PS

Gravitation

Important points:-

1) The phenomenon of gravity was discovered by Sir Isaac Newton.

2) Every body in the universe exerts the gravitational force on every other body in the universe. This force is directly proportional to the product of the masses of the two bodies and inversely proportional to the masses of the two bodies.

3)A force acts on any object moving along a circle and it is directed towards the centre of the circle. This force is called the centripetal force. It can be given by

4)Kepler’s laws:

The orbit of a planet is an ellipse with the Sunn at one of the foci.
The line joning the planet and the Sun sweeps equal areas in equal intervals of time.
The square of period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of a planet from the sun. i.e.

5) Planets revolve around the sun, satellites revolve around the planet, we have high tide and low tide.These are all because of the gravitational force.

6) Earth’s gravitational acceleration (denoted by g) is maximum at ples (9.832 m/s^2) and minimum at equator(9.78 m/s^2).

7) The value of g changes as per :

The radius of earth at that place.
Height of the object from the surface of the earth.
The depth below the surface of the earth. At the centre of the earth g = 0.

8) Mass is the amount of matter present in the object and is constant. Weight is the force with which the earth attracts the object. Mass is a scalar quantity while weight is a vector quantity

Weight = mg and it can change as the value of ‘g’ changes. Unit of mass is kilogram and that of weight is Newton

9) Recently, scientists have detected the gravitational waves which were predicted about 100 years back by Einstein

10) Whenever an object moves under the influence of the force of gravity alone, it is said to be falling freely.

11) The motion of satellites revolving around the earth is also an example of a free fall.

12) At infinite distance from the earth, the gravitational potential energy is zero and for smaller distances i.e. heights the potential energy is negative.

13) If an object is thrown upwards with such an initial velocity that the object is able to overcome the downward pull by the earth and can escape the earth forever and will not fall back on the earth, then that initial velocity is called escape velocity.

Q.1.(A) Match the following and rewrite:

Ans:

(ii)

Ans:

Q.1.(B) Fill in the blanks using correct options:

(i) As per the Kepler’s law the orbit of a planet is …….

(a) circular (b) elliptical (c) spherical (d) linear

(ii) The centripetal force in a circular motion is given by …….

(a) (b) (c) (d)

(iii) If mass of an object is 10 kg, then the weight of that object is ……. N.

(a) 10 (b) 9.8 (c) 98 (d) 20

(iv) A light object and a heavy object are released from the same height in vacuum. Then …….

(a) the heavy object will reach ground first.

(b) the light object willl reach ground first.

(c) both the objects will reach the ground at the same time.

(d) none of the above.

(v) The gravitational acceleration on the moon is one sixth of the gravitational acceleration on the earth. Mass of an object is 12 kg on the earth. So, the mass of same object will be ……. kg on the moon.

(a) 2 (b) 72 (c) 1.2 (d) 12

Ans: (i) (b) ; (ii) (c) ; (iii) (c) ; (iv) (c) ; (v) (d)

Q.1.(C) Answer the following questions:

*(i) What are the effects fo a force acting on an object?
Ans: The effects of a unbalanced force acting on an object are as follows-

(1) It can bring an object at rest in the motion.

(2) It can bring an object in motion at rest.

(3) It can change the direction and speed of the motion. In short, a force creates an acceleration in an object.

*(ii) What types of forces are you familiar with?
Ans: Following are some of the types of forces –

(1) Gravitational force (2) Nuclear force (3) Electro-magnetic force (4) Muscular force.

*(iii) What will be the value of g at the centre of the earth?
Ans: The value of g will be zero at the centre of the earth.

*(iv) Will your weight remain constant as you go above the surface of the earth?
Ans: No. As we go above the surface of the earth, the value of g goes on decreasing. And so, the weight also will go on decreasing as Weight = mg.

*(v) What are (i) free fall (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force ?
Ans: (i) Free fall : The motion of an object under the influence of only one force and that is the force of gravity, is called the free fall. When an object is falling down in a vacuum it can be called as a free fall. A satellite revolving around an earth is another example of free fall.

(ii) Acceleration due to gravity: The earth exerts its gravitational force on the objects near to it. As per Newton’s second law of motion and the force equation F = ma, whenever any force is exerted on an object, it creates some acceleration in it. So, when an object is released from height, an acceleration is created in the object due tho gravitational force and it starts coming down with the acceleration. This acceleration is the acceleratin due to gravity. It is denoted by ‘g’.

(iii) Escape velocity : If we throw an object upwards with certain initial velocity, the velocity of the object goes on decreasing due to gravitational pull. At a certain height, velocity becomes zero and the object starts falling down. If the initial velocity is increased the object will attain the larger height. At a certain value of this initial velocit, the object will overcome the gravitational pull and can escape the earth forever and will not come back. This velocity is called as escape velocity.

(iv) Centripetal force : If an object has certain velocity, it will move in the straight line with the same velocity as per the Newton’s first law of motion. If the object is to perform the circular motion, a force has to be continously exerted on the object in the direction towards the centre. This force is called as the centrepetal force.

Q.2.(A) Answer the following question:

*(i) If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why?
Ans: Let the mass of the heavy object be m.
Its weight = mg
If g becomes twice i.e. changes to 2g, its weight will be
It means its weight will be doubled.
The frictional force between the heavy object and the floor is directly proportional to the weight of the object. So, the frictional force also will be doubled.
The force required for pulling the object is just greater than the frictional force. So, the force required for pulling the object also will be doubled.

*(ii) Explain why the value of g is zero at the centre of the earth.
Ans: When an object is at the centre of the earth, it will be pulled in all the directions by the mass of the earth which is distributed equally around the centre. So, the net gravitational force acting on the object is zero. So, the acceleration due to gravity is zero. i.e. The value of g is zero.

*(iii) Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be
Ans: Let period of revolution be T’ when distance of planet is 2R.
According to Kepler’s law : = constant
If the planet was at a distance of 2R from the star, its period of revolution will be T.

*(iv)What do you know about the gravitational force?
Ans: (1) The gravitational force is of attractive nature and is present between any two bodies in the universe.
(2) This force is directly proportional to the product of masses of the two bodies and it is inversely proportional to the square of the distance between them.
(3) Due to this force, any object released form height falls on the earth.
(4) The satellites revolve around the planet, the planets revolve around the stars because of this force.

*(v) If the area ESF in the following figure of an orbit of a planet is equal to area ASB, what will you infer about EF?
Ans: (1) The time taken to move form point E to F along the orbit is equal to the time taken to move form point A to B along the orbit.
(2) EF < AB.

*(vi) Show that in SI units, the unit of G is
Ans:

*(vii) Is there a gravitational force between two objects kept on a table or between you and your friend sitting next to you? If yes, why don’t the two move towards each other?
Ans: (1) The gravitational force between the two is very very small and it creates a very very small acceleration in them.
(2) The gravitational force is so small that it cannot overcome the frictional force between the objects and the ground. So, they don’t move towards each other.

*(viii) What would happen if there were no gravity?
Ans: If there were no gravity, following things would happen:-
(1) No satellite can rotate around a planet.
(2) No planet can rotate around a star.
(3) The objects released from the height will not fall on the ground.
(4) The objects would have no weight.
(5) The celestial bodies having initial velocity would be travelling in a straight line.
(6) There would be no atmosphere on any of the planet.
(7) Ultimately, there would be no life.

*(ix) What would happen if the value of G was twice as large?
Ans: (1) As , the value of ‘g’ would be the double for a given planet.
(2) As the value of ‘g’ will be double, the weights of the objects would be double weight = mg.
(3) The objects would be falling with double the acceleration i.e. they will reach the ground faster.
(4) The average value of ‘g’ on the surface of the earth would be 9.8.
(5) The escape velocity would be different.

*(x) Will the direction of the gravitational force change as we go inside the earth?
Ans: (1) While going form the surface of the earth to the centre of the earth in a straight line, the direction of the gravitational force will be constant i.e. towards the centre. At the centre, there will be no gravitational force. As we continue to move further in a straight line i.e. from the centre to the surface of the earth, the direction will be exactly reversed. But it will continue to be towards the centre.

*(xi) Suppose you are standing on a tall ladder. If your distance form the centre of the earth is 2R, what will be your weight.
Ans: g = . It means that g is inversely proportional to the square of R. If distance becomes 2R i.e. double then value of g will become one fourth of its original value. So, the weight also will become one fourth of the original one as weight = mg.

*(xii) What would be the value of g on the surface of the earth if its mass was twice as large and its radius half of what it is now?
Ans: g = –(1)
Let the value of acceleration due to gravity be g’ when mass is 2M and radius is .

—(2)
From eq. (1) and (2), we get-
g’ = 8 g
The value of g on the surface of the earth if its mass was twice as large and its radius half of what it is now would be .

*(xiii) According to Newton’s law of gravitation, earth’s gravitational force is higher on an object of larger mass. Why doesn’t that object fall down with higher velocity as compared to an object with lower mass?
Ans: (1) From the formula itself it is clear that the value of ‘g’ is independent of mass of the object.
(2) Though the gravitational force is higher on the object of larger mass, the acceleration is the same. So, the object larger mass does not fall down with higher velocity as compared to an object with lower mass.

Q.2.(B) Solve the following examples.

*(i) An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet.
Ans: Here, t = 5 s, s = 5 m, u = 0 m/s, a = g = ?
By the Newton’s second equation of motion –
s = ut + at
The value of g on the planet is 0.4 m/s.

*(ii) The radius of a planet A is half the radius of planet B. If the mass of A is M, what must e the mass of B so that the value of g on B is half that of its value on A?
Ans: Let the radius of planet A be R and that of planet B be 2R.
Mass of planet A = M. Let the mass of planet B be M/
Let the value of g on A be 2x and that on B be x/
(for planet A) —(1)
(for planet B) —(2)
Dividing eq. (1) by (2) we get –
The mass of B should be 2M.

*(iii) The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.
Ans: On the earth –
m = 5 kg,
Weight = mg = 49 N
On the moon –
m = 5 kg only as the mass does not change and is constant everywhere.
(i) The mass on moon will be 5 kg.
(ii) The weight on moon will be 8.17 N.

*(iv) An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s.
Ans: For upward movement : s = 500 m,
a = g = – 10 m/s, v = 0 m/s, u = ?, t = ?
As per the third equation of motion –
As per the first equation of motion –
For downward movement : s = 500m, u = 0 m/s, g = 10 m/s, t = ?
As per the second equation of motion –
Total time taken to come to the earth = 10 + 10 = 20 s.
(i) Initial velocity = 100 m/s
(ii) It will take 20 s to come back to the earth.

*(v) A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s, calculate its speed on reaching the ground and the height of the table.
Ans: Here, u = 0 m/s, t = 1s, a = g = 10 m/s, v = ?, s = ?
As per the first equation of motion –
As per the third equation of motion –
(i) The speed on reaching the ground = 10 m/s
(ii) Height of the table = 5 m.

*(vi) The masses of earth and moon are kg and kg respectively. The distance between them is km. Calculate the gravitational force of attraction between the two. Use G =
Ans: Here,
Gravitational force F
The gravitational force of attraction =

*(vii) The mass of the earth is kg. The distance between the earth and the Sun is m. If the gravitational force between the two is N, what is the mass of the Sun? Use G =
Ans: Here,
Gravitational force F =
The mass of the Sun is kg.

Q.3. Answer the following questions.

*(i) What is the difference between mass and weight of an object. Will the mass and weight of an object on the earth be same as their values on Mars? Why?
Ans: (1) Mass is the amount of matter present in an object or it is the measure of inertia of an object, while the weight of an object is the force with which the earth attracts that object.
(2) The values of mass will be the same on the earth as well as on the Mars. But the values of weight will be different on the earth and the Mars.
(3) The gravitational force with which the object is attracted to the earth is different form that with which the object is attracted to the Mars. So, their weights are different. The gravitational forces are different because the masses of the earth and the Mars are different.

*(ii) What are Newton’s laws of motion?
Ans: (1) Newton’s first law of motion: An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced force acts on it.
(2) Newton’s second law of motion: The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.
(3) Newton’s third law of motion: Every action force has an equal and opposite reaction force which acts simultaneously.

Q.4. Answer the following questions.

*(i) A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is same as the time taken to come down.
Ans: Let us consider the upward motion of the stone.
Initial velocity = u m/s
Final velocity = 0 m/s
Accleration = – g m/s (negative sign is becaouse acceleration is in the opposite direction of velocity)
Displacement = s = h metre
By the Newton’s third equation of motion –
Let us consider the downward motion of the stone.
Initial velocity = 0 m/s
Let the final velocity be v m/s
Acceleration = g m/s (acceleration is in the direction of velocity)
Displacement = s = h = metre [from (1)]
i.e. Initial velocity of upward motion = Final velocity of downward motion —(2)
Let t be the time taken to go up.
Applying the Newton’s first equation of motion for upward motion –
v = u + at
—(3)
Let t be the time taken to come down.
For downward motion –
v = u + at
[from (2)]
—(4)
From (3) and (4) we get –
i.e. The time taken to go up is same as the time taken to come down.

*(ii) Write the three laws given by Kepler. How die they help Newton to arrive at the inverse square law of gravity?
Ans: Kepler’s laws : (1) The orbit of a planet is an ellipse with the sun at one of the foci.
(2) The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
(3) The square of its period of revolution around the Sun is directly proportional to the cube of the mean distance of a planet form the Sun.
Acceleration in a uniform circular motion = (where v is the speed and r is the radius of the circle)
Centripetal force = (because F = ma)
If we consider a planet of mass m revolving around the Sun in a circular orbit of radius r then the gravitational force between the planet and the Sun is the centripetal force.
Gravitational force = —(1)
v = (where T is period of revolution)
Substituting above value in equation (1), we get –
Gravitational force = m
But as per Kepler’s law: = constant = K
Gravitational force =
But for a given planet = Constant
Gravitational force = Constant
Gravitational force
This is how Newton arrived at the inverse square law of gravity taking help of Kepler’s law.

Periodic classification of elements

Important points:

1) Dobereiner made groups of three elements each having similar chemical properties and called them triads. He arranged the elements in a triad in the increasing order of their atomic masses and found that the atomic mass of the middle element is approximately equal to the mean of the atomic masses of the other two, elements. However he could not classify all the elements known in the triads.

2) Newland’s law of octaves – Newlands arranged all the elements known in the increasing order of their atomic masses and stated that the properties of the eighth element are similar to those of the first element. But this law ‘ could not be applied for all the elements.

3) Mendeleev’s periodic law – Properties of elements are periodic function of their atomic masses.

4) Merits of Mendeleev’s periodic table –

(i) Atomic masses of some elements were revised so as to give them proper place in the periodic table in accordance with their properties. e.g. atomic mass of beryllium was corrected from 14.09 to 9.4.

(ii) Mendeleev kept some vacant places in his periodic table and predicted the properties of the elements yet to be discovered which will fit in the vacant places.

(iii) The noble gases which were discovered later could be fitted into his periodic table in a separate group without disturb. ing the positions of the other elements.

5) Demerits of Mendeleev’s periodic table

(i) There was an ambiguity about the positions of the elements Cobalt and Nickel whose whole number atomic masses were same.

(ii) Isotopes have. same properties but have different atomic masses, so there was a chal~ lenge to fit them in proper places.

(iii) When elements are arranged in an increasing order of atomic masses, the rise in atomic masses does not appear to be uniform. So, it was not pessible to predict the discovery of new ‘ elements between two heavy elements.

(iv) Hydrogen resembles halogens as well as metals, so its place could not be determined correctly.

6) The Modern periodic lawProperties of elements are a periodic function of their atomic numbers.

7) Long form of the modern periodic table has 7 periods. and 18 groups.

8) The entire periodic table is divided into four blocks s-block (groups 1 and 2), p-block (groups 13 to 18), dblock (groups 3 to 12) and f-block which includes lanthanide and actinide series of elements.

9) As we go from left to right in a period, electrons get added in the outer shell and as we go from top to bottom in a group, the number of shells go on increasing.

10) Atomic size goes on decreasing as we go from left ‘ to right in a period where as atomic size goes on increasing as we go from top to bottom in a group.

11) Metallic character goes on reducing as we go from left to right in a period and it goes on increasing as we go from top to bottom in a group.

12) The electrons in the outermost shells are called the valence electrons. They determine the valency of the element.

13) Metals In the first group are called alkali meta|s, those in the second group are called alkaline earth metals. The elements in the 17th group are called as halogens while gases in the 18th group are called as noble gases.

Q.1.(A) Match the following :

Ans:

Q.1.(B) Choose the correct option and rewrite the statement .

*a. The number of electrons in the outermost shell of alkali metals is …….
(i) 1 (ii) 2 (iii) 3 (iv) 7

*b. Alkaline earth metals have valency 2. This means that their position in the modern periodic table is in …….
(i) Group 2 (ii) Group 16 (iii) Period 2 (iv) d-block

*c. Molecular formula of the chloride of an element X is XCI. This compound is a solid having high melting point. Which of the following elements be present in the same group as X.
(i) Na (ii) Mg (iii) AI (iv) SI

*d. In which block of the modern periodic table are the non-metals are found?
(i) s-block (ii) p-block (iii) d-block (iv) f-block

e. Electronic configuration of an element B is (2, 8, 3). The formula of its oxide will be …….
Ans: (a) (i) ; (b) (i) ; (c) (i) ; (d) (ii) ; (e) (iv)

Q.1.(C) Write the name and symbol of the element from the description.
a. The atom having the smallest side.
b. The atom having the smallest atomic mass.
c. The most electronegative atom.
d. The noble gas with the smallest atomic radius.
e. The most reactive non-metal.
Ans: (a) Helium (He) (b) Hydrogen (H) (c) Fluorine (F) (d) Helium (He) (e) Fluorine (F).

Q.1.(D) Write the names from the description.
a. The period with electrons in the shells K, L and M.
b. The group with velency zero.
c. The family of non-metals having valency one.
d.The family of metals having valency one.
e.The family of metals having valency two.
f. The metalloids in the second and third periods.
g. Non-metals in the third period.
h. Two elements having valency 4.
Ans: (a) Period 3 (b) Group 18 (c) Halogens (d)Alkali metals (e) Alkaline earth metals (f) Boron and Silicon (9) Phosphorus, Sulphur, Chlorine, Argon (h) Carbon an Silicon.

Q.1.(E) Answer the following questions:

*a. What are the types of matter?
Ans: The types of matter are (i) Elements (ii) Com pounds and (iii) Mixtures.

*b. What are the types of elements?
Ans: The types of elements are (i) Metals (ii) Non~metals and (iii) Metalloids.

*c. What are the smallest particles of matter called?
Ans: The smallest particles of matter are called atoms.

*d. What is the difference between the molecules of elements and compounds?
Ans : The molecule of an element contains same kind of atoms (e.g. ) while the molecule of a compound contains more than one kinds of atoms (e.g. ).

*.e What are the values of ‘n’ for the shells K, L and M?
Ans: For K: n = 1, For L: n = 2 and for M: n = 3.

Q.1.(F) Write the molecular formulae of oxides of the following elements by referring to the Mendeleev’s periodic table. Na, Si, Ca, C, Rb, P, Ba, CI, Sn, Ca.
Ans:

Q.1.(G) Write the molecular formulae of the compounds of the following elements with hydrogen by referring to the Mendeleev’s periodic table. C, S, Br, As, F, O, N, CI.
Ans:

Q.2.(A) An element has its electron configuration as 2, 8, 2. Now answer the following questions:
a. What is the atomic number of this element?
Ans: The atomic number of this element is 12 (2+8+2).

b. What is the group of this element?
Ans: Group 2.

c. To which period dors this element belong?
Ans: Period 3.

d. With which of the following elements would this element resemble? (Atomic numbers are given in the brackets.) N(7), Be(4), Ar(18), CI(17).
Ans: The element (2, 8, 2) will resemble with be (4), because both are in the group 2.

Q.2.(B) Write down the electronic configuration of the following elements form the given atomic numbers. Answer the following question with explanation.

a. Which of these elements belong to period 3?
Ans: (1) Following is the electronic configuration:

(2) Elements and belog to period 3 because each of them has 3 shells.

b. Whic of these elements belong to the second group?
Ans: (1) Following is the electronic configuration :

(2) The elements and belong to second group as they have 2 valence electrons.

c. Which is the most electronegative element among these?
Ans: (1) Following is the electronic configuration

(2) Be has the smallest atomic radius among above elements because as we go down the group atomic radius increases and as we go from left to right in a period atomic radius decreases. Be is in the second period and to the right of Li.

g. Which of the above elements has the highest metallic character?
Ans: (1) Following is the electronic configuration:

(2) Na has the highest metallic character among above elements because as we go form left to right in a period, metallic character decreases. All above elements are form the third period and Na is to the left of all other elements.

h. Which of the above elements has the highest non-metallic character?
Ans: (1) Following is the electronic configuration:

Q.2.(C) Give scientific reasons:

*a. Atomic radius goes on decreasing while going form left to right in a period.
Ans: (1) While going form left to right within a period, the atomic number increases one by one, meaning the positive charge on the nucleus increases by one unit at a time. However, the additional electron gets added to the same outermost shell.
(2) Due to the increased nuclear charge, the electrons are pulled towards the nucleus to a
greater extent and thereby the size of the atom decreases. Therefore, atomic radius goes on decreasing while going from left to right in a period

*b. Metallic character goes on decreasing while going from left to right in a périods
Ans: (1) Metals have tendency to lose the valence electrons and form the cation. If higher the tendency to lose valence electrons, higher is the metallic character.
(2) In a period, as we go from left to right, the number of valence electrons increases and the pull between the nucleus and the valence electrons increases.
(3) AS a result, the tendency to lose valence electrons decreases. Therefore, metallic character goes oh decreasing while going from teft to right in a period.

*c. Atomic radius goes on increasing down a group.
Ans: (1)Atomic radius is the distance between the centre of the nucleus andthe outer shell.
(2) As we go down the group, one by one shell gets added. So, the distance between the centre of nucleus and the outer shell increases. Therefore, atomic radius goes on increasing down a group.

*d. Elements belonging to the same group have the same velocity.
Ans: (1) In Modern Periodic table, elements having same number of valence electrons or having same outer electronic configuration are placed in the same group.
(2) The valency of an element depends on the number of valence electrons. i.e. for same number of valence electrons, the valency is same. Therefore, elements in the same group have the same valency.

*e. The third period contains only 8 elements even though the electron capacity of the third shell is 18.
Ans: (1) The capacity of the third shell is 18 as per the formula . But the capacity of the last shell can not exceed 8.
(2) The elements in the third period have 3 shells and so their third shell is the last shell. So, its capacity is 8.
(3) The third period contains the elements having 3 shells and the number of electrons in the last shell from 1 to 8. Therefore, the third period contains only 8 elements even though the electron capacity of the third shell is 18.

Q.2.(D) Answer the following questions:

*a. How is the problem regarding the position of Cobalt () and Nickel () in Mendeleev’s periodic table resolved in modern periodic table?
Ans: (1) In modern periodic table. the elements are arranged as per the increasing order of atomic numbers and not the atomic masses.
(2) Cobalt has atomic number 27 and Nickel has atomic number 28. So, there is absolutely no ambiguity and Cobalt appears before Nickel in the modern periodic table.

*b. How did the position of and get fixed In the modern periodic table?
Ans: (1) In modern periodic table, the elements are arranged as per the increasing order of atomic numbers.
(2) Because CI and CI have same atomic number 17, both of them are placed in the same place i.e. in group 17 and period 3.

*c. Can there be an element with atomic mass 53 or 54 in between the two elements, chromium and manganese Mn ?
Ans: (1) In modern periodic table, the elements are arranged as per the increasing order of atomic numbers.
(2) So, chromium (atomic number 24) and manganese (atomic number 25) appear consecutively in the periodic table. Therefore, there cannot be any element between them even though the atomic mass of that element is 53 or 54.

*d. What do you think? Should hydrogen be placed in the group 17 of halogens or group 1 of alkali metals in the modern periodic table?
Ans: (1) In the modern periodic table elements are arranged as per their electronic configuration.
(2) Both hydrogen and alkali metals have one electron each in the last shell.
(3) Halogens have 7 electrons each in their last shell. So, I think that hydrogen should be placed in the group 1 of alkali metals.

*e. What is the maximum number of electrons that can be accommodated in a shell? Write the formula and explain.
Ans: (1) The ‘n’ shell can accommodate maximum 2n number of electrons In it. It means that the maximum capacity of 1 shell is 2, that of 2nd shell is 8, that of 3rd shell is 18 and so on.
(2) But the last shell in any case, cannot have more than 8 electrons in it. It means that if the 3rd shell is the last shell for any atoms, it cannot have 18 electrons in it but it can have maximum 8 number of electrons.

*f. Deduce the maximum electron capacity of the shells K, L and M.
Ans:(1)For shell K, n =1. So, the maximum capacity of it = = 2 x (1) = 2.
(2) For shell L, n= 2. So, the maximum capacity of it = 2n = 2 x (2) = 8.
(3) For shell M, n= 3. So, the maximum capacity of it = 2n = 2 x (3) = 18.

*g. What is the relationship between the electronic configuration of an element and its valency?
Ans: (1) If the number of valence electrons (i.e. electrons in the last shell) is 4 or less than 4, then valency is equal to the number of valence electrons. e.g. Magnesium has 2 valence electrons, So its valency is 2.

(2) If the number of valence electrons is greater than 4, then valency is equal to the difference between 8 and the number of valence electrons. e.g. Oxygen has 6 valence electrons, so its valency = 8 – 6 = 2.
(iii) In case of Helium, though there are 2 valence electrons, its outermost orbit is completely filled. So, its valency is zero.

*h. The atomic number of beryllium is 4 while that of oxygen is 8. Write down the electronic configuration of the two and deduce their valency from the same.
Ans:

*i. What is the periodic trend in the variation of valency while going from left to right within a period? Explain your answer with reference to period 2 and period 3.
Ans: As we go from left to right, within 2nd and 3rd period valency first increases from 1 to 4 and then it reduces from 4 to 0.

*j. What Is the periodic trend in the variation of valency while going down a group? Explain your answer with reference to the group 1, group 2 and group 18.
Ans : (1) in a group, the valency of the elements is the same.
(2) The valency of elements in group 1 is 1, the valency of elements in group 2 is 2 and that in group 18 is 0.

*k. What is the cause of non-metallic character of elements?
Ans : (1) The tendency of an element to form anion or the electronegativity is the non-metallic character of element.
(2) In non-metals there are 4 or more electrons in the last shell. So, electrons are held by the nucleus by a greater attractive force. So, there is a tendency not to lose the electrons but accept the electrons and form the anions. Thus, the greater pull by the nucleus is the cause of non-metallic character of elements.

*l. What is the expected trend in the variation of non-metallic character of elements from left to right in a period.
Ans: (1) The tendency of an element to form anion or the electronegativity is the nonmetallic character of element.
(2) As we go from left to right in a period, the number of electrons in the last shell go on increasing. So, the attractive force between the valence electrons and the nucleus becomes stronger and stronger.
(3) As a result, tendency to lose electrons becomes lesser and lesser and later the tendency to gain electrons and to form an anion becomes greater. So, the non-metallic character of elements goes on increasing from 1eft to right In a period.

*m. What would be the expected trend in the variation of non-metallic character of elements down a group?
Ans : (1) As we go down the group, the shells are getting added.
(2) The distance between the valence electrons and the nucleus increases and the effective nuclear force on valence electrons decreases. So, the electronegativity of the element decreases i.e. the non-metallic character decreases.

n. State the modern periodic law and tell how it is different from the Mendeleev’s periodic law.
Ans: (1) The modern periodic law states that properties of elements are a periodic function of their atomic numbers.
(2) The difference between the modern periodic law and Mendeleev’s periodic law is about the fundamental property of the element. Modern periodic law considers atomic number as the fundamental property while Mendeleev’s periodic law considers atomic mass as the fundamental property.

Q.3.(A) Write short notes:

*a. Mendeleev’s periodic law –
(1) Mendeleev’ periodic law states: Properties of elements are periodic function of their atomic masses.
(2) It means that if the elements are arranged in the increasing order of atomic masses, after certain period, the properties of elements repeat.
(3) Mendeleev prepared his periodic table according to his periodic law. There were many anomalies in his table.
(4) Later, Henry Moseley stated that atomic number and not the atomic mass is the fundamental property of an element and properties of elements are periodic function of their atomic numbers.

*b. Structure of modern periodic table –
(1) In Modern periodic table, elements are arranged in the increasing order of their atomic numbers.
(2) They are arranged in the periods (horizontal rows) and groups (vertical columns) according to their electronic configurations.
(3) In the modern periodic table, there are 7 periods and 18 groups.
(4) The elements in a group show similarities in their properties.
(5) The elements in a period show gradation in the properties.
(6) There are metals to the left side of the modern periodic table and non-metals to the right. Metals and non-metals are separated by a zig-zag line. Along the zig-zag line metalloids are present.
(7) There are two different groups of metals placed below the modern periodic table. They are Lanthanide series of elements (atomic no. 57 to 71) and Actinide series of elements (atomic no. 89 to 103).
(8) All elements In the table are divided In the four blocks. Groups 1 and 2 form s-block, groups 13 to 18 form p-block, groups 3 to 12 form d-block arid Lanthanide and Actinide series elements form the f-block.

*c. Position of isotopes in the Mendeleev’s and the modern periodic table-
(1) Mendeleev arranged the elements according to the increasing order of atomic masses. Isotopes of same element have different atomic masses. So, he placed them in different positions.
(2) But in modern periodic table, the elements are arranged in the increasing order of atomic numbers. Isotopes of same element have same atomic numbers, so they are placed at the same position in modern periodic table.

Q.3.(B) Answer the following questions.

*a. Identify Dobereiner’s triads from the following groups of elements having similar chemical properties: (i) Mg (24.3), Ca (40.1), Sr (87.6) (ii) S (32.1), Se (79.0), Te (127.6) (iii) Be (9.0), Mg (24.3), Ca (40.1).
Ans : Let us check for each group, whether the atomic mass of middle element is equal to the mean of atomic masses of the other two.
(i) Mean atomic mass of other two elements
It is not equal to the atomic mass of middle element Ca (40.1). So, this Is not a Dobereiner’s triad.
(ii) Mean atomic mass of other two elements
It is appeoximately equal to the atomic mass of middle element Se (79.0). So, this is a Dobereiner’s triad.
(iii) Mean atomic mass of other two elements
It is approximately equal to the atomic mass of middle element Mg (24. 3).
So, this iS a Dobereiner’s triad.

*b. There are some vacant places in the Mendeleev’s periodic table. In some of these places the atomic masses are seen to be predicted. Enlist three of these predicted atomic masses along with their group and period.
Ans:

*c. Due to uncertainty in the names of some of the elements, a question mark is indicated before the symbol in the Mendeleev’s periodic table. What are such symbols?
Ans: (1) The symbols before which the question mark was put by Mendeleev are as follows:
Yt, Di, Ce, Er, La,
(2) These were later named as Yttrium (Y ), Didymium [later found to be mixture of praseodymium (Pr) and neodymium (Nd)], Cerium (Ce), Erbium (Er) La (Lanthanum) respectively.

*d. On going through the modern periodic table, it is seen that the elements Li, Be, B, C, N, O, F and Ne belong to the period-2. Write down electronic configuration of all of them. Also write the number of valence electrons and number of shells in case of each element.
Ans:

*e. Look at the elements of the third period. Classify them into metals and non-metals. On which side of the period are the metals? Left or right? On which side of the period did you find the non-metals?
Ans: (1) The elements in the third period are: (i) Metals – Sodium (Na), Magnesium (Mg), Aluminium (AI) (ii) Metalloids – Silicon (Si) and (iii) Non-metals – Phosphorus (P), Sulphur (S), Chlorine (CI) and Argon (Ar).
(2) The metals are on the left side of the period.
(3) Non-metals are on the right side of the period.
(4) Metals and nonmetals are separated by a metalloid in between.

f. What are the merits of Mendeleev’s periodic table?
Ans: (1) Atomic masses of some elements were revised so as to give them proper place in the periodic table in accordance with their properties. e.g. the previously determined atomic mass of beryllium as 14.09 was corrected to 9.4 and beryllium was placed before boron.
(2) Mendeleev kept some vacant places in his periodic table and predicted their atomic masses and other properties. When these elements were discovered later, it was found that the predicted atomic masses and their properties were fairly correct. e.g. he kept vacant places for elements eka-boron, eka-aluminium and eka-silicon. Later the elements Scandium, Gallium and Germanium were discovered which fit into these places respectively.
(3) There was no place for inert gases initially in the Mendeleev’s periodic table. But when these gases were discovered later, they could be fitted in a separate group (zero group) without disturbing the position of other elements.

Q.4. Answer the following questions:

a. Study the figure and answer the questions that follow.

*(1) Write the names of the elements of group 1.
Ans: Hydrogen (H), Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs)and Francium (Fr).

*(2) Write the electronic configuration of the first four elements in this group.
Ans: (i) Hydrogen (1), Lithium (2, 1), Sodium (2, 8, 1) and Potassium (2, 8, 8, 1).

*(3) Which similarity do you find in their configuration?
Ans: They all have one electron in their last shell.

*(4) How many valence electrons are there in each of these elements?
Ans: Each one of them has one valence electron.

(5) Which of the above elements has the smallest atomic radius?
Ans: Hydrogen has the smallest atomic radius among the above elements.

*b. The following table is made on the basis of the modern periodic table. Write in it the symbol (e.g. K), electronic configuration (e.g. 2, 8, 8, 1) and valency (e.g. 1) of the first 20 elements as per shown example.

*c. Following is part of modern periodic table.
Some elements and their atomic radii are given below. Study them and answer the questions that follow.

(1) By referring modern periodic table, find out the groups to which above elements belong.
Ans: They all belong to the first group.

(2) Arrange the above elements vertically downwards in an increasing order of atomic radii.
Ans: Li
Na
K
Rb
Cs

(3) Does this arrangement match with the pattern of the group 1 of the modern periodic table?
Ans: Yes.

(4) Which of the above elements have the biggest and the smallest atom?
Ans: Out of the above elements, Li (Lithium) has the smallest atom and Cs (Cesium) has the biggest atom.

(5) What is the periodic trend observed in the variation of atomic radii down a group?
Ans: As we go down a group the atomic radius increases

d. What are the demerits of Mendeleev’s periodic table and how were they removed in the modern periodic table?
Ans: Demerit: The whole number atomic mass of the elements cobalt (Co) and nickel (Ni) was same. So, there was an ambiguity about their sequence in periodic table.
How it was removed : In modern periodic table, elements are arranged according to their atomic numbers. So, Cobalt (atomic number 27) was rightly placed before Nickel (atomic number 28).
Demerit: The atomic masses of isotopes were different but their properties were same. So, there was a challenge to place the isotopes in proper places.
How it was removed : In modern periodic table, elements are arranged according to their atomic numbers. So, isotopes of an elements are kept at one place as their atomic numbers are same.
Demerit : The elements were arranged in the increasing order of atomic masses. But the increase in atomic masses was not uniform. So, it was not possible to predict as to how many elements could be discovered be tween two heavy elements.
How it was removed: In modern periodic table, elements are arranged accOrding to their atomic numbers. So, the prediction of such elements was possible.
Demerit : Hydrogen resembles metals and halogens. So, there was an ambiguity whether it should be placed in the group of metals or in the group of halogens.
How it was removed : In modern periodic table, elements are arranged according to their atomic numbers and electronic configurations. So, hydrogen was placed in the first group as it has one valence electron.

chemical reaction of elements

Important Points:

Physical Change:
1. Physical change occurs due to changes in temperature and pressure.
2. Composition of the material remain as it IS iri a physical change.
3. It is a temporary change.
4. It is a reversible change. The original substance is recovered as it is after the physical change.

Chemical Change:
1. In a chemical change, A totally new substance is formed after the chemical reaction, between two or more than two compounds. The chemical bonds between reactants break & New chemical bonds are formed to form totally new products.
2. The composition of the product totally changes.
3. it Is a permanent change.
4. Original substance can not be recovered after a chemical change.

Writing a Chemical Equation :
1. A chemical equation is written using symbols for the elements or molecular formulae of compounds. The reactants are written on the left handside of an arrow , headed towards right hand side. The products are written on right hand side of the arrow.
2. A (+) sign is written between two reactants or between two products.
3. The physical state of the reactants or products are mentioned as given below:
aqueous solution (aq), solid (s), liquid (l), gaseous (g)
4. Conditions like temperature, pressure, catalyst etc. are written on the arrow between reactants and products.
5. The reactants and products are always written in their molecular form.
Types of Chemical Reactions:
1) Combination reaction: A+ B à AB
2) Decomposition reaction: AB à A+ B
3) Displacement reaction: AB + C à AC + B
4) Double displacement reaction:  A+B +C+D à A+D+B-C+
5) Endothermic reaction: The reaction in which heat  is absorbed or it has to be supplied externally, that reaction is called as Endothermic reaction.
6) Exothermic reaction: When heat is liberated in a chemical reaction, it is called as Exothermic reaction.
7) Oxidation reaction: When Oxygen is added or hydrogen is removed from a reactant it is an oxidation reaction.
8) Reduction reaction: When hydrogen is added to a reactant or oxygen is removed from the reactant it is a reduction reaction.
9) Redox reaction: When oxidation and reduction both occur at a time in a certain reaction. it is called as a redox reaction.

Q.1.(A) Fill in the blanks by choosing the correct option from given options and justify the statement.

(Oxidation, Reduction, Decompesition, Displacement, Zinc, Copper, Electrolysis, Double Displacement)
1) Water is ……. if electric current is passed through acidified water.
Ans: Water is electrolysed if electric current is passed through acidified water.
When the pure water is acidified using certain acid, water gets ionised into H and O ions.
H+ ions are liberated at the cathode and 0 ions are liberated at the anode. Due to which at the cathode hydrogen gas is liberated and at the anode oxygen gas is liberated.
This decomposition is done with the help of electric current thus it is also called as electrical decomposition or electrolysis.

2) To avoid rusting of Iron Sheet a coating of ……. is applied on it.
Ans: To avoid rusting of iron sheets a coating of zinc is applied on it. it is called as Galvanising. Iron gets corroded if it comes in contact of air moisture or water. To avoid rusting a thin layer of zinc metal which is more electropositive than iron is applied on iron. Corrosion is presented by galvanisation.

3) ……. reaction ocurs when iron nail is placed in aqueous solution of copper sulphate.
Ans: Displacement reaction occurs when iron nail is placed in aqueous solution of copper sulphate.
When more reactive element replaces less reactive element in a compound, the reaction is called as displacement reaction.
Fe, the more reactive element than copper displaces Cu in Cuso4 solution. A thin reddish brown coating of Copper can be seen on iron nail.
This is a displacement reaction.

4) When Magnesium wire is burnt in presence of Oxygen, ……. reaction occurs giving magnesium oxide as a product.
Ans: When Magnesium wire is burnt in presence of Oxygen, Oxidation reaction occurs giving magnesium oxide as a product.
When oxygen is added to a reactant or hydrogen is removed from the reactant, it is an oxidation reaction.
2Mg + –> 2MgO.
Magnesium Oxygen Magnesium oxide.

5) When hydrogen gas is passed over black copper oxide, reddish coloured copper is obtained on ……. of copper oxide.
Ans: When hydrogen gas is passed over black copper, oxide, reddish coloured copper is obtained on reduction of copper oxide.
When oxygen is removed from the reactant and hydrogen is added to the reactant, the chemical reaction is called as reduction.
In the above reaction oxygen from copper oxide is removed giving a reduction reaction, forming red coloured copper.
CuO+H2 -> Cu+H20

6) Conversion of Ferrous Sulphate to Ferric Sulphate is an ……. reaction.
Ans: Conversion of ferrous sulphate to ferric sulphate is an oxidation reaction.
2 FeSO4 –+ Fez (304)
Ionic reaction :
Fez + 8042-‘–§ 2 Fe3+ + 380 42
Fe2+ —i Fe3+
Ferrous Ferric
When one or more than one election is lost it is an oxidation reaction.
When ferrous is converted into ferric, ferrous ion looses 1 electron and gets oxidised.
Oxidation is brought about by Potassium Permanganate.
ZKMnO4+1OFeSO+8HZSO4-JK389342MnSo ‘. . , +5Fe2 ($0420)3+8H
7) When aqueous solutions of Zinc Sulphate and Barium Chloride are mixed together, it is a ……. reaction.
Ans: When aqueous solutions of Zinc Sulphate and Barium Chloride are mixed together it is a double displacement reaction.
BaCl2_+ an04 ._I 83804 i + ZnCI2
Ba.2+ a’nd Zn2+ ions, are replaced with $0 2’ and 20CI respectiveiy.
Negative radicals in both the compounds of reactants are displaced at a time. Thus it is called a double displacement reaction. White precipitate of Barium Sulphate is formed and Zinc Chloride is also formed.

Q.1.(B) Find the correlation between given pair and complete the other pair.
1) Salt Solution: Physical change :: Rusting of iron : ….
2) Vegetable oil + Hmg : Reduction :: 2 Mg + 02 : ….
3) Melting of ice : Endothermic reaction :: Disolution of NaOH: ….
4) Fe –> FeO : Oxidation :: FeO –> Fe : ….
5) NH3 + HCI –> NH4CI : Combination reaction :: ZH2O –> 2H2 + 02 : …….
Ans: 1) Chemical Change; 2) Oxidation; 3) Exotherrnic reaction; 4) Reduction; 5) Decomposition reaction.

Q.1.(C) Identify in the following reactions, which of the reactants are oxidised and which are reduced.

1)2A920–+4Ag+ O2
Ans: Oxygen from silver oxide is removed and silver oxide decomposes to silver and oxygen gas Silver oxide is reduced to silver.

2) 2M9 + O 2′–I2MgO.
When Magnesium is burnt in presence of oxygen. Magnesium oxide is formed. Magnesium is oxidised to magnesium oxide, Oxidation reaction.

3)NiQ+H2-)Ni+H20 « When Nickel oxide is trezated with hydrogen, Nickel is formed.
Addition of hydrogen is reduction, and removal of oxygen is also reduction.
Thus Nickel Oxide is reduced. Hydrogen accepts oxygen and water is formed.
Oxidation and Reduction both occurs at the same time so it is redox reaction.

4) 2Fe + 02 –> 2Fe0.
When iron reacts with oxygen, iron gets oxidised to iron oxide. This is an oxidation reaction.
5) Fe + S –> FeS
This is an oxidation – reduction (redox) reaction.
Fe2+ –> Fe3+ …….. Oxidation
872 –> 2S …….. Reduction
Fe is reducing agent and s in oxidising agent.

Q.1.(D) Complete the given sentences by choosing correct option from the given options.

1) When Magnesium wire is heated it is ……. to form white powder of magnesium oxide.
(a) Reduced (b) Oxidised (c) Decomposed (d) displaced

2) When sugar is heated it gets converted into black coloured carbon and water is formed. This is an example of ……. type of reaction.
(a) decomposition (b) displacement (c) combination (d) double displacement

3) During electrolysis of water ……. is added in it.
(a) salt (b) sugar (c) acid (d) alkali

4) Disolution of Potassium nitrate in water is an ……. reaction.
(a) redox (b) reduction (c) exothermic (d) endothermic

5) Disolution of Sodium hydroxide is an ……. reaction.
(a) Combination (b) Exothermic (c) endothermic (d) displacement

6) When Calcium carbonate is heated, it decomposes releasmg ……. gas.
(a) carbon dioxide (b) Oxygen (c) carbon monoxide (d) Hydrogen Sulphide

7) Ethyl alcohol is oxidised with the help of potassium dichromate and ……. is obtained.
(a) ethyl acetate (b) ethanol (c) methanol (d) Acetic acid.

8) To prevent rancidity of ghee or oil ……. are used.
(a) reducing agents (b) oxidising agents (c) anti oxidants (d) catalyst

9) Ripening of fruit is an example of ……. change.
(a) Chemical (b) temporary (c) physical (d) exothermic

10) Oxidant used for purification of drinking water is …….
(a) Chlorine (b) Sulphuric acid (c) ozone (d) Hydrochloric acid

Ans: 1.(b); 2.(a); 3.(c); 4.(d); 5.(b); 6.(a); 7.(d); 8.(c); 9.(a); 10.(c).

Q.2.(A) Balance the given Chemical reactions and explain in brief.

1) Cu(s) + HNO(aq) –> Cu (N03)2(aq) + N02(g) + H20(l)
Step 1: There are 3 Nitrogen atoms on the product side, thus if we take 3 HNO3 then also hydrogen atoms are not balanced.
Step 2: Cu+4HNO3 –> CU(N032) +2N02 +2H20

2) Ag N0 + NaCl –> Agcn + NaNO
AgN03 + 3NaC| –> AgCl + NaNO3

Reactants = Products

effect of electric current

Important points:-
1) Joules law of heating:
Joules heating also known as Ohmic heating and resistive heating, is the process by which the flow of an electric current through a conductor produces heat.
H = P x t
H = Vlt
H = I x R x I x t
H = IRt
2) In the electric equipment producing heat an alloy such as Nichrome is used because it is resistant to corrosion and has high melting point of about 1400 C. They do not oxidize readily at high temperatures ang is quite suitable at room temperature. Nichrome has very low resistivity and are good conductors of electricity.
3) Magnetic effect of electric current : When an electric current is passed through a conducting wire a magnetic field is produced around it. This phenomenon is known as magnetic effect of electric current
4) Right hand thumb rule: If we hold the current carrying conductor in the right hand such that the outstretched thumb denotes the direction of the current flow then the curled fingers denote the direction of the magnetic field.
5) Solenoid : If a conducting wire is wound around the cylinder in the form of a helix whose cross-section diameter is less compared to the length the coil is called solenoid.
6) The magnetic field lines produced by a bar magnet are similar to the magnetic field lines of a current carrying solenoid. Thus a current carrying solenoid acts just like a bar magnet
7) Electric Motor: An electric motor is a machine which converts electrical energy into mechanical energy.
8) The principle of working of an electric motor is that whenever a current carrying conductor is placed in a magnetic field, it experience a mechanical force. The direction of this force is given by Fleming’s left hand rule.
9) Electromagnetic Induction : Whenever the number of magnetic field lines changes, an electromotive force (emf) is developed at the ends of the conductor and it last as long as there is a change in the magnetic field lines. This phenomenon is known as electro magnetic induction.
10) Galvanometer :
A galvanometer is an electromechanicai instrument used for detecting and indicating electric current. A galvanometer works by producing a deflection of a pointer in response to electric current flowing through a coil in a contstant magnetic field.
11) Flemings Right hand Rule: If the thumb is pointed in the direction of the conductor relative to the magnetic field and the index finger is pointed in the direction of the magnetic field, then the middle finger represents the direction of the induced or generated current within the conductor.
12) Alternating Current: Alternating current is an electric current which periodically reverses its direction.
Direct Current: The electric current flows in a constant direction. The direct current can be increased decreased or stable but it is not oscillating.
An electric generator is a machine which converts mechanical energy into alternating electric energy. Electric generator works according to the Faraday’s laws of electromagnetic induction.

Q. 1. Multiple choice question :

1) The direction of magnetic field lies in the region outside the bor magnet is
a) from the N Pole towards the S Pole of the magnet.
b) from the S Pole towards the N Pole of the magnet.
c) in the direction coming out from both the poles of the magnet.
d) in the direction entering both the poles of the magnet.
Ans: (a) from the N Pole towards the S Pole of the magnet.

(2) which of the following statement is false ?
a)The direction of magnetic field lies from N to S.
b) In a region will magnetic field lies are at a close distance from one another there will be strong magnetic field.
c) The magnetic field lies form closed loops.
d) The magnetic field lies can cross one another.
Ans: (d) The magnetic field lies can cross one another.

(3) Who was the first to observe the magnetic effect of electric current?
a)Faraday b) Oersted c)Volta d)Ampere
Ans: (b) Oersted

(4) With the help of which law can be the direction of magnetic field decided.
a) Faraday’s law
b) Fleming’s right hand rule
c) Right hand thumb rule
d) Flemings left hand rule
Ans: c) Right hand thumb rule

(5) According to the right hand thumb rule direction of what is indicated by the tumb?
a) Electric current b) Magnetic field
c) Magnetic force (d) Motion of conductor
Ans : (a) Electric current.

(6) What is type of field line of a magnetic field Passing through the centre of current carrying circular ring :
a) Circular b) Straight line
c) Ellipse d) Magnetic field is zero at centre.
Ans : (b) Current carrying ring

(7) Which of the following has magnetic field like that of a bar magnet ?
a) Current carrying wire b) Current carrying ring
c) Current carrying solenoid d) Current carrying rectangular loop
Ans: (c) Current carrying solenoid

(8) Who gave the Princnpal of electromagnetic induction ?
a) Faraday b) Oersted
c) Volta d)Ampexe
Ans: (a) Faraday

(9) Which is the direction of niagnetic force acting on a current carrying wire placed in a magnetic field?
a) along magnetic field.
b) along the electric current.
c) perpendicular to magnetic field.
d) opposite to magnetic field.
Ans: (c) perpendicular to magnetic field.

10) Which instrument is used for converting electrical energy into mechanical energy ?
a) Electric generator b) Electric motor
c) Electic Iron d) Electric oven.
Ans: (b) Electric motor.
11) On which principle does the electric generator work ?
a) Electrical energy is converted to mechanical energy.
b) Electrical energy is converted to thermal energy.
c) Mechanical energy is converted to electrical energy.
d) Electrical energy is converted to light energy
Ans: (c) Machanical energy is converted to electrical energy.

(12) The magnitude of AC voltage used in india is ……. and the frequency is …….
a) 110V, 60 HZ b) 110V, 50 Hz
c) 220V, 50 HZ d) 220V, 60 HZ
Ans: (c) c) 220V, 50 H2

(13) Which ofd the following instrument is used to know the presence of electric current ?
a) Fuse b)Galvanometer
c) Voltmeter d) Magnetic needle
Ans:

(14) Afuse wire is …….
a) a conductor b) an insulator
c) a semi-conductor d) made of any material
Ans: a)a conductor

(15) Which type of cUrrent is obtained from a battery.
a)Ac b)DC
c) Both AC and DC d) Depends on the type of battery.
Ans:

(16) Within a bar magnet, the magnetic field lies are
a) from N to S b) from S to N
c) absent d) in both directions
Ans: b) from S to N

17) The strength of magnetic field produced in a current carrying straight conductor is.
a) directly proportional to the distance from wire
b) directly proportional to the magnitude of the current in wire.
c) inversely proportional to the magnitude of the current in th wire.
d) independent of the magnitude of the current in wire.
Ans:

(18) The magnetic field inside a solenoid is
a) strong at N Pole and weak at S Pole.
b) strong at S Pole and weak at N Pole.
c) uniform throughout.
d) Zero
Ans: (c) uniform throughout

(19) What is not true
a) A solenoid is an electromagnet.
b) A solenoid is a temporary magnet.
c) A solenoid is a permanent magnet.
d) A solenoid is a box magnet.
Ans: (c) A solenoid is a permanent magnet.

20) During the situation of short circuit, the current in th circuit,
a) reduces to Zero b) increases greatly
c) does not change d) oscillates
Ans: b) inoreases greatly

Q.2. Match the following :

Ans: (1) – b (2) – c (3) – d (4) – a

Q.3. State and explain Joule’s law of heating.
Ans: Joules heating is also known as ohmic heating and resistive heating,is the process by which the flow of an eletric current through a conductor produces heat.
(ii) P = Eletric Power =
(iii) Consider an eletric circuit as following
(iv) The cell is the source of energy which gives the energy Pxt to the resistor. If is the current following continously through the circuit the heat produced in the resistor in time t will be,
H = P x t = –>(2)
According to Ohm’s law,
= IR –> (3)
H = I x R x I x t
H = I Rt
This is Joule’s law of heating.

Q.4. Write the unit of electric power
P = V x I = Volt x Amp
=

Q.5. Give scientific reasons:

1) In the electric equipment producing heated iron, electric heater, boiler, toaster, etc, an alloy such as Nichrome is used not pure metals.
Ans: (i) Alloys are combination of one or 2 metals Though metals possess useful properties like good conductivity and high strength, alloys combine there properties making a metal useful for a particular application.
(ii) When metals are alloyed properties like their melting point, conductivity etc are affected.
(iii) There are coils inside electrical appliances that take part in electrical heating. Electrical heating is a process in which electrical energy gets converted to heat energy.
(iv) Modern home appliances use nichrome which is wound in coils that produce heat when current passes though the coil.
(v) Nichrome is an alloy of nickel (80%) and chromium (20%) it is silver grey in colour, resistant to corossion, and has high melting point of about 1400°c. They do not oxidize readily at high temperatures and is quite stable at , such a temperature.
(vi) The resistivity of alloys generally differs from its constituent metals. Nichrome has a very low resistivity and are good conductors of electricity.
(vii) The properties like no oxidation at high temperature, low resistivity and high melting point makes nichrome very much suitable for their use as coils in electrical appliances like electric iron toasters etc.

(2) For electric power transmission, copper or aluminium wire is used.
Ans: (i) Copper and aluminium wires are generally used for electricity transmission as they have low resistivity.
(ii) Low resistivity decreases the resistance and hence increases the conductivirty. Hence for electic power transmission, copper or aluminium wire is used.

(3) Tungsten metal is used to make a solenoid type coil in an electric bulb.
Ans: (i) Pure tungsten has some amazing properties including the highest melting pount (3695K), lowest vapour pressure and greatest tensile strength cut of all the metals.
(ii) Because of these properties it is the most commonly used material for light bulb. It can reach high temperature before melting and therefore emit a brighter.
(iii) An electic current can heat the solenoid coil around 2000-3300K, which stays below the melting point of tungsten.
(iv) Hence tungsten is used to make a slolenoid type coil in an electric bulb.

(4) In practice the unit KWh is used for the measurement of electrical energy, rather than Joule ?
Ans: (i) The amount of electrical energy transfered to n appliance depends on its power and the length of time is switched on.
(ii) The amount of mains electrical energy transfered measured in KWh.
(iii) Electricity meters measure the number of units of electricity used in a home or other building units in KWh are used instead of Joules because a Joule is a very small unit of energy.

(5) why is fuse wire used in electrical circuits?
Ans: (i) Fuse wire basically acts on the principle of electrical resistance and heat generation. The thickness of wire is inversely proportionate to the amount of heat generated by a wire by passing current.
(ii) Fuse wire is used to provide overcurrent protection to an electrical circuit. When there is excess current in the circuit, the fuse wire melts and interupts the current flow, saving the equipment.

(6) Why is earthing necessary in electrical systems ?
Ans: (i) Earthing is the process of creating an alternative path for the flow of excessive current safely into the ground in the presence of minimal resistance or impedance.
(ii) The primary purpose of earthing is to reduce the risk of serious electric shock from Current leaking into reinsulated metal parts of an appliance or other electrical devices.
(iii) Earthing also provides protection from large electrical disturbances like lightning strikes and power surges.
iv) Earthing helps minimize such hazards from occurring. Hence good and permanent efficient earthing is an integral part of any electrical system as it is directly related to the safety of human life as well as the equipment.

Q. 6) Explain the following :

1) What is meant by magnetic effect of electric current? Demonstrate by an experiment.
Ans: (i) When an electric current is passed through a conducting wire a magnetic field is produced around it. This phenomenon is known as magnetic effect of electric current.

(ii) When the plug key is kept open, no current passes through the wire. Then the needle points in the N-S direction i.e. along the earth’s magnetic field.
(iii) The needle points along the earth’s magnetic field i.e. the north pole of the magnet points towards the north direction and the south pole points towards the south direction, making the needle parallel to the wire.
(iv) When the key is pressed, the current flows in the wire in the direction from Ato B (i e. from south to north) and the needle deflects towards the west as shown in fig (a)
(v) When the direction of the current in the wire is reversed by reversing the connections at the terminals of the cell. The north pole of the magnet starts deflecting towards the east as shown in fig (b).

(2) State and explain the right hand thumb rule ?
Ans: If We hold the current carrying conductor in the right hand such that the outstreched thumb denotes the direction of the current flow then the curled fingers denote the direction of the magnetic field.

3) What is a solenoid ? Compare the magnetic field produced by a solenoid with a magnetic field of a bar magnet. Draw neat figures and name various components.
Ans: (i) Solenoid: if a conducting wire (eg: copper, wire) is wound around the cylinder in the form of a helix whose cross-section diameter is less compared to the length, the coil is called the solenoid.
(ii) Each turn of a solenoid acts as a circular loop of a w conducting wire. suppose a constant current I is flowing though a solenoid.
(iii) The magnetic field due to the current I in the solenoid is the sum of the magnetic fields produced by each circular turn.
iv) The magnetic field lies at the centre of the solenoid are approximately straight and parallel to the axis of the solenoid is inside the solenoid the magnetic field is uniform.
(v) A strong magnetic field can be obtained by increasing the current in the solenoid.
(vi) If we increase the number of turns in the solenoid of a given length then the magnetic field also increases.
(vii) Comparision between a current carrying solenoid and a bar magnet :
Similarities :
(i) The magnetic field lies produced by a bar magnet are similar to the magnetic field lies of a current carrying solenoid. Thus a current carrying solenoid ,acts just like a bar magnet.
(ii) When a current carrying solenoid is suspended freely it will come to rest with its direction as North-South direction exactly in the same manner as bar magnet does.
(iii) If we take the current carrying solenoid near the iron fillings it attracts the iron fillings behaving like a bar magnet.
Dissimilarities between a current carrying solenoid and bar magnet:
(i) The strength of the magnetic field produced by the solenoid can be changed by changing the current flowing through it whereas the magnetic field produced by the bar magnet can not be changed
(ii)The direction of the magnetic field produced by the bar magnet can not be reversed whereas we can reverse the direction of the magnetic field produced by the solenoid by reversing the direction of the current.

Q. 4. State the fleming’s left hand rule

(i) Fleming’s left hand rule gives the direction of the force acting on a current carrying conductor placed in magnetic field.
(ii) If we stretch the fore fnger, central finger and the thumb of our left hand are mutually prependicular to each other then if the fore finger denotes the direction of the magnetic field and the central finger denotes the direction of the current then the outstretched. thumb denotes the direction of the force acting on it.

Q. 7. Find the odd one out. Give proper explanation.
(a) Fuse wire, bad conductdr, rubber gloves, generator.
Ans: Generator. Not a safety measure

(b) Voltmeter ammeter, galvanometer, thermometer.
Ans: Thermometer – Not related to electricity

(c) Loudspeaker, electric motor, microphone, bar magnet.
Ans: Bar Magnet – No electricity is used.

(d) Armature coil, brushes, commutator, direct current.
Ans: Direct current, Not a part of electric motar.

Q. 8. Electromagnetic induction means:
a) Charging of an electric conductor
b) Production of magnetic field due to a current flowing through a coil.
c) Generation of a current in a coil due to relative molion between the coil and the magnet.
d) Motion of the coil around the axle in an electh motor.
Ans: (c) Generation of a current in a coil due to relative motion between the coil and the magnet.

Q. 9. Explain the construction and working of the following.
a) Electric motor
Ans: Def :.An electric motor is amachine which converts electrica| energy into mechanical energy.
Principle : The principle of working of an electric motor is that whenever a cuurent carrying conductor is placed in a magnetic field, it experiences a mechanical force. The direction of this force is given by Fleming’s left hand rule.
Construction:

(i) Armature coil:
It consist of a single loop of an insulated copper wire wound on iron core in the form of a rectangle.
(ii) Strong magnetic field:
The rectangular copper loop is placed between 2 pole pieces of a strong magnet such as a horse shoe magnet which provides strong magnetic field.
(iii) Split ring commutator:
It consists of 2 halves of a metallic ring. The 2 ends of a rectangular copper coil are connected to these two halves of the ring. Commutator reverses the directions of current in armature coil.
(iv) Brushes:
2 carbon brushes pressed against the commutator and terminal battery.
(v) Battery:
It is connected across the carbon brushes. It supplies current to the armature coil.
Working:
(i) When current flows though coil, arm AB and CD experiences magnetic force.
(ii) The current flows in the arm AB of the loop from A to B though the carbon brushes E and F. Since the direction of the magnetic field is from north pole to the south pole, according to Fleming left hand rule, arm AB of coil experiences force in downward direction. And arm CD
experiences force in upward direction as the current in the arm CD is in a opposite direction to that of the arm AB.
(iii) Both these forces are equal and opposite forces acting at different positions of a armature constitute a couple
(iv) The couple rotates the coil in clockwise direction. After half rotation, the 2 halves of the split ring R and R come in contact with carbon brushes, B and B respectively and the current start flowing in the direction DCBA.
(v) Now a force is exerted on the arm DC in downward direction and on the arm BA in upward direction and the rectangular loop rotates in the clockwise direction.
(vi) The loop and the axle continues to rotate in the clockwise direction, since the current in the rectangular loop is reversed after every half rotation.
(vii) The coil of an electric motor continues to rotate in the same direction. Hence electrical energy is converted into mechanical energy.

(3) Electric Generator:
There are 2 types of electric generator
(A) AC Generator:
(i) An electric i.e. generator is a machine which converts mechanical energy into alternating electric energy.
ii) Electric generator works according to the Faraday’s law of electromagnetic induction i.e. whenever a conductor moves in a magnetic field, an e.m.f. (electromotive force) gets induced across the conductor if the close path is provided to the conductor, induced e.m.f. causes current to flow in the circuit.
Construction:
(i) Armature coil: It consists of a rectangular coil with number of copper wires wounded over an iron dore. This coil is called armature.
(ii) Strong magnet : A strong permanent magnet is placed and the armature is rotated between the magnets where the magnetic field lies are perpendicular to the axis of the armature.
(iii) lip rings: There are 2 slip rings that are connected to the arms of the armature. They are used to provide movable contact.
(iv) Brushes : 2 carbon brushes B and B are connected to pass current from the armature to the slip ring The current is passed through o load resistance which is connected across the slip rings.
Working:
(i) As armature coil rotates the magnetic force line linked with the coil changes, hence induced e.m.f is produced in the armature coil
(ii) Let the armature ABCD is rotating in a such a way that the arm AB moves upward and CD moves downwards
(iii) By Fleming’s right hand rule, electric current is produced in the arms AB and CD in the direction from A->B and from C->D. Thus the induced current flows in the direction A->B->C->D.
In the external circuit current flows from B to B through the galvanometer
(iv) After the armature turns through 180°, the arm AB takes the place of arm CD and the CD takes the place of arm AB i.e. arm AB moves downwards and arm CD moves upwards, again applying Fleming’s right hand rule, we find that the induced current flows in the direction D->C->B->A.
(v) Hence we find that the direction of induced e.m.f. and current in the external circuit is reversed after the armature has rotated though 180°. Hence the current produced is alternating in nature.

(B) DC generator:
construction:
(i) Armature: A large number of turns of insulated copper wire wound on iron core in rectangular shape forms an armature coil.
(ii) Strong magnet : The armature coil is placed in between & poles (N and S) of a strong magnet. This provides a strong magnetic field.
(iii) split ring commutator: It consists of 2 halves (R and R) of a metallic ring. The ends of the armature coil are connected to these rings.
(iv) Brushes : 2 carbon brushes B1 and B2 are used which are in contact with split rings.
(v) Bulb : The output is shown by the glowing bulb connetcted across the carbon brushes.
Working:
(i) When the coil of DC generator rotates in the magnetic field, potential difference is produced in the coil. This gives rise to the flow of current. This is shown by glowing of the blub inthe above figure.
(ii) In DC generator, the flow of current in the circuit is in the same direction as long as the coil rotates in the magnetic field.
(iii) This is because one brush is always in contact with the arm of the armature moving up and other brush is in contact with the arm of the armature moving downward in the magnetic field.

Q. 10. State the Fleming’s left hand rule ?
Ans: If we stretch the fore finger, central finger and the thumb of our left hand mutually prependicular to each other the if the fore figner denotes the direction of the magnetic field and the central finger denotes the directtion of the current them the outstretched. thumb denotes the direction of the force acting on it.

Q. 11. Explain the concept of electromagnetic induction ?
Ans: (i) We know that when an electric current is passed though a conductor a magnetic field is produced around the conductor.
(ii) Faraday thought that as a magnetic field is produced by an electric current carrying conductor, it should be possible to produce an electric current by the magnetic field.
(iii) He performed number of experiments and on the basis of observation he concluded that whenever the number of magnetic field lies changes, an elctromotive force (e.m.f) is developed between the ends of the conductor and it lasts as long as there is a change in the magnetic field lies.
(iv) This phenomenon is known as electromagnetic induction.

Q. 12. What is galvanometer? Write the working principle of galvanometer ?
Ans: i) A galvanometer is an electromechanical instrument used for detecting and indicating electric current.
ii) A galvanometer works by producing a deflection of a pointer in response to electric current flowing though a coil is a constant magnetic field.
iii) Early galvanometers were not caliberated but their later developments were used as measuring instruments called ammeters to measure the current flowing through an electric circuit.
Working Principle of galvanometer:
i) Galvanometer works on the Principle of conversion of electrical energy into mechanical energy. If a current flows in a magnetic field it experiences a magnetic torque.
ii) If it is fue to rotate under the controlling torque it rotates through an angle proportional to the current flowing through it.

Q. 13. Explain the difference:
AC generator and DC generator
Ans: i) Electromagnetic induction occurs when and electrical wire passes though a changing magnetic field. Generators use electromagnetic induction to convert mechanical energy in to electrical energy.
ii) There are several differences between AC and DC gererators. In an AC, or alternating current generator, the electrical current periodically reverses direction. With a DC or direct current generator, the current flows only in one direction.
iii) In AC generators, the coil through which the current flows is fixed, and the magnet is moving. The magnets north and south pales cause the current to flow in opposite directions, producing an alternating current with the DC generators, the coil through which the current flows rotates in a fixed field.
iv) The two ends of the coil attach to a commutator, different halves of a single rotating split ring. Metal brushes connect these split rings to an external circuit,The commutator balances the charges leaving and returning to the generator, resulting in a current that does not change direction.
v) AC and DC generators serve different purposes. Home typically use AC generators to power small motors and common electrical appliances e.g.: vacuum cleaner, food mixers. DC generators power very large electric motors such as those needed for subway systems

Q.14. State and explain Faraday’s law of inductoin ? Demonstrate with and experiment?
Ans: i) When we switch On or Off the current in the solenoid coil a current is produced in the coil.
ii) The induction of current is also observed, when the current is increased or decreased
iii) Now when the electrical current is flowing though, the solenoid coil and the solenoid coil is displaced with respect to the coil, the current is still produced in the coil.
b) when a current is passing through the solenoid coil and the solenoid coil is displaced laterally with respect to the coil.
When the current in the solenoid coil is switched On or Off.
iv) Even if the solenoid coil is kept stationary a change in Current in the solenoidcoil produces a current in the coil.
v) If the solenoid coil is moved away from the coil, We see a deflection in the Galvanometer (fig C) Also’ the faster is the displacement of the selenoid, larger is the diflection of the galvanometer pointer. If the current in the solenoid coil is changed, a current is produced in the coil or if the solenoidcoil is moved towards the coil, then also a current is produced in the coil.
(c) When a current passing through the solenoid coil and the solenoid coil is displaced longitudinally with respect to coil.
Faraday’s Law of induction:
i) From the above experiment it is understood that, if a current is switched On or Off in the solenoid coil, a current it induced in the coil. If we increase or decrease the current in the solenoid coil, the current is again induced in the coil. The current is also induced if there is a relative motion between the solenoid coil and the coil.
ii) it is understood that whenever there is a change in the magnetic lies of force passing through the coil, curient is induced in the coil. This is known as Faraday’s law of induction. And the current produced in the coil is called the induced current.
Define : Magnetic line of force
The path along which the unit north pole moves in a magnetic field is called magnetic line of force or magnetic field line.

Q.15) State Fleming’s right hand rule ?
Ans: i) it shows the direction of induced current when a conductor attached to a circuit moves in a magnetic field.
ii) The right hand is held with the thumb, index finger and middle finger mutually perpendicular to each other as shown in figure.
iii)If the thumb points in the direction of the conductor realtive to the magnetic field, and the index finger is pointed in the direction of the magnetic field, then the middle fIgner represents the direction of the induced or generated current within the conductor.

Q.16. Define AC Current and DC current ? Write the difference between AC and DC currents.
Ans: i) AC is an electric current which periodically reverses its direction DC Current . The electric current flows in a constant direction, distinguishing it from AC.

Numericals
(a) Heat energy is being produced in a resistance in a circuit at the rate if 1000W = The current of 3A is flowing in the curcuit what must be the value of resistance.
Sol: Given: Eletrical Power = 1000 W
Current = 1 = 3A
Resistance = R = ?

(b) 2 tungsten bulbs of wattage 100W and 60W power work on 220V potential difference. If they are connected in parallel, how much current will flow in the main conductor?
Solv: Given:
As they are connected in parallel.

*C) Who will spend more eletrical energy? 500W TV set in 30mins. or 600W TV heater in 20 min ?
Ans:
(d) An eletric iron of 1100W is operated for 2hrs daily what wire be the eletrical consumption expenses for that month of April? The eletric company charges Rs. 5 per unit of energy
Sol: Power =

Summary:
i) Joule’s heating effect: It is the Process by which the flow of eletric current through the conductor produced heat.
Eletric Power =
ii) Magnetic effect of eletric current: When an eletric current is passed through a conducting wire a magnetic field is produced around it.
iii)Solenoid: If a conducting wire e.g. copper wire is wound around the cylinder in the form of a helix is called solenoid. Solenoid behaves like a bar magnet.
iv) Electromagnetic Induction: It is the phenomenon in which generation of an electric current takes place in the coil due to the relative motion between the coil and the magnet.
v) Electic Motor : An electric motor is a machine which converts electrical energy into mechanical energy. it works on the Principle of Fleming’s left hand rule.
vi) Electric generator : An electric generator is a machine which converts mechanicai energy into an alternating or direct electrical energy.
vii) The path along which the unit north pole moves in , a magnetic field is called magnetic lines of force.

heat

Important points:-
1) Heat: Heat Is the transfer of kinetic energy from one medium or object to another or from an energy to a medium of object.
2) Temperature: Temperature is a measure of hotness or coldness experienced in terms of any of several arbitrary scales like celsius and Fahrenheat Temperature is related to how fast the atoms within a substance are moving.
3) Latent heat: It is defined as the heat required to convert a solid into liquid or vapour or a liquid into vapour without change of temperature.
4) Latent heat of fusion: During the transition of ice to water the ice absorbs heat energy but its temperature does not increase. This heat energy is used in weakening the bonds between the atoms of the Ice hence it converts Ice into water.
5) Latent heat of vapourisatien: The amount of heat energy absorbed at constant temperature for the transformation of liquid into. gas is called the latent heat of vapouhsahon
6) Regelation: Regelation is the phenomenon of melting of iée under pressure and freezing again when the pressure is removed.
7) Anomalous Behaviour of Water: Till 4°C density of water gradually increases as you cool it. When you reach 4°C, its density reaches a maximum water expands with .further drop of temperature means the density of water J decreases when you cool it from 4°C to 0°C . This behaviour of water between 0°C to 4°C is caiied as Anomalous behaviour of water.
8) Humidity: Humidity Is the amount of water vapour present in the air.
9) When the air contains maximum possible water . vapour it is said to be saturated with vapour at that temperature.
10) Dew Point : The dew point is the temperature to which an unsaturated air must be cooled. So that the air becomes saturated with vapour. This temperature is called dew point.
11) Absolute humidity : The amount of vapour present in the air is measured using a physical quantity called absolute humidity.
12) The mass per unit volume of air is called absolute humidity. Absolute humidity is measured in kg/m.
13) When the air comes up against the cold outside of a cooled bottle, the air right next to the bottle gets cold. It gets so cold that it cannot hold as much water anymore or comes out of the air onto the bottle.
14) When water within the cracks of the rocks freezes during winter it expands forcing the rock to break into smaller pieces as water expands below 4°C due to its anomalous nature.

*In this subqestion varlety of objective type question like:-Fill in the blanks, odd man out, complete the following , give formula etc. should be given
Q.1. Fill in the blanks and rewrite the sentence.

a) The amount of water vapour in air is determined in terms of its …………
b) Heat is measured in Joule.
c) I Cal = 4.2 Joule
d) The amount of heat required to raise temperature of a substance by PC is called as heat capaclty
e) The amount of heat required to raise the temperature of 1 Kg by I°C is called as specific heat capacity.
f) SI unit of specific heat capacity is j/kg°C
g) The temperature at which liquid changes into vapour is called as boiling point.
h) If objects of equal masses are given equal heat, their final temperature will be different. This is due to differences in their specific heat capacity.
i)During transformation, of liquid phase to solid phase the latent heat is ………. ?

Q. 2. MCQS:

1) 1 gm of ice at 0°C is mixed with l gm of steam at 100°c After thermal equilibrium, the temperature of the mixture is
a) 0°C b) One c) Infinity d) 100°C

2) Why does the bottom of the lake do not freeze in severe winter even when the surface is all frozen?
a) The water has large specific heat.
b) The conductivity of Ice is low
c) The water has large latent heat of fusion.
d) the temperature of the earth at the bottom of the lake is high.
Ans: (b)

3) Specific heat is:-
a) The specific temperature at which the substance is in solid state.
b) The energy needed to increase the temperature of gram of a substance by IC
c) The amount of heat conducted in l min.
d) The heat needed to increase the temperature of 1 gallon of water by 1 degree Fahrenheit
Ans: (b)

4) As a solid undergoes a phase change to a liquid state, it
a) releases heat while remaining at a constant temperature.
b) absorbs heat while remaining at a consta temperature.
c) releases heat as the temperature decrease.
d) absorbs heat as the temperature increase.
Ans: (b)

5) At what temperature are the Celsius and Fahrenheit equal?
a) 40 b) -40° c) -0° ‘ d) +100
Ans: (b)

6) What would happen to a hole in a metal sheet when the sheet is heated ?
a) It decreases in size
b) It increase in size
c) No change is seen
d) First increases and the decreases
Ans: (b)

7) Compared to warm air cool air can held.
a) more water vapour
b) less water vapour
c) The same amount of water vapour
d) temperature is not important
Ans: (b)

8) Which one among the following statements about thermal conductivity is correct ?
a) steel > wood > water
b) steel > water > wood
c) water > steel > wood
d) water > wood > steel
Ans: (b)

9) Body A of mass 2 kg and another body B of mass 4 kg and of same material are kept in the same sunshine for same interval of time. If the rise in temperature is equal for both the bodies, then which one among the following in this regard is correct?
a) Heat absorbed by B is double because its mass is double.
b) Heat absorbed by A is double because its mass is half.
c) Heat absorbed by both A and B are equal because the quantity of heat absorbed does not depend upon mass.
d)All of the above
Ans: (a)

10) In a pressure cooker cooking is faster because the increase in vapour pressure
a) increase the specific heat.
b) decreases the specific heat.
c) decreases the boiling point.
d) increases the boiling point.
Ans: (d)

Q.No.3) Define the following and explain in Short (2 Mks)

1) Heat
Ans: i) Heat is the transfer of kinetic energy from one medium or object to another or from an energy source to a medium or an object.
ii) such energy transfer can occur in 3 ways radiation, conduction and convection.
iii) Example: A good example of heat transfer is heating an ice cube. Initially the ice cube absorbs heat and gets converted into water if we heat it further it gets converted into steam.

2) Latent heat
Ans: i) it Is defined as the heat required to convert a solid into liquid or vapour or a liquid into vapour without change of temperature.

3) Latent heat of fusion
Ans: i) During the transition of ice to water the ice absorbs heat energy but its temperature does not change. This heat energy is used in weakerning the bonds between the atoms of ice hence it converts ice into water.
ii) Thus, latent heat of fusion can be stated as, the heat energy absorbed during the transition of solid into liquid at constant temperature.

4) specific latent heat of fusion
Ans: i) At constant temperature, the amount of heat energy absorbed by unit mass of a solid substance to convert into liquid phase is called the specific latent heat of fusion.

Q.5) What is meant by Regelation? Demonstrate it by an experiment.
Ans: i) Regelation is the phenomenon of melting of ice under pressure and freezing again when the pressure is removed.
ii) It can be demonstrated by looping a fine wire around a block of ice, with a heavy weight attached to it.
iii) The pressure exerted on the ice slowly melts it locally, permitting the wire to gradually penetrate the ice. In sometime the wire comes out, of the lower surface of the ice slab. The wires track will ,refill as soon as the pressure is relieved. So that the ice block will remain solid even after wire passes completely through it.
iv) The melting point of the ice falls by 0.0072°C for each additional atmosphere pressure applied. As soon as the pressure is removed, the melting point is restored to 0°C and water gets converted into ice again.

Q.6. Explain the anomalous behaviour of water?
Ans: i) A common trend seen in substances through out is the fact that all substances expand when they are heated and their density decreases and vice versa on cooling something in contrast and density increases. This is how substance generally react to heat.
ii) if you cool water it follows the general trend till 4°C. Density of water gradually increases as you cool it. When you reach 4°C, its density reaches a maximum density of water decreases when you cool it from 4°C to 0°C.
iii) This behaviour of water between 0°C to 4°C Is called as Anomalus behaviour of water.

Q.7. Observe the following graph considering the change in volume of water as its temperature is raised from 0°C, discuss the difference in the behaviour of Water and other substances.
Ans: i) The given graph is nothing but the anomalous behaviour of water.
ii) A common trend seen in substances is the fact that all substances expand. When they are heated up hence. That is volume of substances increases on heating and decreases in cooling.
iii) But when water is heated from 0°C to 4°C, its volume decreases and Increases only after 4°C as shown in the given graph.

Q.8. Explain the following :

a) What is the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate.
Ans: i) During extremely cold wheather conditions. the atmospheric temprature falls below 0°C resulting in freezing of lake or river water.
ii) Since water at 4°C is the heaviest due to its anomalous behaviour. this water settles at the bottom of the water body and the lightest is the coolest layer accumulates on the top layer. s0 in extremely cold climate, the top of the water is alway’s the first to freeze over and the dissolved oxygen in water is trapped beneath the frozen layer. This phenomenon is responsibie for the survival of the aquatic life.

b) How can you relate the formation of water droplets on the outer surface of a bottle taken out of refrigerator, with formation of dew?
Ans: i) There is a moisture in the air and this is often called ‘humidity‘.
ii) The amount of moisture that can be in the air depends on the temperature of the air. The warmer the air, the more water it can held in it.
iii) When the air comes up against the cold outside of a cold bottle. the air right next to the bottle gets cold. It gets so cold that it cannot held as much water anymore and some of the water condenses or comes out of the air as water droplets onto the bottle.
iv) That cold air next to the bottle is heavier than the air around it (hot air rises and cold air sinks) so the colder air with some water out of it sinks and fresh water fitted air comes in to gets cooled and leave some more condensation.

c) In cold regions In winter, the rocks crack due to the anomalous expansion of water.
Ans: i) Water sometimes finds its way into cracks within the rocks During winter when such water freezes, it expands due to the anomalous expansion of water forcing the rock to break into smaller pieces. Hence the rocks crack.

d) As we go higher than the sea level the boiling points of water decreases?
Ans: i)A liquid at high pressure has a higher boiling point than when that liquid is at atmospheric pressure.
ii) For a given pressure, different liquids boil at different temperatures.
iii) The boiling point of water depends on the atmospheric perssure at that altitude. At high altitudes, the pressure is lower so water cannot boil at 100°C above the sea level.

Q.9. What is meant by specific heat capacities? How will you prove experimentally that different subtances have different specific heat capacities.
Ans: i)Specific heat is the amount of heat needed to raise the temperature of a unit mass of an object by IC
ii) The specific hear capacity is denoted by ‘C’ The SI unit is J/OCkg and CGS unit is cal /gC.
To prove experimentally that different substances have different specific heat capacities
i) Take 3 spheres of different materials. Let the material be iron, copper and lead.
ii) Put all the 3 spheres in boiling water In the beaker for some time so that all the 3 spheres attains the same temperature as 1000C.
iii) Now put them immediately on a thick slab of wax. The depth that each of the Sphere goes into the wax is different for different substance.
iv) The sphere which absorbs more heat from the water will give more heat to the wax, thus the wax will melt more and that sphere will go deeper in the wax.
v) We see that the iron sphere goes deepest in the wax. The lead sphere goes the least and the copper sphere goes to the middle depth.
vi) This shows that the 2 spheres absorbs different amount of heats for the same rise in temperature. Thus the amount of heat energy absorbed by sphere is different for the 3 spheres. This property is called the specific heat capacity.

Q.10. On what basis and how will you determine whether air is satuated with water vapour or not?
Ans: i) There is a limit on how much water vapour the air can contain for a given volume of air at a specific temperature.
ii) When the air contains maximum possible water vapour, it is said to be saturated with vapour at that temperature.
iii) Temperature affects the amount of water vapour the air can hold If the air temperature is low, less water vapours are needed to saturate it. If the air temperature is more, more water vapour is needed to saturate it. Hence saturation of air depends on Its temperature.

Q.11. Define water droplets.
Ans: For a given volume of an at a specific temperature, if amount of water vapour exceeds the certain value, the excess vapour converts into water droplets.

Q.12. Write the difference between heat and temperature?

Q.13. Is the concept of latent heat applicable during transformation of gasseous phase to liquid phase and from liquid phase to solid phase? Where does the latent heat go during these transfromations.
Ans: i) The concept of latent heat is not applicable during the phase change of gas to liquid and the liquid to solid.
ii) The latent heat during thee transformations is released as thermal energy.
Units of specific latent heat:
The S.I. unit of, specific latent heat is Jkg Other common units are calg-1 and kcal kg-1
They are related as,
1Kcal kg -1 = I cal g-1

Numericals:

Q.1. How much heat energy is necessary to raise the temperature of 5 kg of water from 0C to 100C.
Sol: Given m a 5kg , C = 1 kcal/0kgC
and change in temperature
T = 100 – 20 = 80°C
Energy to be supplied to water = Engery gained by water
= mass of water x specific heat of water x change in temp
= m x c x t
= 5 x 80C
= 400k cal

Q.2. How much heat energy is required to melt 5kg of ice? Specific latent heat of ice = 33 6 J g
Sol: Given: m = 5kg = 5000 gm
L = 336 J g
heat energy required = ML
= 5000 x 336 J g
= 168000 J
= 1.68 x 10 J

refraction of light

Important points:
1) Light changes its direction when going from one transparent medium to another transparent medium. This is called the refraction of light.
2) Laws of refraction
(i) The incident ray, refracted ray and normal at the point of incidence are in the same plane.
(ii) The incident ray and refracted ray are on the opposite sides of the normal.
(iii) For a given pair of media, the ratio of sine of angle of incidence and sine of angle of refraction is constant. The ratio is called as the refractive index of the second medium with respect to the first medium.
3) When light travels from rarer medium, it bends toward the normal end when it travels from denser medium to rarer medium it bends away from the normal.
4) Refractive index depends on the velocities of light in the different media.
6) Twinkling of stars, mirage, seeing the sun before the sunrise, seeing the sun after sunset are some examples of refraction seen in the nature.
7) The process of separation of light into its component colours while passing through a medium is called the dispersion of light
8) When light is travelling from a denser medium to a rarer medium the value of angle of incidence for which angle of refraction is 90°, is called the critical angle.
9) Rainbow is seen due to the refraction, dispersion and total internal reflection of light.

Q.1.(A) Fill in the blanks :

*(i) Refractive index depends on the …… of tight.
*(ii) The change in …… of light rays while going from one medium to another is called refraction.
Ans: (i) Velocity (ii) direction

Q.1(B) Choose the correct answer in the following questions:
*(i) What is the reason for the twinkling of stars?
(a) Explosion occurring in stars from time to time.
(b) Absorption of light in the earth’s atmosphere.
(c) Motion of stars.
(d) Changing refractive index of the atmospheric gases.

*(ii) We can see the Sun even when it is little below the horizon because of
(a) Reflection of light (b) Refraction of light
(c) Dispersion of light (d) Absorption of light

*(iii) If the refractive Index of glass with respect to air is 3/2, what is the refractive index of air with respect to glass?
(a) 1/2 (b) 3 (c) 1/3 (d) 2/3

(iv) Refraction of light means
(a) falling on a surface and returning back from it.
(b) splitting into seven colours.
(c) changing its direction when going from one transparent medium to another transparent medium.
(d) travelling in a straight line.

(v) If a ray is incident on a surface at an angle of 90° to the surface, the angle of Incidence is —– .
(a) 90° (b) 0° (c) 45° (d) 270°

Ans: (i) (d); (ii) (b); (iii) (d); (iv) (c); (v) (b)

Q.1.(C) Answer the followmg questions:

*(i) What is meant by reflection of light?
Ans: When light falls on a surface it returns back from me surface. This is called the reflection of light.

*(ii) Will light travel through a glass slab with the same velocity as it travels in air?
Ans: No. As the optical density of glass is more than that of air, light will travel through a glass slab with the slower velocity than that in air.

*(iii) Will the velocity if light be same in all media?
Ans: No. The velocity of light in a media is dependent on the optical density of the media. More the optical density, slower is the speed.

(iv) What is meant by dispersion of light?
Ans: The process of separation of light into its component colours while passing through a medium is called the dispersion of light.

*(v) If a beam of LED light is passed through a prism, will we be able to see the spectrum?
Ans: If the LED light is not monochromatic and is white then we will be able to see the spectrum. But, the spectrum will be very compact.

Q.2.(A) Solve the following examples:

*(i) If the speed of light in a medium is 1.5 x 10 m/s, What is the absolute refractive index of the medium?
Ans: Let v1 be the velocity of light in vacuum. Let v2 bethe velocity is the medium.
Tje absp;ite refractive index of the medium is 2.

*(ii) If the absolute refractive indices of glass and water are 3/2 and 4/3 respectively, what is the refractive index of glass with respect to water?
Ans: Let refractive index of glass with respect to vaccum be n.
Let refractive index of water with respect to vaccum be n.
Let the refractive index of glass with respect to water be n.
If v, v and v are the velocities of light in vaccum, glass and water respectively, then –
n = v/v
3/2 = v/v —(1)
n = v/v
4/3 = v/v —(2)
Dividing eq. (1) by eq. (2)
(3/2)/(4/3) = (v/v)/(v/v)
9/8 = v/v
But v/v = n
v/v = n = 9/8
The refractive index of glass with respect to water is 9/8.

(iii) The absolute refractive index of a medium is 1.5. find the velocity of light in that medium.
Ans: Let v be the velocity of light in the vacuum and v be the velocity in the medium.
If n is the absolute refractive index of the medium, then –
n = v/v
v = v/n
v = (3 x 10)/1.5
v = 2 x 10 m/s
The velocity of light in the medium is 2 x 10 m/s.

(iv) A light ray passes from air to a transparent medium. If the angle of incidence is 30° and angle of reflection is 24°. What is the refractive index of the medium with respect to air? (Take sin 24° = 0.4)
Ans: Refractive index of medium = sin i / sin r
Refractive index of medium = sin 30°/sin 24° = 0.5/0.4 = 1.25
The refractive index of the medium with respect to is 1.25.

Q.2.(B) Answer the following questions:

*(i) Have you seen that objects beyond and above a boil fire appear to be shaking? Why does this happen?
Ans: (1) Due to holi fire, the air above the fire becomes hot and its density is reduced.
(2) Thus, due to changing density, its refractive index also keeps changing.
(3) So the light starting from the objects beyond the holi fire passes through the air of changing refractive index. Its extent of refraction also keeps on changing. Therefore, the objects appear to be shaking.

*(ii) From incident white light how will you obtain white emergent light by making use of two prisms?
Ans: (1)The prisms are kept one after another as shown in the following figure.
(2) Due to the first prism, the white light is dispersed into seven colours.
(3) Due to the second prism, the seven colours combine to give white emergent light.

(iii) What are the laws of refraction?
Ans: (1) incident ray and refracted ray at the point of incidence are on the opposite sides of the normal to the surface.
(2) For a given pair of media, the ratio of sin i to sin r is constant. This ratio is called as the refractive index of the second medium with respect to the first medium.

(v) Why do stars twinkle?
Ans: (1) Stars are at a very large distance from us. So, they appear to be point sources.
(2) Due to the changing density of the atmosphere, the light coming from stars refracts and reaches the observer.
(3) Because of refraction, the apparent position of stars and brightness of stars keeps on continuously changing. Therefore, the stars appear to be twinkling.

Q.3.*(A) Prove the following statements.

a. if the angle of incidence and angle of emergence of a light ray falling on a glass slab are i and e respectively. Prove that i = e.
Ans:

In the above figure, AB is the incident ray, BC is the refracted ray and CD is the emergent ray.
MN and PQ are the normals.
ABM = angle of incidence for the first refraction = i
NBC = angle of refraction for the first refraction = r
BCP= angle of incidence for the second refraction = r (Because NBC and BCP are alternate angles)
QPD= angle of emergence= e
Let velocity of light in air be v and that in glass be v n = v/v and n = v/v
n = 1/n —(1)
Also, n = sin i / sin r and n = sin r/sin e
sin i/sin = 1/ (sin r)/(sin e) .[from (1)]
sin i/sin e = sin e/sin r
sin i = sin e
i = e

b. A rainbow is the, combined effect of the refraction, dispersion and total internal reflection of light.
Ans:
(1) Here small droplets act as prisms.
(2) When sunlight passes through these drops, light gets refracted and dispersed first.
(3) Then it gets totally internally reflected as shown in the figure.
(4) Then it again refracts while coming out of the drop and different colours are seen Thus a rainbow is the combined effect of the refraction, dispersion and total internal reflection.

Q.3.(B) Answer the following questions:

*a. What are the laws of reflection?
Ans: (1) The incident ray, the normal at point of incidence and the reflected ray are in one plane.
(2) The incident ray and the reflected ray are on the opposite sides of the normal.
(3) The angle of incidence and the angle of reflection are equal.

b. Observe the following figure and answer the questions that follow.

(i)What is this phenomenon called?
Ans: This phenomenon is called as mirage.

(ii) Which property of light is responsible for this Phenomenon?
Ans: Refraction of light is responsible for this phenomenon.

(iii) When and where this phenomenon is seen?
Ans: It is seen on the hot surfaces like hot roads or hot deserts when the refractive indices of various layers of atmosphere are different due to the changing temperature.

c. Observe the following figure and answer the questions that follow.

(i) Which phenomenon is this?
Ans: We see the sun before actual sunrise or we continue to see the sun for some time after the actual sun set.

(ii) Which property of light is responsible for this phenomenon?
Ans: Refraction of light is responsible for this phenomen.

(iii) Fill in the blank boxes with suitable tabels.
Ans: (1) Apparent position of the sun
(2) Horizon
(3) Real position of the sun.

Q.4.(A) Observe the following figure and answer questions that follow.

(i) Which phehomenon of light is demonstrated in the figure?
Ans: Total internal reflection.

(ii) What is angle i called?
Ans: Angle i is called as the critical angle.

(iii) Define the angle mentioned above.
Ans: Wheh the light is going from a denser medium to a rarer medium, for a particular value of angle of incidence, the angle of refraction is 90°. This value of angle of incidence called critical angle.

(iv) Write two conditions under which the above phenomenon happens.
Ans: (i) Light must be travelling from denser medium to rarer medium
(ii) The angle of incidence must be greater than critical angle.

(B) Read the following passage and answer the questions given below.

Sir Isaac Newton was the first person to use a glass prism to obtain Sun’s spectrum. When white light is incident on the prism, different colours bend through different angles. Among the seven colours, red bends the least while violet bends the most. Seven colours emerge along different paths and get separated and we get a spectrum of seven colours.

(i) Write the seven colours in the order of increasing refractive index.
Ans: Red, orange, yellow, green, blue, indigo and violet.

(ii) What is this process of separating the colours called?
Ans: Dispersion of light.

(iii) What is the band of seven colours obtained called?
Ans: A spectrum.

(iv) Do we get such a band of seven colours when a beam of light passes through a rectangular glass slab?
Ans: No.

(v) Give an example of a natural phenomenon based on the above property.
Ans: A rainbow.

lenses

Important points:

1) Light rays parallel to the principal axis failing on a convex lens come together at focus i.e. they converge, so convex lens is also called as converging lens.
2) Light rays parallel to the principal axis falling on a concave lens 90 away from one another after refraction in such a way that they appear to be coming from the focus of the lens. So, concave lens is also called as divergent lens or diverging lens.
3) When the incident ray is parallel to the principal axis of a convex lens, the refracted ray passes through the principal focus.
4) When the incident ray passes through the principal focus of a convex lens, the refracted ray goes parallel to the principal axis.
5) When the incident ray passes through the optical centre of a convex lens it goes undeviated.
6) A convex lens can give real as well as virtual images depending upon the position of the object whereas a concave lens always gives virtual and diminished images irrespective of the position of the object.
7) Following are the positions and nature of the images formed by a convex lens.

8) When the incident ray is parallel to the principal axis of a concave lens, the refracted ray when extended backwards, passes through the focus.
9) When the incident ray passes through the focus of a concave lens, the refracted ray goes parallel to the principal axis.
10) When the incident ray passes through the optical centre of a concave lens it goes undeviated.
11) As per the Cartesian sign convention:
(1) Object is always placed to the left of the lens and above the principal axis.
(ii) The distances to the left of the lens are considered to be negative and the distances to the right are considered to be positive.
(iii) Heights measured above the principal axis are taken to be positive and the heights measured below the principal axis are taken to be negative.
(iv) Focal length of a convex lens is positive while that of a concave lens is negative.
12) Lens formula is : 1/v – 1/u = 1/f
13) Magnification is the ratio of the height of the image to the height of the object.
M = h/h
14) Power of a lens is defined as the reciprocal (multiplicative inverse) of the focal length measured in metres. The unit of power is Dioptre(D). More the convergence or divergence more is the power. For a convex lens power is positive while for a concave lens power is negative.
15) The power P of a combination of two lenses of powers P and P2 is given by the formula:
P = P + P
16) Cornea, iris, pupil, eye lens. ciliary muscles, retina are some of the important parts of human eye.
17) The capacity of the lens to change its focal length as per need is called its power of accommodation.
18) In the defect nearsightedness or myopia, the person cannot see the distant objects clearly but can see the nearby objects clearly. The reasons for myopia can be
(i) The curvature of cornea and the eye lens is more
(ii) As the ciliary muscles do not relax sufticientty, the converging power of the eye lens remains high
(iii) The eye ball is elongated so the image of distant object is formed in front of retina.
19) Myopia is corrected using spectacles of concave lenses of suitable power.
20) In the defect farsightedness or hypermetropia, the person cannot see the nearby objects clearly but can see the distant objects clearly. The reasons for hypermetropia can be
(i) The curvature of cornea and the eye lens is less
(ii) As the ciliary muscles are weak, the converging power of the eye lens remains lower than required
(iii) The eye ball is flattened so the image of distant object is formed behind the retina.
21) Hypermetropia is corrected using spectacles of convex lenses of suitable power.
22) The minimum distance of an object from a normal eye, at which it is clearly visible without stress on the eye is called as minimum distance of distinct vision. The position of the object at this distance is called the near point of the eye. For a normal eye the near point is at 25 cm.
23) The farthest distance of an object from a normal eye, at which it is clearly visible without stress on the eye is called as farthest distance of distinct vision. The position of the object at this distance is called the far point of the eye.
24) In old age, the near point of the eye shifts farther. So the person cannot see the nearby objects clearly. This is called as presbyopia. If the person also suffers from the myopia, then he needs spectacles of bifocal lenses. In bifocal lenses, the upper part is concave and the lower part is convex.
25) Concave lenses are used in medical equipment, scanner, CD player, peep hole in door, spectacles, torches, camera, telescope, microscope.
26) Convex lenses are used in simple microscopes, compound microscopes, telescope, spectacles, camera, projector, spectrograph.
27) Persistence of vision when we see an object, the eye lens creates its image on retina. If the object is taken away, the image remains imprinted on our retina for 1/ 16th of a second after the object is removed. That means, the sensation on retina persists for a while. This is called the persistence of vision.
28) Retina is made of rod shaped cells and cone Shaped cells. Rod shaped cells give information about the brightness or dimness of the object to the brain while conical cells give information about the colour of the object.

0.1.(A) Fill in the blanks with the correct option and rewrite the statements.

(i) If the object is placed between F1 and 2F, of a convex lens, the nature of the image will be ……
(a) larger and erect (b) smaller and erect
(c) larger and inverted (d) smaller and inverted

(ii) The focal length of a convex lens is 25 cm. Its power will be …… Dioptre.
(a) 0.25 (b)4 (c) 0.04 (d) -4
(iii) Which of the following cannot be the magnification of a convex lens?
(a) 0.6 (b) -0.6 (c) -1.2 (d) 1.5

(iv) The hole which controls the light entering the eye is called …….
(a) Iris (b) pupil (c) cornea (d) retina

(v) The persons having myopia use the spectacles of …… lenses.
(a) concave (b) convex (c) biconvex (d) bifocal

Ans: (i) (c) larger and inverted; (ii) (b)4; (iii) (a) 0.6; (iv) (b) pupil; (v) (a) concave

Q.1.*(B) Match the columns in the following table.

Ans:

Q.1.(C) Answer the following questions.

*(i) What are real and virtual images?
Ans: The image obtained by a lens when the refracted rays actually meet at a point, is called a real image. The image obtained by a lens when the refracted rays appear to be coming from a point, is called a virtual image.

*(ii) How will you find out whether an image is real or virtual?
Ans: (1) In case of mirrors, the images obtained in front of the mirror are real and the images obtained behind the mirror are virtual.
(2) in case of lenses, the images obtained on the Opposite side of the lens as the object are real and the images obtained on the same side of the lens as the object are virtual.

*(iii) Can a virtual image be obtained on a screen?
Ans: No a virtual image cannot be obtained on a screen.

(iv) Write the lens formula.
Ans: Lensformula:1/v – 1/u = 1/f

(v) Define magnification.
Ans: The ratio of the height of the image (h2) to the height of the object (h) is called the magnification.
]
Magnification = M = h2/h1

(vi) What is the power of a lens?
Ans: The capacity of a lens to converge, or diverge incident rays is called its power (P). Power is the inverse of focal length where the focal length is in meters.
P = 1/f(m)

(vii) What is the unit of power of a lens?
Ans: The unit of the power of a lens is Dioptre.

(viii) Write the relation between the focal length of combination of two lenses (f) and the focal lengths of the individual lenses (f1 and f2).
Ans: 1/f = 1/f1 = 1/f2

Q.2.(A) Give scientific reasons.

*(i) Simple microscope is used for watch repairs.
Ans: (1) If an object is kept within the focal length of a convex lens, a magnified, virtual and erect image is obtained.
(2) If the focal length is shorter the magnification is more. Such a convex lens of small focal length is called a simple microscope. It can give a magnification up to about 20 times.
(3) For watch repairs, its very small parts need to be seen clearly and so, need to be magnified.
Therefore a simple microscope is used for watch repairs.

*(ii) One can sense colours only in bright light.
Ans: (1) The retina is made up of many light sensitive cells.
(2) They are of two types – (i) rod shaped and (ii) cone shaped.
(3) Conical cells respond to different colours and give information about the colours to the brain.
(4) The response of conical cells to the colours is weak in the dim light. Therefore one can sense colours only in bright light.

*(iii) We cannot clearly see an object kept at a distance less than 25 cm from the eye.
Ans: (1) When we see the distant objects, our eye lens becomes flatter and its focal length increases so that the image of the object is formed on the retina.
(2) While seeing the nearby objects the eye lens becomes rounded and its focal length decreases.
(3) This capacity of eye lens to change its focal length according to need is called as power of accommodation.
(4) However, there is a limit to this capacity and if the object is at a distance less than 25 cm from the eye, the eye lens cannot reduce its focal length to the required value.
(5) As a result the clear image of the object cannot be formed on the retina. Therefore we cannot clearly see an object kept at a distance less that 25 cm from the eye.

*(iv) If we rotate fast a burning incense stick along a circle, we see a continuous circle.
Ans: (1) We see an object because its image is formed on the retina. As long as image is there on the retina, we see the object.
(2) When the object is taken away from our sight, the image does not disappear instantaneously. but it lasts on the retina for 1/16th of a second. This is called as the persistence of the vision.
(3) When we rotate the burning incense stick, the position of the object changes very fast the image of the next position is formed before the image of the earlier position disappears. Therefore, we see a continuous circle.

(v) The image formed by a concave lens is always virtual.
Ans: (1)A real image is formed when the rays refracted by a lens meet again at a point.
(2) A concave lens is a diverging lens meaning rays go away from each other after refraction. So, they cannot meet at a point and real image cannot be formed. Therefore, the image formed by a concave lens is always virtual.

Q.2.(B) Solve the following examples
*(i) Doctor has prescribed a lens having power +1.5D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?
Ans:

Focal length of the lens = +67 cm
Focal length of a convex lens is positive. So, the type of the lens is convex. The defect of vision is hypermetropia (farsightedness).

*(ii) 5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.
Ans: Here, u = -25 cm, f = 10 com, v = ?
By lens formula: 1/v – 1/u = 1/f
1/v – 1/-25 = 1/10
1/v = 1/10 – 1/25
1/v = 5-2/50
1/v = 3/50
v = 16.7cm
Now, v/u = h2/h1
16.7/-25 = h2/5
h2 = -3.3cm
Since, the height of image is negative, it is inverted and real.
Image is at a distance of 16.7cm from the lens, size of image = 3.3 com and image is real.

Q.2.(C) Answer the following questions.

*(i) What is the function of iris and the muscles connected to the lens in human eye?
Ans: (1) Iris controls the opening of pupil so that light enters to the eye in desired amount.
(2) The muscles connected to the lens are called biliary muscles and their function is to change the shape of the eye lens and hence to change its focal length as per requirement.

*(ii) How are concave and convex mirrors constructed?
Ans: (1) A hollow glass sphere is cut into two unequal parts in a plane.
(2) The smaller part is converted into a mirror.
(3) If the outer surface of the part is coated with a shiny material, the concave mirror is obtained.
(4) if the inner surface of the part is coated with a shiny material, the convex mirror is obtained.

*(iii) What is the cartesian sign convention for spherical mirrors?
Ans: (1) The pole of the mirror is taken as the origin.
(2) The principal axis is taken as the X-axis of the frame of reference.
(3) The object is always kept on the left of the mirror. Ali distances parallel to the principal axis are measured from the pole of the mirror.
(4) All distances measured towards the right of the pole are taken to be positive, while those measured towards the left are taken to be negative.
(5) Distances measured vertically upwards from the principal axis are taken to be positive.
(6) Distances measured vertically downwards from the principal axis are taken to be negative.
(7) The focal length of a concave mirror is negative while that of a convex mirror is positive.

*(iv) Why do we have to bring a small object near the eyes in order to see it clearly?
Ans: (1) The apparent size of the object we see, depends on the angle subtended by the object at the eye
(2) if the angle subtended with the eye is bigger we see the object bigger.
(3) As the object comes closer and closer to the eye, the angle subtended with the eye increases.
Therefore, we have to bring a small object near the eyes to see it clearly.

*(v) If we bring an object closer than 25 cm from the eyes, why can we not see it clearly even though it subtends a bigger angle at the eye.
Ans: (1) For apparent size of the object to be bigger, it has to be closer to the eyes. But for seeing the object clearly, a clear image has to be formed on retina.
(2) For having clear image on retina it is necessary that the object should be at a distance greater than minimum distance of distinct vision which is 25 cm.
(3) If we bring the object closer than 25 cm from the eyes, the eye lens cannot sufficiently bulge to form a clear image on retina and we cannot see the object clearly.

Q.3.(A) Answer the following questions.

*(i) Draw a figure explaining various terms related to a lens.
Ans:

Centre of curvature – The centres of spheres whose parts form surfaces of the lenses are called centres of curvature. (C1 and C2)
Radius of curvature – The radii of the spheres whose parts form surfaces of the lenses are called radii of curvature. (R and R2)
Principal axis – The imaginary line passing through both centres of curvature is called the principal axis (AB)
Optical centre – The point inside a lens on the principal axis through which light rays pass without changing their path is called the optical centre. (O)
Principal focus – When light rays parallel to the principal axis are incident on a convex lens, they converge to a point on the principal axis. This is called as principal focus. (F and F2)
In case of a concave lens, the light rays parallel to the principal axis which are incident on the concave lens diverge in such a way that they appear to be coming from a point. This point is called as principal focus.
Focal length – The distance between the optical centre and the principal focus is called the focal length of the lens. (OF1 and OF2)

*(ii) At which position will you keep an object in front of a convex lens so as to get a real image of the same size as the object? Draw the figure.
Ans: (1) The object will have to be kept at 2F1 to et a real image of the same size as the object.
(2) Diagram:-

Object – On 2F1
Image – On 2F2

*(iii) Explain the working of an astronomical telescope using refraction of light.
Ans: (1) The astronomical telescope has a larger convex lens which collects the maximum light coming from the heavenly objects. This is called the objective.
(2) Objective forms a real image at its focus as the heavenly objects are at very large distance. This image acts as an object for another convex lens of smaller focal length which is called eye piece.
(3) The distance between the objective and eye piece can be adjusted so that a clear image is obtained.
(4) This distance is adjusted such a way that the real image formed by the objective falls within the focal length of eye piece. It creates a magnified virtual image of the real image.
(5) Thus image of the heavenly object can be seen through the eye piece.

(v) Which are the three rules to be followed to draw a ray diagram of images obtained by convex lenses.
Ans: (1) When the incident ray is parallel to the principal axis, the refracted ray passes through the principal focus.
(2) When the incindent ray passes through the principal focus, the refracted ray is parallel to the principal axis.
(3) When the incident ray passes through the optical centre of the lens, it passes without changing its direction.

(vi) What is presbyopia? How is it corrected?
Ans: (1) Generally the focusing power of the eye lens decreases with age.
(2) The ciliary muscles lose their ability to change the focal length of the lens.
(3) The near point of the eye lens shifts farther from the eye.
(4) Due to all these, old people cannot see the nearby objects clearly. This is called as preSbyopia. (5) If a person having presbyopia is also having myopia or nearsightedness, then he has to use the spectacles of bifocal lenses whose upper part is a concave lens which corrects myopia and the lower part is a convex lens which corrects presbyopia.

(v) Write uses of concave, lenses.
Ans: (1) In medical equipments, scanner and CD player
(2) For peep hole in door
(3) For making spectacles to correct myopia
(4) In torches, to spread the light
(5) In camera, telescope and microscope concave lens is used in front of eye piece to get a good quality image.

(vi) Write uses of convex lenses.
Ans: (1) A convex lens giving magnification upto 20 is used as simple microscope
(2) In compound microscopes, objectives and eye pieces are made up of convex lenses
(3) In retracting telescopes, objectives and eye pieces are made up of convex lenses.

*(vii) How do we perceive different colours?
Ans: (1) The retina is made up of many light sensitive cells.
(2) They are of two types-(i) rod shaped and (ii) cone shaped.
(3) Cone shaped cells respond to the colour and give information about the colour of the object to the brain.
(4) The brain processes all the information received and we see the actual image of the coloured object.
(5) Cone shaped cells respond to the colours in bright light only. So, we can perceive the colours in the bright lights only.

Q.3.(B) Distinguish between:

*(i) Farsightedness and Nearsightedness
Ans:

*(ii) Concave lens and Convex lens.
Ans:

Q.4.(A) Answer the following questions:

(i) Complete the following table of images formed by convex lenses for different positions of the object.

Ans:

(ii) Observe the following figure and name the numbered parts and write their functions.

Ans: (1) Cornea – To refract the incident light.
(2) Iris – To control the opening of pupil.
(3) Eye lens – To form the image of the object on retina.
(4) Ciliary muscles – Change the focal length of eye lens.
(5) Retina – When an image is formed on retina, electric signals are generated and sent to the brain through optic nerves and the object is perceived.

(iii) Observe the following figure and answer the questions given below.

a. Which defect of vision is indicated in the figure?
Ans: The defect is Nearsightedness or myopia.

b. What are the possibie reasons for this defect?
Ans: There are three possible reasons for this defect:
(1) The curvature of the cornea and the eye lens is more than required.
(2) The ciliary muscles do not relax sufficiently, so the converging power of eye lens remains higher.
(3) The eye ball is elongated so that the distance between the lens and the retina is more.

c. How is the.above defect corrected?
Ans: The defect of nearsightedness is corrected by using spectacles of concave lenses of suitable power.

(iv) Observe the following figure and answer the questions given below.

a. Which defect of vision is indicated in the figure?
Ans: The defect is Farsightedness or hypermetmpia.

b. What are the possible reasons for this defect?
Ans: There are three possible reasons for this defect:
(1) The curvature ofthe cornea and the eye lens is less than required.
(2) The ciliary muscles do not contract sufficiently so the converging power of eye lens remains less.
(3) The eye ball is flattened so that the distance between the lens and the retina is less.

c. How is the above defect corrected?
Ans: The defect of farsightedness is corrected by using spectacles of convex lenses of suitable power.

metallurgy

Important points:

1) Metals have lustre. They are ductile and malleable. They are good conductors of electricity and heat.
2) Non-metals do not have above properties in general.
3) Reactive metals combine with oxygen to form their oxides. In general, oxides of metals are basic. Oxides of aluminium and zinc are amphoteric.
4) Metals react with water/ hot water/ steam depending upon their reactivity. In the reaction hydrogen gas is released.
5) The reactive metals react with acids to form the respective salts and hydrogen gas is evolved.
6) Nitric acid being oxidising, it forms oxides of nitrogen when treated with copper and no hydrogen is evolved.
7) More reactive metal displaces less reactive metal from its salt.
8) According to reactivity series the reactivity of common metals in decreasing order is as follows:-
K, Na, Li, Ca, Mg, Al, Zn, Fe, Sn, Pb, Cu, Hg, Ag, Au,
9) Metals form Ionic compounds with non-metals.
10) Non-metals combine with oxygen to form their oxides. The oxides of non-metals are acidic.
11) Metallurgy – The science and technology regarding the extraction of metals from ores and their purification for the use is called metallurgy.
12) Ionic compounds are electrically neutral. They are crystalline. They are hard and brittle and their melting point and boiling point is high. They are generally water soluble.
13) Ionic compounds do not conduct electricity in their solid state but they conduct electricity in their aqueous solution or in their fused state
14) Minerals – The compounds of metals that occur in nature along with the impurities are called minerals.
15) Ores – The minerals from which the metal can be separated economically are called ores
16) Concentration of ores is done by the methods of separation based on gravitation magnetic separation, froth floatation or by teaching depending upon the physical and chemical properties of ores.
17) Extraction of reactive metals is done by electrolytic reduction. For moderately reactive metals, first oxide is obtained by roasting or calcination method and then metal is obtained by reduction. Less reactive metals are mostly available in free state.
18) Concentration of bauxite ore is done by either Bayef‘s method or by Hall’s method Alumina is then electrotytically reduced to get aluminium metal.
19) Metals are puritied electrotytically.
20) Due to various gases present in air like oxygen, carbon dioxide or hydrogen sulphide, metals form various compounds on their surface in the presence of moisture. This is called corrosion.
21) Painting, greasing, varnishing, galvanizing, tinning, anodizatiOn, electroplating, alloying are some of the processes used to prevent corrosion.
22) If one of the metals in an alloy is mercury, the alloy is called amalgam, e.g. the alloy of silver and mercury is called silver amalgam.

*Q.1.(A) Name the following.

a. Alloy of sodium with mercury.
b. Molecular formula of the common ore of aluminium.
c. The oxide that forms salt and water by reacting with both acid and base.
d. The device used for grinding an ore.
e. The non-metal having electrical conductivity.
f. The reagent that dissolves noble metals.
Ans: (a) Sodium amalgam
(b) AI2O3.nH2O
(c) Amphoteric oxide
(d) Ball mill
(9) Graphite (Carbon)
(f) Aqua regia

*Q.1.(B) Make pairs of substances and their properties.

Ans: (a) 2 (b) 4 (c) 1 (d) 3

*Q.1.(C) Identify the pairs of metals and their ores from the following.

Ans: (a) ii (b) iii (c) i

Q.1.(D) Complete the following statement using correct options:
(i) Metal ……. exists in liquid state at room temperature.
(a) Sodium (b) Potassium (c)Calcium (d) Gallium

(ii) Metal ……. has the highest melting point.
(a) Tungsten (b) Sodium (c) Mercury (d) Gallium

(iii) Metals on reaction with dilute sulphuric acid liberate ……. gas.
(a) CO2 (b) H2 (c) SO2 (d) O2

(iv) Which of the following options have the corfect increasing order of reactivity of metals?
(a) K, Li, Al, Fe (b) K, Al, Fe, Li (c) Fe, K, Li, Al (d) Fe, Al, Li, K

(v) In galvanizing, a thin coat of metal ……. is applied on iron or steel.
(a) Tin (b) Chromium (c) Zinc (d) Nickel

Ans: (i) Gallium (ii) Tungsten (iii) H2 (iv) Fe, Al, Li, K (v) Zinc

Q.1.(E) Answer the following questions:

*(i) What is the electronic definition of oxidation and reduction?
Ans: Losing electron/s is oxidation and gaining electron/s is reduction.

*(ii) What are the moderately reactive metals?
Ans: The metals in the middle of the reactivity series are moderately reactive metals. Examples are Zinc, Iron, Tin, Lead, copper etc.

*(iii) In which form do the moderately reactive metals occur in the nature?
Ans: The moderately reactive metals occur in the nature in the form of their sulphides or carbonates.

(iv) What is the difference between roasting and calcination?
Ans: Strongly heating in the presence of sufficient air is called roasting, where as strongly heating in the presence of limited air is called calcination.

*(v) What is meant by corrosion?
Ans: Due to various components of atmosphere, either oxidation of metals takes place or some other salts of metals are produced, consequently resulting into their damage this is called corrosion.

*(vi) Can we permanently prevent the rusting of an iron article by applying a layer of paint on its surface?
Ans: No. If there is a scratch on the painting, that part of Iron gets exposed to air and it starts corroding.

*Q.1.(F) Write chemical equation for the following events.

a. Aluminium came in contact with air.
Ans: 4Al + 3O2 –> 2Al2O3

b. Iron filings are dropped in aqueous solution of ccpper sulphate.
Ans: Fe + CuSO –> FeSO4 + Cu

c. A reaction was brought about between ferric oxide and aluminium.
Ans: 2AI + Fe2O3 –> AI2O3 + 2Fe

d. Electrolysis of alumina is done.
Ans: At anode: 2O2 –> O2 + 4e
At Cathode: Al3 + 3e –> Al
Net reaction: 2Al2O3 –> 4Al + 3O2

e. Zinc oxide is dissolved in dilute hydrochloric acid.
Ans: ZnO + 2HCl –> ZnCl2 + H2O

*Q.2.(A) Explain the terms.

a. Metallurgy b. Ores c. Minerals d. Gangue
Ans: (a) Metallurgy – The science and technology regarding the extraction of metals from ores and their purification for the use is called metallurgy. It also deals with the properties of metals and their alloys.
(b) Ores – Most of the metals are available in the nature in the form of their compounds. These compounds along with the impurities are called as minerals. The minerals from which the metal can be separated economically are called ores. All ores are minerals but all minerals may not be ores.
(c) Minerals Most of the metals are available In the nature in the form of their compounds like oxides, carbonates, sulphides, nitrates etc. These compounds along with the impurities are called as minerals.
(d) Gangue – Ores of metals contain many types of impurities such as soil, sand and rocky material along with the compounds of metals. These impurities collectively are called as gangue.

*Q.2.(B)Give scientific reasons:

a. Lemon or tamarind is used for cleaning copper vessels turned greenish.
Ans: (1) The copper vessels turn green due to the formation of a layer of copper carbonate on their surface
(2) Lemon or tamarind have acids in them (citric acid and tartaric acid respectively)
(3) These acids react with copper carbonate and the greenish layer is removed. Therefore lemon or tamarind is used for cleaning copper vessels turned greenish.

b. Generally the ionic compounds have high melting points.
Ans: (1) During the conversion from solid state to liquid state, intermolecular attractive forces are to be broken.
(2) Ionic compounds have these intermolecular attractive forces very strong. So, it requires considerable amount of energy (heat) to break them.Therefore, ionic compounds have high melting points.

c. Sodium is always kept in kerosene.
Ans: (1) Sodium is a very reactive metal.
(2) It combines vigorously with oxygen from air at room temperature and burns to form sodium oxide. Hence it cannot be exposed to air.
(3) It reacts violently with water to produce hydrogen gas, which may be ignited by the heat of the reaction. It is also lighter than water. Due to these reasons, it cannot be kept under water also.
(4) Sodium doesnot react with kerosene and sinks in it, Therefore to avoid reaction of sodium with oxygen or water vapour in the air it is always kept in kerosene.

d. Pine oil is used in froth floatation.
Ans: (1) Froth floatation process is used for concentration of sulphide ores.
(2) Metal sulphides are hydrophobic and get easily wetted with pine oil and float on the froth.
(3) As they float on the froth due to wetting with pine oil it becomes easier to separate them from the gangue. Therefore, pine oil is used in froth floatation.

e. Anodes need to be replaced from time to time during the electrolysis of alumina.
Ans: (1) During the electrolysis of alumina, following reaction takes place at anode:
2O2- – 4 e- –> O2
(2) The oxygen liberated at anode reacts with carbon anodes to form carbon dioxide due to high temperature.
(3) Thus, carbon anodes get oxidized during electrolysis and get consumed. Therefore, they need to be replaced time to time.

Q.2.(C) Answer the following questions:

*(i) When a copper coin is dipped in silver nitrate solution, a glitter appears on the coin after some time. Why does this happen? Write the chemical equation.
Ans: (1) According to reactivity series, copper is more reactive than silver.
(2) So, a displacement reaction takes place between copper and silver nitrate.
(3) More reactive copper displaces less reactive silver from silver nitrate.
(4) This displaced silver gets coated on the copper coin and a giitter appears on it.
(5) Following reaction take place : Cu + 2AgNO -> Cu(N03)2 + 2 Ag.

*(ii) The electronic configuration of metal ‘A’ is 2, 8, 1 and that of metal ‘B’ is 2, 8, 2. Which of the two metals is more reactive? Write their reaction with dilute hydrochloric acid.
Ans: (1) From the electronic configuration, we know that metal ‘A’ is Sodium and metal ‘B’ is magnesium.
(2) According to the reactivity series sodium is more reactive than magnesium. i.e. metal ‘A’ is more reactive than metal ‘B’.
(3) Their reactions with dilute hydrochloric acid are:
2Na + 2HCL –> 2 NaCl + H2
Mg + 2HCL –> MgCl2 + H2

*(iii) Divide the metals Cu, Zn, Ca, Mg, Fe, Na, Li into three groups, namely reactive metals, moderately reactive metals and less reactive metals.
Ans: Reactive metals: Na, Li, Ca
Moderately reactive metals : Mg, Zn, Fe
Less reactive metals : Cu

(iv) What is aqua regia? What is its peculiarity?
Ans: (1) Aqua regia is a freshly prepared mixture of concentrated hydrochloric acid and concentrated nitric acid in the ratio 3:1.
(2) Its peculiarity is that it can dissolve noble metals like gold and platinum in it.

*(v) In the reaction between chlorine and HBr, a transformation of HBr into Br2 takes place. Can this transformation be called oxidation? Which is the oxidant that brings about this oxidation?
Ans: (1) Yes, this transformation can be cailed as oxidation of HBr. Here, Br gets oxidised to Brz.
(2) Ci2 is the oxidant which brings about this oxidation.

*(vi) Why do silver articles turn blackish while copper vessels tum greenish on keeping in air for a long time?
Ans: (1) Silver reacts with the sulphur compounds like H2S present in air and forms blackish silver sulphide on its surface. So, siiver articles tum blackish on keeping in air for a long time.v
(2) Capper reacts with CQ2 in atmosphere in the presence of moisture and a greenish layer of copper carbonate is formed on its surface. So, copper vessels turn greenish on keeping in air for a long time.

*(vii) Why do pure gold and platinum always glitter?
Ans: (1) Metais have lustre due to which they glitter.
(2) Pure gold and platinum do not get affected by air or water and do not corrode as many other metals do. So, this glitter of gold and platinum does not vanish and they always glitter.

Q.3.(A) Draw a neat labelled diagram.

a. Magnetic separation method.

b. Forth floatation method.

c. Electrolytic reduction of alumina.

d. Hydraulic separation method.

Q.3.(B) Answer the following questions:-

(i) Write any three points of difference between metals and non-metals in relation to their physical properties.
Ans:

(ii) Write the following metals in the descending order of their reactivities in the following of their reactivities in the following boxes:
Mg, K, Cu, Al, Na, Hg, Zn, Li, Ag, Ca, Fe, Pb, Sn.

Ans:

(iii) Write any three chemical properties of non-metals.
Ans: (1) Non-metals combine with oxygen to form acidic oxides. e.g. C + O2 –> CO2
(2) Generally, non-metals do not mad with water except halogens. On dissolving in water. chlorine gives following products C12 + H2O –> HOCL + HCL.
(3) Non-metals combine with hydrogen under mnam conditions of temperature, pressure or presence of catalyst. e.g. (i) S + H2 –> H2S (ii) N2 + 3H2 –> 2NH3

(iv) State any three properties of ionic compounds.
Ans: (1) The attractive force between the positively charged and negatively charged ions is strong. So, ionic compounds exist in solid state and are hard but brittle.
(2) The melting points and boiling points of ionic compounds are high.
(3) Ionic compounds are soluble in water but in soluble in organic solvents like kerosene, petrol, etc.

(v) Name the four general methods of concentration of ores and draw a diagram illustrating any one of the above methods.
Ans: The general methods of concentration of ores are-
(1) Separation based on gravitation
(2) Magnetic separation method
(3) Froth floatation method
(4) Leaching method.
Magnetic separation method (Diagram):

(vi) Explain briefly the Bayer’s method for concentration of bauxite. Give related equations.
Ans: (1) The bauxite ore is ground and leached by heating with concentrated sodium hydroxide at 140 to 150 C under high pressure.
(2) The following reaction takes place and sodium aluminate is formed.
Al2O3.2H2O + 2 NaOH –> 2NaAlO2 + 3H2O
(3) Insoluble iron oxide is separated by filtration.
(4) The filtrate contains sodium silicate obtained by dissolving silica in NaOH.
(5) Aqueous sodium aluminate is diluted by water and cooled upto 50 C. This resluts into precipitation of AI(OH)3. The reaction is as follows:
NaAIO2 + 2H2O –> NaOH + AI(OH)3
(6) Aluminium hydroxide obtained is washed, dired and calcined at 1000 C.
(7) Following reaction takes place and pure alumina is obtained which is obtained which is further electrolysed to get aluminium. 2AI(OH) –> AI2O3 + 3H2O.

(vii) Study the following figure and answer the questions that follow:

a. What is added to molten eletrolyte to lower its melting point?
Ans: Cryolite (Na3AIF6) and fluorapar (CaF2) are added to molten electrolyte to lower its melting point.

b. Write the anode reaction and cathode reaction:
Ans: Anode reaction: 2 O2- – 4e- –> O2 (Oxidation)
Cathode reaction: AI3 + 3e- –> AI (Reduction)

c. At which electrode is molten aluminium collected?
Ans: The molten aluminium is collected at cathode at the bottom of the tank.

*(viii) What are the various alloys used in daily life? Where are those used?
Ans: Following are some of the alloys used in daily life and their uses:
(1) Stainless steel (Fe + C + Cr + Ni)- For kitchen utensils, various machine parts.
(2) Brass (Cu + Zn) – For kitchen utensils, eletrical equipment.
(3) Bronze (Cu + Sn) – For statues, medals etc.
(4) Silver amalgam (Ag + Hg) – in dentistry, for filling the gaps in teeth.
(5) Alnico (AI + Co + Ni) – Used for permanent magnets.
(6) Magnalium (AI + Mg) – Used in airplane bodies.

(ix) What are the properties that the alloy used for minting coins should have?
Ans: (1) The alloy should be wear resistant.
(2) It should have anti-corrosion properties.
(3) It should have excellent striking properties i.e. the plastic flow of material in cold condition must be good.
(4) They should be cheaper so that the cost of material for a coin must be less than its face value.

Q.4.(A) Answer the following questions-

*(i) Complete the following statement using every given options.
During the extraction of aluminium …….
a. Ingredients and gangue in bauxite.
b. Use of leaching during the concentration of ore.
c. Chemical reaction of transformation of bauxite into alumina by Hall’s process.
d. Heating the alumina ore with concentrated caustic soda.
Ans: (1) During the extraction of aluminium the ore bauxite is used. The formula of bauxite is AI2O3.nH2O. It contains the impurites of silica (SiO2), ferric oxide (Fe2O3) and titanium oxide (TiO2).
(2) For the concentration of the ore, in Hall’s process, the powdered ore is leached by heating with aqueous sodium carbonate.
(3) Following reaction takes place:
AI2O3.2H2O + Na2CO3 –> 2NaAIO2 + CO2 + 2H2O.
(4) Soluble sodium aluminate is formed which is filtered to remove insoluble impurities.
(5) The filtrate is warmed and neutralised by passing CO2 gas through it.
(6) Following reaction takes place:
2NaAIO2 + 3H2O + CO2 –> 2AI(OH)3 + Na2CO3
(7) In Bayer’s process, powdered ore is leached by heating with concentrated solution of NaOH at 140 to 150C under high pressure. Following reaction takes place: AI2O3.2H2O + 2NaOH –> 2NaAIO2 + 3H2O.
(8) Iron oxide is filtered out but the filtrate contains sodiumm silicate obtained by dissolving silica in NaOH.
(9) Aqueous sodium aluminate is diluted by water and cooled upto 50C. This results into precipitation of AI(OH)3. The reaction is as follows:
NaAlO2 + 2H2O –> NaOH + Al(OH)3
(10) Aluminium hydroxide obtained by either Bayer’s process or by Hall’s process is washed, dried and calcined at 1000C.
(11) Following reaction takes place and pure alumina is obtained which is further electrolysed to get aluminium. 2AI(OH)3 –> AI203 + 3H20.

(ii) Write chemical properties of metals. Give one example each.
Ans: (1) Metals combine with oxygen to form their oxides. Generally, oxides of metals are basic. However, oxides of aluminium and zinc are amphoteric.
e.g. 4Na + O2 –> 2Na2O
(2) Metals react with water, hot water or steam as per their reactivity. Hydrogen gas is evolved and hydroxide or oxide of the metal is formed. e.g. (i) 2K + H2O –> 2KOH + H2 + heat (ii) 2Al 43H2O –> Al2O3 + 3H2.
(3) Metals react with dilute hydrochloric acid or sulphuric acid to give corresponding salt of metal and hydrogen is released. e.g. Mg + 2HCI –> MgCi2 + H2
(4) Metals react with nitric acid to give nitrate salts and oxides of nitrogen are formed depending upon the concentration of nitric acid. e.g. Cu + 4HNO3 (Conc) –> Cu(NO3)2 + 2NO2 + 2H2O
(5) More reactive metals displace less reactive metals from the aqueous solution of theit salts.
e.g. Fe + CuSO –> FeSO + Cu

*(iii) Write the measures to be taken to stop the corrosion of metallic articles.
Ans: (1) Steel or iron articles are coated with paint, oil, grease or varnish. e.g. window grills
(2) Iron or steel parts are given a coating of zinc. This process is called as galvanizing. e.g. Steel sheets, pipes
(3) Brass or copper vessels are given a coating of tin. This process is called as timing or ‘kalhaee’. e.g. Kitchen utensils
(4) Anodization – Articles of aluminium or copper are given a coating of their respective oxide by the process of electrolysis. e.g. Pressure cookers, pans
(5) Electroplating – In this method, less reactive metal is coated on a more reactive metal by electrolysis. e.g. silver plating ,on copper spoons, Chromium plating on steel articles.
(6) Alloying – Alloys have less tendency to corrode. So, metals are mixed with other metals or non-rnetals with definite proportion to form such alloys. e.g. Stainless steel is an alloy of iron. carbon and chromium and it does not rust.

(iv) Observe the following figure and answer the questions that follow:

a. Which process is depicted by the figure?
Ans: The figure depicts ‘Forth floatation method’.

b. Which type of ores are concentrated by this method?
Ans: The sulphide ores are concentrated by this method.

c. Which property of such ores is made use of in the process?
Ans: Metal sulphides are hydrophobic and gangue particles are hydrophilic.

d. Where are the particles of concentrated ore are obtained in the process?
Ans: The particles of sulphide ore float on the top along with the froth (foam).

e. Name one oil which is used in this process.
Ans: Pine oil.

carbon compounds 

Important Points:
1) The bonds formed due to the sharing of electrons are called covalent bonds. Carbon compounds contain covalent bonds.
2) Covalent bonds can be single, double or triple.
3) Carbon has tetra valency. It can form long chains with other carbon atoms by forming strong covalent bonds with them. Carbon can form straight chains, branched chains or closed chains. Due to all these reasons, there are about 10 million of carbon compounds known. This number is greater than all other compounds put together.
4) Catenation Carbon has a unique property of forming strong covalent bonds with other carbon atoms and hence forming big molecules. This property is called catenation.
5)The compounds which contain only carbon and hydrogen in them are called hydrocarbons.
6) Saturated hydrocarbons have only single covalent bonds between carbon atoms. Unsaturated hydrocarbons have atleast one double covalent bond or triple covalent bond between carbon atoms.
7) The saturated hydrocarbons having general formula CnH2n+2 are called alkanes. Unsaturated hydrocarbons having general formula C H2 are called alkenes, They have one double bond. Unsaturated hydrocarbons having general formula CnH22 2222 are called alkynes. They have one triple bond.
8) The atom of an element which substitutes hydrogen atom in a hydrocarbon is called hetero atom.
9) The hetero atoms or the group of atoms containing hetero atoms due to which a carbon compound acquires specific chemical properties are called functional groups.
10) Homologous series The series of compounds formed by joining the same functional group in the place of a particular hydrogen atom on the chains having sequentially increasing length is called homologous series. e.g. alkenes, alkenes, alkynes, alcohols are some of the examples of homologous series.
11) The members of a homologous series have a general molecular formula.
12) For naming carbon compounds systematically, the nomenclature system by IUPAC is accepted world wide.
13) Carbon compounds under go one or more of the following types of reactions Combustion, Oxidation, Addition, Substitution.
14) Saturated hydrocarbons under go substitution reactions while unsaturated hydrocarbons under go addition reactions.
15) Ethanol, ethanoic acid are some of the important compounds.
16) Esters are formed by the reaction between a carboxylic acid and an alcohol. They have sweet odour.
17) The giant carbon molecules formed from hundreds of th0usands of atoms are called macromolecules.
18) A macromolecule formed by regular repetition of a small unit (monomer) is called polymer. Polyethylene, polystyrene, polyvinyl chloride are some of the examples of polymers.
19) Polysaccharides, cellulose, proteins etc. are the examples of natural polymers.

Q.1.(A) Match the pairs

Ans:

*Q.1.(B) Draw an electron dot structure of the following molecules.(Without showing the circles)

a. Methane
Ans: Methane (CH4)

b. Ethene
Ans: Ethene (C2H4)

c. Methanol
Ans: Methanol (CH3OH)

(Note: Lone pairs of electrons of Oxygen shown smaller)
d. Water
Ans: Water (H2O)

(Note: Lone pairs of electrons of Oxygen shown smaller)

*Q.1.(C) Write the IUPAC names of the following structural formulae.

a. CH3 – CH2 – CH2 – CH3
Ans: Butane

b. CH3 – CHOH – CH3
Ans: Propan -2-ol

c. CH3 – CH2 – COOH
Ans: Propanoic acid

d. CH3 – CH2 – NH2
Ans: Ethanamine

e. CH3 – CHO
Ans: Ethanal

f. CH3 – CO – CH2 – CH3
Ans: Butan -2-one

*Q.1.(D) Identify the type of the following reactions of carbon compounds.

a. CH3 – CH3 – CH3 – OH –> CH3gCH3 – COOH
Ans: Oxidation

b. CH3 – CH3 – CH3 –> 3CO3 + 4H3O
Ans: Combustion

c. CH3 – CH = CHCH3 + Br3 –> CH3 – CHBr – CHBr – CH3
Ans: Addition

d. CH3 – CH3 + CI3 –> CH3 – CH3 – CI + HCL
Ans: Substitution

e. CH3 – CH3C – H3 – CH3 – OH –> CH3 – CH3 – CH = CH3 + H30
Ans: Dehydration

f. CH3 – CH3 – COOH + NaOH –> CH3 – CH3 -COO-Na+ + -H30
Ans: Neutralization

9. CH3 – COOH + CH 3-0H –> CH33 – COO – CH + H30
Ans: Esterification

Q.1.(E)Answer the following questions.

*(i) What are the types of compounds?
Ans: The compounds are of two types:
(1) ionic com pounds (ii) Covalent compounds

*(ii) Objects in everyday use suoh as foodstuff, fibres, paper., medicines, wood, fuels are made of various compounds. Which constituent elements are common in these compounds?
Ans: The most common elements in these objects are carbon (C), hydrogen (H) and oxygen (O).

*(iii) Atomic number of chlorine is 17. What is the number of electrons in the valence shell of chlorine?
Ans: The number of electrons in the valence shell of chlorine is 7.

*(iv) Molecular formula of chlorine is CI2. Draw electron-dot and line structure of a chlorine molecule.
Ans:

(Note: Lone pairs of electrons are shown smaller)

*(v) The molecular formula of water is H2O. Draw electron-dot and line structuresj for this traitomic molecule. (Use dots for electrons of oxygen atom and crosses for electrons of hydrogen atoms)
Ans:

(Note: Lone pairs of electrons of Oxygen shown smaller)

*(vi) The molecular formula of ammonia is NH3. Draw electron-dot and line structures for ammonia molecule.
Ans:

*(vii) With which bond C atom in CO2 is bonded to each of the O atoms?
Ans: C atom is bonded to each of the O atom with a double bond each.

*(viii) The molecular formula of carbon dioxide is CO2. Draw electron-dot (without showing circle) and line structures for carbon dioxide molecule.
Ans:

(Note: Lone pairs of electrons of Oxygen shown smaller)

*(xi) Molecular formula of propane is C3H8. From this draw its structural formula.
Ans: Propane (C3H8)

*(x) The molecular formula of ethyne is C2H2. From this, draw its structural formula and eletron-dot structrue.
Ans: Ethyne(C2H2)

*(xi) By how many -CH2-(methylene) units do the formulae of any two consecutive members of a homologous series differ?
Ans: Any two consecutive members of a homologous series differ by one -CH2-(methylene) unit.

*(xii)which is the component of biogas that makes it useful as fuel?ns: Any two consecutive members of a homologous series differ by one -CH2-(methylene) unit.

Ans: Methane.

*(xiii) Which product is formed by the combustion of elemental carbon?
Ans: Carbon dioxide (CO2).

*(xiv) Is the biogas combustion reaction endothermic or exothermic?
Ans: Exothermic.

*(xv) Which one of ethanoic acid and hydrochloric acid is stronger?
Ans: Hydrochloric acid is stronger than ethanoic acid.

*(xvi) Which indicator paper out of blue litmus paper and pH paper is useful to distinguish between ethanoic acid and hydrochloric acid?
Ans: pH paper will be useful for distinguishing between ethanoic acid and hydrochloric acid because it, will tell us the pH of both acids. The acid with lower pH is hydrochloric acid and that with higher pH is ethanoic acid.

*(xvii) What is meant by a chemical bond?
Ans: A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds.

*(xviii) What is the number of chemical bonds that an atom of an element forms called?
Ans: Valency of the element.

*(xix) What are the two important types of chemical bonds?
Ans: (1) Ionic bond or electrovalent bond (2) Covalent bond

*Q.2.(A) Explain the following terms with example.

a. Structural isomerism
Ans: (1) The compounds having different structural formulae but having the same molecular formula are called as isomeric compounds or isomers. This property is called structural isomerism.
(2) Following are the two isomers of butane (C4H10).

b. Covalent bond.
Ans: (1) The chemical bond formed by sharing of two valence electrons between the two atoms is called covalent bond.
(2) H (1) has one eIectron in the outermost shell. It would attain the stability by having the
configuration of nearest inert gas i.e. He (2).(3) CI (2,8,7) has seven electrons in its outermost shell. It would attain the stability by having the configuration of nearest inert gas i.e. Ar (2, 8, 8).
(4) If both of them share one electron each H gets a configuration of (2) and Cl gets a configuration of (2, 8, 8) and a covalent bond is formed between them. This is how a covalent compound HCI is formed having a covalent bond between H and CI.
(5)A covalent bond is represented by a line and the structure of HCI is shown as H CI.

c. Hetero atom in a carbon compound
Ans: (1) Hydrocarbons are the compounds which contain only carbon and hydrogen.
(2) They are supposed to be the basic carbon compounds.
(3) Many more compounds can be derived from them by replacing one or more hydrogen atoms by other atoms like chlorine, bromine, oxygen etc. or by group of atoms. Such atoms are called hetero atoms.
(4) e.g. From methane (CH). chloromethane (CH3OI) can be obtained by replacing one hydrogen atom by a chlorine atom. Here, Chlorine atom is a hetero atom.

d. Functional group
Ans: (1) The carbon compounds acquire specific chemical properties due to hetero atoms or a group of atoms that contains hetero atoms irrespective of the length and nature of the carbon chain in that compound.
(2) Such hetero atoms or groups of atoms containing hetero atoms are called functional groups.
(3) Chloro, (Cl), Alcohol (OH), Aldehyde (-CHO), Ketone (-CO-) are some examples of functional groups.
(4) Methanol (CH3OH), ethanol (C2HSOH) and Propanol (C3H7OH) are some alcohol shaving functional group -OH.

e.Alkane
Ans: (1) The hydrocarbons with general formula CnHm2 are called alkanes.
(2) They are saturated hydrocarbons and they contain only single covalent bonds.
(3) Methane (CH), Ethane (C2HG) Propane (C H8) are some examples of alkanes.

f. Unsaturated hydrocarbon
Ans: (1) A hydrocarbon which contains at least one double bond or triple bond is called an unsaturated hydrocarbon. (2) Ethene (CH2 = CH2), Propene (CH3cH = CH2), ,Ethyne (CHECH) are some examples of unsaturated hydrocarbons.

g. Homopolymer
Ans: (1) A macromolecule formed by regular repetition of a small unit is called a polymer.

(2) The polymers which are formed by repetitioon of single monomer are called homopolymers.
(3) Poiyethylene is a polymer of a single monomer CH2 = CH2, so it is a homopolymer.

h. Monomer
Ans: (1) A macromolecule formed by regular repetition of a small unit is called a polymer.
(2) The small unit that repeats regularly to form a polymer, is called a monomer.
(3) The polymer polyvinyl chloride (PVC) is formed by the repetition of vinyl chloride (CI-CH=CH2). So, vinyl chloride is caooed the monomer of PVC.

i. Reduction
Ans: (1) When a carbon atom in a carbon compound gains a bond to hydrogen and loses a bond to a hetero atom or to another carbon atom, that carbon compound is said to be reduced.
(2) This reaction is also called as hydrogenation.
(3) The conversion of ethene into ethane in the presence of a catalyst Ni is an example of reduction. CH2 = CH2 + H2 –> C2H6.
(4) Conversion of vegetables oil into vegetables ghee is another example of reduction.

j. Oxidant
Ans: (1) An oxidant is a reactant that oxidizes or removes electrons from other reactants during a redox reaction.
(2) It is also called as an oxidiser or oxidising agent.
(3) The conversion of ethanol (C2H5OH) into ethanoic acid (CH3-COOH) by addition of potassium permanganate (KMnO4) is an oxidation reaction and KMnO4 acts as an oxidant.

*Q.2.(B) Write structural formulae for the following IUPAC names.

a. pent-2-one
Ans:



b. 2-chlorobutane
Ans:



c. propan-2-ol
Ans:



d. methanol
Ans:



e. butanoic acid
Ans:



f. 1-bromopropane
Ans:



g. ethanamine
Ans:



h. butanone
Ans:

Q.2.(C) Answer the following questions

*(i) What is a catalyst? Write any one reaction which is brought about by use of catalyst?
Ans: (1) A substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change is called as a catalyst.
(2) In the conversion of ethene into ethane, the catalyst Ni is used. CH2=CH2 + H2_, Ni C2 H6.

*(ii) To which group in the periodic table does the element carbon belong? Write down the electronic configuration of carbon and deduce the valency of carbon.
Ans: (1) Carbon belongs to 14th group of the periodic table.
(2) The atomic number of carbon is 6 and its electronic configuration is (2,4).
(3) The valency of carbon is 4.

*(iii) The molecular formula of sulphur is S8 in which eight sulphur atoms are bonded to each other to form one ring. Draw the electron-dot structure for S8. without showing the circles.
Ans:


(Note- The lone pairs of electrons of sulphur are shown smaller)

*(iv) Draw electron-dot structure of cyclohexane.
Ans:



*(v) Propane is one of the combustible components of L.P.G. Write down the reaction for Propane(C3H8):
Ans: Combustion of propane (C3H8):
C3H8 + 5O2 –> 3 CO2 + 4 H2O

*(vi) Explain the ihaction when lime water turns milky on passing CO2.
Ans: When CO2 is passed through lime water [Ca(OH)2], calcium carbonate is formed which is insoluble in water and white in colour. Due to this, the lime water turns milky. Following is the equation:
Ca(OH)2 + C02 –> CaCO3+ H2O
(Note Lime water test is used to identify carbon dioxide gas.)

*(vii) Explain the reaction that would take place when a piece of sodium is dropped in ethanoic acid.
Ans: When sodium is dropped in ethanoic acid (acetic acid), sodium ethanoate (sodium acetate) will be formed and hydrogen gas will be released. The equation is as follows:
2 CHaCOOH + 2 Na –> 2 CHsCOONa + H2

*(viii) Two test tubes contain two colourless liquids ethanol and ethanoic acid. Explain by writing reaction which chemical test you would perform to tell which substance is present in which test tube.
Ans: (1) Add sodium bicarbonate in both the test tubes.
(2) Sodium bicarbonate does not react with ethanol.
(3) Sodium bicarbonate reacts with ethanoic acid and gives effervescence of carbon dioxide.
CH3COOH + NaHCO3 –> CH3COONa + H2O + CO2

*(ix) When fat is heated with sodium hydroxide solution, soap and glycerin are formed. Which functional groups might be present in fat and glycerin?
Ans: (1) Fat might be having -COOH functional group a d glycerin might be having -OH functional group.

*(x) Structural formulae of some monomers are given below. Write the structural formulae of the homopolymers formed from them.


Ans:
structural formulae of homopolymers



*(xi) From the given structural formula of polyvinyl acetate, that is used in paints and glues, deduce the name and structural formula of the corresponding monomer.



Ans: (1) The name of monomer is vinyl acetate.
(2) The structural formula is as follows:

Q.3.(A) Answer the following questions:

*(i) Draw all posible structural formulae of compounds from their molecular formula given below.
a. C3H8
Ans: Only one polymer is possible.


b. C4H10
Ans: Following two polymers are possible:



(ii) What causes the existence of very large number of carbon compounds?
Ans: (1) Carbon has a property of catenation i.e. it can form covalent bonds with another carbon atom.
(2) The bonds between two carbon atoms are strong and stable so that carbon can form huge chains of its atoms.
(3) Carbon can form straight chains as well as branched chains and cyclic chains.
(4) Carbon has tetra valency. Due to all these reasons a large number of carbon compounds exist.

*(iii) Saturated hydrocarbons are classified into three types. Write these names giving on example each.
Ans:





*(iv) Give names of three functional groups containing different hetero atoms. Write name and structural formula of one example each.
Ans:





*(v) Fill in the following boxes:



Ans:



*(vi) What is meant by vinegar and gasohol? What are their uses?
Ans: (1) Vinegar is 5-8% aqueous solution of ethanoic acid (acetic acid). It is used as a preservative. It is also used in various recipes. It is used for cleaning purpose.
(2) Gasohol is a mixture of petrol (gasoline) and ethanol (about 10%). It is used as a fuel in vehicles.

(vii) Write a note on homologous series.
Ans: (1) The compounds having same functional group in the place of a particular hydrogen item, but having sequentially increasing length of carbon chains form a homologous series.
(2) Each member of a homologous series is called a homologue.
(3) The members of a homologous series can be represented by a general formula.
(4) The consecutive members of a series differ by a methylene group (-CH2-) and their molecular mass differs by 14u.
(5) The chemical properties of compounds in a homologous series are very much similar due to the presence of the same functional group.
(6) Alkanes (general formula: CnH2n+2), Carboxylic acids (general formula: CnH2n+1COOH), Aldehydes (CnH2n+1CHO) are some examples of homologous series. Methane, ethane, propane etc. are the members of alkane series. Methanoic acid, ethanoic acid, propanoic acid etc. are the members of carboxylic acids series. Methanal, ethanal, propanal etc. are the members of aldehyde series.

(viii) Complete the following chart.



Ans:



(Note- R and R’ are alkyl groups)

*(ix) Draw three structural formulae having molecular formula C5H12 and name them as n-pentane, i-pentane and neo-pentane.
Ans:

Q.4. Answer the following questions:

*(i) Give any four functional groups containing oxygen as the hetero atom in it. Write name and structural formula of one example each.
Ans: (1) Alcohol e.g. Ethanol



(2) Carboxyl e.g. Ethanoic acid



(3) Aldehyde e.g. Ethanal



(4) Ketone e.g. Propanone



(ii) Write the equation of one example each of the following types of reactions.




Ans:

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