Class IX Solution of GRAVITATION

I. Very short answer type questions:

1. Write one difference between gravitation and gravity.
Ans. Gravitation is the force of attraction between any two bodies and gravity is the force of attraction between a body and the Earth.

2. How is the acceleration due to the gravity of the falling body related to its mass?
Ans. Acceleration due to the gravity of a falling body doesn't depend on the mass of the body.

3. Why G is called universal constant?
Ans. The value of G remains constant at all the places in the universe, so it a called universal constant.

4. Define the mass and weight of an object.
Ans. Mass is the quantity of matter contained in the body and the weight of the body is the force with which Earth attracts the body.

5. Can the mass of an object be zero?
Ans. No.

6. What do you understand by the gravitational force of earth?
Ans. Gravitational force is the force with which earth attracts the other body.

7. How does acceleration due to gravity change with the shape of the earth?
Ans. The earth is not a perfect sphere. As the radius of the earth increases from the pole to the equator, the value of g becomes smaller at the equator and greater at the pole.

8. Is the value of G dependent on the medium present between the two bodies?

Ans. No.

9. At what place on the earth's surface is the weight of a body be maximum and minimum?

Ans. The weight of a body is maximum at the pole and minimum at the equator of the earth.

10. If the mass of a body is 8 Kg and weight is 12 N then what will be its mass and weight in the moon? 

Ans. Mass of the body will be 8 kg and weight will be one-sixth of 12 means 2N (12/6). Because the mass of the body does not change but the weight of a body is different in different places.  

11. State the factors on which acceleration due to gravity 'g' and 'G' depends.

Ans. Acceleration due to gravity g depends on the mass of the earth and G which is universal constant depends on the distance or radius and mass of the planet.

12. On what factors does the weight of a body depend?

Ans. Mass(m) and gravity (g) are the two factors on which weight depends.

 As Weight = mg.

13. How does the weight of a rocket change as it moves from earth to moon?
Ans. The weight of a rocket decreases as it moves from earth to moon because weight is directly related to g and at the moon, weight becomes one-sixth.

14. What is the relation between the weight of an object on the moon and on the earth?
Ans. The weight of an object on the moon is one-sixth the weight of the object on the earth. 

II. Short answer type questions

1. Give reasons for the following:
(a) A sheet of paper falls slower than when it is crumpled into a ball.
(b) What is meant by the statement that acceleration due to gravity is 9.8 m/s²?

Ans. (a)A sheet of paper falls slower than when it is crumpled into a ball because the resistance provided by the air is more for a sheet of paper than the crumpled ball of paper. In a vacuum, both will touch the ground together.

(b) Acceleration due to gravity is 9.8 m/s²b means, every second an object which is in free fall,  the velocity of the object to increase 9.8 m/s due to gravity. 

2. What is meant by free fall? A stone and a small coin are dropped from the same height at the same time, which of the two will touch the ground first?

Ans. When a body falling from a height towards the earth under the gravitational force of the earth and no other force acting on it is called free fall.

When stone and a small coin are dropped from the same height at the same time, both the stone and coin will touch the ground together because the acceleration due to the gravity of the falling body doesn't depend on the mass of the body.

3. (i) Write the units of the following:
(a) Gravitational constant G (b) Mass (c) Weight.
(ii) Why a body weighs more at the pole than at equator?

Ans. (i) (a) Unit of gravitational constant G is Nm²kg^-2

(b) The unit of mass is kg.
(c) The Unit of weight is N.
(ii) A body weighs more at the pole than at the equator because the radius of the earth is smaller at the pole than the equator and weight is inversely proportional to the radius of the earth.

4.Explain what happens to the force between two objects if:
(a) The mass of one object is doubled?
(b) The distance between the two objects is tripled?
Ans. (a) The forces between two objects become double
when the mass of one object is doubled. Force is directly proportional to mass.
(b) The forces between two objects become one-ninth when the distance between the two objects is tripled.
F₁ = m₁m₂/r² and when r is triple,new force
 F₂ = m₁m₂/(3r)² 
= m₁m₂/9r²
F₂ = F₁/9 

5. What is the importance of the universal law of Gravitation? Or State any four natural phenomena explained by the universal law of Gravitation.
Ans. The universal law of gravitation successfully explained several phenomena which are:
(i) the force that binds us to the earth;
(ii) the motion of the moon around the
earth;
(iii) the motion of planets around the Sun;
and
(iv) the tides due to the moon and the Sun.

6. How does the force of gravity between two objects change when the distance between them is reduced to half?
Ans. The force of gravity between two objects becomes 4 times when the distance between them is reduced to half.
The force of gravity:
F₁ = m₁m₂/r² 
And when the distance between them is reduced to half the force of gravity: 
F₂ = m₁m₂/(r/2)² 
= 4 m₁m₂/r² .

7. Stone and the earth attract each other with an equal and opposite force. Why then we see only the stone falling towards the earth but not the earth rising towards the stone? 
Ans . According to the third law of motion, stone and the earth attract each other with an equal and opposite force. But according to the second law of motion, for a given force, acceleration is inversely proportional to the mass of an object. The mass of stone is negligibly small compared to that of the earth. So, we see only the stone falling towards the earth but not the earth rising towards the stone. 

8. Write three differences between 'g'(gravitational acceleration) and 'G'(gravitational constant).
Ans. g (gravitational acceleration)      
 1. g is the acceleration due to gravity acquired by the body due to the earth gravitational pull. 
 2. g is a vector quantity.
3. It is different in different places on the earth's surface.  
G (gravitational constant)  
1. G is numerically equal to the force of attraction between two masses of 1kg which are separated by a distance 1m.
2. G is a scaler quantity.
3. Its value is the same everywhere in the universe. earth surface.

9. Give the relation between mass and weight. What kind of quantity is the weight?
Ans. Mass is the quantity of matter contained in the body and the weight of the body is the force with which Earth attracts the body.
The relation is: W, is the product of the mass m of the object and the magnitude of the gravitational acceleration g 
 = weight/acceleration (W/g)
Weight is a vector quantity.

10. A man of mass 60 Kg is standing on the floor holding a stone weighing 40N. What is the force with which the floor is pushing him up?
Ans. The mass of the man is 60kg.
Weight of man = mg= 60 x 9.8 =588N
Weight of stone = 40N
According to the newton's third law of motion total force exerted by the man to the ground = the force with which the floor is pushing him up=588 +40 = 628N
The force with which the floor is him up= 628N.

11. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate:

(a) The maximum height to which it reaches.
(b) The total time it takes to return to the earth's surface.
Ans. (a) Let the max. height to which it reaches is h.
We have  v²-u²= 2gh
h = v²-u²/2g  where v = 0 and u = - 49 m/s
h = -(49)²/2 x (-9.8) = 2401/19.6 = 122.5 m.
(b) Let the total time it takes to return to the earth's surface s t.
We have v = u +gt
and t = (- u) /-g
 t = -49/-9.8 = 5 s.

12. A body of weight 600N rests on a lift. If the lift begins to fall freely under the gravity, what is the force with which the body presses on the floor?
Ans. When the lift is falling freely under the gravity then the body of weight 600N kept in it also falls freely under the action of gravity. In this case, the reaction force of the floor of the lift on the body is zero. So the action force of the body on the floor of the lift should also be zero.(by Newton's III Law of motion). Hence no force is exerted by the body on the floor of the lift when the lift is falling freely under the gravity. The body is weightless in such a situation.

13. A sphere of mass 40 kg is attracted by another sphere of mass 80 kg. by a force of 2.5 x10^-6 N and when their centers are 300 mm apart Find the value of G.

Ans. The force of gravitation exerted by Earth or any other planet on the objects on its surface is called gravity. Every object attracts every other object due to the force of gravitation. So gravity is a special case of Gravitation.
F = GMm/r²
2.5 x10^-6 = G (80 x 40)/(300 x 10^-3)²

2.5 x 10^-12 = G (3200)/90000
G = (2.5 x 90000 x 10^-12)/3200 = (25 x 90 x 10^-12)/3200
= 2250 x 10^-12/3200 
= 22500 x 10^-11/3200
G = 7.02135 x 10^-11

14. A ball is thrown upwards from the ground at a tower with a speed of 20 meters per second. There is a window in the tower at a height of 15 m from the ground. How many times and when will the ball pass a window?
Ans. Initial velocity = u = 20 m/s.
The maximum height the ball will reach h =? 
Using equation  v²- u²= 2gh  
At maximum height v = 0 m/s
h= (v²- u²)/2g = - u²/2g = - 20²/2 x 10 = 400/20 = 20m.
h = 20m
The ball will reach the height of 20m.
The ball will reach the window two times when the ball goes upward and then comes back downward. 
Now to calculate the time:
We have     h = ut - 1/2gt²                  15 = 20t - 0.5 x10t² 
                   15 = 20t - 5t²
  5t²- 20t + 15 = 0
  t² -  4t + 3 = 0
  t² - 3t - t + 3 = 0
  t(t - 3) - 1(t -3) = 0
 t = 1 , 3 s
Therefore ball pass the window at 1 sec. and 3 sec. respectively.
   
15. (a) Seema byes a few grams of gold at the pole as per the instruction of one of her friends. She hands over the same when she meets her friend at the equator. Will the friend agree with the weight of gold bought? If not why?
(b) If the moon attracts the earth, why does the earth not move towards the moon?
Ans. (a) No. The weight of the same amount of gold will be less on the equator because weight is the force of attraction with which the earth attracts the body and weight = mg. Therefore weight is directly proportional to g and the value of g is less at the equator than the pole.
(b) The moon attracts the earth but the earth does not move towards the moon because the mass of the moon is very less as compared to the earth.

16. Suppose a planet exists whose mass and radius both are half those of earth. Calculate the acceleration due to gravity on the surface of this planet.
Ans. If g_e and g_p  are the acceleration due to gravity on the earth and planet surface then 
 g_e= GM/R² and  g_p= Gm/r²,   
Where R = radius of the earth 
r =  radius of the planet.
m = mass of the planet and
M = mass of the earth.
According to question m = 1/2M and r = 1/2R
Therefore g_p = Gm/r² 
= G x M  x 4 
      2 x R² 
= 2g_e 
=2 x 9.8 = 19.6m/s²
                                             

III. Long answer type questions:.

1. What happens to the magnitude of the force of gravitation between two objects if:

(a) Mass of one of the objects is tripled?
(b) The distance between the objects is doubled?
(c) The mass of both objects is doubled.

Ans. (a) The force of gravitation between two objects is given by 'Universal gravitational Law'. It is numerically stated as:

F = Gm₁m₂/r²

(a) Mass of one of the objects is tripled: F = G3m₁m₂/r² 

F = 3 Gm₁m₂/r²  

Force will be tripled.

(b) The distance between the objects is doubled:

F = Gm₁m₂/(2r)²

F = 1/4 Gm₁m₂/r²

Force will be one-fourth of its previous value.

(c) The mass of both objects is doubled:

F = G2m₁2m₂/r² 

F = 4Gm₁m₂/r²

Force will be four times its previous value.

3. (I) Write the name of the scientist:

(a) Who discovered that force is the cause of motion?
(b) Who formulated the universal law of gravitation?
(c) Who found out the value of gravitational constant 'G’.

(II) (a) Explain why the value of 'g' is greater at poles than the equator.

(b) Why does a body reach the ground quicker at poles than equator when dropped from the same height?

Ans. (a) Galileo Galilei discovered that force is the cause of motion.

(b) Sir Issac Newton formulated the universal law of gravitation.

(c) Henry Cavendish found out the value of gravitational constant 'G’.

(II) (a) The value of g directly proportional to, the radius of the earth and radius of the earth is less at the pole than the equator, so the value of 'g' is greater at poles than the equator.

(b) A body reaches the ground quicker at poles than the equator when dropped from the same height because g is greater at the pole than the equator. 

4. Derive an expression for acceleration due to gravity in terms of mass M of earth and radius R.

Ans. According to the Newtons third law of motion, the force of attraction between the earth and the earth and body is given by F = GMm/r²
Where m = mass of the object, M = mass of the earth, r = distance between earth and object and G =universal constant.
And according to Newton's second law:
F = mg.  (where g is the acceleration due to gravity)
From eq. 1 and 2 we get
mg = GMm/r²              g = GM/r². 
If the body is located on the earth's surface of the earth then r = R.
Therefore g = GM/R²
This expression gives acceleration due to gravity at the surface of the earth where the value of g is 9.8 m/s²

5. A stone is dropped from the edge of the roof. It passes a window 2m high, in 0.1sec. How far is the roof above the window? 
Ans. Given :  
 When a stone is dropped from the edge of the roof, 
its initial velocity = 0 .
The stone passes the window of 2m high in 0.1s.
Let the distance between the roof and that window be 'h' m and the time taken to reach that "h"m is t. 
Therefore h = ut + 1/2gt² = 1/2gt²........eq (1)
A/Q Stone passes the window in 0.1 seconds. 
So it travels h+2 m in t + 0.1seconds.
h + 2 = 1/2g(t+ 0.1)² 
1/2gt² + 2 = 1/2g(t²+0.01 + 0.2t)  From eq ..(1)
1/2gt² - 1/2gt² + 2 = 0.005 g + 0.1gt 
2 = 0.005 x 10 + 0.1 x 10 x t 
2= 0.05 + t 
2 - 0.05 = t 
t = 1.95 s.
Therefore, The stone takes 1.95 seconds to travel from roof to top of the window.
The distance between the roof and top of the window (h) 
h = 1/2 x 10 x t²
h = 5t² 
h= 5x 1.95²
  = 19.0125 m.
Therefore, the height of the roof above the top of the window is 19.0125m

6. State the Universal law of Gravitation? Derive an expression for gravitational force between two bodies.
Ans. The universal law of gravitation states that the force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Let the two bodies A and B be of masses m₁ and m₂ respectively.
            According to universal law:
                        F  ∝ m₁m₂........eq. (1)
                        F  ∝ 1/r²............eq. (2)
                        From eq...1 & 2 we get
                         F ∝ m₁m₂/r²
                         F = G m₁m₂/r²
  Where G is the universal constant.

7. A stone is dropped from a height of 50 m on earth. At the same time, another stone is thrown vertically upwards from the ground with a velocity of 50m/s. At what height from the ground will the two stones meet (g = -10m/s²)
Ans. Let the distance covered by stone A which is dropped from a height be h. 
h = 1/2gt²   ( u = 0 m/s).....1
Height of the Stone B which is thrown vertically upwards
= 50 - h
= ut  - 1/2gt²
50 - h = 50 t - 1/2 g t²
50  - 1/2 gt² = 50 t - 1/2 g t²  ( from eq.  1)
50 = 50t
t = 1s
Now distance from the ground = 50 - 1/2gt² = 50 - 1/2 x 10 x1 = 45 m

8. A ball is thrown vertically upwards. The speed of the ball was 10m/s when it had reached one half of its maximum height. How high does the ball rise? 
Ans. (a) When a ball is thrown vertically upward, the speed of the ball at half of the max.height =v = 10 m/s.
Let the maximum height be H and the final velocity is zero at height this height.
Now distance = H.
We have v² - u² = 2x g x H
             0² - u² = 2 x -10 x H
             - u² = -20H
               u² = 20H
When the ball reaches half the maximum height, the final velocity is 10 m/s, and the distance traveled is H/2.
We have v² - u² = -2gH/2
            10² - 20H = 2 x -10 x H/2
            100 - 20H = -10H
             10H = 100
              H   = 10m
Therefore the maximum height is 10 m.

9. If the mass of one of the objects is doubled and the mass of others remains the same and if the distance between them is halved then how does the gravitational force change?
Ans. Let the masses of two objects are m₁ and m₂. and the distance between them = r  
Therefore the gravitational force F₁ = m₁m₂/r².
Now, when the mass of one of the object is doubled and the mass of others remains the same and if the distance between them is halved then:
Force F₂ = 2m₁m₂/(r/2)².
        F₂ =  8 m₁m₂/r².
        F₂ =  8F₁ 
 If the mass of one of the objects is doubled and the mass of others remains the same and if the distance between them is halved then the gravitational force becomes eight times.

10. The radius of the earth is about 6370 Km. An object of mass 30 Kg is taken to a height of 230 km above the surface of the earth.
(a) What is the mass of the body?
(b) What is the acceleration due to gravity at this height?
(c) What is the weight of the body at this height?
Ans.  (a) The of the will not change because mass is constant in everywhere.
(b) We have g = g^'(1+h/R)²
Where g = acceleration due to gravity.
h= height of the body aabove the surface of the earth =230km
R= radius of the earth = 
 g^' = acceleration due to gravity at the height of 230km.
g = g^'(1+ h/R)²
9.8 =g^'(1 + 230/ 6370)²= g^'(6370 + 230)/ 6370)²
9.8 =  g^'(6600/ 6370)² = g^'(1.04)² =  g^'x1.08
 g^'= 9.8/1.08 = 9.07m/s²
(c) The weight of this body = mg = 30 x 9.07 = 272.1 kg. 

11. What is the magnitude of the gravitational force exerted by a 15Kg mass on a 25 Kg mass separated by a distance of 25 cm? What is the acceleration produced on each mass?
Ans. Let the two masses m and n and the gravitational force between them is F then according to law 
F= Gmn/r², where r is the distance between them.
F = (6.674×10^-11. x 15 x 25)/0.25 x 0.25
F = 40044x 10^-11N = 4 x 10^-7N
Acceleration produced in 15 kg mass a = F/m = 4 x 10^-7N/15kg
a= 2.66 x10^-8  m/s²
Acceleration produced in 25 kg mass a = F/a =  4 x 10^-15N/25kg
a = 1.6 x 10^-8  m/s²

12. A boy on a cliff 49 m high drops a stone, one second later he throws another stone vertically downwards. The two stones hit the ground at the same time. What was the velocity with which the second stone was thrown?
Ans. For the first stone:
h= 49 m, u = 0 m/s, g = 9.8 m/s²
Then v² - u² = 2gh
 v²= 2gh + u²  = 2 x 9.8 x 49 + 0 = 960.4m/s
v = 30.99 or 31m/s
time t = (v - u)/g = (31 - 0)/9.8 = 3.16s
Now for second stone :
h = 49 m, g = 9.8 m/s², time t = 3.16 -1= 2.16s
We have to find u.
 h = ut + 1/2 at²
49 =2.16u + 1/2 x 9.8 x 2.16 x 2.16 = 2.16u + 22.86
2.16u = 49 - 22.86 = 26.14
u = 26.14/2.16=12.10m/s 
The velocity with which the second stone was thrown is 12.10m/s 

13. A ball is thrown vertically upward and it returns after 6 sec. Find:
(a) The velocity with which it was thrown up.
(b) The max.height it reaches.
(c) Its position after 4 sec.
Ans. (a) Let the time for aximum height be t = 6/2 = 3 s.
At the maximum height v = 0 
g = - 9.8 m/s²
(a) We have v = u + gt = u - 9.8 × 3
                   0 = u - 29.4
 u = 29.4 m/s
The velocity with which it was thrown up = 29.4m/s²

(b) We have 2aS = v² - u²
h = v²- u²/2g 
= 0 - 29.4 × 29.4/- 2× 9.8
= 44.1 m
Thus, Maximum height it reaches = 44.1 m.

14.[Do yourself 👉 (a) A ball is dropped from the roof of a building, it takes 10 seconds to reach the ground to find the height of the building.
(b) A stone is released from the top of a tower of height 19.6 m. Calculate its velocity just before touching the ground]

15. (a) Determine the magnitude of the gravitational force between a planet of mass 6 x 10²⁴.kg and a 1 kg object on its surface. Let the radius of the planet be 6 x 10⁶m G = 6.67 x 10^-11Nm²kg.
(b) To estimate the height of the bridge over a river, a stone is dropped freely in the river from the bridge. The stone takes 2 sec. to reach the water surface in the river. Calculate the height of the bridge from the water level.
Ans. (a) According to the law of gravitation F = GMm/R²
Where R = 6 x 10⁶m 
F= (6.67 x 10^-11 x 6 x 10²⁴ x 1 )/(6 x 10⁶)² = 40.02 x 10¹³ /36 x 10¹² 
 = 1.11 x 10 N
F = 11.1N.
(b) Let the height be h.
 g = 9.8m/s²
 u = 0 m/s
 h= ut + 1/2gt²
 h = 1/2 x 9.8 x 2 x 2 = 19.6m

16. An astronaut carried a pot containing soil waiting for 60 Newton from the earth to the surface of the motion moon he kept it there and just before returning journey from Moon to Earth he waited for the soil there on the surface of the moon and found that it was only 10 Newton where did the rest of soil go and how much mass of soil has lost?
Ans. We know that wt of the soil on the moon = 1/6th of the weight of the soil on the earth and it is due to,  g of moon = 1/6th g of earth.
Given that weight of the soil on the earth surface = 60 N
Therefore the weight of the soil on the moon surface = 1/6 x 60 = 10 N.
As mass always remains constant therefore there is no loss in mass of soil on the moon surface and a decrease in weight is due to the difference in gravity.

17. [Do yourself 👉 (a)  Find the weight of an 80 kg man on the surface of the moon? What should be his mass on the earth and on the Moon?
(b) An object of mass 8 units weight 300 Newton on the surface of earth what would be its mass and weight on the surface of the moon?]

18. There are 2 planets A and B masses m₁ and m₂ respectively. They revolve around the sun in a circular orbit. The orbital radius of A from the sun is twice that of orbital radius of B. The mass of A is twice that of the mass of B.Compare the forces of gravitation between A to Sun and B to Sun.
Ans. Given: Mass of the planet A = 2x Mass of the planet B and  
the orbital radius of A from the sun = 2 x  The orbital radius of B from the sun.
Therefor  m₁= 2m₂ ....... eq (1)
and          r₁ = 2r₂ ........ eq (2)
According to the law of gravitation :
F₁ = GMm₁/r₁²     and  F₂ = GMm₂/r₂²
F₁ =  GM 2m₂/(2r₂)². (m₁= 2m₂) 
       =GM2m₂/4(r₂)^2.  ( r₁ = 2r₂  )
        = GMm₂/2(r₂)²  
F₁ =  1/2 F₂
The force of gravitation between April to Sun is half of the force of gravitation between B to Sun.

19. (a) Calculate the force of gravitation between two objects of mass 50 kg and 120 kg., Kept at a distance of 10 m from one another.
(b) A stone is dropped from a height of 10 m on an unknown planet having g = 20 m/s².  Calculate the speed of the stone when it hits the surface of the planet. Also, calculate the time it takes to fall through this height.
Ans. (a) Given  m₁=50kg,  m₂= 120kg and r = 10 m.
G = 6.67 x 10^-11 Nm²kg.h
The force of gravitation between two objects F = Gm₁m₂/r²
 F = (6.67 x 10^-11 x 50 x 120)/10²
F = 400.20 x 10^-11N
or F = 4 x  10^-9 N
(b) Given height h= 10m, g = 20 m/s² and u = 0 m/s
We have v² - u² = 2gh
              v² = 2 x 20 x 10 = 400. ( u = 0 )
              v = 20 m/s
To find t:
 We have v = u + gt 
             20 = 0 + 20 x t =20t
              t = 20/20 = 1 s
              t = 1 s .       
             

Class IX Solution of STRUCTURE OF ATOMS

I. Very short answer type questions:

1. Define atoms. 

Ans. Atoms are the tiny particles and the word atom first used by Democritus. Atoms are the building blocks of matter.

2. Write the name of particles present in atoms or Write the name of subatomic particles.

Ans. The name of particles present in atoms are electrons, protons, and neutrons.

3. Define atomic number.

Ans. The number of protons present in the nucleus of the atom of an element is called the atomic number.

4. Define mass numbers.

Ans. The sum of the number of protons and the number of neutrons present in the nucleus of the atom of an element is called the mass number of that element.

5. Define electronic configuration.

Ans. Arrangement of electrons around the nucleus of an atom is called electronic configuration.

6. Write the formula for calculating the maximum no.of electrons in an orbit?
Ans. The maximum number of electrons in each shell or orbit is determined by a formula 2n². Where n is the number of shells.

7. What happens to an element Z if its atom gains three electrons?
Ans. When the atom of an element Z gains three electrons, it develops three negative charges and called negative ions.

8. J Chadwick discovered a subatomic particle that has no charge and has mass nearly equal to that of a proton. Name the particle and give its location in the atom?
Ans. The Particle is neutron and is present in the nucleus.

9. Which one of the Na⁺ and He has completely filled L and K shells? Give the reason to support your answer.
Ans. Na⁺ has completely filled L shell and He has completely filled K shells. 
Reason: Atomic no. of Na = 11     E. C = 2,8,1 
after loosing one electron E.C of Na = 2,8 , L shell is completely filled.
Atomic no. of He = 2  and E.C = 2 so its K shell is completely filled. 

10. The atomic numbers of atoms of two elements are 18 and 20 respectively and their mass number is 40. 
What is the relation between such pairs of atoms? Will they have the same property?
Ans.  Atoms of different elements with different atomic numbers, which have the same mass number, are known as isobar. They have different chemical properties due to the different number of electrons and the same in physical properties.

11. Which elements have zero combining capacity or valency?
Ans. The atoms of inert elements, having a completely filled outermost shell show little chemical activity and their combining capacity or valency is zero. 

12. How many electrons can be accommodated in the outermost shell of an atom?
Ans. Eight electrons can be accommodated in the outermost shell of an atom. 

13. If the electronic configuration of an atom is 2, 8, 1, calculate the total number of electrons and write the name of the element.
Ans. The total number of electrons is 11 and this element is sodium.

14. What is the valency of oxygen if the atomic number is 8?
Ans. The atomic number of oxygen is 8 and electronic configuration is - 2, 6
So valency is = 8 - 6 = 2

15. Who concluded that the size of the nucleus is very small compared to the size of an atom?
Ans. Rutherford concluded that the size of the nucleus is very small compared to the size of an atom.

16. Define polyatomic ions.
Ans. A group of atoms carrying a charge is known as a polyatomic ion.
Ex - Ammonium NH₄⁺ , Nitrate NO₃^-, carbonate HCO₃^- etc

II. Short answer type questions:

1. Write the properties of electron, proton, and neutron. 

Ans. Electrons protons and neutrons are the subatomic particles of the atom.

Electrons - 

. Electrons are negative in charge.

. The mass of an electron is 1.672 x 10^-24 

. They revolve in a fixed orbit around the nucleus. 

Protons

. Protons are positive in charge.

. The mass of a proton is 9.108 x 10^-19  

. They are located inside the nucleus at the center of the atom 

Neutrons

. Neutrons are electrically neutral in charge.

. The mass of a neutron is 1.676 x 10^-24  

. They are located inside the nucleus at the center of the atom.

2. An atom of an element has three electrons in its 3rd orbit which is the outermost shell and Write
(i) the electronic configuration 
(ii) atomic number
(iii) number of protons, electrons, neutrons 
(iv) valency.
Ans. Three electrons are present in the 3rd or last orbit, 
(i) The electronic configuration is E.C = 2,8,3
(ii) The atomic number is =13
(iii) The no. of protons = 13, no. of electrons = 13 ,and no.of neutrons = 14
(i) Valency = 3 

3. There are 16 protons and 15 Neutrons are present in an element. Calculate its atomic number and its atomic mass no. 
State its valency.
Ans. Atomic number = No. of protons = 16
Atomic mass no. = No. of protons + No. of neutrons = 16 + 15 = 31 
E.C = 2,8,6, Valency = 8 - 6 = 2

4. If an element X is available in the form of say, two isotopes ⁷⁹Br₃₅ (49.7 %) and  ⁸¹Br₃₅ (50.5 %), calculate the average atomic mass of the atom of X.
Ans. Atomic mass = 49.7/100 x 79 =39.26 g
Atomic mass = 50.5/100 x 81 = 40.91
Av. atomic mass = 39.26 + 40.91 = 80.17 or 80g

5. An atom of an element is represented as ¹⁹Y₉ .

Write the
(a)  Atomic number
(b) The electronic configuration
(c) No. of protons and
(d) Valency.
Ans. In an atom of element ¹⁹Y₉ 
(i) Atomic number = 9
(ii) The atomic number of Y is = 9
E.C = 2,7  
(iii) No. of protons = 9
(iv)  Valency = 8 - 1 = 1

6. The electronic configuration of an element X  is 2 , 8 , 2 :
(a) Find the number of electrons present in the atom of element X.
(b) Write its atomic number and
(c) Its atomic mass.
Ans. (a) No. of electrons present in X = 12  ( 2+ 8 + 2 = 12)
(b) Atomic number = No. of electrons = 12, It is magnesium Mg.
(c) Atomic mass = 24

7. List Bhor-bury rules for the distribution of electrons in different shells. 
Ans. According to Neils Bohr rules: the distribution of electrons in different orbits or shells are represented by the letters K, L, M, N,… or the numbers, n=1,2,3,4,….formula 2n², where ‘n’ is the orbit number or energy level index, 1,2,3,…. 
The maximum number of electrons in different shells are as follows:
K - 2, L - 8, M - 18, N - 32 .....and so on
The maximum number of electrons that can be accommodated in the outermost orbit is 8. 
Electrons are not accommodated in a given shell unless the inner shells are filled.

8. Draw the atomic structure of an atom with atomic number 11.
Ans. Atomic number - 11
Element is Sodium Na.
Electronic Configuration - 2,8,1
Atomic structure







9. Composition of the nuclei of two atomic species X and Y are given below.
Element    X   Y
Proton      8    8
Neutron    8    10
i. Give the mass number.
ii. What is the relationship between the two elements?
Ans. Mass number of X = Proton number + Neutron number

Mass number = 8 + 8 = 16 g
Mass number of Y = Proton number + Neutron number = 8 + 10 = 18 g
The atomic number of both the atoms is 8 and mass number is different so they are isotopes.

10. Give one word of the following;
(i) Positively charged atom.
(ii) A group of atoms carrying a charge?
Ans. (i) Positively charged atom - Cation.
(ii) A group of atoms carrying a charge - Polyatomic ion.

11. (a) J. Chadwick discovered a subatomic particle that has no charge and has mass nearly equal to that of the proton. Name the particle and give its location in the atom. 
(b) K and L shells of an atom completely filled electrons, then what would be 
(i) The total number of electrons in the atom and 
(ii) It's valency?
Ans. (a) The particle is neutron and it is present in the nucleus of the atom.
(b) If k and L shells of an atom are completely filled electrons, then what would be 
(i) The total number of electrons in the atoms in which K and L shells are completely filled electrons is 10.
(ii) Valency is = 0 
Reason : E .C = 2,8 , It is an inert element.

12.  State three features of the nuclear model of an atom put forward by Rutherford. 
Ans. Rutherford put forward the nuclear model of an atom, which had the following features: 
(i) There is a positively charged center in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.
(ii) The electrons revolve around the nucleus in well-defined orbits. 
(iii) The size of the nucleus is very small as compared to the size of the atom.

13. Mention the postulates Niels Bohr put forward to overcome the objection raised against Rutherford's atomic model.
Ans. In order to overcome the objections raised against Rutherford’s model of the atom, Neils Bohr put forward the following postulates about the model of an atom: 
(i) Only certain special orbits are known as discrete orbits of electrons are allowed inside the atom. 
(ii) While revolving in discrete orbits the electrons do not radiate energy. 


14. An element X has mass number 27 and it contains 13 protons.
(a) Write the symbolic representation of the element.
(b) Find the number of Neutrons in the element.
(c) Write the name of the element.
Ans. (a) The atomic number of an element is 13 so the symbolic representation of the element X will be Al. 
(b) The number of Neutrons in the element = mass number - atomic number 
= 27 - 13 = 14
(c) The name of the element is aluminium.

15. Sulphur dioxide is a colorless pungent-smelling gas and is a major air pollutant.
(a) Write the electronic configuration of its constituent element Sulphur and Oxygen. (Given - ³²S₁₆, ¹⁶O₈ )
(b) Write the valency of Sulphur and Oxygen.
(c) Are sulphur and Oxygen isotope? Explain your answer.
Ans. (a) The electronic configuration of S whose atomic number is 16 = 2, 8, 6
The electronic configuration of O whose atomic number is 8 = 2,6(b) The valency of both sulpher and oxygen  = 8 - 6 = 2 
(c) No, sulpher and oxygen are not isotopes because their atomic number is not same.

16. Calculate the average atomic mass of chlorine if it exists commonly in two isotopes. ³⁵Cl₁₇ (75%)   and   ³⁷Cl₁₇ (25%)
Ans. The average atomic mass of chlorine if it exists commonly in two isotopes ³⁵Cl₁₇ (75%)   and   ³⁷Cl₁₇ (25%)
 (75 x 35)/100 + (25 x 37)/100 = 26.25 + 9.25 =35.50

17. (a) What are the canal rays?  
(b) State three characteristics of canal rays.

Ans. (a) Canal rays - Canal rays are α-particles which are doubly-charged helium ions. Since they have a mass of 4 u, the fast-moving canal ray or α-particles have a considerable amount of energy.
(b) Three characteristics of canal rays - 
i. Canal rays are positively charged particles ii. fast-moving ray and deflected by the electric and magnetic field. iii.α-particles are much heavier than the protons. 

18. If ¹⁴X₆  is an isotope of element Y and has 2 neutrons less than X.
(a) Write the symbol of element Y showing its mass number and atomic number.
(b) Write the valency for these elements. What relation is given for the chemical properties of  X and Y.
Ans. (a) If ¹⁴X₆  is an isotope of element Y and has 2 neutrons less than X  then the atomic number of Y = 6 and atomic mass = 14 - 2 = 12.
(b) Electronic configuration of X and Y = 2, 4 ( At. no. = 6)
The chemical properties of X and Y are the same.

19. An element ¹⁴X₁₂ loses two electrons to form a cation which combines with the anion of element  ³⁵X₁₇ formed by gaining an electron.
(a) Write the electronic configuration of element X
(b) Write the electronic configuration of the anion of Y
(c) Write the formula of the compound formed by the combination of X and Y.
Ans. (a) The atomic number of X is 12.
E.C = 2,8,2   Valency = +2 , Cation
(b) The atomic number of Y is 17
E.C = 2,8,7   Valency = -3 , Anion
(c) The formula of the compound will be = X₃Y₂.

21. (i) ²⁴Mg₁₂ and ²⁶Mg₁₂  are symbols of two isotopes of Mg. Compare atom of these isotopes with respect to
(a) Composition of their nuclei.
(b) Electronic configuration and valency.
(ii) Give the reason why two isotopes of magnesium have different mass numbers?
Ans. (i) (a) The atomic number of two isotopes of Mg are 24 and 26 
In ²⁴Mg₁₂ - Number of Protons = 12, Number of Electrons = 12 and the  
Number of Neutrons = 24 - 12 = 12
In ²⁶Mg₁₂ - Number of Protons = 12, Number of Electrons = 12 and the  
Number of Neutrons = 26 - 12 = 14.
(b) E.C of ²⁴Mg₁₂ - 2,8,2  ( At. No. = 12)  Valency = 2
 E.C of ²⁶Mg₁₂ - 2,8,2 ( At. No. = 12)  Valency = 2
(ii) Two isotopes of magnesium have different mass numbers due to different in neutron numbers. 

22. Two elements X and Y combine in a ratio of 3 : 8  by mass and the compound Z is formed. Z is one of the essential components of photosynthesis to take place. If Z is also one of the greenhouse gases:
(a) Identify X,Y, and Z.
(b) Write the electronic configuration of X and Y.
(c) Write the atomicity of the molecule of Z. 
Ans. Carbon dioxide is an essential component of photosynthesis to take place and it is also one of the greenhouse gases.
So two elements X and Y which are C and O combine in a ratio of 3 : 8  by mass and the compound Z that is CO₂ is formed.
(b) E.C of C = 2,4  ( At.No. = 6) and E.C of O = 2,6  (At. No.= 8)
(c) The atomicity of  CO₂ = 3  ( One atom of carbon and two atoms of oxygen)

23. If the mass no. of an element is 23 and its atomic no.is 11 then
(a) Write its electronic configuration and its valency.
(b) Find the no. of neutrons in its nucleus.
(c) Mention the types of the ion formed by it.
Ans. Mass no. 23 and At. No. = 11 then
(a) E.C = 2,8,1   and  Valency = 1
(b) No. of neutrons = Mass No. -  Atomic number = 23 - 11 = 12 
(c) It formes negative ion meas anion.


24. State the difference between isotopes and isobar. Give an example of each.
Ans. Atoms of the same elements with the same atomic numbers, which have different mass numbers, are known as isotopes. They have different physical properties due to the different number of neutrons and the same in chemical properties. Ex - ¹²C₆ , ¹³C₆ ,¹⁴C₆ .
Atoms of different elements with different atomic numbers, which have the same mass number, are known as isobar. They have different chemical properties due to the different number of electrons and the same in physical properties. Ex- ²³Ne₁₀   ²³Na₁₁ 

25. Name the scientists who discovered Proton,  Electron, and Neutron. What is the charge and mass of a neutron?
Ans. In 1920 Rutherford discovered Proton, J.J. Thomson (1856- 1940), a British Physicists discovered electrons and in 1932, J. Chadwick discovered neutron which is another subatomic particle which had no charge and a mass nearly equal to that of a proton. 
The charge of the neutron is zero and the mass is 

1.674929 x 10^-27 kg, nearly equal to that of a proton. 

III. Long answer type questions:

1. Give reasons for the following:
(a) An atom is electrically neutral.
(b) Noble gases show the least reactivity.
(c) The nucleus of an atom is heavy and positively charged.
(d) Ions are more stable than the atom.
Ans. (a) An atom is electrically neutral because atom has the same number of electrons (negatively charged ) and protons ( positively charged). After losing or gaining electrons atom becomes an ion.
(b) Nobel gases show the least reactivity because noble gases are stable elements as they have a filled valence shell of electrons and no need to lose or gain electrons to gain a stable electronic configuration. 
(c) The nucleus of an atom is heavy and positively charged due to the presence of neutrons and protons.
(d) Ions are more stable than the atom because ions are formed by gaining or losing electrons due to which they gain stable electronic configuration.    

2. (a) Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have?
(b)The atomic mass of an element X is 16.2 u. What is the % age of isotope X having atomic number 8 and mass numbers 16 and 18?
Ans. (a) The atomic mass of Helium is 4u.
Atomic mass no. = no. of protons + no. of neutrons 
The number of neutrons = Atomic mass no. - no. of protons
The number of neutrons = 4 - 2 = 2
(b) The atomic mass of X = 16.2 u 
Let the % of the isotope of X with mass number 16 be = a and 
the % of the isotope of X with mass number 18 be =100 - a
A/Q   16.2 = a/100 x 16 + (100 - a)/100 x 18
         16.2 = 16a/100 + 1800/100 - 18a/100
        18a/100 - 16a/100 = 18 -16.2 = 1.8
        2a/100 = 1.80
         a        = 1.80 x 100/2 = 180/2 = 90%
      100 - a = 10 %
 The % age of isotope X is,having atomic number 8 and mass numbers 16 and 18 10% and 90%.

3. How many electrons, protons and neutrons are there in an element  ¹⁹ X ₉ ? What will be the valency?
(a) Define isotope and isobar with examples.
(b) What do the species ⁴B₂ and ⁴C₃ represent?
Ans. (a) Isotope - Atoms of the same elements with the same atomic numbers, which have different mass numbers, are known as isotopes. They have different physical properties due to the different number of neutrons and the same in chemical properties. Ex - ¹²C₆ , ¹³C₆ ,¹⁴C₆ .
Isobar - Atoms of different elements with different atomic numbers, which have the same mass number, are known as isobar. They have different chemical properties due to the different number of electrons and the same in physical properties. Ex- ²³Ne₁₀   ²³Na₁₁ 

(b) ⁴B₂ represents Boron with atomic mass number 4 and atomic number 2 means the number of electrons 2 and the number of protons 2.

  ⁴C₃ represents Carbon with atomic mass number 4 and atomic number 3 means the number of electrons 3 and the number of protons 3.

Both have the same atomic mass and different atomic number, so they are isobar.

They are different in chemical properties but the same in physical properties.

4. (a) List three subatomic particles of an element with its charge.
(b) A helium atom has an atomic mass 4 u and it's atomic no. is 2. How many neutrons does it have?
Ans. (a) Three subatomic particles are:
(i) Electron- Negative charge
(ii) Proton - Positive charge and 
(iii) Neutron - Neutral. 
(b) At. mass of He = 4 u and At. No = No. of protons = 2
No. of neutrons - At. mass no. - No. of electrons = 4 - 2 = 2
No. of neutrons = 2

5. Study the following table and answer the questions:

Elements	Protons	Neutrons	Electrons
A	2,	3,	2
B	10	9,	10
C	17	20	16
D	17	17	17

E	18	19	18
F	17	20	17

 (a) What is the mass no. of elements A and B?
(b) What is the atomic no. of element B?
(c) Which two elements represent a pair of isotope and why?
(d) What is the valency of B and D?
(e) Identify the cation.
(f)  Identify a pair of isobar.
Ans. (a) The mass no. of elements A = No. protons + No. of neutrons
= 2 + 3 = 5
The mass no. of elements B = No. protons + No. of neutrons
= 10 + 9 = 19
(b)The atomic no. of element B is = No of protons = No. of electrons = 10
(c) D and F represent a pair of isotope because they have the same no. of atomic no. that is 17 ( At. no = no. of protons = no. of electrons )
(d) At. no of B = 10 and E.C = 2,8. It is a stable element, so valency is zero.
At. no. of D = 17 and E.C = 2,8,7. Valency = 8 - 7 = 1.
(e) Cation - Cation is a positively charged ion, that would be attracted to the cathode in electrolysis.
(f) A pair of isobar = ⁴⁰K, and ⁴⁰Ca.
Isobars are atoms of different chemical elements that have the same number of nucleons.  
                                                   
6. List three differences between electrons, protons, and neutrons.
What observation in alpha-particle scattering experiment is the Rutherford to make the following:
a) Most of the space in an atom is empty. 
b) The positive charge of the atom occupies very little space.
c) The mass of the atom is concentrated in a very small volume within the atom.
Ans. Three differences between electrons, protons, and neutrons:
 Electron - 
Following observations were made: 
a) Most of the space in an atom is empty - Most of the fast-moving α-particles passed straight through the gold foil. 
b) The positive charge of the atom occupies very little space - Very few particles were deflected from their path.
c) The mass of the atom is concentrated in a very small volume within the atom - A very small fraction of α-particles were deflected by 180 degrees.

7. (a) Which isotope is used in the treatment of cancer?
(b) the particle contains 11 protons and 10 electrons. Write the symbol and name of the particles.
(c) What do you think would be the observation if the alpha-particle 
scattering experiment is carried out using a foil of a metal other than gold.
Ans. (a) An isotope of cobalt is used in the treatment of cancer.
(b) Number of protons = 11 and number of electrons = 10
The atomic number is 11 so the element is sodium and the symbol is Na.
(c) If the alpha particle scattering experiment is carried out using a foil of a metal other than gold, there would be no change in the observation. Rutherford selected a gold foil because he wanted as thin a layer as possible. Gold is malleable and a thin foil of gold can be easily made.

8. (a) State the limitations of J.J. Thomson’s model of an atom.
(b) Define valency by taking the example of Mg and O₂
(c) S^-2 has completely filled K, L and M shells. Find its atomic number.
Ans. (a) According to Rutherford the orbital revolution of the electron is not expected to be stable. But any particle in a circular orbit would undergo acceleration. During acceleration, charged particles would radiate energy and lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know. 
(b) Valency - Valency of an atom is the number of electrons lost, gained or shared by an atom during the course of the chemical reaction. 
Atomic number of Mg = 12 and electronic configuration is - 2,8,2 
Valence electron is 2. 
Mg  - 2e → Mg⁺² Magnesium atom loses 2 electrons so valency is 2.