Solution
HOME WORK
Class 09 - Science
Section A
1. The earth does not fall into the sun because the earth remains in its circular orbit due to the gravitational force acting on it.
2. Both will reach the ground at the same time because acceleration due to gravity is independent of the masses of freely falling
bodies.
3. We know that,
Weight of object on the moon = ( ) × its weight on the earth.
1
That is,
We
Wm = =
10
N
tre
6 6
= 1.67 N
Thus, the weight of the object on the surface of the moon would be 1.67 N.
4. The Earth and the Moon experience equal gravitational forces from each other.However the mass of the Earth is much larger than
en
the mass of the moon.Hence it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth.For this reason
Earth does not move towards the moon.
5. We have given
gc
Mass, m = 10 kg
Acceleration due to gravity, g = 9.8 ms-2
W=m×g
W = 10 kg × 9.8 ms-2 = 98 N
hin
Thus, the weight of the object is 98 N.
Section B
6. Mass of object (m) = 10 kg
ac
Acceleration due to gravity on earth (ge) = 9.81 ms-2
Acceleration due to gravity on moon (gm)
Co
= 9.81
6
ms-2
Weight of the object on the earth = mge
= 10 × 9.81 = 98.1 N
Weight of the object on the moon = mg m
ya
10×9.81
= 6
= 16.35 N
7. A weight of an object is directly proportional to the mass of the earth (M) and inversely proportional to the square of the radius of
the earth (R), i.e.,
ag
A weight of object ∝ M
2
R
Original weight, Wo = mg = mG M
2
..........(1)
R
Pr
R
According to question, hypothetically, Mass of earth,M' = 4 M and Radius of earth, R' = 2
.
′
M
Now, New weight an object =Wn =mg' = mG ′2
..........(2)
R
Now, put the values of M' and R' in equation (2), we get.
Weight of an object, Wn = mG = (16 mG)
4M
2
= 16 × Wo M
2
R R
( )
2
The weight will become 16 times.
8. Initial velocity of stone (u) = 0
Final velocity of stone (v) = ?
Height attained (S) = 19.6 m
Acceleration due to gravity (g) = + 9.8 ms-2
We know : v2 = u2 + 2 gS
v2 = (0)2 + 2 × 9.8 × 19.6
v2 = 19.6 × 19.6
1/3
−−−−−−−−−
v = √19.6 × 19.6
v = 19.6 ms-1
9. For both the stones, Their initial velocity, u = 0
Acceleration in downward direction = g
Using equation, S =ut + 1
2
at2
Now, h = ut + 1
2
gt2
⇒ h=0+ 1
2
gt2
⇒ h= 1
2
gt2
−−
⇒ t=√ 2h
Since time period depends upon the height and acceleration due to gravity. Here, both stones fall from same height and planet.
Therefore, Both stones will take the same time to reach the ground.
10. It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the
tre
pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area the larger wold be the
pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence pressure exerted on the shoulder is
very large.
en
Section C
11. As u = 0, using equation, S= ut + 1
2
at2 , We have,
gc
h1 = 1
2
gt
2
1
⇒ h2 = 1
2
gt
2
2
,
−−
t1 h1
⇒
t2
=√ h2
hin
Since, time period does not depends upon the mass and size. Therefore, Ratio will not change in either case because acceleration
remains the same.
12. Height of coconut (S) = 15 m
Acceleration due to gravity (g) = -10 ms-2
ac
Time (t) = ?
Initial velocity (u) = 20 ms-1
Co
Using S = ut + 1
2
at2 we have
15 = 20 × t + 1
2
× (-10)t2 . This is a quadratic equation in time.
Simplifying we have
5t2 - 20 t + 15 = 0 or t2 - 4t + 3 = 0
ya
2
−(−4)± √(−4) −4×1×3
t= 2
= 4±2
2
= 1 or 3 s = 1 or 3 s
Thus the projectile will pass the coconut 1s after it is thrown while moving in the upward direction and 3 s while moving in the
ag
downward direction on its return journey.
13. Given, the mass of the mars, Mm = 6.39 × 1023 kg
Pr
The mass of the Jupiter, Mj = 1.89 × 1027 kg
The distance between the mars and jupiter,
d = 7.49 × 105 m
G = 6.7 × 10-11 Nm2 /Kg2
The force exerted by the Jupiter on the mars,
= 1.44 × 1029 N
Mm × Mj −11 23 27
F=G 2
= 6.7× 10 ×6.39× 10
2
×1.89× 10
d 5
(7.49× 10 )
Thus, the force exerted by the Jupiter on the mars is 1.44 × 1029 N.
14. Case (i) = For the first ball
Initial velocity (u) = 0
Final velocity (v) = ?
Acceleration due to gravity (g) = 10 ms-2
Distance of fall (S) = 20 m
2/3
Time of fall (t) = ?
Using S = ut + at we have time of fall of the first ball
1
2
2
1 2
20 = 0 × t + × 10 × t
2
2
=> t = 4
=> t = 2 s
Since the second ball is thrown 1 second later. Therefore for the two balls to reach the ground at the same time, the second ball
should in motion for t = 2 - 1 = 1 second
1
Therefore using the equation S = ut + 2
at
2
we have
1 2
20 = u × 1 + × 10 × 1
2
−1
=> u = 15ms
15. As we know,
Gm1 m2
F =
2
[symbols have their usual meanings]
r
Gm1 m2
i. r′
= 3r ⇒ F
′
= =
F
[force decreases by 9 times]
tre
2 9
9r
4Gm1 m2
ii. m ′
1
= 2m1 and m
′
2
= 2m2 ⇒ F
′
=
2
= 4F [force increases by 4 times]
r
4Gm m
iii. m ′
1
= 2m1 and m
′
2
= 2m2 r' ⇒ F ′
=
1 2
= F [force remains unchanged]
4r2
en
Section D
16. i. Let the two bodies have masses m1 and m2 and they are placed at the same distance R from the centre of the earth. According
to the question, if the same force acts on both of them, then
gc
GMm1
F1 = 2
....(i)
R
GMm2
and F2 = 2
....(ii)
R
As, F1 = F2
hin
GMm1 GMm2
Hence, 2
= 2
R R
So, m1 = m2, their masses will be the same.
GM
ii. Mathematically, g =
ac
2
R
Where, g = acceleration due to gravity
G = universal gravitational constant, M = mass of the earth and R = radius of the earth
Co
iii. G is known as the universal gravitational constant because its value remains the same all the time everywhere in the universe,
applicable to all bodies whether celestial or terrestrial.
17. The value of g on the earth is 9.8 m/s2
i. a. g on the moon is given by
ya
m/s2
g 9.8
g' = 6
= 6
= 1.63
b. Mass of the person on the moon = 110.84
1.63
= 68 kg
c. Mass will be constant and does not change from place to place. Hence the mass of the person on the earth is the same that
ag
on the moon.
Weight of person on the earth = mg = 68 × 9.8 = 666.4 N
ii. According to the Newton's law of gravitation, the force of attraction between earth and the body is given by
Pr
F= ...(i)
GMm
2
R
where, M = mass of the earth, R = radius of the earth, m = mass of person and G = 6.67 × 10-11 N-m2/kg2
Force produces an acceleration 'g'. So from Newton's second law, F = mg ....(ii)
Equating (i) and (ii) we get,
mg = GMm
2
R
GM
∴ g= 2
R
3/3
