Solution
UNIT TEST III SET - B
Class 11 - Physics
Section A
1.
(d) 32 N
Explanation:
Mm
Fsurface = G 2
Re
F Rc = G
Mm
2
= 4
9
× F
surface
=
4
9
× 72 = 32 N
Re + Re
2
( )
2
2.
(d) increases by 2%
Explanation:
g= GM
2
R
For constant G and M
Δg
g
× 100 = −2
ΔR
R
× 100 = -2(-1)% = + 2%
The value of g increases by 2%.
3. (a) g ∝ r
Explanation:
Inside the earth g ′
=
4
3
πρGr
∴ g ∝ r
4.
(b) towards m2 on m1 and towards m on m 1 2
Explanation:
Since the gravitational force is attractive in nature, one Force will be on m1 which will be directed towards m 2. i.e.,
→ →
m1 m2 m1 m2
F 12 = G
2
and other force will be on m2 which will be directed towards m 1 i.e., F 21 = −G
2
r21 r12
→ →
Clearly, It can be seen that F 12 = −F 21 .
5. (a) Both A and R are true and R is the correct explanation of A.
Explanation:
Gm1 m2
According to Newton's law of gravitation, F = 2
r
G(2m1 )(2m2 )
When m1, m2 and r all are doubled, F = 2
(2r)
Gm1 m2
2
, i.e. F remains the same.
r
6.
(c) A is true but R is false.
Explanation:
Variation of g with depth from surface of earth is given by g' = g (1 − d
R
)
At the centre of earth, d = R,
∴ g' = g (1 − R
R
) =0
∴ Apparent weight of body = mg' = 0
Section B
7. according to the relation between g and G is
acceleration due to gravity is g
1/5
radius = r
mass = M
and Gravitational constant =G
4 3
Gρ πr
GM 3 4π
g = = = Gρr
r2 r2 3
g ∝ r
g r1
1
=
g r2
2
OR
Radius of earth = R
density of earth = p
GM
acceleration due to gravity = g = 2
R
4 3
M = πR ρ
3
4
g = πGRρ
3
acceleration due to gravity on the planet
′ 4 ′ ′
g = πGR ρ
3
The gain in PE at the highest point will be same in both cases. Hence
mg h = mgh
′ ′
4
m× πGRρh
mgh 3
′
h = =
mg ′ 4 ′ ′
m πG R ρ
3
′ ′
Rρh 3R ×4ρ ×1.5
= ′
= ′
′ R×ρ
R ρ
= 18 m
8. i. Since the acceleration due to gravity reduces by 36%, the value of acceleration due to gravity there is
100 - 36 = 64%
It means, g' = g. if h is the height of location above the surface of earth, then
64
100
2
g' = g R
2
(R+h)
or 64
100
g=g R
2
(R+h)
10
=
R+h
R
or 8R + 8h = 10R
= 1.6 × 106m.
6
⇒ h= 2R
8
=
R
4
=
6.4×10
ii. The radius of the earth
R = 6400 km
Now, calculate the height from the surface of the earth at which the value of g is reduced by 36 from the value at the surface of
earth.
g' = 36% of g ...(ii)
And now comparing equations
36% of g = g (1 − 2h
R
)
36 2h
⇒ ( ) g = g (1 − )
100 R
2h
⇒ 0.36 = 1 - R
⇒
2h
R
= 1 - 0.36
⇒
2h
R
= 0.64
⇒ 2h = (0.64) × R
⇒ h=( 0.64
2
) × R
⇒ h = 0.32 × R
⇒ h = 0.32 × 6400
⇒ h = 2048 km
Hence, 2048 km height.
GMm
9. Gravitational PE of mass m in orbit of radius R = U = − R
GMm
∴ Ui =
2R
GMm
Uf = −
3R
2/5
Energy required = Potential energy of the Earth(mass system when mass is at distance 3R) – Potential energy of the Earth (mass
system when mass is at distance 2R)
1 1
ΔU = Uf − Ui = GM m [ − ]
3 2
GMm
=
6R
Section C
10. Let A and B be the positions of the two spheres and P be the midpoint of AB.
Gravitational field at P due to mass at A,
G × 100
E1 =
GM
2
=
2
, along PA
r (0.5)
Gravitational field at P due to mass at B,
G × 100
E2 =
2
, along PB
(0.5)
As E1 and E2 have equal magnitudes but opposite directions, so the resultant gravitational field at P is zero.
As gravitational potential is a scalar quantity, so total potential at P is
GM GM 2GM
V = VA + VB = − r
−
r
= −
r
−11
= -2.668 × 10-8 J kg-1
2 × 6.67 × 10 × 100
=− 0.5
11. The work obtained in bringing a body from infinity to a point in gravitationa field is called gravitational potential energy.
Force of attraction between the earth and the object when an object is at the distance a from the center of the earth.
GmM
F =
2
x
Consider the dW is the small amount of work done in bringing the body without acceleration through the very small distance dx.
dW = Fdx
And total work done to bring the body from infinity to point p which is at distance r from the center of the earth.
r
GmM GMm
w = ∫ dx = −
x2 r
∞
Hence, the gravitational potential energy = − GMm
12. a. T = 7h 39min = 459 × 60s , orbital radius is given by R and
R = 9.4 × 10 km = 9.4 × 10 m , M = ?
3 6
m
∴ Mass of mars is given by, M
2 3
4π R
= ⋅ m
G T 2
2 6 3
4×(3.14) × (9.4× 10 )
= −11 2
6.67× 10 ×(459×60)
= 6.48 × 10 23
kg
2 3
Tm R
b. Using Kepler's third law, =
MS
2 3
T R
ES
where RMS is the mass-sun distance RES is the earth-sun distance
3/2
∴ Tm = (1.52) × 365 = 684 days
OR
The projectile is acted upon by two mutually opposing gravitational forces of the two spheres. The neutral point N (see Fig.) is
defined as the position where the two forces cancel each other exactly. If ON = r, we have
GMm 4GMm
=
2 2
r (6R−r)
(6R - r)2 = 4r2
6R - r = ± 2r
r = 2R or - 6R
The neutral point r = – 6R does not concern us in this example. Thus ON = r = 2R. It is sufficient to project the particle with a
speed that would enable it to reach N. Thereafter, the greater gravitational pull of 4M would suffice. The mechanical energy at the
surface of M is
1 2 GMm 4GMm
El = mv − −
2 R 5R
At the neutral point N, the speed approaches zero. The mechanical energy at N is purely potential.
GMm 4GMm
EN = − −
2R 4R
3/5
From the principle of conservation of mechanical energy
1 2 GM 4GM GM GM
v − − = − −
2 R 5R 2R R
or
2 2GM 4 1
v = ( − )
R 5 2
1/2
3GM
v = ( )
5R
A point to note is that the speed of the projectile is zero at N, but is nonzero when it strikes the heavier sphere 4M.
Section D
13. Read the text carefully and answer the questions:
The figure shows the schematic drawing of Cavendish's experiment to determine the value of the gravitational constant. The bar
AB has two small lead spheres attached at its ends. The bar is suspended from a rigid support by a fine wire. Two large lead
spheres are brought close to the small ones but on opposite sides as shown. The name of G from this experiment came to be
6.67 × 10-11 N-m2/kg2
(i) (b) equal and opposite
Explanation:
equal and opposite
(ii) (b) zero
Explanation:
zero
(iii) (b) Both zero and non-zero
Explanation:
Both zero and non-zero
(iv) (b) restoring torque of the wire equals the gravitational torque
Explanation:
restoring torque of the wire equals the gravitational torque
Section E
14. Let the gravitational field due to the two stars be zero at some point O lying at a distance x from the centre of smaller star.
(16M)m
Then G Mm
2
= G
2
x (10a − x)
or 1
2
=
16
2
x (10a − x)
or 16x2 = (10a - x)2
or 4x = ± (10a - x)
The negative sign is inadmissible, so x = 2 a
The body of mass m, when fired from point P lying on the surface of heavier star, must cross the threshold (the point O),
otherwise, it would return back.
The gravitational potential energies when the body of mass m lies at positions P and O are given by
GMm G × 16M × m 65GMm
Up = − − = −
8a 2a 8a
GMm G × 16M × m 5GMm
UO = − − = −
2a 8a 2a
4/5
∴ Increase in potential energy,
ΔU = Up - UO = − +
5GMm
2a
65GMm
8a
=
45GMm
8a
If v is the minimum speed with which the body is fired from P so as to reach O, then
1 2 45GMm
mv = ΔU =
2 8a
−−−−− −−−−
45 GM 3 5GM
or v = √ 4 a
=
2
√
a
OR
The force (F) of gravitational attraction on a body of mass m due to earth of mass M and radius R is given by
F = G ...(i)
mM
2
R
We know from Newton's second law of motion that the force is the product of mass and acceleration.
F = ma
But the acceleration due to gravity is represented by the symbol g. Therefore, we can write
F = mg ...(ii)
From the equation (i) and (ii), we get
mg = G or g =
mM
...(iii)
2
GM
2
R R
When body is at a distance 'r' from centre of the earth then g = GM
2
.
R
a. The acceleration due to gravity on the earth is given by
g =
GM
2
i.e. GM = R2g ...(1)
R
The acceleration due to gravity at height 'h' from the surface of the earth is given by
i.e. GM = (R + h)2gh ...(2)
GM
gh =
2
(R+h)
From (1) and (2) we have,
gh 2
R
=
g 2
(R+h)
gh 2h
= (1 − )
g R
b. The acceleration due to gravity on the surface of the earth is given by.
GM
g= 2
...(i)
R
let 'Q' be the density of the material of the earth.
Now, mass = volume × density
M = πR × ρ 4
3
3
Substituting in equation (1) we get
G 4 3 4
g = × πR × ρ = πGRρ
2 3 3
R
g =
4
3
πGRρ ...(ii)
gd =
4
3
πG(R − d)ρ ...(iii)
dividing equation (iii) by (ii) we get
g R−d
d d
= = (1 − )
g R R
d
gd = g (1 − )
R
5/5
