Gravitation The Law Of Falling(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

GRAVITATION THE LAW OF FALLING MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21. Determine the escape velocity of a rocket on the far side of a moon of a planet. The radius of the moon is

2.64 \times 10^{6}

m and its mass is

1.495 \times 10^{23}

. The mass of the planet is

1.9 \times 10^{27}

kg, and the distance between planet and the moon is

1.071 \times 10^{9}

m. Include the gravitational effect of planet and neglect the motion of the planet and the moon as they rotate about their CM.

Determine The Escape Velocity Of A Rocket On The Far Side Of...

1.560×102ms−1
1.560×103ms−1
2.8×104ms−1
1.560×104ms−1

Discuss Question

Answer: Option D. -> 1.560×104ms−1
:
D
Total potential energy of the rocket is

U = - G [\frac{M_{p} m}{(d + R_{m})} + \frac{M_{m} m}{R_{m}}]

v_{e}

is the escape velocity,we can write

\frac{1}{2} m v_{e}^{2} = - U

v_{e}^{2} = 2 G (\frac{M_{p}}{(d + R_{m})} + \frac{M_{m}}{R_{m}})

2 \times 6.67 \times 10^{- 11} (\frac{1.90 \times 10^{27}}{1.071 \times 10^{9} + 2.64 \times 10^{6}} + \frac{1.495 \times 10^{23}}{2.64 \times 10^{6}})

2.436 \times 10^{8}

v_{e} = 1.560 \times 10^{4} m s^{- 1}

Question 22. If W1, W2 and W3 represents the work done in moving a particle from A to B along three different paths 1, 2 and 3, respectively (as shown) in the gravitational field of a point mass m, find the correct relation between W1, W2 and W3.

W1 > W2 > W3
W1 = W2 = W3
W1 < W2 < W3
W2 > W1 > W3

Discuss Question

Answer: Option B. -> W1 = W2 = W3
:
B
Gravitational field is conservative in nature hence the work done does not depend on the path followed but only on initial and final positions

Question 23. In some region, the gravitational field is zero. The gravitational potential in this region

Must be variable
Must be constant
Cannot be zero
Must be zero

Discuss Question

Answer: Option B. -> Must be constant
:
B

I = \frac{- d V}{d r}

, If I = 0 then V = constant

Question 24. The mean radius of the earth is R, its angular speed on its own axis is

ω

and the acceleration due to gravity at earth's surface is g. The cube of the radius of the orbit of a geostationary satellite will be

R2gω
R2ω2g
Rgω2
R2gω2

Discuss Question

Answer: Option D. -> R2gω2
:
D

O r b i t a l v e l o c i t y v_{0} = \sqrt{\frac{G M}{r}} = \sqrt{\frac{g R^{2}}{r}} a n d v_{0} = r ω T h i s g i v e s r^{3} = \frac{R^{2} g}{ω^{2}}

Question 25. If

v_{e}

and

v_{o}

represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R, then

ve=vo
√2vo=ve
ve=vo√2
ve and vo are not related

Discuss Question

Answer: Option B. -> √2vo=ve
:
B

v_{e} = \sqrt{2 g R} a n d v_{0} = \sqrt{g R} ∴ \sqrt{2} v_{o} = v_{e}

Question 26. An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy

E_{0}

. Its potential energy is

−E0
1.5E0
2E0
E0

Discuss Question

Answer: Option C. -> 2E0
:
C
Potential energy =

2 \times (T o t a l e n e r g y) = 2 E_{0}

Because we know =

U = \frac{- G M m}{r} a n d E_{0} = \frac{- G M m}{2 r}

Question 27. A rocket of mass M is launched vertically from the surface of the earth with an initial speed V. Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is

R(gR2V2−1)
R(gR2V2−1)
R(2gRV2−1)
R(2gRV2−1)

Discuss Question

Answer: Option C. -> R(2gRV2−1)
:
C

Δ K . E = Δ U \Rightarrow \frac{1}{2} M V^{2} = G M_{e} M (\frac{1}{R} - \frac{1}{R + h}) \dots \dots (i) A l s o g = \frac{G M_{e}}{R^{2}} \dots \dots (i i) O n s o l v i n g (i) a n d (i i) h = \frac{R}{(\frac{2 g R}{V^{2}} - 1)}

Question 28. A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is

Discuss Question

Answer: Option A. -> R3
:
A
If body is projected with velocity v

v < v_{e}

then height up to which it will rise,

h = \frac{R}{\frac{v_{e}^{2}}{v^{2}} - 1}

v = \frac{v_{e}}{2} (g i v e n) ∴ h = \frac{R}{(\frac{v_{e}}{\frac{v_{e}}{2}})^{2} - 1} = \frac{R}{4 - 1} = \frac{R}{3}

Question 29. Four particles each of mass M, are located at the vertices of a square with side L. The gravitational potential due to this at the centre of the square is

−√32GML
−√64GML2
Zero
√32GML

Discuss Question

Answer: Option A. -> −√32GML
:
A

Four Particles Each Of Mass M, Are Located At The Vertices O...

Potential due to a point mass M at a distance r is

\frac{- G M}{r}

Potential at the centre due to single mass =

\frac{- G M}{\frac{L}{\sqrt{2}}}

Potential at the centre due to all four masses

= - 4 \sqrt{2} \frac{G M}{L} = - \sqrt{32} \times \frac{G M}{L}

Question 30. The value of

g

at a particular point on the earth surface is

9.8 m s^{- 2}

. Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass. The value of ‘g’ at the same point (assuming that the distance of the point from the centre of earth does not shrink) will now be: