Gravitation Test

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Gravitation Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $$V$$. Due to the rotation of planet about its axis the acceleration due to gravity $$g$$ at equator is $$\dfrac{1}{2}$$ of $$g$$ at poles. The escape velocity of a particle on the pole of planet in terms of $$V$$.

A
$$V_e = 2V$$

B
$$V_e = V$$

C
$$V_e = V/2$$

D
$$V_e = \sqrt{2}V$$

Solution
Question 2
1 / -0

At what height over the earth's pole does the freefall acceleration decreases by 1% ?

A
$$64 km$$

B
$$16 km$$

C
$$8 km$$

D
$$32 km$$

Solution

$$mg'=\dfrac{GMm}{R^2+h^2}$$

$$mg=\dfrac{GMm}{R^2}$$

$$\dfrac{g'}{g}=\dfrac{R^2}{R^2+h^2}$$
$$\dfrac{g'}{g}=(R^2)(R^2+h^2)^{-2}$$
$$\dfrac{g'}{g}=(1+\dfrac{h^2}{r^2})^{-2}$$

Expanding using binomial expansion and neglecting higher terms,
$$\dfrac{g'} {g}= \displaystyle \left ( 1 - \dfrac{2h}{R} \right ) = \dfrac{99}{100} or h = 32 km$$
Question 3
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The condition for a uniform spherical mass $$m$$ of radius $$r$$ to be a black hole is [$$G= $$gravitational constant, $$g=$$ acceleration due to gravity].

A
$$\displaystyle\left [ \dfrac{2Gm}{r} \right ]^{1/2}\leq c$$

B
$$\displaystyle\left [ \dfrac{2gm}{r} \right ]^{1/2}= c$$

C
$$\displaystyle\left [ \dfrac{2Gm}{r} \right ]^{1/2}\geq c$$

D
$$\displaystyle\left [ \dfrac{gm}{r} \right ]^{1/2}\geq c$$

Solution

For a uniform spherical mass to be a black hole, total energy of the "particle + spherical mass" system $$\leq 0$$
Let the mass of particle be $$m_1$$ moving with its maximum speed $$c$$.
$$\therefore$$ $$\dfrac{1}{2}m_1 c^2 + (-\dfrac{Gmm_1}{r}) \leq 0$$
$$\implies \ c\leq\bigg[\dfrac{2Gm}{r}\bigg]$$
Question 4
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A body is projected up with a velocity $$2$$ times the escape velocity $$v_{e}$$ from the surface of the earth. The velocity of the body at a point far away the earth's attraction is

A
$$v_{es}$$

B
$$\sqrt{7} v_{es}$$

C
$$\sqrt{2} v_{es}$$

D
$$\sqrt{3} v_{es}$$

Solution

$$v_e = (\dfrac{2GM}{R})^{0.5}$$
now, $$V = 2v_e = 2(\dfrac{2GM}{R})^{0.5} = (\dfrac{8GM}{R})^{0.5}$$

By conservation of energy, $$ \dfrac{-GMm}{R} +\dfrac{8GMm}{2R} = 0 + \dfrac{mv^2}{2}$$

solving this, we get $$v = (\dfrac{3 \times 2GM}{R})^{0.5} = 3^{0.5}v_e$$
Question 5
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A particle of mass $$10 g$$ is kept on the surface of a uniform sphere of mass $$100 kg$$ and radius $$10 cm$$. Find the work to be done against the gravitational force between them to take the particle far away from the sphere

A
$${\displaystyle 13.34\times 10^{-10}J}$$

B
$${\displaystyle 13.33\times 10^{-10}J}$$

C
$${\displaystyle 6.67\times 10^{-9}J}$$

D
$${\displaystyle 6.67\times 10^{-10}J}$$

Solution

$$W=\Delta PE=\left ( \dfrac{-GMm}{\infty } \right )-\left ( \dfrac{-GMm}{R} \right )=\dfrac{6.67\times10^-11 \times 100\times 10\times 10^{-3}}{1}= 6.67\times10^{-10} J$$
Question 6
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The change in the value of $$g$$ at a height $$h$$ above the surface of the earth is the same as at a depth d below the surface of the earth.When both $$h$$ and d are much smaller than the radius of earth,then which one of the following is true?

A
$$a=h/2$$

B
$$d=3h/2$$

C
$$d=2h$$

D
$$h=d$$

Solution

$$\displaystyle{{g}'=g \left ( 1-\frac{2h}{R} \right )=g \left ( 1-\frac{d}{R}\right )\therefore d=2h}$$
Question 7
1 / -0

A $$75 kg$$ astronaut is repairing Hubble telescope at a height of $$600 km$$ above the surface of the earth. Find his weight there.

A
$$740 N$$

B
$$700 N$$

C
$$650 N$$

D
$$610 N$$

Solution

$$\displaystyle mg' = mg \left ( 1 - \dfrac{2h}{R} \right )$$
$$= 75 \times 10 (1 - 0.185) = 610 N$$
Question 8
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If the orbital speed of moon is increased by $$41.4\%$$ then moon will

A
leave its orbit and will escape out.

B
fall on earth.

C
attract all bodies on earth towards it.

D
have time period equal to 27 days.

Solution

Velocity of the planet is $$v=\sqrt{\dfrac{GM}{a}}$$
Kinetic Energy is $$K=\dfrac{1}{2}m{v}^{2}=\dfrac{GMm}{2a}$$
Potential Energy is $$U=-{\dfrac{GMm}{a}}$$
$$Total\ energy= K+U= -{\dfrac{GMm}{2a}}$$
$${v}= \sqrt{\dfrac{GM}{a}}$$
So if$$,\quad v=1+0.414 v=1.414v= \sqrt{2} v= \sqrt{\dfrac{2GM}{a}}$$
Escape Velocity is $$v_e=\sqrt{\dfrac{2GM}{a}}$$
So the moon will escape.
Question 9
1 / -0

A spaceship is released in a circular orbit near the Earth's surface. How much additional velocity will have to be given to the spaceship in order to escape out of this orbit?

A
$$3.28 \quad m/s$$

B
$$3.28 \times 10^3 m/s$$

C
$$3.28 \times 10^7 m/s$$

D
$$3.28 \times 10^{-3} m/s$$

Solution

Here, orbital velocity and escape velocity are
$$v_0 = \sqrt{gR}$$
$$v_e = \sqrt{2gR}$$
Hence, required additional velocity will be,
$$v = v_e - v_o$$
$$v = \sqrt{2gR} - \sqrt{gR}$$
$$v = (\sqrt 2 - 1) \sqrt{gR}$$
$$v = (1.414 - 1)\sqrt{(10)(6400 \times 10^3)}$$..........$$($$since, $$g = 10 ms^{-2}$$ and $$R = 6400 \times 10^3 m)$$
$$v = (0.414)(8) \times 10^3$$
$$v = 3.3 \times 10^3 ms^{-1}$$
Question 10
1 / -0

Find the work done to take a particle of mass m from surface of the earth to a height equal to $$2R$$.

A
$$2$$ $$mg R$$

B
$$\displaystyle \frac{mg R}{2}$$

C
$$3 mg R$$

D
$$\displaystyle \frac{2mgR}{3}$$

Solution

$$Work\quad Done\quad =\quad \Delta \quad PE\\ PE=\quad \dfrac { GMm }{ r } \\ \Delta PE=GMm(\dfrac { 1 }{ R } -\dfrac { 1 }{ 3R } )=\dfrac { 2GMm }{ 3R } =\dfrac { 2 }{ 3 } mgR$$

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Gravitation Test - 30

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Gravitation Test - 30

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SHARING IS CARING

Question 1

$$V_e = 2V$$

$$V_e = V$$

$$V_e = V/2$$

$$V_e = \sqrt{2}V$$

Solution

Question 2

$$64 km$$

$$16 km$$

$$8 km$$

$$32 km$$

Solution

Question 3

$$\displaystyle\left [ \dfrac{2Gm}{r} \right ]^{1/2}\leq c$$

$$\displaystyle\left [ \dfrac{2gm}{r} \right ]^{1/2}= c$$

$$\displaystyle\left [ \dfrac{2Gm}{r} \right ]^{1/2}\geq c$$

$$\displaystyle\left [ \dfrac{gm}{r} \right ]^{1/2}\geq c$$

Solution

Question 4

$$v_{es}$$

$$\sqrt{7} v_{es}$$

$$\sqrt{2} v_{es}$$

$$\sqrt{3} v_{es}$$

Solution

Question 5

$${\displaystyle 13.34\times 10^{-10}J}$$

$${\displaystyle 13.33\times 10^{-10}J}$$

$${\displaystyle 6.67\times 10^{-9}J}$$

$${\displaystyle 6.67\times 10^{-10}J}$$

Solution

Question 6

$$a=h/2$$

$$d=3h/2$$

$$d=2h$$

$$h=d$$

Solution

Question 7

$$740 N$$

$$700 N$$

$$650 N$$

$$610 N$$

Solution

Question 8

leave its orbit and will escape out.

fall on earth.

attract all bodies on earth towards it.

have time period equal to 27 days.

Solution

Question 9

$$3.28 \quad m/s$$

$$3.28 \times 10^3 m/s$$

$$3.28 \times 10^7 m/s$$

$$3.28 \times 10^{-3} m/s$$

Solution

Question 10

$$2$$ $$mg R$$

$$\displaystyle \frac{mg R}{2}$$

$$3 mg R$$

$$\displaystyle \frac{2mgR}{3}$$

Solution

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