Gravitation Test

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Gravitation Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

The gravitational P.E. of a rocket of mass 100 kg at a distance of $$10^{7}$$ m from the earths centre is $$-4\times 10^{9}$$ J. The weight of the rocket at a distance of $$10^{9}$$ m from the centre of the earth is :

A
$$4\times 10^{-2}$$ N

B
$$4\times 10^{-9}$$ N

C
$$4\times 10^{-6}$$ N

D
$$4\times 10^{-3}$$ N

Solution

Gravitational potential energy is given by
$$PE=-\dfrac {GMm}{R}=-4\times 10^9$$

But $$R = 10^7$$ and $$PE=-4\times 10^9$$ J

$$\Rightarrow GMm=4\times10^9\times10^7=4\times10^{16}$$
Now the weight of the satellite at a radius of $$10^9m$$ from earth's centre is given by,
$$mg=\dfrac {GMm}{R^2}=\dfrac {4\times10^{16}}{10^{18}}=4\times10^{-2}$$ N
Question 2
1 / -0

A sky laboratory of mass $$2\times 10^{3}Kg$$ is raised from a circular orbit of radius 2R to a circular orbit of radius 3R. The work done is (approximately):

A
$$10^{16}J$$

B
$$2\times 10^{10}J$$

C
$$10^{6}J$$

D
$$3\times 10^{10}J$$

Solution

$$W=U_{2}-U_{1}=\displaystyle \frac{+Mm}{3R}(-\frac{GMm}{2R})=\frac{GMm}{6R}=\frac{mgR}{6}$$
$$= W=\displaystyle \frac{2\times 10^{3}\times 9.8\times 6.4\times 10^{6}}{6}=2\times 10^{10}J$$
Question 3
1 / -0

The kinetic energy needed to project a body of mass $$m$$ from the earth surface to infinity is

A
$$\dfrac{1}{2}$$mgR

B
$$2 mgR$$

C
$$mgR$$

D
$$\dfrac{1}{4}$$mgR

Solution

Kinetic energy required $$=\displaystyle \frac{1}{2}mV_{e}^{2}$$$$=\displaystyle \frac{1}{2}m(\sqrt{2gR})^{2} = mgR$$
Question 4
1 / -0

If $$V_{e}$$ is the escape velocity of a body from a planet of mass M and radius R. Then the velocity of the satellite revolving at height h from the surface of the planet will be:

A
$$V_{e}\sqrt{\dfrac{R}{R+h}}$$

B
$$V_{e}\sqrt{\dfrac{2R}{R+h}}$$

C
$$V_{e}\sqrt{\dfrac{R+h}{R}}$$

D
$$V_{e}\sqrt{\dfrac{R}{2\left ( R+h \right )}}$$

Solution

Escape velocity, $$v_e = \sqrt{\dfrac{2GM_e}{R}}$$
and orbital velocity, $$v_o = \sqrt{\dfrac{GM_e}{(R+h)}}$$
On dividing both equations, $$v_o = v_e \sqrt{\dfrac{R}{2(R+h)}}$$
Question 5
1 / -0

A body of mass 'm' is taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be -

A
mg 2R

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

C
3 mgR

D
$$\displaystyle \frac{1}{3}mgR$$

Solution
Question 6
1 / -0

The ratio of Earth's orbital angular momentum (about the sun) to its mass is $$4.4 \times 10^{15}\ m^{2}\ s^{-1}$$. The area enclosed by the earth's orbit is approximately

A
$$1 \times 10^{22}\ m^{2}$$

B
$$3 \times 10^{22}\ m^{2}$$

C
$$5 \times 10^{22}\ m^{2}$$

D
$$7 \times 10^{22}\ m^{2}$$

Solution

Area enclosed by a body in an orbit is given by: $$\dfrac {dA}{dt}=\dfrac {L}{2m}$$
Total orbital area enclosed by the earth is the area covered in one revolution i.e 365days.
$$\therefore A=\dfrac {L}{2m}365\times 24\times 3600=\dfrac {4.4\times10^{15}}{2}365\times 86400=6.938\times 10^{22}$$ m$$^2$$
Question 7
1 / -0

At what height, is the value of g half that on the surface of earth? (R = radius of the earth)

A
0.414R

B
R

C
2R

D
3.5R

Solution

We know, g/g' = $$r^2$$/$$R^2$$, where g is gravitational force on earth, g' is gravitational force at a height.
Given g' = g/2
$$g' = \dfrac{gR^2}{r^2} = \dfrac{g}{2}$$
$$\Rightarrow r^2 = 2R^2$$
$$\Rightarrow r = R\sqrt{2} = R+h$$
$$\Rightarrow h = R(\sqrt{2}-1)$$
$$ = R(1.414-1)$$
$$ = 0.414 R$$
Question 8
1 / -0

Experimental value of G is

A
$$6.67\times 10^{-11} Nm^2/kg^2$$

B
$$66.7\times 10^{-11} Nm^2/kg^2$$

C
$$667\times 10^{-11} Nm^2/kg^2$$

D
None of these

Solution
Question 9
1 / -0

Maximum weight of a body is

A
at the center of the earth.

B
inside the earth.

C
on the surface of the earth.

D
above the surface of earth.

Solution

$$\textbf{Hint:}$$ Gravitational acceleration vary with height or depth.

$$\textbf{Step1:Weight of body}$$
Weight of a body at any place is given by,
$$W = mg$$
Where,
$$m$$ is mass of the body,
g is the value of acceleration due to gravity at the place where body is kept.

$$\textbf{Step2:Variation in gravitational acceleration}$$
Let the value of acceleration due to gravity at surface of Earth is $$g$$.
$$\bullet$$ If radius of Earth is $$R$$, the value of acceleration due to gravity at height $$h$$ from the surface of Earth is given by the formula,
$$g'=g\left( 1-\dfrac{2h}{R} \right)$$
$$\bullet$$ The value of acceleration due to gravity at depth $$h$$ from surface of Earth is given by the formula,
$$g'=g\left( 1-\dfrac{h}{R} \right)$$
$$\bullet$$ Therefore, the value of acceleration due to gravity is maximum at the surface of Earth.
Thus, the weight of the body $$( W=mg )$$ is maximum at the surface of Earth.
Question 10
1 / -0

The weight of an object in the coal mine, sea level and at the top of the mountain are respectively $$\omega_1 , \omega_2, \omega_3$$ then

A
$${\omega}_1 < {\omega}_2 > {\omega}_3$$

B
$${\omega}_1 = {\omega}_2 = {\omega}_3$$

C
$${\omega}_1 < {\omega}_2 < {\omega}_3$$

D
$${\omega}_1 > {\omega}_2 > {\omega}_3$$

Solution

For position $$(1)$$:
$$g_1=\dfrac{GM}{R^3}(R-d)$$, $$w_1=mg_1$$
For position $$(2)$$:
$$g_2=\dfrac{GM}{R^2}$$, $$w_2=mg_2$$
For position $$(3)$$:
$$g_3=\dfrac{GM}{(R+h)^2}$$, $$w_3=mg_3$$
Clearly we can see that $$g_1 < g_2 > g_3$$
Thus $$w_1 < w_2 > w_3$$ (A).

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Gravitation Test - 27

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Gravitation Test - 27

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Question 1

$$4\times 10^{-2}$$ N

$$4\times 10^{-9}$$ N

$$4\times 10^{-6}$$ N

$$4\times 10^{-3}$$ N

Solution

Question 2

$$10^{16}J$$

$$2\times 10^{10}J$$

$$10^{6}J$$

$$3\times 10^{10}J$$

Solution

Question 3

$$\dfrac{1}{2}$$mgR

$$2 mgR$$

$$mgR$$

$$\dfrac{1}{4}$$mgR

Solution

Question 4

$$V_{e}\sqrt{\dfrac{R}{R+h}}$$

$$V_{e}\sqrt{\dfrac{2R}{R+h}}$$

$$V_{e}\sqrt{\dfrac{R+h}{R}}$$

$$V_{e}\sqrt{\dfrac{R}{2\left ( R+h \right )}}$$

Solution

Question 5

mg 2R

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

3 mgR

$$\displaystyle \frac{1}{3}mgR$$

Solution

Question 6

$$1 \times 10^{22}\ m^{2}$$

$$3 \times 10^{22}\ m^{2}$$

$$5 \times 10^{22}\ m^{2}$$

$$7 \times 10^{22}\ m^{2}$$

Solution

Question 7

0.414R

R

2R

3.5R

Solution

Question 8

$$6.67\times 10^{-11} Nm^2/kg^2$$

$$66.7\times 10^{-11} Nm^2/kg^2$$

$$667\times 10^{-11} Nm^2/kg^2$$

None of these

Solution

Question 9

at the center of the earth.

inside the earth.

on the surface of the earth.

above the surface of earth.

Solution

Question 10

$${\omega}_1 < {\omega}_2 > {\omega}_3$$

$${\omega}_1 = {\omega}_2 = {\omega}_3$$

$${\omega}_1 < {\omega}_2 < {\omega}_3$$

$${\omega}_1 > {\omega}_2 > {\omega}_3$$

Solution

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