Gravitation Test

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Gravitation Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

The distances from the centre of the earth, where the weight of a body is zero and one-fourth that of the weight of the body on the surface of the earth are (assume $$R$$ is the radius of the earth)

A
$$0$$, $$\displaystyle \frac { R }{ 4 } $$

B
$$0$$, $$\displaystyle \frac { 3R }{ 4 } $$

C
$$\displaystyle \frac { R }{ 4 } $$, $$0$$

D
$$\displaystyle \frac { 3R }{ 4 } $$, $$0$$

Solution

The weight of the body at the centre of the earth is equal to zero because
$$g_{centre}=g\left(1-\dfrac{d}{R}\right)=g\left(1-\dfrac{R}{R}\right)=0$$
$$\dfrac{g_{1}}{g}=\left(1-\dfrac{d}{R}\right)=\dfrac{1}{4}\Rightarrow d=\dfrac{3R}{4}$$
So from the centre, $$d=\dfrac{R}{4}$$
Question 2
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In a relation $$\displaystyle \sqrt{\frac{2GM}{R}} \geq C$$, G stands for universal gravitational constant, M for mass of a very very dense material, R for a small radius and C for velocity of light. Then,

A
the above relation indicates that orbital velocity is equal to velocity of light.

B
the above relation indicates something like a black hole.

C
the above relation indicates that the light is escaping.

D
the above relation indicates that any object starting from the given mass with speed more than light will return to it.

Solution

$$(\dfrac{2GM}{R})^{0.5} V_e$$ is the escape velocity of planet with mass $$M$$ and radius $$R$$. If, $$V_e$$ is greater than speed of light, it means that mass is concentrated in Schwarzschild radius. So, it behaves as the black hole where even light cannot escape its gravitation.
Question 3
1 / -0

If $${ g }$$ is the acceleration due to gravity on the earth's surface, the change in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth is

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

B
$$\displaystyle { 2mgR }$$

C
$$\displaystyle { mgR }$$

D
$$\displaystyle { -mgR }$$

Solution

$$\text{Gain in potential energy}= \triangle U=U_2-U_1$$
$$\triangle U=(-\dfrac{GMm}{2R})-(-\dfrac{GMm}{R})$$

$$\triangle U=\dfrac{GMm}{R}-\dfrac{GMm}{2R}$$

$$\triangle U=\dfrac{GMm}{2R}$$

$$\triangle U=\dfrac{mgR^2}{2R} As,GM=gR^2$$

$$\triangle U=\dfrac{mgR}{2}$$
Question 4
1 / -0

Two equals masses each $$m$$ and $$m$$ are hung from a balance whose scale pans differ in vertical height by $$'h'$$. The error in weighing in terms of density of the earth $$\rho $$ is:

A
$$\displaystyle \pi G \rho m h$$

B
$$\cfrac { 1 }{ 3 } \pi G \rho m h$$

C
$$\cfrac { 8 }{ 3 } \pi G \rho m h$$

D
$$\cfrac { 4 }{ 3 } \pi G \rho m h$$

Solution

we know, $$g_h=g[1-\dfrac{2h}{R}]$$
$$g-g_h=\dfrac{2gh}{R}$$
$$g-g_h=2\times\dfrac{GM}{R^3}h$$
$$g-g_h=\dfrac{2Gh}{R^3}\times\dfrac{4}{3}\pi R^3\rho$$
or, $$g-g_h=\dfrac{8}{3}\pi G\rho h$$
or, $$mg-mg_h=\dfrac{8}{3}\pi G\rho hm$$
Question 5
1 / -0

If $$g$$ is acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of earth to a height equal to the radius $$R$$ of the earth is

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

B
$${ 2mgR }$$

C
$${ mgR }$$

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

Solution

$$\text{The gain in potential energy}=\triangle U=U_2-U_1$$
$$\triangle U=(-\dfrac{GMm}{2R})-(-\dfrac{GMm}{R})=\dfrac{GMm}{2R}$$

$$\triangle U=\dfrac{1}{2}(\dfrac{GM}{R^2})mR \, \text{As,} \ g= \dfrac{GM}{R^2}$$

$$\triangle U=\dfrac{1}{2}mgR$$
Question 6
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A body of mass $${ m }$$ rises to a height $${ h }=\dfrac{R}{5}$$ from the earth's surface where $${ R }$$ is the earth's radius. If $$g$$ is acceleration duem to gravity at the earth's surface, the increase in potential energy is

A
$${ mgh }$$

B
$$\cfrac { 4 }{ 5 } { mgh }$$

C
$$\cfrac { 5 }{ 6 } { mgh }$$

D
$$\cfrac { 6 }{ 7 } { mgh }$$

Solution

$$\text{The gain in potential energy}\quad \triangle U=U_2-U_1$$ $$=\left(-\dfrac{GM_em}{R+h}\right)-\left(-\dfrac{GM_em}{R}\right)$$
$$\triangle U=\dfrac{GM_em(R+h-R)}{(R+h)R}$$
$$\triangle U=\dfrac{mgR\times h}{R+h}$$
$$\text{Substitute}\quad R=5h$$
$$\triangle U=\dfrac{5mgh^2}{6h}=\dfrac{5}{6}mgh $$
Question 7
1 / -0

The value of $${ g }$$ at a certain height $${ h }$$ above the free surface of the earth is $${ x }/{ 4 }$$ where $${ x }$$ is the value of $${ g }$$ at the surface of the earth. The height $${ h }$$ is

A
$${ R }$$

B
$${ 2R }$$

C
$${ 3R }$$

D
$${ 4R }$$

Solution

Acceleration due to gravity at some distance $$r$$ from center of the earth=$$\dfrac{GM}{r^2}$$.
It is given that $$x=\dfrac{GM}{R^2}$$
and $$\dfrac{x}{4}=\dfrac{GM}{r^2}$$
$$\implies r=2R$$
which is the distance from center of earth.
Therefore distance from surface$$=h=2R-R=R$$
Question 8
1 / -0

The change in the value of $$g$$ at a height $$h$$ above the surface of earth is the same as at a depth $$d$$ below the earth. When both $$d$$ and $$h$$ are much smaller than the radius of earth, then which one of the following is correct?

A
$$d=\cfrac { h }{ 2 } $$

B
$$d=\cfrac { 3h }{ 2 } $$

C
$$d=2h$$

D
$$d=h$$

Solution

$$\textbf{Step1:Acceleration due to gravity}$$
$$\bullet$$ Acceleration due to gravity at a height $$g'_{h} = g\left (1 - \dfrac {2h}{R}\right )$$
$$\bullet$$ Acceleration due to gravity at a depth $$g_{d}' = g \left (1 - \dfrac {d}{R}\right )$$

$$\textbf{Step2:Relation between d and h}$$
According to question,
$$ g_{h}' = g_{d}'$$
$$\Rightarrow g\left (1 - \dfrac {2h}{R}\right ) = g' \left (1 - \dfrac {d}{R}\right )$$
where d : depth in earth surface
R : radius of earth
h : height above the earth
$$\Rightarrow d = 2h$$

Hence option (C) is correct.
Question 9
1 / -0

If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is

A
$$2R$$

B
$$\cfrac {R}{2}$$

C
$$R$$

D
$$\cfrac{R}{4}$$

Solution

$$\dfrac{1}{2}\sqrt{\dfrac{2GM}{R}}=\sqrt{\dfrac{GM}{R+h}}$$

$$\dfrac{1}{4}\times\dfrac{2GM}{R}=\dfrac{GM}{R+h}$$

$$\dfrac{1}{2}\times\dfrac{1}{R}=\dfrac{1}{R+h}$$

$$2R=R+h$$

or $$h=R$$
Question 10
1 / -0

The gravitational potential due to earth at infinite distance from it is zero. Let the gravitational potential at a point $$P$$ be $$-{ 5 }{ J }{ kg }^{ -1 }$$. Suppose, we arbitrarily assume the gravitational potential at infinity to be $$+{ 10 }{ J }{ kg }^{ -1 }$$, then the gravitational potential at $$P$$ will be

A
$$-{ 5 }{ J }{ kg }^{ -1 }$$

B
$$+{ 5 }{ J }{ kg }^{ -1 }$$

C
$$-{ 15 }{ J }{ kg }^{ -1 }$$

D
$$+{ 15 }{ J }{ kg }^{ -1 }$$

Solution

According to the problem as the potential at $$\infty$$ increases by $$+10\ J\ kg^{-1}$$, hence potential will increases by the same amount everywhere (potential gradient will remain constant). Hence, potential point $$P=10-5=+5\ J\ kg^{-1}$$

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Re-Attempt Weekly Quiz Competition

Gravitation Test - 33

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Gravitation Test - 33

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Question 1

$$0$$, $$\displaystyle \frac { R }{ 4 } $$

$$0$$, $$\displaystyle \frac { 3R }{ 4 } $$

$$\displaystyle \frac { R }{ 4 } $$, $$0$$

$$\displaystyle \frac { 3R }{ 4 } $$, $$0$$

Solution

Question 2

the above relation indicates that orbital velocity is equal to velocity of light.

the above relation indicates something like a black hole.

the above relation indicates that the light is escaping.

the above relation indicates that any object starting from the given mass with speed more than light will return to it.

Solution

Question 3

$$\displaystyle \dfrac { mgR }{ 2 } $$

$$\displaystyle { 2mgR }$$

$$\displaystyle { mgR }$$

$$\displaystyle { -mgR }$$

Solution

Question 4

$$\displaystyle \pi G \rho m h$$

$$\cfrac { 1 }{ 3 } \pi G \rho m h$$

$$\cfrac { 8 }{ 3 } \pi G \rho m h$$

$$\cfrac { 4 }{ 3 } \pi G \rho m h$$

Solution

Question 5

$$\cfrac { 1 }{ 2 } { mgR }$$

$${ 2mgR }$$

$${ mgR }$$

$$\cfrac { 1 }{ 4 } { mgR }$$

Solution

Question 6

$${ mgh }$$

$$\cfrac { 4 }{ 5 } { mgh }$$

$$\cfrac { 5 }{ 6 } { mgh }$$

$$\cfrac { 6 }{ 7 } { mgh }$$

Solution

Question 7

$${ R }$$

$${ 2R }$$

$${ 3R }$$

$${ 4R }$$

Solution

Question 8

$$d=\cfrac { h }{ 2 } $$

$$d=\cfrac { 3h }{ 2 } $$

$$d=2h$$

$$d=h$$

Solution

Question 9

$$2R$$

$$\cfrac {R}{2}$$

$$R$$

$$\cfrac{R}{4}$$

Solution

Question 10

$$-{ 5 }{ J }{ kg }^{ -1 }$$

$$+{ 5 }{ J }{ kg }^{ -1 }$$

$$-{ 15 }{ J }{ kg }^{ -1 }$$

$$+{ 15 }{ J }{ kg }^{ -1 }$$

Solution

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