Gravitation Test

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Gravitation Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

The height at which the weight of a body becomes $$\dfrac {1}{16}th$$, its weight on the surface of earth (radius $$R$$), is

A
$$3R$$

B
$$4R$$

C
$$5R$$

D
$$15R$$

Solution

weight w = mg
m = mass (constent)
$$ \dfrac{w_{1}}{w_{2}} = \dfrac{g_{1}}{g_{2}}; g_{1} = \dfrac{GM}{R^{2}}$$
$$ 9_{2} = \dfrac{GM}{(R+h)^{2}}$$ ;h = height
So$$ \dfrac{w_{1}}{w_{2}} = \dfrac{(R+h)^2}{R^{2}}$$
putting $$ \dfrac{w_{1}}{W_2} = 16 = \dfrac{(R+h)^{2}}{R^{2}}$$
$$ \Rightarrow \dfrac{R+h}{R} = 4 \Rightarrow 1+\dfrac{h}{2} = 4 \Rightarrow \boxed{h = 3R }$$
Question 2
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At what height above the surface of earth acceleration due to gravity becomes $$1 \ mm/s^2$$?
[Take $$R$$ as radius of earth and acceleration due to gravity at the surface of earth $$g = 10 m/s^2$$]

A
$$101 \ R$$

B
$$100 \ R$$

C
$$99 \ R$$

D
$$199 \ R$$

Solution
Question 3
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If $$'R'$$ is the radius of the earth, then the height at which the weight of a body becomes $$\dfrac{1}{4}^{th}$$ of its weight on the surface of the earth is:

A
$$2R$$

B
$$R$$

C
$$\dfrac{3R}{8}$$

D
$$\dfrac{R}{4}$$

Solution

we know that acceleration due to gravity varies height h and the surface of the earth
$${ g }^{ \prime }=g\left( 1-\dfrac { 2h }{ R } \right) $$
hence the weight of the body
$$m{ g }^{ \prime }=mg\left( 1-\dfrac { 2h }{ R } \right) $$

$$w^{ \prime }=w\left( 1-\dfrac { 2h }{ R } \right) $$

$$w^{ \prime }=w\left( 1-\dfrac { 2h }{ R } \right) $$

$$\dfrac { 1 }{ 4 } =\left( 1-\dfrac { 2h }{ R } \right) $$

$$\dfrac { 2h }{ R } =1-\dfrac { 1 }{ 4 } =\dfrac { 3 }{ 4 } $$

$$h=\dfrac { 3R }{ 8 } $$
Question 4
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A lift is descending with a constant velocity $$\ 'V'$$, A man in the lift drops a coin experiences an acceleration towards the floor equal to:

A
$$g+v$$

B
$$g-v$$

C
$$g$$

D
$$zero$$

Solution
Question 5
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Two planet have radii in the ratio $$1:2$$ and densities in the ratio $$1:3$$. The value of the ratio between acceleration due to gravity on the first planet to that of the second planet is:

A
$$6$$

B
$$3$$

C
$$1/4$$

D
$$1/6$$

Solution

$$g=\dfrac{GM_p}{r^2}$$

$$M_p=Density\times Volume=Density\times\dfrac{4\pi r^3}{3}$$
$$g=\dfrac{G\times density\times 4\pi r^3}{3r^2}$$

Thus g$$\propto r$$ and g$$\propto Density$$

So,$$\dfrac{g_1}{g_2}=\dfrac{1\times 1}{2\times 3}=\dfrac{1}{6}$$
Question 6
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A body weighs $$900$$N on the earth. Find its weight on a planet whose density is $$\dfrac{1}{3}^{st}$$ the density of earth and radius is $$\dfrac{1}{4}^{th}$$ that of the earth.

A
$$75$$N

B
$$500$$N

C
$$62$$N

D
$$320$$N

Solution

Weight of object on earth $$W = {\dfrac{GMm}{R^2}}$$

Now, $$M = {\rho}V$$ =$${\dfrac{4\pi \rho R^{3}}{3}}$$

$$\therefore W = {\dfrac{G \rho \pi R^{3} G m}{3 R^{2}}} = {\dfrac{4 \pi \rho R G m}{3}}$$

$${\dfrac{W_P}{W_E}} = {(\dfrac{\rho_P}{\rho_E})}{(\dfrac{R_P}{R_E})}$$

$$ = {\dfrac{1}{3}} \times {\dfrac{1}{4}}$$ = $${\dfrac{1}{12}}$$

$${W_P} = {\dfrac{900}{12}} = 75 N$$
Question 7
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The depth at which the value of acceleration due to gravity becomes $${\dfrac{1}{n}}$$ times the value at the surface is (R be the radius of the earth):

A
$${\dfrac{R}{n}}$$

B
$${\dfrac{R}{n^2}}$$

C
$${\dfrac{R(n-1)}{n}}$$

D
$${\dfrac{Rn}{(n-1)}}$$

Solution

option C
$${ g }^{ 1 }=g\left[ 1-\dfrac { d }{ R } \right] $$

$$\Longrightarrow \dfrac { g }{ n } =g\left[ 1-\dfrac { d }{ R } \right] \Longrightarrow d=\left[ \dfrac { n-1 }{ n } \right] R$$
Question 8
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A particle when thrown. moves such that it passes from same height at $$2$$ and $$10 s$$, the height is:

A
$$g$$

B
$$2g$$

C
$$5g$$

D
$$10g$$

Solution

If t1 and t2 are the time when the body is at the same height Then $$h = \dfrac{1}{2} \times g\times {t}_1\times {t}_2$$ $$h = \dfrac{1}{2} \times g\times {2}\times {10}$$h = 10 g
So that the correct option is D
Question 9
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The escape velocity of an object from the earth depends upon the mass of earth $$(M)$$, its mean density $$(\rho)$$, its radius $$(R)$$ and gravitational constant $$(G)$$, thus the formula for escape velocity is:

A
$$v=R\sqrt{\dfrac{8\pi}{3}G\rho}$$

B
$$v=M\sqrt{\dfrac{8\pi}{3}G\rho}$$

C
$$v=\sqrt{2GMR}$$

D
$$v=\sqrt{\dfrac{2GM}{R}}$$

Solution
Question 10
1 / -0

If the gravitational potential energy of two point masses infinitely away is taken to be zero then gravitational potential energy of a galaxy is

A
Zero

B
Positive

C
Negative

D
Can have any value

Solution