Gravitation Test

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Gravitation Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

The variation of acceleration due to gravity $$g$$ with distance $$d$$ from centre of the earth is best represented by ($$R=$$Earths radius):

A

B

C

D

Solution

As we know that,
Gravity inside the earth is given as
$$g=\cfrac{GMm}{R^3}$$
While gravity outside the earth is given as
$$g=\cfrac{GMm}{r^2}$$
Hence option $$A$$ is correct.
Question 2
1 / -0

The height at which the acceleration due to gravity becomes $$g/9$$ (where $$g=the$$ acceleration due to gravity on the surface of the earth) in terms of $$R$$, the radius of the earth, is

A
$$R/2$$

B
$$\sqrt { 2 } R$$

C
$$2R$$

D
$$\dfrac { R }{ \sqrt { 2 } }$$

Solution

Acceleration due to gravity at a height $$'h'$$ is given by:-

$$'g'=g\left ( \dfrac{R}{R+H} \right )^{2}$$

where $$g\rightarrow $$ on the surface of the earth.

$$R\rightarrow $$ Radius of earth.

As $$g'=g/9,$$ we get :

$$\Rightarrow \dfrac{g}{9}=g\left ( \dfrac{R}{R+H} \right )^{2}$$

$$\Rightarrow \dfrac{R}{R+H}=\dfrac{1}{3}\Rightarrow h=2R$$
Question 3
1 / -0

The fractional change in the value of free-fall acceleration $$'g'$$ for a particle when it is lifted from the surface to an elevation $$h. (h < < R)$$ is

A
$$h/ R$$

B
$$-2h/ R$$

C
$$2h/ R$$

D
None of these

Solution

$$g = \dfrac {GM}{R^{2}} .....(i)$$
$$\dfrac {dg}{dR} = \dfrac {-2GM}{R^{3}}$$
Putting $$dR = h$$ we obtain
$$\Rightarrow \dfrac {dg}{h} = \dfrac {-2GM}{R^{2}}\dfrac {1}{R} ..... (ii)$$
From (i) and (ii)
$$\Rightarrow \dfrac {dg}{g} = -2\left (\dfrac {h}{R}\right )$$
Change is $$-ve$$.
That means $$g$$ decreases.
Hence (B) is correct.
Question 4
1 / -0

At what depth below the surface of the earth acceleration due to gravity $$'g'$$ will be half of its value at $$1600\ km$$ above the surface of the earth?

A
$$1.6\times 10^{6}m$$

B
$$2.4\times 10^{6}m$$

C
$$3.2\times 10^{6}m$$

D
$$4.8\times 10^{6}m$$

Solution

Variation of $$g$$ outside the earth
$${ g }^{ 1 }=g\left( 1-\dfrac { 2h }{ R } \right) $$
$${ g }^{ 1 }=10\left( 1-\dfrac { 2\times 1600 }{ 6400 } \right) $$
$${ g }^{ 1 }=5m/{ s }^{ 2 }$$
Variation of $$g$$ inside the earth
$${ g }^{ 11 }=g\left( 1-\dfrac { h }{ R } \right) $$
Now, it is given that,
$${ g }^{ 11 }=\dfrac { { g }^{ 1 } }{ 2 } $$
$$2.5m/{ s }^{ 2 }=10\left( 1-\dfrac { h }{ 6400 } \right) $$
$$\dfrac { 1 }{ 4 } =1-\dfrac { h }{ 6400 } $$
$$h=\dfrac { 3 }{ 4 } \times 6400$$ Km
$$h=4.8\times { 10 }^{ 6 }$$ m
Question 5
1 / -0

Weight of a body on earth's surface is $$W$$. At a depth half way to the centre of the earth, it will be (assuming uniform density in earth).

A
$$W$$

B
$$W/2$$

C
$$W/4$$

D
$$W/8$$

Solution

The variation of g with depth is given by

$${{g}^{'}}=\dfrac{g}{1-\dfrac{d}{R}}\,$$ where d is depth

At a depth half way to the centre of the earth $$d=\dfrac{R}{2}$$

$$ {{g}^{'}}=\dfrac{g}{1-\dfrac{R}{2R}} $$

$$ g'=\dfrac{g}{2} $$

On multiplying both sides by m

$$ mg'=\dfrac{mg}{2} $$

$$ {{W}^{'}}=\dfrac{W}{2} $$

At the depth half way, weight becomes half.
Question 6
1 / -0

A body of mass 'm' is raised from the surface of the earth to a height 'nR' (R- radius of earth). Magnitude of the change in the gravitational potential energy of the body is (g-acceleration due to gravity on the surface of earth)

A
$$\left( \dfrac { n }{ n+1 } \right) mgR$$

B
$$\left( \dfrac { n-1 }{ n } \right) mgR$$

C
$$\dfrac { mgR }{ n } $$

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

Solution

At surface gravitaional potential energy $$U_1=-\dfrac{GMm}{R}$$
At height $$nR, U_2=-\dfrac{GMm}{R+nR}$$
$$\left| { U }_{ 2 }-{ U }_{ 1 } \right| =\left| \left( \dfrac { -GMm }{ R+nR } \right) -\left( \dfrac { -GMm }{ R } \right) \right| \\ =\dfrac { GMm }{ R } -\dfrac { GMm }{ R+nR } \\ \Rightarrow \dfrac { GMm }{ R } \left[ 1-\dfrac { 1 }{ n+1 } \right] \Rightarrow \left( \dfrac { n }{ n+1 } \right) \dfrac { GMm }{ R } \\ \Rightarrow \left( \dfrac { n }{ n+1 } \right) mgR$$
Question 7
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The radius of a planet is $$4$$ times the radius of the earth. The time period of revolution of the planet will be:

A
$$1\ yr$$

B
$$2\ yr$$

C
$$4\ yr$$

D
$$8\ yr$$

Solution

According to Kepler's law
$${ T }^{ 2 }\alpha { R }^{ 3 }\quad \rightarrow \left( I \right) $$
Now,
$${ R }_{ 1 }=4R$$
So, $${ T }_{ 1 }^{ 2 }\alpha { R }_{ 1 }^{ 3 }$$
$${ T }_{ 1 }^{ 2 }\alpha { 64R }^{ 3 }\quad \rightarrow \left( II \right) $$
on dividing both equations, we get,
$$\dfrac { { T }^{ 2 } }{ { T }_{ 1 }^{ 2 } } =\dfrac { 1 }{ 64 } $$
$${ T }_{ 1 }=8T$$
i.e, It will take $$8$$ year to complete a revolution.
Question 8
1 / -0

In a double star system one of mass $$m_{1}$$ and another of mass $$m_{2}$$ with a separation $$d$$ rotate about their common centre of mass. Then rate of sweeps of area of star of mass $$m_{1}$$ to star of mass $$m_{2}$$ about their common centre of mass is

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

B
$$\dfrac {m_{2}}{m_{1}}$$

C
$$\dfrac {m_{1}^{2}}{m_{2}^{2}}$$

D
$$\dfrac {m_{2}^{2}}{m_{1}^{2}}$$

Solution
Question 9
1 / -0

If potential at the surface of a planet is taken as zero, the potential at infinity will be $$(M$$ and $$R$$ are mass and radius of the planet).

A
Zero

B
$$\infty$$

C
$$\dfrac {GM}{R}$$

D
$$-\dfrac {GM}{R}$$

Solution

Potential at surface of planet $$=\dfrac { -GM }{ R } $$. When infinity is taken as reference but here potential will be $$+\dfrac { GM }{ R } $$ because some positive work must be done to take the particle from infinity to the surface.
Question 10
1 / -0

Weight of a body decreases by $$1.5$$%, when it is raised to a height $$h$$ above the surface of the earth. When the same body is taken to same depth $$h$$ in a mine, its weight will show

A
$$0.75$$% increase

B
$$3.0$$% decrease

C
$$0.75$$% decrease

D
$$1.5$$% decrease

Solution

Given weight,
$$1.5=\dfrac { \dfrac { GM }{ { R }^{ 2 } } -\dfrac { GM }{ { (R+h) }^{ 2 } } }{ \dfrac { GM }{ { R }^{ 2 } } } \\ =\dfrac { { R }^{ 2 }+2Rh+{ h }^{ 2 }-{ R }^{ 2 } }{ { \left( R+h \right) }^{ 2 } } \\ \Rightarrow \dfrac { 2Rh+{ h }^{ 2 } }{ { \left( R+h \right) }^{ 2 } } =\dfrac { 1.5 }{ 100 } $$
At depth $$'h', g'=g(1-\dfrac{h}{R})$$
SO g' decreases, $$\dfrac{g-g'}{g}=\dfrac{h}{R}$$
$$\Rightarrow \dfrac { \dfrac { 2h }{ R } +{ \left( \dfrac { h }{ R } \right) }^{ 2 } }{ { \left( 1+\dfrac { h }{ R } \right) }^{ 2 } } =\dfrac { 1.5 }{ 100 } \\ \dfrac { 2h }{ R } =\dfrac { 1.5 }{ 100 } \\ \dfrac { h }{ R } =\dfrac { 0.75 }{ 100 } $$
But since $$h<<<R$$
At depth $$'h'$$ there will be a decrese of $$g$$ by $$0.75$$%