Gravitation Test

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Gravitation Test - 71

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Weekly Quiz Competition

Question 1
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The work done in shifting a particle of mass $$m$$ from the centre of earth to the surface of the earth is

A
$$-mgR$$

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

C
$$zero$$

D
$$mgR$$

Solution

Given,The mass of the particle be $$m$$,
Initilly it is on the centre of the earth, so
The work done in moving a particle from center of earth to its surface$$\ W=\int_0^{R}Fdr$$
$$=\int_0^R(-\dfrac{G(\rho\dfrac{4}{3}\pi r^3)m}{r^2})dr$$
$$=\int_0^R(-\dfrac{4G\rho \pi mr}{3})dr$$
$$=\dfrac{GMm}{2R},$$we know that $$F=\dfrac{GMm}{R^2}=mg$$
$$=\dfrac{1}{2}mgR$$
(Since $$g=\dfrac{GM}{R^2}$$)
So, The work done in shifting a particle of mass from the centre of the earth to the surface of the earth is $$=\dfrac{mgR}{2}.$$
Question 2
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The work done by an external agent to shift a point mass from infinity to the centre of the earth is W. Then choose the correct relation.

A
W$$=$$0

B
W > 0

C
W < 0

D
W$$\leq $$ 0

Solution

$$W_{ext}=\Delta U=U_{f}-U_{i}=U_{f}-0=U_{f}$$
$$ W_{ext}=\displaystyle \frac{-3GMm}{2R} \rightarrow W_{ext}<0$$
Question 3
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A particle hanging from a massless spring stretches it by 2 cm at earths surface. How much will the same particle stretch the spring at a height of 2624 km from the surface of the earth? (Radius of earth $$=$$ 6400 km).

A
$$1 cm$$

B
$$2 cm$$

C
$$3.3 cm$$

D
$$4 cm$$

Solution

$$g= \dfrac{GM}{R^{2}}$$ and $$g_{h}=\displaystyle \dfrac{GM}{(R+h)^{2}}$$
$$ \Rightarrow \dfrac{g_{h}}{g}=\left(\dfrac{R}{R+h}\right)^{2}=\left(\dfrac{6400}{6400+2624}\right)^{2}=\dfrac{1}{2}$$
$$\Rightarrow g_{h}=\displaystyle \dfrac{g}{2}$$
At the surface of earth, $$mg = kx$$
At the height $$\mathrm{h}' \Rightarrow mg_{h}=kx^{'}$$
$$\displaystyle \Rightarrow\dfrac{g_{h}}{g}=\frac{x^{'}}{x}\Rightarrow x^{'}=\left(\dfrac{g_{h}}{g}\right)x=\left(\frac{1}{2}\right)(2cm)=1cm$$
Question 4
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If $$v_{e}$$ is escape velocity and $$v_{0}$$ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by:

A
$$v_{0}=v_{e}$$

B
$$v_{e}=\sqrt{2v_{0}}$$

C
$$v_{e}=\sqrt{2}v_{0}$$

D
$$v_{0}=\sqrt{2}v_{e}$$

Solution
Question 5
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A spherical $$\Uparrow \mathrm{o}\mathrm{l}\mathrm{e}$$ of radius $$\mathrm{R}/2$$ is excavated from $$\mathrm{t}\Uparrow \mathrm{e}$$ asteroid of mass
$$\mathrm{M}$$ as shown in fig. $$\mathrm{T}\Uparrow \mathrm{e}$$ gravitational acceleration at a point on $$\mathrm{t}\Uparrow \mathrm{e}$$
surf ace of $$\mathrm{t}\Uparrow \mathrm{e}$$ asteroid just above $$\mathrm{t}\Uparrow \mathrm{e}$$ excavation is :

A
$$\mathrm{G}\mathrm{M}/\mathrm{R}^{2}$$

B
$$\mathrm{G}\mathrm{M}/2\mathrm{R}^{2}$$

C
$$\mathrm{G}\mathrm{M}/8\mathrm{R}^{2}$$

D
$$7\mathrm{g}\mathrm{W}8\mathrm{R}^{2}$$

Solution

The energy at centre of hole is energy due to empty part $$\left( { \varepsilon }_{ 1 } \right) $$ plus energy due to remaining mass $$\left( { \varepsilon }_{ 2 } \right) $$.
The energy due to empty mass $$=0$$ at center of the empty portion
$$E={ E }_{ 1 }+{ E }_{ 2 }$$
$${ E }_{ 2 }=E-{ E }_{ 1 }$$
$$=\dfrac { 1 }{ 8 } \dfrac { GM }{ { \left( \dfrac { R }{ 2 } \right) }^{ 2 } } $$
$$E=\dfrac { GM }{ { 2R }^{ 2 } } $$
Question 6
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A spherical ball is dropped in a long column of viscous liquid. Which of the following graphs represent the variation of
i) The gravitational force with time
ii) The viscous force with time
iii) The net force acting on the ball with time

A
Q, R, P

B
R, Q, P

C
P, Q, R

D
R, P, Q

Solution
Question 7
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A body of mass m is placed on earth surface which is taken from earth surface to a height of $$h = 3R$$ then change in gravitational potential energy is:

A
$$\dfrac{mgR}{4}$$

B
$$\dfrac{2}{3}mgR$$

C
$$\dfrac{3}{4}mgR$$

D
$$\dfrac{mgR}{2}$$

Solution

change in gravitational potential energy is given by $$\triangle U=\cfrac { gmhR }{ (R+h) } =\cfrac { mg3R.R }{ \quad R+3R } $$ [$$given that h=3R$$]
=$$\cfrac { 3 }{ 4 } mRg$$
Question 8
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A particle starts from rest at a distance $$\mathrm{R}$$ from the centre and along the axis of a fixed ring of radius $$\mathrm{R}$$ a mass M. Its velocity at the centre of the ring is:

A
$$\sqrt{\frac{\sqrt{2}\mathrm{G}\mathrm{M}}{\mathrm{R}}}$$

B
$$\sqrt{\frac{2\mathrm{G}\mathrm{M}}{\mathrm{R}}}$$

C
$$\sqrt{(1-\frac{1}{\sqrt{2}})\frac{\mathrm{G}\mathrm{M}}{\mathrm{R}}}$$

D
$$\sqrt{(2-\sqrt{2})\frac{\mathrm{G}\mathrm{M}}{\mathrm{R}}}$$

Solution
Question 9
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Directions For Questions

The earth does not have a uniform density; it is most dense at its centre and least dense at its surface. Anapproximation of its density is $$p(r ) =(A - Br )$$ , where $$A = 12,700kg/m^{3},B=1.5\times 10^{-3}kg/m^{4}$$ and $$r$$ is the distance from the centre of earth. Use $$R=6.4\times 10^{6}m$$ for the radius of earth approximated as a sphere, imagine dividing the earth into concentric, elementary spherical shells. Each shell has radius $$r$$, thickness $$dr$$ ,volume $$dV$$ and mass $$dm = p(r)dV$$. By integrating $$dm$$ from zero to $$R$$ the mass of earth can be found. Knowing the fact that a uniform spherical shell gives no contribution to acceleration due to gravity inside it,we can also find $$g$$ as a function of $$r$$.

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Acceleration due to gravity as a function of $$r$$ is given by

A
$$\dfrac{4}{3}\pi Gr(A-BR)$$

B
$$4\pi Gr(A-BR)$$

C
$$\dfrac{4}{3}\pi Gr(A-\dfrac{3}{4}BR)$$

D
$$\dfrac{4}{3}\pi Gr\left ( A-\dfrac{4}{3}BR \right )$$

Solution

Mass of the volume enclosed in radius r will be

$$M=\int _{ 0 }^{ r }{ \rho (r)dV } =\int _{ 0 }^{ r }{ 4\pi { r }^{ 2 }(A-Br)dr } =4\pi (A\dfrac { { r }^{ 3 } }{ 3 } -B\dfrac { { r }^{ 4 } }{ 4 } )=\dfrac { 4 }{ 3 } \pi { r }^{ 3 }(A-\dfrac { 3 }{ 4 } Br)$$

Since no contribution due to mass outside sphere of radius r. Hence, Acceleration due to gravity will be

$$g=\dfrac { GM }{ { r }^{ 2 } } =\dfrac { 4 }{ 3 } \pi r(A-\dfrac { 3 }{ 4 } Br)$$
Question 10
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Find the distance between the centre of gravity and centre of mass of a two-particle system attached to the ends of a light rod. Each particle has the same mass. Length of the rod is $$R$$, where $$R$$ is the radius of the earth.

A
$$R$$

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

C
zero

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

Solution

The centre of gravity of the Earth lie at its centre (Point p in picture)
The centre of mass of the rod is $$\cfrac { mR+m.0 }{ 2m } =\cfrac { R }{ 2 } \\ $$
The distance between CM of the rod and CG of the Earth is=$$\cfrac { R }{ 2 } $$ [$$PS=\cfrac { R }{ 2 } $$ in the picture]