MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 13

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).

0%
$1 cm$
0%
$2 cm$
0%
$3.3 cm$
0%
$4 cm$

Explanation

$g= \dfrac{GM}{R^{2}}$ 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

$\mathrm{h}' \Rightarrow mg_{h}=kx^{'}$

$\displaystyle \Rightarrow\dfrac{g_{h}}{g}=\frac{x^{'}}{x}\Rightarrow x^{'}=\left(\dfrac{g_{h}}{g}\right)x=\left(\frac{1}{2}\right)(2cm)=1cm$

$v_{e}$ 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:

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

Three planets of same density and with radii

$\mathrm{R}_{1}, \mathrm{R}_{2}$ and

$\mathrm{R}_{3}$ , such that

$\mathrm{R}_{1}=2\mathrm{R}_{2}=3\mathrm{R}_{3},$ have gravitation fields on the surfaces

$\mathrm{E}_{1}, \mathrm{E}_{2}, \mathrm{E}_{3}$ and escape velocities

$\mathrm{v}_{1}, \mathrm{v}_{2}, \mathrm{v}_{3}$ respectively, Then

0%
$\displaystyle \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{2}$
0%
$\displaystyle \frac{\mathrm{E}_{1}}{\mathrm{E}_{3}}=3$
0%
$\displaystyle \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=2$
0%
$\displaystyle \frac{\mathrm{v}_{1}}{\mathrm{v}_{3}}=\frac{1}{3}$

Explanation

$E=\dfrac{GM}{R^{2}}$ $=\dfrac{G\left ( \dfrac{4}{3}\pi R^{3}\rho \right )}{R^{2}}$

$\Rightarrow E\propto R$

$\Rightarrow \dfrac{E_{1}}{E_{2}}=\dfrac{R_{1}}{R_{2}}$ $=2$

$\Rightarrow (A)$ is not true

$\dfrac{E_{1}}{E_{3}}=\dfrac{R_{1}}{R_{3}}$ $=3$

$\Rightarrow (B)$ is true

$v=\sqrt{gR}\Rightarrow v\alpha R$

$\dfrac{v_{1}}{v_{2}}=\dfrac{R_{1}}{R_{2}}=2$

$\Rightarrow (C)$ is true

$\dfrac{v_{1}}{v_{3}}=\dfrac{R_{1}}{R_{3}}=3$

$\Rightarrow (D)$ is not true

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 :

0%
$\mathrm{G}\mathrm{M}/\mathrm{R}^{2}$
0%
$\mathrm{G}\mathrm{M}/2\mathrm{R}^{2}$
0%
$\mathrm{G}\mathrm{M}/8\mathrm{R}^{2}$
0%
$7\mathrm{g}\mathrm{W}8\mathrm{R}^{2}$

Explanation

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

$E={ E }_{ 1 }+{ E }_{ 2 }$

${ E }_{ 2 }=E-{ E }_{ 1 }$

$=\dfrac { 1 }{ 8 } \dfrac { GM }{ { \left( \dfrac { R }{ 2 } \right) }^{ 2 } }$

$E=\dfrac { GM }{ { 2R }^{ 2 } }$

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:

0%
$\dfrac{mgR}{4}$
0%
$\dfrac{2}{3}mgR$
0%
$\dfrac{3}{4}mgR$
0%
$\dfrac{mgR}{2}$

Explanation

change in gravitational potential energy is given by

$\triangle U=\cfrac { gmhR }{ (R+h) } =\cfrac { mg3R.R }{ \quad R+3R }$ [

$given that h=3R$ ]

$\cfrac { 3 }{ 4 } mRg$

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:

0%
$\sqrt{\frac{\sqrt{2}\mathrm{G}\mathrm{M}}{\mathrm{R}}}$
0%
$\sqrt{\frac{2\mathrm{G}\mathrm{M}}{\mathrm{R}}}$
0%
$\sqrt{(1-\frac{1}{\sqrt{2}})\frac{\mathrm{G}\mathrm{M}}{\mathrm{R}}}$
0%
$\sqrt{(2-\sqrt{2})\frac{\mathrm{G}\mathrm{M}}{\mathrm{R}}}$

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

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Q, R, P
0%
R, Q, P
0%
P, Q, R
0%
R, P, Q

$\mathrm{W}_{1},\ \mathrm{W}_{2}$ and

$\mathrm{W}_{3}$ represent the work done in moving a particle from A to

$\mathrm{B}$ along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass

$\mathrm{m}$ , find the correct relation between

$\mathrm{W}_{1},\ \mathrm{W}_{2}$ and

$\mathrm{W}_{3}$ .

0%
$\mathrm{W}_{1}>\mathrm{W}_{2}>\mathrm{W}_{3}$
0%
$\mathrm{W}_{1}=\mathrm{W}_{2}=\mathrm{W}_{3}$
0%
$\mathrm{W}_{1}<\mathrm{W}_{2}<\mathrm{W}_{3}$
0%
$\mathrm{W}_{2}>\mathrm{W}_{1}>\mathrm{W}_{3}$

Acceleration due to gravity as a function of

$r$ is given by

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$\dfrac{4}{3}\pi Gr(A-BR)$
0%
$4\pi Gr(A-BR)$
0%
$\dfrac{4}{3}\pi Gr(A-\dfrac{3}{4}BR)$
0%
$\dfrac{4}{3}\pi Gr\left ( A-\dfrac{4}{3}BR \right )$

Explanation

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

$g=\dfrac { GM }{ { r }^{ 2 } } =\dfrac { 4 }{ 3 } \pi r(A-\dfrac { 3 }{ 4 } Br)$

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.

0%
$R$
0%
$\dfrac{R}{2}$
0%
zero
0%
$\dfrac{R}{4}$

Explanation

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]

Consider the following statements about acceleration due to gravity on earth and mark the correct statement(s) :

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The value of $g$ is constant throughout
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$g' = g(1 - \dfrac {d}{r})$
0%
$g$ is slightly less (by about $1$ %) when distance $< 200 m$
0%
$g$ is slightly greater when distance $< 200 m$

Explanation

Acceleration due to gravity at depth d,

$g'=g(1-\dfrac{d}{r})$ , where

$r$ is the radius of earth.

Thus

$g$ is not constant throughout.

Also for

$d=200 m$

$g'= g(1-\dfrac{200}{6400\times 1000})$

Now

$\dfrac{g-g'}{g} \times 100= .003$ %

Thus

$g$ is much more less than

$1$ %.

The angular velocity of rotation of the earth in order to make the effective acceleration due to gravity equal to zero at equator should be

$g=10m/s^2, R=6400 km$ .

0%
$2.15\times 10^3$ rad/sec
0%
$2.25\times 10^{-3}$ rad/sec
0%
$1.25\times 10^3$ rad/sec
0%
$1.25\times 10^{-3}$ rad/sec

Explanation

$g'=g-\omega^2R cos^2\theta$
At equator

$g'=0$

$cos\theta =0$

$g=\omega^2R$

$\omega^2=\frac {g}{R}$

$\omega=\sqrt {\dfrac {g}{R}}=\sqrt {\dfrac {10}{6400\times 1000}}$

$=\sqrt {\dfrac {1}{640000}}=\dfrac {1}{800}=1.25\times 10^{-3} rad/sec.$

The ratio of the radii of the planets

$P_{1}$ and

$P_{2}$ is

$k_{1}.$ The corresponding ratio of the acceleration due to the gravity on them is

$k_{2}.$ The ratio of the escape velocities from them will be

0%
$k_{1}k_{2}$
0%
$\sqrt{k_{1}k_{2}}$
0%
$\sqrt{(k_{1}/k_{2})}$
0%
$\sqrt{(k_{2}/k_{1})}$

Explanation

$v_{e}=\sqrt{2gR}$

$\therefore \dfrac{(v_{e})p_{1}}{(v_{e})p_{2}}=\dfrac{\sqrt{g_{1}R_{1}}}{\sqrt{g_{2}R_{2}}}=\sqrt{\left \{ \left ( \dfrac{g_{1}}{g_{2}} \right )\times \left ( \dfrac{R_{1}}{R^{2}} \right ) \right \}}=\sqrt{k_{1}k_{2}}$

A planet is revolving in an elliptical orbit around the sun as shown in figure.The areal velocity (area swapped by the radius vector with respect to sun in unit time) is :

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$\dfrac{1}{4}\left ( r_{1}+r_{2} \right )v_{1}$
0%
$\dfrac{1}{2}r_{2}v_{2}$
0%
$\dfrac{1}{2}\dfrac{v_{1}r{_{1}}^{2}}{r_{2}}$
0%
dependent on the position of planet from sun

Explanation

Kepler's law of periods says that the area swept by a planet for a certain time interval is constant.
i.e

$\dfrac {dA}{dt}=\dfrac {L}{2m}$ ; where

$L$ is the angular momentum and is constant.

Therefore,

$\dfrac {dA}{dt}$ is constant.

$\dfrac {dA}{dt}=\dfrac {L}{2m}=\dfrac {mv_2r_2}{2m}=\dfrac {1}{2}v_2r_2$

Acceleration due to gravity as a function of

$r$ is given by :

0%
$\dfrac {4}{3} \pi Gr (A - Br)$
0%
$4 \pi Gr (A - Br)$
0%
$\dfrac {4}{3} \pi Gr (A - \dfrac {3}{4} Br)$
0%
$\dfrac {4}{3} \pi Gr (A - \dfrac {3}{2} Br)$

Explanation

$g = \dfrac {GM(r)}{r^2} = \dfrac {G.\dfrac{4}{3}\pi r^3 (A - \dfrac {3}{4}Br)}{r^2} = \dfrac {4 \pi Gr}{3} \left[A - \dfrac {3}{4} Br \right]$

A planet of radius

$R =1/10 \times {(radius \quad of \quad earth)}$ has the same mass density as Earth. Scientist dig a well of depth

$\dfrac{R}{5}$ on it and lower a wire of the same length and of linear mass density

$10^{-3} kgm^{-3}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in places is (take the radius of earth

$= 6\times 10^6$ and the acceleration due to gravity on earth is

$10ms^{-2}$ )

0%
$96 N$
0%
$108 N$
0%
$120 N$
0%
$150 N$

Explanation

$R_e$ is the radius of the earth

$g_e=\dfrac{GM_e}{R_e^2}=G\dfrac{4}{3}\pi R_e\rho$
Now g at the mentioned planet is given as

$g_p=\dfrac{GM}{r^2}$ , here

$m=\dfrac{4}{3}\pi r^3\rho$
Thus

$g_p=G\dfrac{4}{3}\pi r\rho$
Thus we get

$F=\displaystyle\int^R_{4R/5}\lambda g_p dr$

$F= \lambda \times G\dfrac{4}{3}\pi \rho\times \dfrac{9R^2}{50}$

Solving, we get

$F= 108N$

The ratio of the earth's orbital angular momentum (about the Sun) to its mass is

$4.4\times 10^{15} m^2s^{-1}$ . The area enclosed by the earth's orbit is approximately

$\underline{\hspace{0.5in}} m^2$ .

0%
$6.94\times 10^{22} m^2$
0%
$4.24\times 10^{22} m^2$
0%
$4.92\times 10^{22} m^2$
0%
$5.54\times 10^{22} m^2$

Explanation

Areal velocity of a planet around the Sun is constant and is given by

$\dfrac {dA}{dt}=\dfrac {L}{2m}\Rightarrow dA=\dfrac {L}{2m}dt$
Integrating both sides

$\int dA=\dfrac {L}{2m}\int dt\Rightarrow A=\dfrac {L}{2m}\cdot T$
where

$L=$ angular momentum of the planet (earth) about the Sun and

$m=$ mass of planet (earth).
Hence,

$A=\dfrac {L}{2m}T=\dfrac {1}{2}\times 4.4\times 10^{15}\times 365\times 24\times 3600 m^2$

$Area=6.94\times 10^{22}m^2$

If the change in the value of

$g$ at a height

$h$ above the surface of the earth is the same as at a depth

$x$ below it when both

$x$ and

$h$ are much smaller than the radius of the earth, then

0%
$x = h$
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$x = 2h$
0%
$x = \dfrac{h}{2}$
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$x = \dfrac{h}{3}$

Explanation

$\textbf{Hint}$ : Apply the formula of acceleration due to gravity above and below the earth surface.

$\textbf{Step 1}$ :

We know acceleration due to gravity on the earth's surface is given by

$g=\dfrac{GM}{R^2}$ where

$G$ is gravitational constant,

$R$ is the radius of the earth and

$M$ is the mass of the earth.

Given that the change in value of

$g$ at a height

$h$ above the surface of the earth is the same as at a depth

$x$ below it.

$g \propto \dfrac{1}{R^2}$

$g_h \propto \dfrac{1}{(R+h)^2}$

$\dfrac{g_h}{g}=\dfrac{R^2}{(R+h)^2}$

$\implies \dfrac{g_h}{g}=\dfrac{1}{(1+\dfrac{h}{R})^2}$

$\implies \dfrac{g_h}{g}=(1-\dfrac{2h}{R})$

$\implies g_h=g- \dfrac{2gh}{R}$

$\implies g-g_h=\dfrac{2gh}{R}$

$\textbf{Step 2}$ :

The acceleration due to gravity at depth

$x$ is given by

$g_x=\dfrac{4}{3}G_p(R-x)$

So,

$g_x \propto (R-x)$

$\dfrac{g_x}{g}=\dfrac{R-x}{R}$

$\dfrac{g_x}{g}=1-\dfrac{x}{R}$

$g-g_x=\dfrac{gx}{R}$

Given that the change in value of

$g$ at a height

$h$ above the surface of the earth is the same as at a depth

$x$ below it.

$g-g_h=g-g_x$

$\dfrac{2gh}{R}=\dfrac{gx}{R}$

$x=2h$

Thus option B is correct

Calculate energy needed for moving a mass of

$4kg$ from the centre of the earth to its surface (in joule). If radius of the earth is 6400 km and acceleration due to gravity at the surface of the earth is

$g = 10 m/sec^2$

0%
$1.28 \times 10^8 J$
0%
$1.28 \times 10^6 J$
0%
$2.56 \times 10^8 J$
0%
$2.56 \times 10^10 J$

Explanation

$W = U_R - U_C = [- \dfrac {GMm}{R} + \dfrac {3GMm}{2R}]$

$\Rightarrow \dfrac {GMm}{2R} = \dfrac {mgR^2}{2R}$

$\Rightarrow mgR/2 = 2gR = 2 \times 10 \times 6400 \times 10^3 = 1.28 \times 10^8 J$

The mass of the Jupiter is

$1.9\times 10^{27}\ kg$ and that of the sun is

$1.99\times 10^{38}\ kg$ . The mean distance of the Jupiter from the sun is

$7.8\times 10^{11}\ m$ . Speed of the Jupiter is (assuming that Jupiter moves in a circular orbit around the sun)

0%
$1.304\times 10^4\ m/sec$ .
0%
$13.04\times 10^4\ m/sec$ .
0%
$1.304\times 10^6\ m/sec$ .
0%
$1.304\times 10^2\ m/sec$ .

Explanation

$F = \dfrac{GMm}{r^2}$

$= \dfrac{6.67\times 10^{-11}\times (1.99\times 10^{30})\times (1.9\times 10^{27})}{(7.8\times 10^{11})^2}$

$= 4.14\times 10^{23} N$

Now let its speed of jupiter be

$v$ , then

$F = \dfrac{mv^2}{R}$

$\Rightarrow v = \sqrt{\dfrac{FR}{m}}$

$= \sqrt{\dfrac{(7.8\times 10^{11})\times (4.10\times 10^{23})}{(1.9\times 10^{27})}}$

$= 1.304\times 10^4 m/sec$ .

A space ship is launched into a circular orbit close to the surface of the earth. The additional velocity now imparted to the spaceship in the orbit to overcome the gravitational pull is

0%
${ 11.2 }km{ s }^{ -1 }$
0%
${ 8 }km{ s }^{ -1 }$
0%
${ 3.2 }km{ s }^{ -1 }$
0%
${ 1.414 }\times { 8 }km{ s }^{ -1 }$

Explanation

Given:

$R= 6400 km$

$g=9.8 m/s^2$

Orbital velocity of space ship revolving in an orbit of radius r,

$v_o= \sqrt{\dfrac{gR^2}{r}}$

For

$r= R$

$\implies v_o= \sqrt{gR}= \sqrt{9.8 \times 6400\times 1000} \approx 7.919 km/s$

Escape velocity,

$v_e= \sqrt{2gR} = \sqrt{2(9.8) \times 6400\times 1000}= 11.2 km/s$

Thus additional velocity imparted

$= 11.2- 7.919 \approx 3.29 km/s$

Find approximately the third cosmic velocity

$v_3$ in km/s i.e. the minimum velocity that has to be imparted to a body relative to the Earth's surface to drive it out of the Solar system. It is given that the rotation of the Earth about its own axis is to be neglected.

0%
$16.6$
0%
$20$
0%
$10$
0%
None of these

Explanation

Assume first that the attraction of the earth can be neglected. Then the minimum velocity, that must be imparted to the body to escape from the Sun's pull, is, as in

$1.230$ , equal to,

$(\sqrt 2-1)v_1$

$v_1^2=\gamma M_s/r$ ,

$r$ is radius of the earth's orbit,

$M_s$ is mass of the Sun.

In the actual case near the earth, the pull of the Sun is small and does not change much over distances, which are several times the radius of the Earth. The velocity

$v_3$ in question is that which overcomes the earth's pull with sufficient velocity to escape the Sun's pull. Thus

$mv_3^2/2-\gamma M_E/R=m(\sqrt 2-1)^2v_1^2/2$

$R$ is radius of the earth,

$M_s$ is mass of the earth.

$v_1^2=\gamma M_E/R$ ,

so,

$v_3=\sqrt{2v_2^2+(\sqrt 2-1)^2v_1^2}=16.6\ km/s$

A particle would take time

$t_1$ to move down a straight tube from the surface of earth (supposed to be homogeneous sphere) to its centre. If gravitational acceleration were to remain constant, time would be

$t_2$ . The ratio

$t/t'$ will be

0%
$\displaystyle \frac{\pi}{2 \sqrt{2}}$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{2\pi}{3}$
0%
$\displaystyle \frac{\pi}{\sqrt{3}}$

Explanation

Lets consider a distance of dr at a distance of r from center of earth.

now,

$dt_1 = \dfrac{dr}{-v}$ [since the velocity is towards the center of earth]

Energy is conserved , Initial = final energy

So,

$\dfrac{-GMm}{R} = \dfrac{mv^2}{2} + \dfrac{-GMm}{2R^3}(3R^2 - r^2)$

solving this, we get v =

$(\dfrac{GM(R^2 - r^2)}{R^3})^{0.5}$

Integrating on both sides with limits from R to -R

$t_1 = \pi (\dfrac{R^3}{GM})^{0.5} = \pi \times (\dfrac{R}{g})^{0.5}$

for

$t_2$ , use S =

$ut + \dfrac{a(t)^{2}}{2}$

So,

$R = 0 + \dfrac{g(t)^2}{2} \rightarrow t = (\dfrac{2R}{g})^{0.5}$

But,

$t_2 = 2t = 2(\dfrac{2R}{g})^{0.5}$

Taking ratio,

$\dfrac{t_1}{t_2} = \dfrac{\pi \times (\dfrac{R}{g})^{0.5}}{2 (\dfrac{2R}{g})^{0.5}} = \dfrac{\pi}{8^{0.5}}= \dfrac{\pi}{2 \sqrt2}$

Which of the following is correct?

0%
The value of g is constant throughout
0%
$\displaystyle g \propto \frac{1}{r^2}$
0%
$g$ is slightly less (by about 1%) when distance $< 200 m$
0%
$g$ is slightly greater when distance $< 200 m$ .

Explanation

The gravitational field inside earth is given by

$g=\dfrac { GM }{ { r }^{ 2 } } (\dfrac { { r }^{ 3 } }{ { R }^{ 3 } } )=\dfrac { GMr }{ { R }^{ 3 } }$

Hence, at a small depth h,

$g'=\dfrac { GM(R-h) }{ { R }^{ 3 } }$

Thus,

$\Delta g=\dfrac { GMh }{ { R }^{ 3 } } =g\dfrac { h }{ R }$

$\dfrac { \Delta g }{ g } =\dfrac { h }{ R }$

$\dfrac { \Delta g }{ g } =\dfrac { 200 }{ 6400000 } <1 percent$

Answer is option C.

The largest and the shortest distance of the earth from sun are a and b, respectively. The distance of the earth from sun when it is at a point where perpendicular drawn from the sun on the major axis meets the orbit is

0%
$\displaystyle \frac{ab}{a + b}$
0%
$\displaystyle \frac{ab}{2(a + b)}$
0%
$\displaystyle \frac{2ab}{a+b}$
0%
$\displaystyle \frac{a+b}{2ab}$

Explanation

$\dfrac{y}{r}=1-ecos\theta$ (here

$e$ is the eccentricity)

$r=a$ ,

$\theta=0$

$\dfrac{y}{a}=1-e$

$r=b$

$\theta=180^o$

$\dfrac{y}{b}=1+e$

Thus

$\dfrac{y}{a}+\dfrac{y}{b}=2$

$y=\dfrac{2ab}{a+b}$

The maximum vertical distance through which a fully dressed astronaut can jump on the earth is

$0.5 m$ . If mean density of the Moon is two-third that of the earth and radius is one quarter that of the earth, the maximum vertical distance which he can jump on the Moon and the ratio of the time of duration of the jump on the Moon to hold on the earth are

0%
$3 m$ , $6:1$
0%
$6 m$ , $3:1$
0%
$3 m$ , $1:6$
0%
$6 m$ , $1:6$

Explanation

The work that the astronaut can do is same on both earth and the moon.

Hence,

$mg_{e}h_{e}=mg_{m}h_{m}$

$\implies m(\dfrac{GM_{e}}{R_{e}^2})h_{e}=m(\dfrac{GM_{m}}{R_{m}^2})h_{m}$

Using

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

$h_{m}=h_{e}(\dfrac{\rho_{e}R_{e}}{\rho_{m}R_{m}})=3m$

Again since work done has be to same. initial kinetic energy would same at both the places, hence the initial velocity.

$t_{jump}=\dfrac{2v_{0}}{g}$

$\implies \dfrac{t_{m}}{t_{e}}=\dfrac{\rho_{e}R_{e}}{\rho_{m}R_{m}}=6:1$

$V_e$ is the escape velocity of a body from a planet of mass

$M$ and radius

$R$ . Then, the velocity of satellite revolving at height

$h$ from the surface of planet will be

0%
$V = V_e \sqrt{\dfrac{R}{(R + h)}}$
0%
$V = V_e \sqrt{\dfrac{2R}{(R + h)}}$
0%
$V = V_e \sqrt{\dfrac{(R + h)}{R}}$
0%
$V = V_e \sqrt{\dfrac{R}{ 2 (R + h)}}$

Explanation

The expression of escape velocity is

$\displaystyle V_e=\sqrt{\dfrac{2GM}{R}} ..................................(1)$
Let

$V$ be the velocity of satellite revolving at height

$h$ from the surface of planet.
here,

$\displaystyle \dfrac{mV^2}{(R+h)}=\dfrac{GMm}{(R+h)^2}$
or

$\displaystyle V=\sqrt{\dfrac{GM}{R+h}} .......................(2)$

$\displaystyle (2)/(1), V=V_e\sqrt{\dfrac{R}{2(R+h)}}$

A tunnel is dug along a chord of the earth at a perpendicular distance

$\displaystyle \frac{R}{3}$ from the earth's centre. Assume wall of the tunnel is frictionless. Find the force exerted by the wall on mass m at a distance x from the centre of the tunnel

0%
$\displaystyle \frac{mg \sqrt{\frac{R^2}{9} + x^2}}{R}$
0%
$\displaystyle \frac{mg x}{\sqrt{R^2 / a + x^2}}$
0%
$\displaystyle \frac{mg}{3}$
0%
$\displaystyle \frac{mg x}{R}$

Explanation

Mass enclosed in a sphere of radius

$(r=R/3)$ is

$\dfrac{M}{\dfrac{4}{3}\pi R^3}.\dfrac{4}{3}\pi r^3=\dfrac{Mr^3}{R^3}$

Thus force on a wall of mass m at a distance r from center=

$\dfrac{G(M.\dfrac{r^3}{R^3})m}{r^2}=\dfrac{GMmr}{R^3}$

For

$r=R/3$ ,

$F=\dfrac{GMm}{3R^2}=\dfrac{mg}{3}$

Four equal masses (each of mass

$M$ ) are placed at the corners of a square of side

$a$ . The escape velocity of a body from the centre

$O$ of the square is

0%
$\displaystyle \sqrt[4] {\frac {2GM}{a}}$
0%
$\displaystyle \sqrt {\frac {8\sqrt {2}GM}{a}}$
0%
$\displaystyle \frac {4GM}{a}$
0%
$\displaystyle \sqrt {\frac {4\sqrt {2}GM}{a}}$

Explanation

Potential energy of particle at the centre of square

$\displaystyle =-4 \left ( \dfrac {GMm}{\dfrac {a}{\sqrt {2}}} \right )$

$\displaystyle \therefore -4\left ( \dfrac {GMm}{a/\sqrt {2}}\right )+\dfrac {1}{2}mv^2=0 \Rightarrow v^2 = \dfrac {8\sqrt {2}GM}{a}$

Let

$V$ and

$E$ denote the gravitational potential and gravitational field at a point, respectively. It is possible to have

0%
$V=0$ and $E=0$
0%
$V=0$ and $E\ne 0$
0%
$V\ne 0$ and $E=0$
0%
$V\ne 0$ and $E\ne 0$

Explanation

We know the relations

$E=-\dfrac { dV }{ dx }$ and hence

$V=-\int { E.dx }$ .

All the options can satisfy both the equations since the parameters are considered only for a point and we know that

$\dfrac { dy }{ dx }$ can be 0 without

$y=0$ for a given point.

Answer is ABCD.

Which of the following statements are true about acceleration due to gravity?

0%
$g$ decreases in moving away from the centre if $r>R$
0%
decreases in moving away from the centre if $r<R$
0%
$g$ is zero at the centre of the earth
0%
$g$ decreases if earth stops rotating on its axis

Explanation

Moving away from the center, g varies as

$g=\dfrac { GM }{ { r }^{ 2 } }$ . Thus, g decreases in moving away from the centre if r>R.

Inside the earth, g varies as

$g=\dfrac { GMr }{ { R }^{ 3 } }$ . Hence at the center of earth, r=0 and hence g=0.

If earth stops rotating, the gravitational force will increase since there will be no centrifugal force.

Answer is A, C.

A body of mass

$m$ is taken from earth's surface to the height equal to the radius of earth, the change in potential energy will be of

0%
$\displaystyle mgR_e$
0%
$\displaystyle mgR_e/2$
0%
$\displaystyle 2mgR_e$
0%
$\displaystyle mgR_e/4$

Explanation

The gravitational potential energy at a height

$h$ is

$\displaystyle U_h=\frac {-GM_em}{R_e+h}$
If

$\displaystyle h=R_e \Rightarrow U_h=\frac {-GM_em}{+2R_e}= \frac {-mgR_e}{2}$
& the gravitational potential energy on the surface of earth is

$\displaystyle U_s=\frac {-GM_em}{R_e}$
So change in potential energy

$\displaystyle \bigtriangleup U=U_h-U_s$

$\displaystyle =\frac {-GM_em}{2R_e}+ \frac{GM_em}{R_e}= \frac {GM_em}{2R_e}$

$\displaystyle \bigtriangleup U=\frac{mgR_e}{2}$
It means that to move a body of a mass

$m$ from the surface of earth to a height

$h=R_e$ , the energy required is

$\displaystyle \frac {GM_em}{2R_e}=\frac{mgR_e}{2}$

A tunnel is dug along a chord of the earth at a perpendicular distance

$R/2$ from the earth's centre. The wall of the tunnel may be assumed to be frictionless. A particle is released from one end of the tunnel. The pressing force by the particle on the wall, and the acceleration of the particle vary with

$x$ (distance of the particle from the centre) according to

Explanation

Let the particle of mass

$m$ is at point P which lies at a distance

$x$ from the centre of the earth.

$\therefore x^2 = b^2 + \dfrac{R^2}{4}$ ..................(1)

where

$R$ is the radius of earth.

Let

$\rho$ and

$M$ be the density and mass of earth respectively.

Mass of part of earth having radius

$x$

$M' = \rho \bigg(\dfrac{4\pi}{3}\pi x^3\bigg) = \dfrac{M}{\frac{4\pi}{3}R^3} \times \dfrac{4\pi}{3}x^3$

$\therefore$

$M'= M \dfrac{x^3}{R^3}$

Now net force experienced by the particle

$F = \dfrac{G M' m}{x^2} = \dfrac{GMm}{R^3}x$

Pressing force

$F_{\perp} = F cos\theta = \dfrac{GMm x}{R^3} \times \dfrac{\frac{R}{2}}{x} = \dfrac{GMm}{2 R^2} = constant$

Let

$a$ be the acceleration of the particle.

$\therefore$

$ma = F_{\parallel} = F sin\theta$

Thus

$m a = \dfrac{GMm}{R^3} x \times \dfrac{b}{x}$

$\implies a = \dfrac{GM}{R^3} b = \dfrac{GM}{R^3} \sqrt{x^2 - \dfrac{R^4}{4}}$

A tunnel is dug along a chord of the earth at a perpendicular distance

$x$ (distance of the particle from the centre) according to :

Explanation

Net force towards centre of earth $\displaystyle ={mg}'=\dfrac {mgx}{R}$

Normal force $\displaystyle N={mg}'sin \theta$

Thus pressing force $\displaystyle N=\dfrac {mgx}{R}\dfrac {R}{2x}$

$\displaystyle N=\dfrac {Mg}{2}$ constant and independent of $x$ . Hence $(b)$ .

Tangential corce. $\displaystyle F=ma={mg}'cos \theta$

$\displaystyle Q={g}'cos \theta=\dfrac {gx}{R} \dfrac {\sqrt {\dfrac {R^2}{4}-x^2}}{x}$

$\displaystyle a=\dfrac {gx}{R}\sqrt {R^2-4x^2}$

Curve is parabolic and at $\displaystyle x=\dfrac {R}{2},a=0$ , Hence $(C)$

A (nonrotating) star collapses onto itself from an initial radius

$R_1$ with its mass remaining unchanged. Which curves in figure best describes the gravitational acceleration

$a_g$ on the surface of the star as a function of the radius of the star during the collapse

0%
$\displaystyle a$
0%
$\displaystyle b$
0%
$\displaystyle c$
0%
$\displaystyle d$

Suppose, the acceleration due ot gravity at the earth's surface is

$10 m s^{-2}$ and at the surface of Mars it is

$4.0ms^{-2}$ . A

$60$ kg passenger goes from the Earth to the Mars in a spaceship moving with a constant velocity. Neglect all other objects in the sky. Which part of figure bests represents the weight (net gravitational force) of the passenger as a function of time?

0%
$\displaystyle A$
0%
$\displaystyle B$
0%
$\displaystyle C$
0%
$\displaystyle D$

A tunnel is dug along a diameter of the earth. If

${ M }_{ e }$ and

${ R }_{ e }$ are the mass and radius, respectively, of the earth, then the force on a particle of mass

${ m }$ placed in the tunnel at a distance

${ r }$ from the centre is :

0%
$\displaystyle \cfrac { { G }{ M }_{ e }{ m } }{ { R }_{ e }^{ 3 } } { r }$
0%
$\displaystyle \frac { { G }{ M }_{ e }{ m } }{ { R }_{ e }^{ 3 }{ r } }$
0%
$\displaystyle \frac { { G }{ M }_{ e }{ m }{ R }_{ e }^{ 3 } }{ r }$
0%
$\displaystyle \cfrac { { G }{ M }_{ e }{ m } }{ { R }_{ e }^{ 2 } } { r }$

Explanation

Let us first calculate the mass of inner solid sphere of radius r.

Mass of inner sphere

$M'=\dfrac{M_e}{\dfrac{4}{3}\pi {R_e}^3}\times\dfrac{4}{3}\pi r^3=\dfrac{M_e}{{R_e}^3}r^3$

Now,

$g=\dfrac{GM_er^3}{{R_e}^3}\times\dfrac{1}{r^2}$

$g=\dfrac{GM_er}{{R_e}^3}$

Force on particle of mass m

$=mg=\dfrac{GM_emr}{{R_e}^3}$

Note that those layers of the sphere which have radii larger than r do not have any contribution to the gravitational force.

A person brings amass of

$1 kg$ from infinity to a point A lnitially, the mass was at rest but it moves at a speed of

$3 m/s$ as it reaches A. The work done by the person on the mass is

$- 5.5 J$ The gravitational potential at

$A$ is

0%
$-1J/kg$
0%
$-4.5 J/kg$
0%
$-5.5J /kg$
0%
$-10J/kg$

Explanation

Using work energy theorem:

$W_{person}+W_{gravity} = \Delta K.E.$

$-5.5+m (V_A-V_{infinity}) = \dfrac{1}{2}mv^2- 0 = \dfrac{1}{2}\times 1 \times 3^2$

$-5.5+ V_A -0= 4.5$

$V_A= 10 J/Kg$

The ratio of SI units to CGS units of g is.

0%
$10^2$
0%
10
0%
$10^{-1}$
0%
$10^{-2}$

Escape velocity for a projectile at earth's surface is

$v_e$ . A body is projected form earth's surface with velocity

$2_{v_e}$ . The velocity of the body when it is at infinite distance from the centre of the earth is

0%
$v_e$
0%
$2v_e$
0%
$\sqrt 2 v_e$
0%
$\sqrt 3 v_e$

Explanation

Let velocity of body when it is at infinity be v and also P.E is zero at infinity.

Apply conservation of energy at surface and infinity-

$\dfrac{1}{2}m(2v_e)^2 - \dfrac{GMm}{R}= \dfrac{1}{2}mv^2$

$\dfrac{1}{2}m\bigg[2\sqrt{\dfrac{2GM}{R}}\bigg]^2 - \dfrac{GMm}{R}= \dfrac{1}{2}mv^2$

$\dfrac{3GMm}{R}= \dfrac{1}{2}mv^2$

$v^2= 3 \dfrac{2GM}{R} = 3 v^2_e$

$\implies v= \sqrt{3} v_e$

If the gravitational acceleration at surface of earth is

$g$ , then increase in potential energy in lifting an object of mass

$m$ to a height equal to the radius

$R$ of earth will be.

0%
$\dfrac {mgR}{2}$
0%
$2mgR$
0%
$mgR$
0%
$\dfrac {mgR}{4}$

Explanation

Gravitational petential energy at r,

$P.E= -\dfrac{GMm}{r}$

$GM= gR^2 \implies P.E= -\dfrac{mgR^2}{r}$

For

$r=R$ ,

$P.E= -mgR$

For

$r=2R$ ,

$P.E'= -\dfrac{mgR}{2}$

Gain in P.E=

$P.E'- P.E=\dfrac{mgR}{2}$

Maximum weight of a body is.

0%
At the centre of the earth
0%
Inside the earth
0%
On the surface of the earth
0%
Above the surface of earth

Read the assertion and reason carefully to mark the correct option out of the options given below :

Assertion : At height

$h$ from ground and at depth

$h$ below ground, where h is approximately equal to

$0.62 R$ , the value of

$g$ acceleration due to gravity is same.

Reason : Value of

$g$ decreases both sides, in going up and down.

0%
If both assertion and reason are true and the reason is the correct explanation of the assertion
0%
If both assertion and reason are true but reason is not the correct explanation of the assertion
0%
If assertion is true but reason is false
0%
If assertion is false but reason is true

Explanation

Inside the earth's surface, gravitational field varies as

$g=\dfrac { GM }{ { r }^{ 2 } } \dfrac { { r }^{ 3 } }{ { R }^{ 3 } } =\dfrac { GMr }{ { R }^{ 3 } }$

Outside the earth's surface,

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

For gravity at height and depth h to be same,

$\dfrac { GM(R-h) }{ { R }^{ 3 } } =\dfrac { GM }{ { (R+h) }^{ 2 } }$

One solution is

$h=0.$ Solving this, we get

$h\approx 0.62R$

It is clear from both the expressions that the gravitational field decreases as one moves away from earth in any direction.

Answer is option A.

The density of the core of a planet is

$P_{1}$ and that of the outer shell is

$P_{2}$ the radii of the core and that of the planet are

$R$ and

$2R$ respectively. The acceleration due to gravity at the surface of the planet is same as at depth

$R$ . Find the ratio of

$\displaystyle \frac{P_{1}}{P_{2}}$

0%
$7:3$
0%
$3:7$
0%
$7:4$
0%
$4:7$

Explanation

Radius of hollow sphere is $\displaystyle \dfrac{R}{2}$ , so mass in this hollow portion would had been, $\displaystyle \dfrac{M}{8}$ .

Now net force on m due to whole sphere $=$ force due to remaining mass $+$ force due to cavity mass.

$\therefore$ Force due to remaining mass=force due to whole sphere-force due to cavity mass

$\displaystyle =\dfrac{GMm}{d^{2}}-\dfrac{GMm}{8\left ( d-R/2 \right )^{2}}$

$\displaystyle =\dfrac{GMm}{d^{2}}\left [ 1-\dfrac{1}{8\left ( 1-\dfrac{R}{2d} \right )^{2}} \right ]$

There are two planets. The ratio of radius of the two planets is

$k$ but ratio of acceleration due to gravity of both planets is

$g$ . What will be the ratio of their escape velocity?

0%
$(kg)^{1/2}$
0%
$(kg)^{-1/2}$
0%
$(kg)^{2}$
0%
$(kg)^{-2}$

Explanation

The equation of escape velocity is

$v_e=\sqrt{2gr}$

so,

$\dfrac{v_e1}{v_e2}=\dfrac{\sqrt{2g_1r_1}}{\sqrt{2g_2r_2}}=\sqrt{kg}$

Suppose a vertical tunnel is dug along the diameter of the earth, assumed to be a sphere of uniform mass density

$\rho$ . If a body of mass

$m$ is thrown in this tunnel, its acceleration at a distance

$y$ from the centre is given by:

0%
$\displaystyle \frac{4\pi}{3}G\rho ym$
0%
$\displaystyle \frac{3}{4}\pi \rho y$
0%
$\displaystyle \frac{4}{3}\pi \rho y$
0%
$\displaystyle \frac{4}{3}\pi G\rho y$

Explanation

HINT: Acceleration due to gravity is directly proportional to distance from centre inside the earth.

$\textbf{STEP 1: Acceleration due to gravity inside the earth}$

If a vertical tunnel is dug along the diameter of the earth, whose mass density is

$\rho$ , then the value of g or the acceleration due to gravity would vary along with the tunnel.

The value of acceleration due to gravity on the surface of earth is

$g = \dfrac{GM}{R^2}$

The value of acceleration due to gravity inside the earth is

$g_1 = \dfrac{GMy}{R^3}$

where,

$g =$ is acceleration due to gravity

$G =$ universal gravitational constant

$M =$ mass of earth

$R =$ radius of earth

$\textbf{STEP 2:Gravitational acceleration in terms of density}$

According to the given conditions, if the mass density of earth is uniform, then

$M = \dfrac 43 \pi R^3\rho\;\;\rightarrow(1)$

where,

$\rho$ is the uniform mass density of earth

Write g in terms of

$\rho$ we have

$g = \dfrac{4\pi GR\rho}3$

And, at a distance of

$y$ from the center the value is

$g_1 = \dfrac{4\pi G\rho y}3$

Thus, option D is correct.

The gravitational potential difference between the surface of a planet and a point

$20 m$ above it is

$16 J/kg$ . Then the work done in moving a

$2 kg$ mass by

$8 m$ on a slope

$60$ degree from the horizontal, is:

0%
$11.1 J$
0%
$5.55 J$
0%
$16 J$
0%
$27.7 J$

Imagine a spacecraft going from the earth to the moon. How does its weight vary as it goes from the earth to the moon?

0%
Initially decreases then becomes zero and again increases.
0%
Initially increases and again increases.
0%
Initially decreases again increases.
0%
zero throughout the path

Explanation

When a spacecraft is going from Earth to Moon,as long it is under Earth's gravitation,the acceleration due to gravity like

${ g }^{ \prime }={ g }(1-\cfrac { h }{ r } )$ [g=acceleration due to gravity at the surface of the Earth]

Then, when it goes out of the Earth's gravitation,its weight will be zero.

After that,it will enter to the moon's gravitational field.

Then its acceleration due to Gravity will be

${ g }^{ \prime }={ g }(1-\cfrac { (-h) }{ r } )\quad$

i.e,weight will be increasing

At what height in km over the earth's pole the free fall acceleration decreases by one percent? (Assume the radius of the earth to be

$6400\ km$ )

0%
$32$
0%
$64$
0%
$80$
0%
$1.253$

Explanation

Acceleration on the earth's surface is

$g=\dfrac{GM}{R^2}$ and the acceleration at height h will be

$g_h=\dfrac{GM}{(R+h)^2}$

So,

$\dfrac{g_h}{g}=\dfrac{R^2}{(R+h)^2}$

According to question,

$\dfrac{g_h}{g}=\dfrac{99}{100}$

so,

$\dfrac{99}{100}=\dfrac{R^2}{(R+h)^2}$

$\dfrac{R}{(R+h)}=0.995$ or

$R=0.995R+0.995h$

$h=\left(\dfrac{0.005}{0.995}\right )R=\left(\dfrac{0.005}{0.995}\right)(6400)=32.16\simeq 32$ km

A body hanging from a massless spring stretches it by

$3$ cm on Earth's surface. At a place

$800$ km above the Earth's surface, the same body will stretch the spring by : (Radius of Earth

$=6400$ km)

0%
$\left(\displaystyle\frac{34}{27}\right)$ cm
0%
$\left(\displaystyle\frac{64}{27}\right)$ cm
0%
$\left(\displaystyle\frac{27}{64}\right)$ cm
0%
$\left(\displaystyle\frac{27}{34}\right)$ cm
0%
$\left(\displaystyle\frac{35}{81}\right)$ cm

Explanation

Acceleration due to gravity,

$g=\displaystyle\frac{GM}{r^2}$

$\therefore g\propto \frac{1}{r^2}$
When the body is hanged on a spring

$F=-kx=mg$ ............(i)
Spring force

$F$ acts on it, where x is extension in spring. Let

$800$ km above the Earth's surface, the stretch in the length of spring is

$x$ ' and value of

$g$ is

$g'$ .

$g=\displaystyle\frac{GM}{R^2}$ and

$g'=\displaystyle\frac{GM}{(R+h)^2}$
So,

$g'=\displaystyle\frac{g}{\left(\displaystyle 1+\frac{h}{R}\right)^2}$ .........(ii)
where, R=radius of Earth
From Eq. (i), we get

$kx=mg$

$\Rightarrow x\propto g$ .........(iii)
Therefore,

$\displaystyle\frac{x'}{x}=\frac{g'}{g}$
From Eqs. (ii) and (iii), we get

$\displaystyle\frac{x'}{x}=\frac{g}{g\left(\displaystyle 1+\frac{h}{R}\right)^2}=\frac{g}{\displaystyle g\left(1+\displaystyle\frac{800\times 10^3}{6400\times 10^3}\right)^2}$

$=\displaystyle \frac{g}{g\left(1+\displaystyle\frac{1}{8}\right)^2}$

$=\displaystyle\frac{g}{g\left(\displaystyle \frac{g}{8}\right)^2}=\frac{g\times 64}{g\times 81}=\frac{64}{81}$

$x'=\displaystyle\frac{64}{81}x$

$[x=3cm]$

$\Rightarrow x'=\displaystyle\frac{64}{81}\times 3=\frac{64}{27}$

$\Rightarrow x'=\displaystyle\frac{64}{27}$ cm.

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 13 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 13 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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