MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 10

Assuming the earth as a sphere of uniform density, the acceleration due to gravity half way toward the centre of the earth will be

0%
$0.75g$
0%
$0.5g$
0%
$0.25 g$
0%
$0.125 g$

An ice cube of size

$a = 10\ cm$ is floating in a tank

$(base = 50cm\times 50cm)$ partially filled with water. The change in gravitational potential energy, when ice completely melts is ________ [Density of ice is

$900\ kg\ m^{-3}$ and

$g = 10\ ms^{-2}]$ .

0%
$-0.0455\ J$
0%
$-0.016\ J$
0%
$-0.24\ J$
0%
$-0.072\ J$

Explanation

Relative density of ice is

$0.9$ hence

$90$ % of volume of ice is immersed in water when melts completely level of water does not change.

Therefore change in

$CG=$ { GG }^{ 1 }=0.5cm$$

Change in potential energy

$\Delta V=-mgh$

$={ -\left( 0.1 \right) }^{ 3 }\left( 900 \right) \left( 0.5\times { 10 }^{ -2 } \right) 5$

$=-0.0455$

$W$ is the weight of a man on the surface of the earth. What would be his weight

$(W)$ if he goes to a height

$(h)$ equal to radius of the earth, from the surface of the earth?

0%
$W=\dfrac{W}{2}$
0%
$W=\dfrac{W}{3}$
0%
$W=\dfrac{W}{4}$
0%
$W=2W$

Explanation

Given that,

Weight =

$W$

Height =

$R$

We know that,

Weight is inversely proportional to square of distance between man and center of earth

Then, initial distance is

$R$ and initial weight is

$W$ and when it is at height

$R$ then distance is

$2R$

Now,

$W\propto \dfrac{1}{{{d}^{2}}}$

Now,

$\dfrac{{{W}_{1}}}{{{W}_{2}}}={{\left( \dfrac{{{d}_{2}}}{{{d}_{1}}} \right)}^{2}}$

$\dfrac{{{W}_{1}}}{{{W}_{2}}}={{\left( \dfrac{2R}{R} \right)}^{2}}$

$\dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{4{{R}^{2}}}{{{R}^{2}}}$

$\dfrac{W}{{{W}_{2}}}=4$

${{W}_{2}}=\dfrac{W}{4}$

Hence, the weight at height is

$\dfrac{W}{4}$

When a small mass

$m$ is suspended at lower end of the elastic wire having upper end fixed with ceiling. There is loss in gravitational potential energy. let it be x, due to extension of wire, mark correct option,

0%
The lost energy can be recovered
0%
The lost energy can be irrecoverable
0%
only $\dfrac{x}{2}$ amount of energy recoverable
0%
only $\dfrac{x}{3}$ amount of energy recoverable

Explanation

Work energy theorem

$\begin{array}{l}{w_{spring}} + {w_{gravity}} + {w_{loss}} = 0\\ - \dfrac{1}{2}k{x^2} + mgx + {w_{loss}} = 0\\ - \dfrac{1}{{2mgx}} + mgx + {w_{loss}} = 0\\{w_{loss}} = - \dfrac{{mgx}}{2}\end{array}$

Since there is a loss so not recoverable
half of it is lost rest is recoverable

Given that mass of earth is

$M$ and its radius

$R$ . body is dropped from a height equal to the radius of the earth above the surface of the earth. When it reaches the ground velocity of body will be

0%
$\dfrac {GM}{R}$
0%
$\left(\dfrac {GM}{R}\right)^{1/2}$
0%
$\dfrac {2GM}{R}$
0%
$\left(\dfrac {2GM}{R}\right)^{1/2}$

Explanation

This question can be very easily tackled using conservation of energy with a bit care,

Total energy of body at height 2R =

$m_b\dfrac{ GM_e} {4R_e^2}2R$ =Total energy of body at surface of earth=

$\dfrac{m_bv^2} {2}$

From these equations we get v=

$[\dfrac{GM_e}{R_e}] ^\dfrac{1} {2}$

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:

0%
$6$
0%
$3$
0%
$1/4$
0%
$1/6$

Explanation

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

$M_p=Density\times Volume=Density\times\dfrac{4\pi r^3}{3}$

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

Thus g

$\propto r$ and g

$\propto Density$

So,

$\dfrac{g_1}{g_2}=\dfrac{1\times 1}{2\times 3}=\dfrac{1}{6}$

If radius of earth contracted by 0.1% its mass remaining same,then weight of the body at earth's surface will increase by

0%
0.1%
0%
0.2%
0%
0.3%
0%
Remains same

Explanation

If radius is decreased by 0.1%

$r'=r-0.001 r=0.999 r$

$W'=\dfrac{GMm}{(0.999r)^2}=\dfrac{GMm}{(0.999)^2r^2}$

$=\dfrac{W}{0.999^2}$

$=1.002 W$

$\Delta W=W'-W=0.002$

$0.002 \times 100=0.2$ % increase.

The mean distance of Earth from the Sun is

$149.6 \times 10^6$ km and the mean distance of Mercury from the Sun is

$57.9 \times10^6$ km. The period of Earth’s revolutions is 1 year, what is the period of Mercury’s revolution?

0%
Infinite
0%
0.24 year on earth
0%
3.2 year on earth
0%
2.1 year on earth

Explanation

$\cfrac { { T }_{ 1 }^{ 2 } }{ { r }_{ 1 }^{ 3 } } =\cfrac { { T }_{ 2 }^{ 2 } }{ { r }_{ 2 }^{ 3 } }$

$1=$ Earth;

$2=$ Mercury

$\cfrac { { 1 }^{ 2 } }{ { \left( 149.6\times { 10 }^{ 6 } \right) }^{ 3 } } =\cfrac { { T }_{ 2 }^{ 2 } }{ { \left( 57.9\times { 10 }^{ 6 } \right) }_{ }^{ 3 } }$

$\Rightarrow \cfrac { 1 }{ 3348071.936\times { 10 }^{ 18 } } =\cfrac { { T }_{ 2 }^{ 2 } }{ 194104.539\times { 10 }^{ 18 } } \Rightarrow { T }_{ 2 }^{ 2 }=0.057975\Rightarrow { T }_{ 2 }=\sqrt { 0.057975 }$

${T}_{2}=0.24$ year on Earth

$1$ year on Earth

$=365$ days

At which height above the surface of the earth of radius

$R$ , will the acceleration due to gravity decrease by

$0.1\%$ ?

0%
$\dfrac {R}{100}$
0%
$\dfrac {R}{200}$
0%
$\dfrac {R}{1000}$
0%
$\dfrac {R}{2000}$

An astronaut orbiting in a spaceship round the earth has a centripetal acceleration of

$2.45m/s^{2}$ . The height of spaceship from earth's surface is (

$R=$ radius of earth)

0%
$3R$
0%
$2R$
0%
$R$
0%
$R/2$

Explanation

Given that,

Acceleration $a=2.45\,m/{{s}^{2}}$

Mass of earth $M=6\times {{10}^{24}}\,kg$

Radius of earth $R=6\times {{10}^{6}}\,m$

Gravitational constant $G=6.67\times {{10}^{-11}}\,N{{m}^{2}}/kg$

Now centripetal acceleration is

${{a}_{c}}=\dfrac{{{v}^{2}}}{R}$

${{a}_{c}}=\dfrac{{{\left( \sqrt{\dfrac{GM}{R}} \right)}^{2}}}{R}$

${{a}_{c}}=\dfrac{GM}{{{R}^{2}}}$

${{a}_{c}}=\dfrac{GM}{{{\left( R+h \right)}^{2}}}$

Now, put the values

$2.45=\dfrac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{{{\left( 6\times {{10}^{6}}+h \right)}^{2}}}$

${{\left( 6\times {{10}^{6}}+h \right)}^{2}}=\dfrac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{2.45}$

${{\left( 6\times {{10}^{6}}+h \right)}^{2}}=\dfrac{40.02\times {{10}^{13}}}{2.45}$

${{\left( 6\times {{10}^{6}}+h \right)}^{2}}=1.6334\times {{10}^{14}}$

$6\times {{10}^{6}}+h=\sqrt{1.6334\times {{10}^{14}}}$

$h=1.278\times {{10}^{7}}-6\times {{10}^{6}}$

$h=\left( 12.78-6 \right)\times {{10}^{6}}$

$h=6.78\times {{10}^{6}}$

$h=6.8\times {{10}^{6}}\,m$

So, $h\cong R$

Hence, the height of spaceship is $R$

The height at which the weight of a body becomes

$\dfrac {1}{16}th$ , its weight on the surface of earth (radius

$R$ ), is

0%
$3R$
0%
$4R$
0%
$5R$
0%
$15R$

Explanation

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

$\dfrac{w_{1}}{w_{2}} = \dfrac{(R+h)^2}{R^{2}}$

putting

$\dfrac{w_{1}}{W_2} = 16 = \dfrac{(R+h)^{2}}{R^{2}}$

$\Rightarrow \dfrac{R+h}{R} = 4 \Rightarrow 1+\dfrac{h}{2} = 4 \Rightarrow \boxed{h = 3R }$

1187794_1105826_ans_6ffa93fa047a46d899747115208249be.jpg

$'R'$ 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:

0%
$2R$
0%
$R$
0%
$\dfrac{3R}{8}$
0%
$\dfrac{R}{4}$

Explanation

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

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

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

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

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

$h=\dfrac { 3R }{ 8 }$

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

$g = 10 m/s^2$ ]

0%
$101 \ R$
0%
$100 \ R$
0%
$99 \ R$
0%
$199 \ R$

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:

0%
$g+v$
0%
$g-v$
0%
$g$
0%
$zero$

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.

0%
$75$ N
0%
$500$ N
0%
$62$ N
0%
$320$ N

Explanation

Weight of object on earth

$W = {\dfrac{GMm}{R^2}}$

Now,

$M = {\rho}V$ =

${\dfrac{4\pi \rho R^{3}}{3}}$

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

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

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

${\dfrac{1}{12}}$

${W_P} = {\dfrac{900}{12}} = 75 N$

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):

0%
${\dfrac{R}{n}}$
0%
${\dfrac{R}{n^2}}$
0%
${\dfrac{R(n-1)}{n}}$
0%
${\dfrac{Rn}{(n-1)}}$

Explanation

option C

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

$\Longrightarrow \dfrac { g }{ n } =g\left[ 1-\dfrac { d }{ R } \right] \Longrightarrow d=\left[ \dfrac { n-1 }{ n } \right] R$

A particle when thrown. moves such that it passes from same height at

$2$ and

$10 s$ , the height is:

0%
$g$
0%
$2g$
0%
$5g$
0%
$10g$

Explanation

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

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:

0%
$v=R\sqrt{\dfrac{8\pi}{3}G\rho}$
0%
$v=M\sqrt{\dfrac{8\pi}{3}G\rho}$
0%
$v=\sqrt{2GMR}$
0%
$v=\sqrt{\dfrac{2GM}{R}}$

The mass and radius of the sun are

$1.99\ \times\ 10^{30}\ kg$ and

$R=6.96\ \times\ 10^{8}\ m$ . The escape velocity of rocket from the sun is =......

$km/s$

0%
$11.2$
0%
$12.38$
0%
$59.5$
0%
$618$

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

0%
Zero
0%
Positive
0%
Negative
0%
Can have any value

Three particles each of mass

$'m'$ are at the vertices of an equilateral triangle of side length

$'l'$ . Then the work done is moving one particle to infinity is:

0%
$\dfrac{3Gm^2}{l}$
0%
$\dfrac{Gm^2}{2l}$
0%
$\dfrac{2Gm^2}{l}$
0%
$\dfrac{4Gm^2}{l}$

The value of

$g$ at the surface of earth is

$9.8\ m/s^{2}$ then the value of

$'g'$ at a place

$480\ km$ above the surface of the earth will be nearly (radius of the earth is

$6400\ km)$ .

0%
$9.8\ m/s^{2}$
0%
$7.2\ m/s^{2}$
0%
$8.3\ m/s^{2}$
0%
$4.2\ m/s^{2}$

Explanation

Given that,

$g=9.8\,m/{{s}^{2}}$

$h=480\,km$

$R=6400\,km$

We know that,

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

$g'=9.8\left( 1-\dfrac{2\times 480}{6400} \right)$

$g'=8.33\,m/{{s}^{2}}$

Hence, the value of g’ is $8.33\,m/{{s}^{2}}$ .

The value of acceleration due to gravity will be

$1\%$ of its value at the surface of earth at a height of

$(R_{e} = 6400\ km)$ .

0%
$6400\ km$
0%
$57600\ km$
0%
$2560\ km$
0%
$6100\ km$

The SI unit of the universal gravitational constant G:

0%
$Nm{{Kg}^{ - 2}}$
0%
$N{m^2}{{Kg}^{ - 2}}$
0%
$N{m^2}{{Kg}^{ - 1}}$
0%
$Nm{{Kg}^{ - 1}}$

Explanation

Force

$=\dfrac{G\ M_{1}\ M_{2}}{R^{2}}$

$[G]=\dfrac{[Force][R]^{2}}{[M_{1}][M_{2}]}$

$=\dfrac{N\ m^{2}}{(kg)^{2}}$

$=N\ m^{2}\ kg^{-2}$

Option

$B$ .

At what height from the surface of the earth, the value of g is reduced by 36% from its value at the surface of the earth of radius R=6400 km?

0%
1200 km
0%
1600 km
0%
2000 km
0%
2200 km

A body weights

$63 N$ on the surface of the Earth. At a height

$h$ above the surface of Earth, its weight is

$28 N$ while at a depth

$h$ below the surface Earth, the weight is

$31.5 N$ . The value of

$h$ is:

0%
$0.4 \ R$
0%
$0.5 \ R$
0%
$0.8 \ R$
0%
$R$

At what height above the surface of earth the value of "g" decreases by 2%? [radius of the earth is 6400 km]

0%
32 km
0%
64 km
0%
128 km
0%
1600 km

Explanation

The effective value of a above height h from eaarth is given by :

$g_{eff}=\frac{g}{\left ( 1+\frac{h}{R} \right )^{2}}$ ( R = radius of Earth )

Now given

$g_{eff}=0.98g$ (As value of g decreases by 2 %)

$\Rightarrow \left ( 1+\frac{h}{R} \right )^{2}=\frac{g}{0.98g}=1.020$

$\Rightarrow 1+\frac{h}{R}=\sqrt{1.020}=1.010$

$\Rightarrow h=0.010\times R=\frac{10}{1000}\times 6400km= 64 km$

The period of a simple pendulum on the surface of earth is

$T$ . At an altitude of half of the radius of earth from the surface, its period will be

0%
$\sqrt{\dfrac{3}{2}}T$
0%
$\dfrac{3T}{2}$
0%
$\dfrac{2T}{3}$
0%
$\sqrt{\dfrac{2}{3}}T$

Explanation

Time Period of pendulum is given by:

$T=2\pi\sqrt{\dfrac{L}{g}}\rightarrow (1)$ (L

$=$ length of pendulum, g

$=$ acceleration of gravity)

$h=\dfrac{R}{2}$ the acceleration due to gravity is given by:

$g_{eff}=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^2}=\dfrac{g}{\left(1+\dfrac{1}{2}\right)^2}=\dfrac{4g}{9}$

So new time period of pendulum will be

$T_{new}=2\pi\sqrt{\dfrac{L}{\dfrac{4g}{9}}}=\sqrt{\dfrac{3}{2}}\times 2\pi\sqrt{\dfrac{L}{g}}$

$T_{new}=\sqrt{\dfrac{3}{2}}T$ .

At a certain height above the earth's surface, the gravitational acceleration is

$4$ % of its value at the surface of the earth. Find the height. (

$R$ is the surface of the earth)

0%
$2R$
0%
$4R$
0%
$R$
0%
$R/2$

Explanation

$\textbf{Hint: Use formula of acceleration due to gravity at height h from earth's surface}$

$\textbf{Step1:Expression of acceleration due to gravity at a height h}$

The acceleration due to gravity at a height h above Earth's surface is:

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

$\textbf{Step2: Find height h}$

$g_h$ is

$4\%$ of g so

$g_h=0.04$ g

$\Rightarrow 0.04g=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^2}$

$\Rightarrow \left(1+\dfrac{h}{R}\right)^2=\dfrac{1}{0.04}=25$

$\Rightarrow 1+\dfrac{h}{R}=5$

$\Rightarrow \dfrac{h}{R}=4\Rightarrow h=4R$ .

$\textbf{Hence option B correct}$

The value of

$'g'$ at a certain height above the surface of the earth is

$16$ % of its value on the surface. The height is

$(R = 6300 \ km)$

0%
$10500 \ km$
0%
$12500 \ km$
0%
$2688 \ km$
0%
$9450 \ km$

A rocket is fired from the earth to the moon. The distance between the earth and the moon is

$r$ and the mass of the earth is

$81$ times the mass of the moon. The gravitational force on the rocket will be zero, when its distance from the moon is then

0%
$\dfrac{r}{20}$
0%
$\dfrac{r}{15}$
0%
$\dfrac{r}{10}$
0%
$\dfrac{r}{5}$

Explanation

Net force on the rocket is zero. i.e. force due to the earth and force due to moon is same but in opposite direction.
$Fe=Fm$
$\dfrac{GMe Mr}{re^2}=\dfrac{GMmMr}{rm^2}$
$r$ is the total distance
$r= re+ rm$
$Me=81 Mm$
$\dfrac{a 81 Mm Mr}{re^2}=\dfrac{G M m Mr}{rm^2}$
$r_m=\dfrac{r_e}{g}$
$g rm +rm =r \Rightarrow 10 r_m=r$
$\Rightarrow r_m=\dfrac{r}{10}$

The depth at which the value of acceleration due to gravity

$\dfrac {1}{n}$ times the value at the surface is: (

$R=$ radius of the earth).

0%
$\dfrac {R}{n}$
0%
$R{\dfrac {(n-1)}{n}}$
0%
$\dfrac {R}{n^{2}}$
0%
$R\left(\dfrac {n}{n+1}\right)$

A pendulum that beats seconds on the surface of the earth were taken to a depth of (1/4)th the radius of the earth.What will be its time period of oscillation?

0%
5.107 sec
0%
1.208 sec
0%
3.40 sec
0%
2.309 sec

A body weight 144 N at the surface of earth. When it is taken to a height of h = 3R,where R is radius of earth it would weight.

0%
48 N
0%
36 N
0%
16 N
0%
9 N

Explanation

Given that,

Weight of body $W=144\,N$

Height $h=3R$

Radius of earth $=R$

We know that,

Acceleration due to gravity at the surface of the earth $=g$

Acceleration due to gravity at height $g'=3R$

Now,

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

$\dfrac{g}{g'}={{\left( \dfrac{R}{R+3R} \right)}^{2}}$

$\dfrac{g}{g'}=\dfrac{1}{16}$

Now, weight at the height

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

$\dfrac{w'}{w}=\dfrac{1}{16}$

$w'=\dfrac{w}{16}$

$w'=\dfrac{144}{16}$

$w'=9\,N$

Hence, the weight at the height is $9\ N$

When work done by of gravity is negative, which of the following will definitely occur

0%
Potential energy increases
0%
kinetic energy decreases
0%
Gravitational potential energy increase
0%
Gravitational potential energy decreases

If a tunnel is dug along the diameter of the earth and a body is dropped into it, then the time taken by it to cross the tunnel once is:-

0%
$2 \pi \sqrt{\dfrac{Re}{g}}$
0%
$\sqrt {2gr}$
0%
$2 \pi \sqrt{\dfrac{g}{Re}}$
0%
$\pi \sqrt{\dfrac{g}{Re}}$

Two planets are at mean distance

$d_1$ and

$d_2$ from the sun and their periods are

$T_1$ and

$T_2$ respectively, Then:

(n is the frequency)

0%
$n_1^2d_1^2=n_2^2d^2_2$
0%
$n_2^2d_2^3=n_1^2d^3_1$
0%
$n_1d_1=n_2d_2$
0%
$n_1^2d_1=n_2^2d_2$

Explanation

According to kepler's 3rd law of planetary motion-

$T^2 \propto r^3$ [Here,

$r=d$ ]

As we know that,

$T = \dfrac{1}{n}$

$\left(\dfrac{1}{frequency}\right)$

$\dfrac{1}{n^2} \propto d^3 \Rightarrow \dfrac{n^2_2}{n_1^2} = \dfrac{d^3_1}{d^3_2}$

$\Rightarrow n_2^2 d_2^3 = n_1^2 d_1^3$

option B is correct.

2030495_1140232_ans_d9526aaebee5443ba91fdaefaaffdbc3.png

The value of acceleration due to gravity at a height R from surface of the earth is then(R=radius of the earth and g=acceleration due to gravity on earth surface )

0%
$0$
0%
$\sqrt{g}$
0%
$\dfrac{g}{4}$
0%
$\dfrac{g}{2}$

Explanation

$g'=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^2}$

$g'=\dfrac{g}{\left(1+\dfrac{R}{R}\right)^2}=\dfrac{g}{(1+1)^2}$

$g'=\dfrac{g}{4}$ .
option

$C$

The value of gravitational acceleration at a height equal to the radius of the earth is

0%
Equal to its value at the earth's surface
0%
50% of its value at the surface
0%
25% of its value at the surface
0%
75% of its value at the surface

At what height from the surface of the earth will the value of acceleration due to gravity be reduced by

$36\%$ from the value at the surface?
(Radius of earth=

$6400\ km$ )

0%
$1500\ km$
0%
$1200\ km$
0%
$1000\ km$
0%
$1600\ km$

The value of 'g' at a certain height above the surface of the earth is 27% of its value on the earth's surface.The height is : (R=6300 km)

0%
2100 km
0%
13031 km
0%
14031 km
0%
14700 km

If the gravitational potential on the surface of earth is

$V_0$ then potential at a point at height half of the radius of earth is

0%
$\dfrac{V_0}{2}$
0%
$\dfrac{2}{3}V_0$
0%
$\dfrac{V_0}{3}$
0%
$\dfrac{3V_0}{2}$

Explanation

${u_r} = \frac{{ - GMm}}{r}\left( {Gravitational\,\,potential} \right)$

$= {v_0}$ at surface

at height

$h \sim \frac{r}{2}$

${u_{{r^2}}} \sim \frac{{ - GMm}}{{\frac{{3r}}{2}}} = \frac{2}{3}{V_0}$

Hence,

option

$(B)$ is correct answer.

At what temperature will the root mean square velocity of oxygen molecule be sufficient so as to escape from the earth? Escape velocity from the earth is

$11.0\ km/s$ and the mass of the one molecule of oxygen

$5.34\times 10^{-25}\ kg$ (Boltzmann constant

$k=1.38\times 10^{-23}\ J/K$ ):

0%
$3\times 10^{5}\ K$
0%
$3.5\times 10^{5}\ K$
0%
$1.56\times 10^{5}\ K$
0%
none of these

The amount of work done in lifting a body of mass 'm' from the surface of the earth to a height equal to twice the radius of the earth is

0%
$\dfrac{2GMm}{3R}$
0%
$\dfrac{3GMm}{2R}$
0%
$\dfrac{5GMm}{3R}$
0%
$\dfrac{3GMm}{5R}$

The loss in wight of a body taken from earth's surface to a height

$h$ is 1%. The charge in weight taken into a mine of depth

$h$ will be

0%
$1$ % loss
0%
$1$ % gain
0%
$0.5$ % gain
0%
$0.5$ % loss

The weights of an object in a coal mine, at the surface and at the top of a mountain at the same piece on the earth are

$W_1, W_2, W_3$ respectively then

0%
$W_1 > W_2 > W_3$
0%
$W_3 > W_2 > W_1$
0%
$W_1 = W_2 = W_3$
0%
$W_1 < W_2 > W_3$

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

0%
$R/2$
0%
$\sqrt { 2 } R$
0%
$2R$
0%
$\dfrac { R }{ \sqrt { 2 } }$

The decrease in trhe value of g on going to a height above the earth's surface will be :-

0%
$g/2$
0%
$\dfrac{5g}{9}$
0%
$\dfrac{4g}{9}$
0%
$\dfrac{g}{3}$

Explanation

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

$g$ at height

$h$

${ g }_{ h }=\dfrac { GM }{ { \left( R+h \right) }^{ 2 } } \quad \quad \left( h=R/2 \right)$

${ g }_{ h }=\dfrac { GM }{ { \left( R+\dfrac { R }{ 2 } \right) }^{ 2 } } =\dfrac { 4GM }{ 9{ R }^{ 2 } } =\dfrac { 4g }{ 9 }$

Decrease value of

$g$

$g-\dfrac { 4g }{ 9 } =\dfrac { 5g }{ 9 }$

The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be

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

A particle of mass

$m_1$ lies inside a spherical shell of mass

$m_2$ and radius R at a distance r from the centre. The gravitational potential energy of the system is

0%
$-\dfrac{G(m_1+m_2)}{r}$
0%
$\dfrac{Gm_1m_2}{R^2}$
0%
$+\dfrac{Gm_1m_2}{r^2}$
0%
$+\dfrac{Gm_1m_2}{R}$

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 10 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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