MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 9

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

A projectile is fired vertically upwards from the surface of the earth with a velocity

$kv_e$ where

$v_e$ is the escape velocity and

$k < 1$ . If R is the radius of the earth, the maximum height to which it will rise measured from the centre of earth will be (neglect air resistance).

0%
$\dfrac{1-k^2}{R}$
0%
$\dfrac{R}{1-k^2}$
0%
$R(1-k^2)$
0%
$\dfrac{R}{1+k^2}$

A man weighs

$80$ kg on the surface of earth of radius R. At what height above the surface of earth his weight will be

$40$ kg?

0%
$2637.6km$
0%
$3.675km$
0%
$8765km$
0%
$(\sqrt{2}+1)R$

Explanation

On surface of earth

$g$ is

$9.81m/{ s }^{ 2 }$ . The weight is given by

$80kg$ .

The gravitational force

$\left( g \right) =\dfrac { G\times { m }_{ 1 }\times { m }_{ 2 } }{ { d }^{ 2 } }$ where

$G$ is gravitational constant distance between two objects that weight

${ M }_{ 1 }$ and

${ M }_{ 2 }$ . Considering weight on earth as

${ M }_{ 1 }$ .

$\left( g \right) =\dfrac { h\times { m }_{ 1 } }{ { d }^{ 2 } }$

Radius of earth is

$6371km$ which will be halved

$2637.6km$ .

Therefore for a person going on earth to weight

$40kg$ . He has to move

$2637.6km$ above earth's surface.

Which of the following laws are conserved, if the areal acceleration is zero

0%
Law of conservation of angular velocity
0%
Law of conservation of angular momentum
0%
Law of conservation of angular acceleration
0%
Law of conservation of angular displacement

The value of universal gravitational constant on earth for a particle of mass 5 kg is

0%
$6.67 \times 10^{-11}$
0%
$6.67 \times 10^{-7}$
0%
$5 \times 6.67 \times 10^{-11}$
0%
$6.67 \times 10^{-23}$

Explanation

Gravitational constant depends only on the mass of the earth and not on the mass of the particle on the earth. Hence its values remains constant =

$6.67 \times 10^{-11}$

The correct option is (a)

Two identical particles of mass

$m$ are placed at a distance

$r$ from each other. If their separation is doubled, then the effect on gravitational constant will be

0%
Gravitational constant remains same
0%
Gravitational constant becomes quadrupled
0%
Gravitational constant becomes 1/4th of the actual one
0%
Gravitational constant becomes doubled

Explanation

According to the Newton's law of gravitation, gravitational force between two objects is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

$F \propto \dfrac{m_1m_2}{r^2}$

Hence,

$F = \dfrac{Gm_1m_2}{r^2}$ , where

$G$ is the proportionality constant , known as the Universal Gravitational constant.

Gravitational constant does not depend on masses or the distance between them

Changing the distance between the masses, changes the force between them.

Hence, The correct option is (A)

The areal velocity of an object of mass m=2 kg revolving around another object is given by

$2m^2/s$ , what is the angular momentum of the particle

0%
$6 kg-m^2/s$
0%
$8 kg-m^2/s$
0%
$4 kg-m^2/s$
0%
$2 kg-m^2/s$

Explanation

The areal velocity of a planet is related to its Angular momentum by the equation areal velocity = 2m x areal velocity

Substituting the values, we get, areal velocity=

$8 kg-m^2/s$

The correct option is (b)

If the gravitational constant is expressed in terms of

$dynes\ m^{-2} kg^{2}$ , how will the value of G change:

0%
$6.67 \times 10^{-11} dynes \ kg^2 m^{-2}$
0%
$6.67 \times 10^{-8} dynes \ kg^2 m^{-2}$
0%
$6.67 \times 10^{-6} dynes \ kg^2 m^{-2}$
0%
$6.67 \times 10^{-3} dynes \ kg^2 m^{-2}$

Explanation

'G' is expressed as

$G=6.67\times { 10 }^{ -11 }N.{ m }^{ -2 }.{ kg }^{ +2 }\\ \therefore 1N={ 10 }^{ 5 }dyne\\ 1m={ 10 }^{ 2 }cm\\ 1kg={ 10 }^{ 3 }g.$

$G=6.67\times { 10 }^{ -11 }\times { 10 }^{ 5 }dynes.kg.{ m }^{ -2 }$ [As,

$1N={ 10 }^{ 5 }$ dyne]

A particle of mass M is placed at origin and a small mass m is placed at A, at a distance of r from M. A force F is applied to m to make it move from A to a nearby point B. When the force becomes zero, it is observed that the mass m moves from B back to A. This is due to the reason

0%
Potential of B is larger than potential of A
0%
Objects starts moving in gravitational field until constant potential difference exist
0%
The line B to A is equipotential surface
0%
The mass moves from B to A, since A is nearer to origin

Explanation

The potential of a particle is given by

$V = -GM/R$ , M is the mass of the object that exerts gravitational potential

A flow is constituted only, if it is present along a potential gradient and this flow will take place in the direction from a region of higher potential to a lower one,

As seen from the formula,

$r_B >r_A \implies V_B>V_A$ (because of the negative sign in the potential). Thus, particle moves from B to A

The correct option is (a)

Linear momentum of the planet is

0%
Different for different points of the orbit
0%
Conserved
0%
Not conserved
0%
None of these

The weight (W) of an object at a depth of R/4 from the surface of the earth will be (R is the radius of the earth)

0%
Zero
0%
W/4
0%
W/2
0%
W

Explanation

The apparent acceleration due to gravity at a depth d is given by

$g'=g(1-d/R)$ .Substituting d=R/4, we get g' = g/4

Thus the apparent weight becomes W/4

The correct option is (b)

A spring balance is graduated on sea level. If a body of mass 5 kg is weighed with this balance and the balance is taken to a height of 360kms and the object is weighed again, then the weight of the object

0%
Will be more than 5 kg
0%
Will be less than 5 kg
0%
Will remain same
0%
Will first increase and then decrease

Explanation

The apparent acceleration due to gravity is given by

$g'=g(1-2h/R)$ . As h increases, g' decreases. Thus weigh will go on decreasing continously

The option (b) is correct

At what height over the earths pole, the free fall acceleration decreases by one percent (assume the radius of earth to be 6400 km)

0%
32 km
0%
80 km
0%
1.293 km
0%
64 km

Explanation

Acceleration due to gravity at poles =

$9.8 m/s^2$

The apparent acceleration due to gravity at a height h is given by

$g'=g(1-2h/R)$ .

Rearranging the terms, we get,

$(g'-g)/g=2h/R \implies h= (R/2) (\delta g/g) = (6400/2) \times (1/100) = 32 km$ .

The correct option is (a)

The gravitational potential energy is the

0%
work done in bringing an object from infinity to radius r
0%
work done in moving an object around the earth
0%
work done by an object in attaining an object's acceleration equal to $9.8 m/s^2$
0%
work done in moving an object between two points horizontally

The variation of acceleration due to gravity

$g$ with distance

$d$ from centre of the earth is best represented by (

$R=$ Earths radius):

Explanation

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.

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

Explanation

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 :

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

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

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

0%
$h/ R$
0%
$-2h/ R$
0%
$2h/ R$
0%
None of these

Explanation

$g = \dfrac {GM}{R^{2}} .....(i)$

$\dfrac {dg}{dR} = \dfrac {-2GM}{R^{3}}$
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.

Gravitational potential energy is

0%
the product of force and velocity
0%
the product of force and momentum
0%
the product of weight and height lifted by an object
0%
the product of force and displacement, if the particle is moving in a cirlce

Explanation

Explanation:

The product of weight and the height to which it has been elevated above the surface of the earth is gravitational potential energy.

The expression of gravitational potential energy is defined as,

$U=mgh$

The acceleration due to gravity at a height

$\left (\dfrac {1}{20}\right )^{th}$ the radius of earth above earth's surface is

$9\ m/s^{2}$ . Its value at a point at an equal distance below the surface of earth is ________

$m/s^{2}$ .

0%
$55\ m/s^{2}$
0%
$9.5\ m/s^{2}$
0%
$12\ m/s^{2}$
0%
$15\ m/s^{2}$

Explanation

The variation of acceleration due to gravity outside the earth.

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

${ g }^{ 1 }=g\left( 1-\dfrac { 1 }{ 10 } \right)$

$g=g\left( \dfrac { 9 }{ 10 } \right)$

$g=10m/{ s }^{ 2 }$

Now, for inside point,

${ g }^{ 11 }=g\left( 1-\dfrac { d }{ R } \right)$

${ g }^{ 11 }=10\left( 1-\dfrac { 1 }{ 20 } \right)$

${ g }^{ 11 }=9.5m/{ s }^{ 2 }$

Escape velocity from the surface of moon is less than that from the surface of earth because the moon has no atmosphere while earth has a very dense one.

0%
True
0%
False

Explanation

As escape velocity depends on the acceleration due to gravity. So, escape velocity for moon must be less than that for earth. And due to less escape velocity the gas molecules easily escape from the planet as it become less than

$rms$ velocity of gas molecules resulting no atmosphere.

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?

0%
$1.6\times 10^{6}m$
0%
$2.4\times 10^{6}m$
0%
$3.2\times 10^{6}m$
0%
$4.8\times 10^{6}m$

Explanation

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

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).

0%
$W$
0%
$W/2$
0%
$W/4$
0%
$W/8$

Explanation

The variation of g with depth is given by

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

$d=\dfrac{R}{2}$

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

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

On multiplying both sides by m

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

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

At the depth half way, weight becomes half.

The earth, moving around the sun in a circular orbit, is acted upon by a force and hence work must be done on earth by this force.

0%
True
0%
False

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

0%
$\dfrac {m_{1}}{m_{2}}$
0%
$\dfrac {m_{2}}{m_{1}}$
0%
$\dfrac {m_{1}^{2}}{m_{2}^{2}}$
0%
$\dfrac {m_{2}^{2}}{m_{1}^{2}}$

The radius of a planet is

$4$ times the radius of the earth. The time period of revolution of the planet will be:

0%
$1\ yr$
0%
$2\ yr$
0%
$4\ yr$
0%
$8\ yr$

Explanation

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.

If a planet gets inflated, keeping its density constant, then

$g$ will increase.

0%
True
0%
False

Explanation

Due to increase in radius the expression.

$g=\dfrac { GM }{ { r }^{ 2 } } =\dfrac { G\left( density\times volume \right) }{ { r }^{ 2 } }$

$g\alpha r$

So, flation leads to increase in acceleration due to

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).

0%
Zero
0%
$\infty$
0%
$\dfrac {GM}{R}$
0%
$-\dfrac {GM}{R}$

Explanation

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.

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

0%
$0.75$ % increase
0%
$3.0$ % decrease
0%
$0.75$ % decrease
0%
$1.5$ % decrease

Explanation

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

$0.75$ %

The ratio of acceleration due to gravity at a depth

$h$ below the surface of earth and at a height

$h$ above the surface for

$h<< R$

0%
constants only when $h<< R$
0%
increases linearly with $h$
0%
increases parabolically with $h$
0%
decreases

Explanation

Above the surface,

$g'_{above}=\dfrac{GM}{(R+h)^2}$ (at depth h)

Inside the earth,

$g'_{below}=\dfrac{GM}{R^2}(1-\dfrac{h}{R})$

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

But for

$\dfrac { h }{ R } <<<1$

$\dfrac{g'_{below}}{g'_{above}}$ =

$(1-\dfrac { h }{ R } ){ \left( 1+\dfrac { h }{ R } \right) }^{ 2 }$

$=(1-\dfrac { h }{ R } ){ \left( 1+\dfrac { 2h }{ R } +{ \left( \dfrac { h }{ R } \right) }^{ 2 } \right) }\\ =1+\dfrac { 2h }{ R } +{ \left( \dfrac { h }{ R } \right) }^{ 2 }-\dfrac { h }{ R } -2{ \left( \dfrac { h }{ R } \right) }^{ 2 }-{ \left( \dfrac { h }{ R } \right) }^{ 3 }\\ =1+\dfrac { h }{ R }$

It increases linearly with

$'h'$

If R is the radius of the earth, then the ratio of acceleration due to gravity on surface of the earth to acceleration due to gravity at height nR is:

0%
$\left( {n + 1} \right)^{ - 1}$
0%
$\left( {n + 1} \right)^2$
0%
$\left( {n + 1} \right)^{ - 2}$
0%
$\left( {n + 1} \right)^3$

Explanation

Given

Radius of Earth

$=R$

Acceleration due to gravity

$g=\cfrac{GM}{R}$

Also

$g^{\prime}=g(\cfrac{1-2h}{R})$

Given height

$=nR$

$g^{\prime}=g(\cfrac{1-2nR}{R})$

When body is raised by

$nR$

$g^{\prime}=\cfrac{GM}{(R+nR)^2}\\g^{\prime}=\cfrac{GM}{R(1+n)^2}$

Now ratio of

$g$ and

$g^{\prime}$

$\Rightarrow(n+1)^2$

The value of

$g$ on the earths surface is

$980\ cm/{sec}^{2}$ . Its value at a height of

$64\ Km$ from the earth's surface is

0%
$960.40\ cm/sec ^{ 2 }$
0%
$984.90\ cm/sec ^{ 2 }$
0%
$982.45\ cm/sec ^{ 2 }$
0%
$977.55\ cm/sec ^{ 2 }$

Explanation

At earth's surface.

$g=\dfrac{GM}{R^2}=9.8 m/s^2$

$\dfrac{GM}{(6400 \times 10^3)^2}=9.8 m/s^2$ ...(1)

At a height 64 kms from earth's surface,

$\dfrac{GM}{((6400+64) \times 10^3)^2}=g'$ ...(2)

(1)

$\div$ (2)

$(\dfrac{6464}{6400})^2=\dfrac{9.8}{g'}$

$g'=(\dfrac{6464}{6400})^2 \times 9.8=9.6069 m/s^2$

$=960.69 cm/s^2$

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)

0%
$\left( \dfrac { n }{ n+1 } \right) mgR$
0%
$\left( \dfrac { n-1 }{ n } \right) mgR$
0%
$\dfrac { mgR }{ n }$
0%
$\dfrac { mgR }{ (n-1) }$

Explanation

At surface gravitaional potential energy

$U_1=-\dfrac{GMm}{R}$

At height

$nR, U_2=-\dfrac{GMm}{R+nR}$

$\left| { U }_{ 2 }-{ U }_{ 1 } \right| =\left| \left( \dfrac { -GMm }{ R+nR } \right) -\left( \dfrac { -GMm }{ R } \right) \right| \\ =\dfrac { GMm }{ R } -\dfrac { GMm }{ R+nR } \\ \Rightarrow \dfrac { GMm }{ R } \left[ 1-\dfrac { 1 }{ n+1 } \right] \Rightarrow \left( \dfrac { n }{ n+1 } \right) \dfrac { GMm }{ R } \\ \Rightarrow \left( \dfrac { n }{ n+1 } \right) mgR$

The height of the point vertically above the earths surface at which the acceleration due to gravity becomes

$1\%$ of its value at the surface is (

$R$ is the radius of the earth)

0%
$9R$
0%
$10R$
0%
$8R$
0%
$20R$

At what height from the ground will be the value of

$g$ be the same as

that in

$10 km$ deep mine below the surface of earth.

0%
20 km
0%
7.5 km
0%
5 km
0%
2.5 km

Explanation

Same changes in value of

$'g'$ can be observe at a depth

$'x'$

and height

$'2x'$

given that the depth

$'d'=x=10km$

$\therefore$ the same charge can be observed at a height

$2x=20km$

Hence,

the correct option is

$A.$

A particle hanging from a spring stretches it by 1 cm at earth's surface. Radius of the earth is 6400 km. At a place 800 km above the earth's surface, the same particle will stretch the spring by

0%
0.79 cm
0%
1.2 cm
0%
4 cm
0%
17 cm

Explanation

The extension is

$x=\dfrac{mg}{k}$

At earth's surface,

$x=\dfrac{1}{100}=\dfrac{m \times \dfrac{GM}{R^2} }{k}$ ..(1)

At 800 km,above the earth's surface,

$x'=\dfrac{n\times \dfrac{GM}{(R+800)^2}}{k}$ ..(2)

(2)

$\div$ (1)

$\dfrac{x'}{\dfrac{1}{100}}=\dfrac { \dfrac { GM }{ { \left( R+800 \right) }^{ 2 } } }{ \dfrac { GM }{ { R }^{ 2 } } } =\dfrac { { R }^{ 2 } }{ { \left( R+800 \right) }^{ 2 } }$

$\Rightarrow x'\times 100=\dfrac { x' }{ { \left( 1+\dfrac { 800 }{ R } \right) }^{ 2 } } =\dfrac { x' }{ { \left( 1+\dfrac { 800 }{ 6400 } \right) }^{ 2 } } =\dfrac { 1 }{ 1.265 } \\ =0.79\times { 10 }^{ -2 }\quad m\\ =0.79\quad cm$

The value of acceleration due to gravity at a point P inside the earth and at another point Q outside the earth is g/2 .(g being acceleration due to gravity at the surface of the earth). Maximum possible distance in terms of radius of earth R between P and Q is:

0%
$2R$
0%
$2R(\sqrt{2}+1)$
0%
$\dfrac{R}{2} (2\sqrt{2}-1)$
0%
$\dfrac{R}{2} (2\sqrt{2}+1)$

Explanation

In above the surface,

$\dfrac{g}{2} = \dfrac{GM}{(R+h)^2} ....1$

And for below the surface,

$\dfrac{g}{2} = \dfrac{GM}{(R-d)^2} ....2$

Equating both the above equation then we get,

$R^2 = h^2 + 2Rh$

And,

$R^2 = d^2 – 2Rd$

By solving quadratic for both hand d terms as:

$h = (\sqrt2 – 1)R$

And,

$d = (\sqrt2 + 1)R$

Thus,

$h + d = (\sqrt2 + 1)R + (\sqrt2 - 1)R$

$=> 2\sqrt{2}R$ meter.

A body weight W Newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be :

0%
$\dfrac{W}{2}$
0%
$\dfrac{2W}{3}$
0%
$\dfrac{4W}{9}$
0%
$\dfrac{W}{4}$

Explanation

Given,

$R=6400km,w=mg\Rightarrow g=\dfrac{w}{m}$

Now the weight of the body at the height is

$\dfrac{1}{2}R$

$g'=g(1-\dfrac{d}{R})$

So the weight of the body at the

$\dfrac{1}{2} R$

$W'=W(1-\dfrac{R}{2R})=W\dfrac{R}{2R})=\dfrac{W}{2}$

The height at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth. (Radius of the earth = 6400 km) is (approximately) :

0%
2630
0%
2640
0%
2650
0%
2660

Explanation

The new value of

$gravity$ at a height

$h$ is given as

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

As visible from the formula that incresing

$h$ results in decresing gravity.

So just put

$g\prime=g/2$ i.e. 50% less than

$g$ where

$g$ is gravity at surface. so we get

$g/2= \dfrac{gR^2}{(R+h)^2}$ or

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

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

$\sqrt[2]{2}= \dfrac{R+h}{R}=1+ h/R$ putting values we get

$\dfrac{h}{R}= 1.414-1=0.414$

$h=0.414R=2649.6Km$

Option C is correct.

At what altitude (h) above the earth's surface would the acceleration due to gravity be one fourth of its value at the earths surface?

0%
$h=R$
0%
$h=4R$
0%
$h=2R$
0%
$h=16R$

A particle weights 120N on the surface of the earth. At what height above the earth's surface will its weight be 30N? Radius of the earth is 6400 km.

0%
6000
0%
6400
0%
5800
0%
7000

A particle is taken to a height

$R$ above the surface, where

$R$ is the radius of the earth. The acceleration due to gravity there is:

0%
$2.45\ m/s^{2}$
0%
$4.9\ m/s^{2}$
0%
$4.8\ m/s^{2}$
0%
$19.6\ m/s^{2}$

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere ?(given : Mass of oxygen

$(m) =2.76\times 10^{-26}$ kg Boltzmann's constant

$k_B=1.38 \times 10^{-23}JK^{-1})$ :

0%
$2.508 \times 10^4K$
0%
$8.360 \times 10^4 K$
0%
$5.016 \times 10^4 K$
0%
$1.254 \times 10^4 K$

Explanation

Given,

$mass=2.7\times10^{-26}kg.k=1.38\times10^{-23}j/k$

So,

$V_{es}=V_{rms}$

$11.2\times10^3=\sqrt{\dfrac{3kT}{m}}\Rightarrow T=\dfrac{[11.2\times10^3]^2}{3k}=\dfrac{[11.2\times10^3]^2}3\times1.38\times10^{-23}=8.360\times10^4$

The altitude at which the weight of a body is only

$64\%$ of its weight on the surface of the earth is (Radius of the earth

$6,400km$ )

0%
$1600 M$
0%
$2Km$
0%
$160 Km$
0%
$1600 Km$

Explanation

Given,

$R=6400km$

So,

$g'=g(1-\dfrac{2h}{R})=g-\dfrac{2gh}{R}\Rightarrow g-g'=\dfrac{2gh}{R}$

Percentage decreases in weight,

$=\dfrac{mg-mg'}{mg}\times100=\dfrac{g-g}{g}\times100=\dfrac{2gh}{gR}\times100=\dfrac{2h}{R}\times100\Rightarrow 64=\dfrac{2\times h}{6400}\times100\Rightarrow h=2.048km$

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%
$15R$

If the radius of the earth is

$6400km$ , the height above the surface of the earth ,where the value of acceleration due to gravity will be

$1\%$ of its value from the surface of the earth is

0%
$6400km$
0%
$64km$
0%
$711km$
0%
$5760km$

A body of mass m is taken from earth's surface to q a height equal to radius of earth . the change in potential will be

0%
$mg$
0%
$\dfrac{1}{2}mg R$
0%
$2mgR$
0%
$\dfrac{1}{4}mgR$

Explanation

Given,

Mass

$=m,$

Height

$h=R$

Potential energy at surface

$U_s = \dfrac{-GMm}{R} = -mgR$

$(gR^2 = GM)$

Potential energy at height

$h = R$ ,

$U_h = -\dfrac{GMm}{R+R} = \dfrac{-mgR}{2}$

Change in potential energy

$\Delta U = \dfrac{-mgR}{2}-(-mgR) = \dfrac{mgR}{2}$

At what height above the surface of earth, acceleration due to gravity is

$\Bigg \lgroup \frac{1}{8} \Bigg \rgroup^{th}$ of its value at the surface of earth: (

$R_e$ = Radoius of earth)

0%
$\sqrt{3}R_e$
0%
$2R_e$
0%
$2\sqrt{2} R_e$
0%
$(\sqrt{3} - 1) R_e$

Earth can be considered as uniform solid sphere of radius

$R. g_{1}$ and

$g_{2}$ be acceleration due to gravity at points

$R/2$ below the surface and

$R/2$ above the surface of earth respectively. Then

$\dfrac {g_{1}}{g_{2}}$ equals to

0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{2}$
0%
$\sqrt {2}$
0%
$\dfrac {9}{8}$

If 'R' is radius of the earth, the height above the surface of the earth where the weight of a body is

$36\%$ less than its weight on the surface of the earth is

0%
$\dfrac{4R}{5}$
0%
$\dfrac{R}{5}$
0%
$\dfrac{R}{6}$
0%
$\dfrac{R}{4}$

How much deep inside the earth radius should a an go ,so that his wight become on fourth of that one the earth's surface/

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

Explanation

If the weight has to be

$\dfrac{1}{4}$ of the original, the new weight is

$\dfrac{mg}{4}$

Therefore, the gravitational force

$=\dfrac{g}{4}=\dfrac{mg}{\dfrac{4}{m}}$

Now,

$\dfrac{g}{4}=g(1-\dfrac{h}{R})\Rightarrow1-\dfrac{h}{R}=\dfrac{1}{4}\Rightarrow R-\dfrac{h}{R}=\dfrac{1}{4}\Rightarrow 4(R-h)=R\Rightarrow h=\dfrac{3}{4}R$

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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