MCQ Questions for Class 11 Engineering Physics Gravitation Quiz 5

An object lying at the equator of the earth will fly off the surface (will feel weightlessness), if the length of the day becomes

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$1.00 h$
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$1.41 h$
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$17.0 h$
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$17.0 min$

Explanation

Escape velocity at earth radius

$R$ is given by

$V = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}$
Now, time period

$T=\dfrac{2\pi R}{\sqrt{2gR}} = \pi \sqrt{\dfrac{2R}{g}} = 3.14 \times \sqrt{\dfrac{2 \times 6.4 \times 10^6}{9.8}} = 1\ hr(approx.)$
A is the right answer.

In a relation

$\displaystyle \sqrt{\frac{2GM}{R}} \geq C$ , G stands for universal gravitational constant, M for mass of a very very dense material, R for a small radius and C for velocity of light. Then,

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the above relation indicates that orbital velocity is equal to velocity of light.
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the above relation indicates something like a black hole.
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the above relation indicates that the light is escaping.
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the above relation indicates that any object starting from the given mass with speed more than light will return to it.

Explanation

$(\dfrac{2GM}{R})^{0.5} V_e$ is the escape velocity of planet with mass

$M$ and radius

$R$ . If,

$V_e$ is greater than speed of light, it means that mass is concentrated in Schwarzschild radius. So, it behaves as the black hole where even light cannot escape its gravitation.

${ g }$ is the acceleration due to gravity on the earth's surface, the change in the potential energy of an object of mass

$m$ raised from the surface of the earth to a height equal to the radius

$R$ of the earth is

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$\displaystyle \dfrac { mgR }{ 2 }$
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$\displaystyle { 2mgR }$
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$\displaystyle { mgR }$
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$\displaystyle { -mgR }$

Explanation

$\text{Gain in potential energy}= \triangle U=U_2-U_1$

$\triangle U=(-\dfrac{GMm}{2R})-(-\dfrac{GMm}{R})$

$\triangle U=\dfrac{GMm}{R}-\dfrac{GMm}{2R}$

$\triangle U=\dfrac{GMm}{2R}$

$\triangle U=\dfrac{mgR^2}{2R} As,GM=gR^2$

$\triangle U=\dfrac{mgR}{2}$

Two equals masses each

$m$ and

$m$ are hung from a balance whose scale pans differ in vertical height by

$'h'$ . The error in weighing in terms of density of the earth

$\rho$ is:

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$\displaystyle \pi G \rho m h$
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$\cfrac { 1 }{ 3 } \pi G \rho m h$
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$\cfrac { 8 }{ 3 } \pi G \rho m h$
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$\cfrac { 4 }{ 3 } \pi G \rho m h$

Explanation

we know,

$g_h=g[1-\dfrac{2h}{R}]$

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

$g-g_h=2\times\dfrac{GM}{R^3}h$

$g-g_h=\dfrac{2Gh}{R^3}\times\dfrac{4}{3}\pi R^3\rho$
or,

$g-g_h=\dfrac{8}{3}\pi G\rho h$
or,

$mg-mg_h=\dfrac{8}{3}\pi G\rho hm$

A spring balance is calibrated at sea level. If this balance is used to measure the weight of a body at successive increasing heights from the surface of the earth, then the weight indicated by spring balance will

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decrease continuously
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increase continuously
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first decrease, then increase
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remains constant

Explanation

The acceleration due to gravity on or above earth is given by g =

$\dfrac {GM}{r^2}$ , where r is distance from earth's center.
i.e.

$g \alpha \dfrac{1}{r^2}$
now as u increase the heights, r values takes

$r_e, r_e + h , r_e + 2h$ etc.
We can observe that, r is increasing and this implies that

$g$ is decreasing accordingly.

A small mass

$m$ is moved slowly from the surface of the earth to a height

$h$ above the surface. The work done (by an external agent) in doing this is

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$mgh$ , for all values of $h$
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$mgh$ , for $h<< R$
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$1/2mgR$ , for $h=R$
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$-1/2mgR$ , for $h=R$

Explanation

Work done=Gain in potential energy

$=\dfrac{GMm}{R}-\dfrac{GMm}{R+h}$

For

$h<<r$ ,

Work done

$=\dfrac{GMmh}{R^2}=mgh$

For

$h=R$

Work done=

$\dfrac{GMm}{R}-\dfrac{GMm}{2R}=\dfrac{GMm}{2R}=\dfrac{1}{2}mgR$

If both the mass and radius of the earth decrease by

$1$ % the value of

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acceleration due to gravity would decrease by nearly $1$ %
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acceleration due to gravity would increase by $1$ %
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escape velocity from the earth's surface would decrease by $1$ %
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the gravitational potential energy of a body on earth's surface will remain unchanged

Explanation

The acceleration due to gravity is: $g=\dfrac{GM}{R^2}$

The new value of g would be: $g'=\dfrac{G(0.99M)}{(0.99R)^2}\simeq1.01\dfrac{GM}{R^2}=1.01g$

Thus,g would increase by about 1%.The new escape velocity would be:

${V_e}'=\sqrt{\dfrac{2\times0.99M\times G}{0.99R}}=\sqrt{\dfrac{2MG}{R}}=V_e$

Thus,the escape velocity will remain unchanged.The potential energy of a body of mass m on earth's surface would be

$-\dfrac{GM(0.99M)}{(0.99R)}=-\dfrac{GmM}{R}$

Thus,the potential energy will also remain unchanged.Hence,the correct choices are (b) and (d).

A body of mass

${ m }$ rises to a height

${ h }=\dfrac{R}{5}$ from the earth's surface where

${ R }$ is the earth's radius. If

$g$ is acceleration duem to gravity at the earth's surface, the increase in potential energy is

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${ mgh }$
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$\cfrac { 4 }{ 5 } { mgh }$
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$\cfrac { 5 }{ 6 } { mgh }$
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$\cfrac { 6 }{ 7 } { mgh }$

Explanation

$\text{The gain in potential energy}\quad \triangle U=U_2-U_1$

$=\left(-\dfrac{GM_em}{R+h}\right)-\left(-\dfrac{GM_em}{R}\right)$

$\triangle U=\dfrac{GM_em(R+h-R)}{(R+h)R}$

$\triangle U=\dfrac{mgR\times h}{R+h}$

$\text{Substitute}\quad R=5h$

$\triangle U=\dfrac{5mgh^2}{6h}=\dfrac{5}{6}mgh$

The value of

${ g }$ at a certain height

${ h }$ above the free surface of the earth is

${ x }/{ 4 }$ where

${ x }$ is the value of

${ g }$ at the surface of the earth. The height

${ h }$ is

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${ R }$
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${ 2R }$
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${ 3R }$
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${ 4R }$

Explanation

Acceleration due to gravity at some distance

$r$ from center of the earth=

$\dfrac{GM}{r^2}$ .

It is given that

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

and

$\dfrac{x}{4}=\dfrac{GM}{r^2}$

$\implies r=2R$

which is the distance from center of earth.

Therefore distance from surface

$=h=2R-R=R$

The change in the value of

$g$ at a height

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

$d$ below the earth. When both

$d$ and

$h$ are much smaller than the radius of earth, then which one of the following is correct?

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$d=\cfrac { h }{ 2 }$
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$d=\cfrac { 3h }{ 2 }$
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$d=2h$
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$d=h$

Explanation

$\textbf{Step1:Acceleration due to gravity}$

$\bullet$ Acceleration due to gravity at a height

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

$\bullet$ Acceleration due to gravity at a depth

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

$\textbf{Step2:Relation between d and h}$

According to question,

$g_{h}' = g_{d}'$

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

where d : depth in earth surface

R : radius of earth

h : height above the earth

$\Rightarrow d = 2h$

Hence option (C) is correct.

If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is

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$2R$
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$\cfrac {R}{2}$
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$R$
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$\cfrac{R}{4}$

Explanation

$\dfrac{1}{2}\sqrt{\dfrac{2GM}{R}}=\sqrt{\dfrac{GM}{R+h}}$

$\dfrac{1}{4}\times\dfrac{2GM}{R}=\dfrac{GM}{R+h}$

$\dfrac{1}{2}\times\dfrac{1}{R}=\dfrac{1}{R+h}$

$2R=R+h$

$h=R$

An object is taken from a point

$P$ to another point

$Q$ in a gravitational field:

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assuming the earth to be spherical, if both $P$ and $Q$ lie on the earth's surface, the work done is zero
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if $P$ is on the earth's surface and $Q$ above it, the work done is minimum when it is taken along the straight line $PQ$
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the work done depends only on the position of $P$ and $Q$ and is independent of the path along which the particle is taken
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there is no work done if the object is taken from $P$ to $Q$ and then brought back to $P$ along any path

Explanation

If P and Q both lie on the earth's surface, this means both have same

$P.E$ that implies same mechanical energy(as K.E is Zero). Thus no work is done in moving an object from P to Q.

As gravitational field is a conservative field and thus work done by gravitational force depends only on the position of P and Q and is independent of path taken.

Also work done by a conservative force along a closed loop is Zero.

$g$ is acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass

$m$ raised from the surface of earth to a height equal to the radius

$R$ of the earth is

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$\cfrac { 1 }{ 2 } { mgR }$
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${ 2mgR }$
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${ mgR }$
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$\cfrac { 1 }{ 4 } { mgR }$

Explanation

$\text{The gain in potential energy}=\triangle U=U_2-U_1$

$\triangle U=(-\dfrac{GMm}{2R})-(-\dfrac{GMm}{R})=\dfrac{GMm}{2R}$

$\triangle U=\dfrac{1}{2}(\dfrac{GM}{R^2})mR \, \text{As,} \ g= \dfrac{GM}{R^2}$

$\triangle U=\dfrac{1}{2}mgR$

The gravitational potential due to earth at infinite distance from it is zero. Let the gravitational potential at a point

$P$ be

$-{ 5 }{ J }{ kg }^{ -1 }$ . Suppose, we arbitrarily assume the gravitational potential at infinity to be

$+{ 10 }{ J }{ kg }^{ -1 }$ , then the gravitational potential at

$P$ will be

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$-{ 5 }{ J }{ kg }^{ -1 }$
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$+{ 5 }{ J }{ kg }^{ -1 }$
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$-{ 15 }{ J }{ kg }^{ -1 }$
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$+{ 15 }{ J }{ kg }^{ -1 }$

Explanation

According to the problem as the potential at

$\infty$ increases by

$+10\ J\ kg^{-1}$ , hence potential will increases by the same amount everywhere (potential gradient will remain constant). Hence, potential point

$P=10-5=+5\ J\ kg^{-1}$

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

$40 kg$ ?

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

Explanation

Let

$r$ be the distance from center of earth where it weighs 40 kg.

$W_{surface}=\dfrac{GMm}{R^2}$

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

Since

$W_{r}=\dfrac{1}{2}W_{p}$ ,

$r=\sqrt{2}R$ .

Hence the height of this point

$=r-R=\sqrt{2}R-R=(\sqrt{2}-1)R$

Let

$V$ and

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

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$\displaystyle V=0$ and $E=0$
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$\displaystyle V=0$ and $E \neq 0$
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$\displaystyle V\neq 0$ and $E = 0$
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All of the above

Explanation

A) At

$\infty$ both V and E are zero.
B) Let,

$V_ \infty = \dfrac{GM}{R}$ for an unit mass.

So,

$V_R = 0$ i.e. at the radius

$R$ of solid sphere(mass

$M$ ) and

$E_R = \dfrac{GM}{R^2}$
C) Inside a spherical shell V =

$\dfrac{-GM}{R}$ and E =0
Thus, all the above(D) is correct.

Which of the following are not correct?

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The escape velocity for the Moon is $6km$ ${s}^{-1}$
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The escape velocity from the surface of Moon is $v$ . The orbital velocity for a satellite to orbit very close to the surface of Moon is $v/2$
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When an earth satellite is moved from one stable orbit to a further stable orbit, the gravitational potential energy increases
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The orbital velocity of a satellite revolving in a circular path close to the planet is independent of the density of the planet.

Explanation

A.The escape velocity for the Moon is $2.4kms^{-1}$

D. $v_0=\sqrt{\dfrac{2GM}{R}}=\sqrt{\dfrac{2G}{R}\times\dfrac{4}{3}\pi {R^3}\rho}$

or ${v_0}\propto {\sqrt{\rho}}$

so,the incorrect choices are A and D.

The escape velocity of a projectile from the earth is approximately

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$\displaystyle 7$ $km/sec$
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$\displaystyle 112$ $km/sec$
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$\displaystyle 11.2$ $km/sec$
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$\displaystyle 1.1$ $km/sec$

Explanation

The escape velocity of projectile from earth is

$\displaystyle v_{e}=\sqrt {2g{R_e}}$ , where

$\displaystyle R_{e}$ is radius of earth
since

$\displaystyle g=9.8m/sec^2, R_{e}=6.4\times10^6$ metre

$\displaystyle \Rightarrow v_{e}=11.2km/sec$

$g$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass

$m$ raised from the surface of the earth to a height equal to the radius

$R$ of the earth is

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$\cfrac { 1 }{ 2 } mgR$
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$2mgR$
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$\cfrac { 1 }{ 4 } mgR$
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$mgR$

Explanation

Gain in potential energy

$=(-\dfrac{GMm}{2R})-(-\dfrac{GMm}{R})$

$=\dfrac{GMm}{R}-\dfrac{GMm}{2R}$

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

$=\dfrac{mgR^2}{2R} As, GM=gR^2$

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

There is no atmosphere on the moon because.

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It is closer to the earth
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It revolves round the earth
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It gets light from the sun
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The escape velocity of gas molecules is lesser than their root mean square velocity

The escape velocity of an object projected from the surface of a given planet is independent of

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Radius of the planet
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The direction of projection
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The mass of the planet
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None of these

Explanation

The escape velocity of an object from any planet is

$\displaystyle v_{escape}=\sqrt {2gR} = \sqrt {2GM/R}$
Where

$R$ &

$M$ are the radius & mass of the planet

The weight of a body at the centre of the earth is

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Zero
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Infinite
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Same as on the surface on earth
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None of these

Explanation

Hint:
The magnitude of the gravitational force on a body equals the weight of the body.

Solution:
Step 1:
Concept used:
The force that pulls a body towards the center of the earth is its weight.
$W =mg$ .......(1)

where 'g' is the acceleration due to gravity,
W is the weight of a body of mass 'm'.

Step2:
A test mass always experiences an acceleration towards the center of the earth so it always tries to go towards the center of the earth. therefore, if the test mas is placed at the center of the earth then it will not feel any acceleration towards any other point, therefore the acceleration at the center of the earth is zero.

Calculating the weight of a body at the center of the earth:
The force exerted by any portion of the Earth at its center will be neutralized by the force exerted by the opposite portion. When the value of g is substituted in equation (1) as zero, the total weight will be zero.

$v_e$ and

$v_0$ represent the escape velocity and orbital velocity of a satellite corresponding to circular orbit of radius

$R$ , then

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$\displaystyle v_e=v_o$
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$\displaystyle v_e=\sqrt 2 v_o$
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$\displaystyle v_e=(1/\sqrt {2})v_o$
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$\displaystyle v_e$ and $\displaystyle v_0$ are not related

Explanation

escape velocity

$v_e=\sqrt{GM/R}$

Orbital velocity

$v_o=\sqrt{GM/2R}$

so,

$v_e=\sqrt 2v_o$

Taking the gravitational potential at a point infinte distance away as zero, the gravitational potential at a point

$A$ is

$-5$ unit. If the gravitational potential at a point infinite distance away is taken as

$+10$ units, the potential at a point

$A$ is

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$\displaystyle -5$ unit
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$\displaystyle +5$ unit
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$\displaystyle +10$ unit
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$\displaystyle +15$ unit

Explanation

The gravitational potential

$V$ at a point distance

$'r'$ from a body a mass

$m$ is equal to the amount of work done in moving a unit mass from infinity to that point

$\displaystyle V_r-V_\infty =-\int_{\infty}^{r} \bar{E}.d \bar{r}=-GM (1/r-1/ \infty)$

$\displaystyle = \frac {-GM}{r} \left [ As \bar{E}= \frac {-dV}{dr} \right ]$

$(i)$ In the first case
when

$\displaystyle V_{\infty}=0, V_{r}=\frac {-GM}{r}= -5$ unit

$(ii)$ In the second case

$V_{\infty}=+10$ unit

$\displaystyle V_{r}-10=-5$
or

$\displaystyle V_{r}=+5$ unit

The mass of the moon is

$1/81$ of earth's mass and its radius

$1/4$ that of the earth. If the escape velocity from the earth's surface is

$11.2$ km/sec. its value from the surface of the moon will be

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$\displaystyle 0.14 kms^{-1}$
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$\displaystyle 0.5 kms^{-1}$
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$\displaystyle 2.5 kms^{-1}$
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$\displaystyle 5.0 kms^{-1}$

Explanation

$\displaystyle v_e = \sqrt {\frac{2GM_e}{R_e}}$

$v_m = \sqrt {\dfrac{2G \dfrac{M_e}{81}}{\dfrac {R_e}{4}}}= \dfrac {2}{9}v_e$

$\displaystyle =2/9 \times 11.2 kms^{-1}=2.5 kms^{-1}$

The kinetic energy needed to project a body of mass

$m$ from the earth surface (radius

$R$ ) to infinity is

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$\displaystyle mgR/2$
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$\displaystyle 2mgR$
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$\displaystyle mgR$
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$\displaystyle mgR/4$

Explanation

If any object through with escaping velocity then the object will never come back at the ground, So we can assume it will reach at infinite distance.

$\displaystyle KE=\frac {1}{2}mv_{esc}^{2} = \frac {1}{2}m (\sqrt {2gR})^2= mgR$

The ratio of the radii of the planets

$R_1$ and

$R_2$ is

$k$ . The ratio of the acceleration due to gravity is

$r$ . The ratio of the escape velocities from them will be

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$\displaystyle kr$
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$\displaystyle \sqrt {kr}$
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$\displaystyle \sqrt {(k/r)}$
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$\displaystyle \sqrt {(r/k)}$

Explanation

We know that,

$\displaystyle v_e=\sqrt {(2gR)}$

$\displaystyle \therefore \frac {(v_e)p_1}{(v_e)p_2}=\frac {\sqrt {(2g_1R_1)}}{\sqrt {(2g_2R_2)}}=\sqrt {\left ( \frac {g_1}{g_2} \right)} \times \sqrt {\left ( \frac {R_1}{R_2} \right )}=\sqrt {kr}$

The radius of a planet is

$1/4^{th}$ of

$R_e$ and its acc, due to gravity is

$2g$ . What would be the value of escape velocity on the planet, if escape velocity on earth is

$v_e$

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$\displaystyle \frac {v_e}{\sqrt {2}}$
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$\displaystyle v_e \sqrt {2}$
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$\displaystyle 2v_e$
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$\displaystyle \frac {v_e}{2}$

Explanation

The escape velocity on the earth is defined as

$\displaystyle v_e = \sqrt {2g_eR_e}$
Where

$R_e$ &

$g_e$ are the radius & acceleration due to gravity of earth
Now for planet

$\displaystyle g_p=2g_e, R_p=R_e/4$
So

$\displaystyle v_p =\sqrt {2g_pR_p}=\sqrt {2 \times 2g_e \times R_e/4}=\frac {v_e}{\sqrt {2}}$

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

$20 m$ above the surface is

$2$ joule/kgs. If the gravitational field is uniform, then the work done in carrying a

$5$ kg body to a height of

$4 m$ above the surface is

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$\displaystyle 2 J$
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$\displaystyle 20 J$
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$\displaystyle 40 J$
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$\displaystyle 10 J$

Explanation

For uniform gravitational field,

$\displaystyle g =-\dfrac {V}{r}= - \frac {-2}{20}=\frac {1}{10}ms^{-2}$

Now,

$\displaystyle W= mgh = 5 \times \frac {1}{10}\times 4 = 2J$

Consider earth to be a homogeneous sphere. Scientist

$A$ goes deep down in a mine and scientist

$B$ goes high up in a balloon. The gravitational field measured by

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$A$ goes on decreasing and that by $B$ goes on increasing
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$B$ goes on decreasing and that by $A$ goes on increasing
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Each decreases at the same rate
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Each decreases at different

Explanation

$Consider\quad Earth\quad to\quad be\quad a\quad uniform\quad solid\quad sphere.\\ Inside\quad the\quad sphere\quad g\quad \propto \quad R,\quad while\\ outside\quad the\quad sphere\quad g\propto \quad \dfrac { 1 }{ { R }^{ 2 } } \\ Hence,\quad the\quad observatvion\quad of\quad each\quad scientist\quad will\quad differ.$

The weight of an object in the coal mine, sea level and at the top of the mountain, are respectively

$W_1$ ,

$W_2$ and

$W_3$ then

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$\displaystyle W_1 < W_2 > W_3$
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$\displaystyle W_1 = W_2 = W_3$
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$\displaystyle W_1 < W_2 < W_3$
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$\displaystyle W_1 > W_2 > W_3$

Explanation

At the surface of earth, the value of

$g=9.8m/sec^2$ . If we go towards the centre of earth or we go above the surface of earth, then in both the cases the value of

$g$ decreases.
Hence

$\displaystyle W_1=mg_{mine}, W_2=mg_{sea level}, W_3=mg_{mountain}$
So

$\displaystyle W_1<W_2>W_3$ (

$g$ at the sea level

$=g$ at the surface of earth)

The escape velocity of a body depends upon its mass as

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$\displaystyle m^0$
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$\displaystyle m^1$
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$\displaystyle m^2$
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$\displaystyle m^3$

The largest and the shortest distance of the earth from the sun is

$r_1$ and

$r_2$ . Its distance from the sun when it is at perpendicular to the major axis of the orbit drawn from the sun:

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$\displaystyle (r_1+r_2)/4$
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$\displaystyle (r_1+r_2)/(r_1-r_2)$
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$\displaystyle 2r_1r_2/(r_1+r_2)$
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$\displaystyle (r_1+r_2)/3$

Explanation

Using the ellipse property is :

$2/R=1/r_1 + 1/r_2$

$R=2r_1r_2/(r_1+r_2)$

Escape velocity when a body of mass

$m$ is thrown vertically from the surface of the earth is

$v$ , what will be the escape velocity of another body of mass

$4 m$ if thrown vertically

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$\displaystyle v$
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$\displaystyle 2v$
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$\displaystyle 4v$
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None of these

The total mechanical energy E possessed by a body of mass 'm', moving with a velocity 'v' at a height 'h' is given by :

$E\, =\, \displaystyle \frac{1}{2}\, mv^{2}\, +\, mgh.$ Make 'm' the subject of formula.

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$m\, =\, \displaystyle \frac{2E}{v^{2}\, +\, gh}$
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$m\, =\, \displaystyle \frac{2E}{v^{2}\, +\, 2gh}$
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$m\, =\, \displaystyle \frac{3E}{v^{2}\, +\, 2gh}$
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$m\, =\, \displaystyle \frac{E}{v^{2}\, +\, 2gh}$

Explanation

$E= \frac{1}{2}mv^{2}+mgh$

Multiply both side 2:

$2E= mv^{2}+2mgh$
Or

$2E= m\left ( v^{2}+2gh \right )$
Or

$m= \dfrac{2E}{v^{2}+2gh}$

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is

$V$ . Due to the rotation of a planet about its axis the acceleration due to gravity

$g$ at equator is

$\dfrac{1}{2}$ of

$g$ at poles. The escape velocity of a particle on the pole of a planet in terms of

$V$ is

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$\displaystyle V_e=2V$
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$\displaystyle V_e=V$
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$\displaystyle V_e=\dfrac{V}{2}$
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$\displaystyle V_e=\sqrt {3}V$

Explanation

$\displaystyle V=\omega R$

$\displaystyle g=g_0-\omega^2R \quad \left( g=at equator, g_0=at poles \right)$

$\displaystyle \dfrac {g_0}{2}=g_0-\omega^2R;\quad \omega^2R=\dfrac {g_0}{2}; \quad V^2=\dfrac {g_0R}{2}$

$\displaystyle v_e=\sqrt {2g_0R}=\sqrt {4V^2}=2V$

A satellite of mass '

$m$ ', moving around the earth in a circular orbit of radius

$R$ , has angular momentum

$L$ . The areal velocity of satellite is :

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$\displaystyle L/2m$
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$\displaystyle L/m$
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$\displaystyle 2L/m$
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$\displaystyle 2L/m_e$

Explanation

According to kepler's second law, areal velocity is constant .
i.e.

$A_v = \dfrac{rv}{2} = \dfrac{\omega r^2}{2} = \dfrac{L}{2m}$

In the region of only gravitational fields of mass '

$M$ ' a particle is shifted from

$A$ to

$B$ via three different paths of length 5m, 10m and 25m. The work done in different paths is

$W_1, W_2, W_3$ respectively then:

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

Explanation

The gravitational field is a conservative field i.e. work done by gravitational field to displace a body is independent of its path. In other words, work done is determined by difference in potential energy at initial and final positions of body.
Therefore,

$W = P_f - P_i$ .(

$P_f$ =final potential energy,

$P_i$ = Initial potential energy )
So,

$W_1 = P_B - P_A$ and similarly

$W_2 = P_B - P_A$

$W_3 = P_B - P_A$
Thus

$W_1 = W_2 = W_3$

The amount of work done in lifting a mass '

$m$ ' from the surface of the earth to height

$2R$ is

0%
$\displaystyle 2mgR$
0%
$\displaystyle 3mgR$
0%
$\displaystyle \frac {3}{2}mgR$
0%
$\displaystyle \frac {2}{3}mgR$

Explanation

Work done

$\displaystyle = U_f-U_i$

$\displaystyle =-\dfrac {GMm}{2R+R}-\left ( -\dfrac {Gmm}{R} \right )=\dfrac {2}{3}\dfrac {GmM}{R}=\dfrac {2}{3}mgR$

At what height from the surface of earth will the value of g be reduced by

$36$ % from the value at the surface?

$R = 6400km$ .

0%
$400 km$
0%
$800 km$
0%
$1600 km$
0%
$3200 km$

Explanation

${g}'=0.64g=\displaystyle\dfrac{g}{\left ( 1+\dfrac{h}{R} \right )}^{2}$

$\Rightarrow 1+\dfrac{h}{R}=\dfrac{5}{4}$

$h=\dfrac{R}{4}=1600\ km$

The change in potential energy, when a body of mass

$m$ is raised to a height

$nR$ from the earth's surface is

$(R=$ radius of earth

$)$

0%
$\displaystyle mgR \left ( \frac {n}{n-1}\right )$
0%
$\displaystyle nmgR$
0%
$\displaystyle mgR \left ( \frac {n^2}{n^2+1}\right )$
0%
$\displaystyle mgR \left ( \frac {n}{n+1}\right )$

Explanation

$\displaystyle \bigtriangleup U=U_f-U_i=-\dfrac {GMm}{nR+h}-\left (-\dfrac {GMm}{R}\right )$

$\displaystyle =\dfrac {n}{n+1}.\dfrac {GMm}{R}=\dfrac {n}{n+1}mgR$

The escape velocity from a planet is

$v_e$ . A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

0%
$\dfrac {u}{\sqrt {2}}$
0%
$u$
0%
$\dfrac {u}{2\sqrt {2}}$
0%
$\dfrac {u}{{2}}$

Weight of a body of mass

$m$ decreases by

$1%$ when it is raised to height h above the earth's surface. If the body is taken to a depth

$h$ in a mine, then in its weight will

0%
decrease by $0.5%$
0%
decrease by $2%$
0%
increase by $0.5%$
0%
increase by $1%$

Explanation

As,

$\dfrac{g'-g}{g}=-\dfrac{2h}{R_e}=-1%$

$\dfrac{g''-g}{g}=-\dfrac{h}{R_e}=-\dfrac{1}{2}%=-0.5%$

so, Weight in mne will decrease by

$0.5%$ .

A missile is lauched with a velocity less than the escape velocity. Sum of its kinetic energy and potential energy is.

0%
positive
0%
negative
0%
may be negative or positive depending upon its

initial velocity
0%
zero

Explanation

At the surface,

$T.E= \dfrac{1}{2}mv^2 + (-\dfrac{GMm}{R})$

For

$v= escape velocity (v_e)= \sqrt{\dfrac{2GM}{R}}$

$\implies T.E=0$

Thus for

$v< v_e$

$T.E <0$

Which of the following statement is wrong for acceleration due to gravity.

0%
$g$ decreases on going above the surface of earth
0%
$g$ increases on going below the surface of earth
0%
$g$ is maximum at pole
0%
$g$ increases on going from equator to poles

Explanation

Acceleration due to gravity at a depth x from the surface,

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

Thus, g decreases on going below the surface of earth.

The value of '

$g$ ' reduces to half of its value at surface of earth at a height '

$h$ ', then

0%
$h = R$
0%
$h = 2R$
0%
$h \equiv ( \sqrt 2 +1)R$
0%
$h \equiv ( \sqrt 2 -1)R$

Explanation

$g'=g\bigg[1+\dfrac{h}{R}\bigg]^{-2}$

$g'=g/2$

$\dfrac{g}{2}=g\bigg[1+\dfrac{h}{R}\bigg]^{-2}$

$\implies 1+ \dfrac{h}{R}= \sqrt{2}$

Hence

$h= (\sqrt{2}-1)R$

$v_{e}$ is the escape velocity for earth when a projectile is fired from the surface of earth. Then the escape velocity if the same projectile is fired from its centre is

0%
$\displaystyle \sqrt{\frac{3}{2}}v_{e}$
0%
$\displaystyle \frac{3}{2}v_{e}$
0%
$\displaystyle \sqrt{\frac{2}{3}}v_{e}$
0%
$\displaystyle \frac{2}{3}v_{e}$

Explanation

To calculate gravitational potential at the center of earth, take a spherical shell element at radius r.

$dV=-\dfrac { GM }{ r } \dfrac { 4\pi { r }^{ 2 }dr }{ \dfrac { 4 }{ 3 } \pi { R }^{ 3 } } =-\dfrac { 3GMrdr }{ { R }^{ 3 } }$

Integrating this, we get

$V=-\dfrac { 3 }{ 2 } \dfrac { GM }{ R }$

The potential at the surface of earth is

$-\dfrac { GM }{ R }$

Since escape velocity is proportional to the square root of potential energy and hence square root of gravitational potential, escape velocity at centre of earth will be

$\sqrt { \dfrac { 3 }{ 2 } } { v }_{ e }$

The gravitational potential energy of a body at a distance

$r$ from the centre of earth is

$U$ . Its weight at a distance

$2r$ from the centre of earth is

0%
$\displaystyle\frac{U}{r}$
0%
$\displaystyle\frac{U}{2r}$
0%
$\displaystyle\frac{U}{4r}$
0%
$\displaystyle\frac{U}{\sqrt{2}r}$

Explanation

$U=\displaystyle\frac{-GMm}{r}$

$\therefore -GM=\displaystyle\frac{Ur}{m}$

$E=-\displaystyle\frac{GM}{(2r)^{2}}=\displaystyle\dfrac{\left(\dfrac{Ur}{m}\right)}{4r^{2}}=\dfrac{U}{4mr}$

$F=mE=\displaystyle\dfrac{U}{4r}$

If the radius of moon is

$1.7 \times 10^{6}\ m$ and its mass is

$7.34\times 10^{22}\ kg$ . Then its escape velocity is

0%
$2.4\times 10^{3}\ ms^{-1}$
0%
$2.4\times 10^{2}\ ms^{-1}$
0%
$3.4\times 10^{3}\ ms^{-1}$
0%
$3.4\times 10^{2}\ ms^{-1}$

Explanation

$V_{e}=\sqrt{\displaystyle\frac{2GM}{R}}$

$=\sqrt{\displaystyle\frac{2(6.67\times10^{-11})(7.34\times10^{22})}{1.7\times10^{6}}}$

$=2.4\times 10^{3}$ m/s

Acceleration due to gravity at earth's surface is '

$g$ ' m/s

$^2$ . Find the effective value of acceleration due to gravity at a height of

$32 km$ from sea level :(

$R_e = 6400 km$ ).

0%
$0.5 g ms^{2}$
0%
$0.99 g ms^{2}$
0%
$1.01 g ms^{2}$
0%
$0.90 g ms^{2}$

Explanation

Acceleration due to gravity at a height h from the earth surface,

$g'=g\bigg[1+\dfrac{h}{R}\bigg]^{-2}$

Given

$h= 32 km \quad R= 6400 km$

Using bionomial expansion,

$(1+x)^n= 1+ nx$ for

$x<< 1$

Thus as

$h<< R$

$\implies g'=g\bigg[1-\dfrac{2h}{R}\bigg]$

$g'=g\bigg[1-\dfrac{2(32)}{6400}\bigg]$

$\implies g'= .99 g m/s^2$

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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