Gravitation Test

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Gravitation Test - 74

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Question 1
1 / -0

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?

A
$$(kg)^{1/2}$$

B
$$(kg)^{-1/2}$$

C
$$(kg)^{2}$$

D
$$(kg)^{-2}$$

Solution

The equation of escape velocity is $$v_e=\sqrt{2gr}$$

so, $$\dfrac{v_e1}{v_e2}=\dfrac{\sqrt{2g_1r_1}}{\sqrt{2g_2r_2}}=\sqrt{kg}$$
Question 2
1 / -0

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

A
$$\displaystyle mgR_e$$

B
$$\displaystyle mgR_e/2$$

C
$$\displaystyle 2mgR_e$$

D
$$\displaystyle mgR_e/4$$

Solution

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 $$\displaystyle \frac {GM_em}{2R_e}=\frac{mgR_e}{2}$$
Question 3
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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

A
$$\displaystyle a$$

B
$$\displaystyle b$$

C
$$\displaystyle c$$

D
$$\displaystyle d$$
Question 4
1 / -0

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:

A
$$\displaystyle \frac{4\pi}{3}G\rho ym$$

B
$$\displaystyle \frac{3}{4}\pi \rho y$$

C
$$\displaystyle \frac{4}{3}\pi \rho y$$

D
$$\displaystyle \frac{4}{3}\pi G\rho y$$

Solution

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.
Question 5
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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?

A
$$\displaystyle A$$

B
$$\displaystyle B$$

C
$$\displaystyle C$$

D
$$\displaystyle D$$

Solution

g is inversely proportional to square of R and also, at some point force due to earth and mass cancels each other, and gravitational force can't be negative.
Hence, C is the correct graph.
Question 6
1 / -0

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

A
$$10^2$$

B
10

C
$$10^{-1}$$

D
$$10^{-2}$$

Solution
Question 7
1 / -0

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

A
$$v_e$$

B
$$2v_e$$

C
$$\sqrt 2 v_e$$

D
$$\sqrt 3 v_e$$

Solution

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-
$$\dfrac{1}{2}m(2v_e)^2 - \dfrac{GMm}{R}= \dfrac{1}{2}mv^2$$
$$\dfrac{1}{2}m\bigg[2\sqrt{\dfrac{2GM}{R}}\bigg]^2 - \dfrac{GMm}{R}= \dfrac{1}{2}mv^2$$
$$ \dfrac{3GMm}{R}= \dfrac{1}{2}mv^2$$
$$v^2= 3 \dfrac{2GM}{R} = 3 v^2_e$$
$$\implies v= \sqrt{3} v_e$$
Question 8
1 / -0

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.

A
$$\dfrac {mgR}{2}$$

B
$$2mgR$$

C
$$mgR$$

D
$$\dfrac {mgR}{4}$$

Solution

Gravitational petential energy at r, $$P.E= -\dfrac{GMm}{r}$$
As $$GM= gR^2 \implies P.E= -\dfrac{mgR^2}{r}$$
For $$r=R$$, $$P.E= -mgR$$
For $$r=2R$$, $$P.E'= -\dfrac{mgR}{2}$$
Gain in P.E= $$P.E'- P.E=\dfrac{mgR}{2}$$
Question 9
1 / -0

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

A
$$7:3$$

B
$$3:7$$

C
$$7:4$$

D
$$4:7$$

Solution

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

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

$$\displaystyle =\dfrac{GMm}{d^{2}}\left [ 1-\dfrac{1}{8\left ( 1-\dfrac{R}{2d} \right )^{2}} \right ]$$
Question 10
1 / -0

Suppose, the acceleration due to gravity at the Earth's surface is $$10m s^{-2}$$ and at the surface of Mars it is $$4.0 m s^{-2}$$. 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 best represents the weight (net gravitational force) of the passenger as a function of time.?

A
A

B
B

C
C

D
D

Solution

$$g\propto \displaystyle\frac{1}{R^2}$$ so we will not get a straight line.
Also, $$F=0$$ at a point where Force due to Earth=Force due to Mars.