Gravitation Test

Result

Gravitation Test - 70

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The depth at which the effective value of acceleration due to gravity is $$\dfrac{g}{4}$$ is

A
$$R$$

B
$$\dfrac{3R}{4}$$

C
$$\dfrac{R}{2}$$

D
$$\dfrac{R}{4}$$

Solution

Value of g at depth 'd'
$$g'= g \bigg( 1-\dfrac{d}{R} \bigg) $$ Given $$g'=\dfrac {g}4$$

$$\implies \dfrac{g}{4} = g\bigg(1-\dfrac{d}{R} \bigg)$$

$$ \implies d= \dfrac 34 R$$
Question 2
1 / -0

Universal gravitational constant, G depends:

A
on the nature of the particle

B
on the medium present between the particles

C
on time

D
does not depend on any of these factors.

Solution

Universal gravitational constant, $$G$$ is independent of the nature of the particle, medium between the particles, and time. Its value is constant anywhere in the Universe, and hence it's called 'Universal'.
Question 3
1 / -0

The acceleration of a body due to the attraction of the earth (radius $$R$$) at a distance $$2R$$ from the surface of the earth is ($$g= $$ acceleration due to gravity at the surface of the earth)

A
$$\dfrac{g}{9}$$

B
$$\dfrac{g}{3}$$

C
$$\dfrac{g}{4}$$

D
$$g$$

Solution

Value of g at hight h, $$g'= g\bigg( \dfrac{R}{R+h} \bigg)^2 $$

h=2R (Given)

$$g'= g \bigg( \dfrac{R}{R+2R} \bigg)^2 = \dfrac{g}{9}$$
Question 4
1 / -0

Practically the value of G for the first time was measured by ...........

A
Newton

B
Cavendish

C
Archimedes

D
Galileo

Solution

The Cavendish has performed an experiment to calculate the gravitation force between two masses. and he gave the value of gravitational constant $$G$$, which is equal to $$6.75\times 10^{-11}\ \ Nm^2kg^{-2}$$
Question 5
1 / -0

An object moves from earths surface to the surface of the moon. The acceleration due to gravity on the earths surface is $$10 m/s^{2}$$. Considering the acceleration due to gravity on the moon to be $$1/6th$$ times of that of earth. If $$R$$ be the earths radius and its weight be $$W$$ and the distance between the earth and the moon is $$D$$. The correct variation of the weight $$W'$$ versus distance $$d$$ for a body when it moves from the earth to the moon is

A

B

C

D

Solution

At the earths surface the weight of body is, $$W = mg$$
At the moon the distance of body from the earth's surface is $$d = D$$
At moons surface the value of acceleration due to gravity is $$g' = \dfrac {g}{6}$$
$$\Rightarrow W' = \dfrac {W}{6}$$
From the relation of acceleration due to gravity at height $$d$$ above earth's surface,
$$g' = \dfrac {gR^{2}}{(R + d)^{2}} ..... (i)$$
For $$d = 0, \Rightarrow g' = g$$ or $$W' = W$$
At $$d = R$$,
$$\Rightarrow g' = \dfrac {gR^{2}}{4R^{2}} = \dfrac {g}{4}$$
Or, $$W' = \dfrac {mg}{4} = \dfrac {W}{4}$$
At distance $$d = D$$, when the body reaches surface of moon
$$W' = \dfrac {mg}{6} = \dfrac {W}{6}$$
Since, Eq. (i) suggests that variation of $$g$$ with distance $$(d)$$ is non-linear, hence the graph of $$W'$$ vs $$d$$ will be non-linear as well. Also $$d\uparrow, W'\downarrow$$
Thus the correct variation is represented by following graph.
Question 6
1 / -0

A boy can jump to a height $$h$$ from ground on earth . What should be the radius of a sphere of density $$\delta $$ such that on jumping on it, he escapes out of the gravitational field of the sphere?

A
$$\sqrt{\dfrac{4\pi G\delta }{3gh}}$$

B
$$\sqrt{\dfrac{4\pi gh }{3G\delta }}$$

C
$$\sqrt{\dfrac{3 gh }{4 \pi G\delta }}$$

D
$$\sqrt{\dfrac{3 G\delta }{4 \pi gh }}$$

Solution

While jumping he converts his initial $$kinetic$$ $$energy$$ to $$potential$$ $$energy$$
$$\dfrac{mv^2}{2}=mgh$$
$$v=\sqrt{2gh}$$
Now this $$ velocity$$ has to be equal to $$escape$$ $$velocity$$ on the sphere
$$Escape$$ $$velocity=\sqrt{\dfrac{2GM_s}{R}}=\sqrt{\dfrac{8 \pi G R_s^2 \delta}{3}}$$
By comparing both the velocities we get $$Radius(R_s)=\sqrt{\dfrac{3gh}{4 \pi G \delta}}$$
Question 7
1 / -0

The value of $$g$$ at a height $$h$$ above the surface of the earth is the same as at a depth $$d$$ below the surface of the earth. When both $$d$$ and $$h$$ are much smaller than the radius of earth, then which one of the following is correct

A
$$d= \dfrac{h}{2}$$

B
$$d= \dfrac{3h}{2}$$

C
.$$ d = 2h$$

D
$$d= h$$

Solution

The value of gravity at a height $$h$$ above earth's surface is given by
$$g_h=g\left(1-\dfrac {2h}{R}\right)$$
and at a depth $$d$$ is given by
$$g_d=g\left(1-\dfrac {d}{R}\right)$$
Since these two values are same, we can see that $$d=2h$$
Question 8
1 / -0

At what distance above the surface of earth, the gravitational force will be reduced by $$10\%$$, if the radius of earth is $$6370$$ Km.

A
$$750 Km$$

B
$$650 Km$$

C
$$450 Km$$

D
$$344 Km$$

Solution

Gravitational force on the earth surface,

$$F_{G_s}= \dfrac{GMm}{R^{2}}$$

Gravitational force at a height 'h' above,

$$F_{G_{h}}= \dfrac{GMm}{(R+h)^{2}}$$

G.T. $$F_{G_{h}}= \dfrac{90}{100}(F_{G_{S}})$$

$$\Rightarrow \dfrac{GMm}{(R+h)^{2}}=\dfrac{9}{10}(\dfrac{GMm}{R^{2}})$$

$$\Rightarrow R+h=\sqrt{\dfrac{10}{9}}(R)$$

$$\Rightarrow 6370+h= \dfrac{3.162}{3} (6370)$$

$$\Rightarrow h=\dfrac{0.162 \times 6370}{3}=344$$ km
Question 9
1 / -0

A particle of mass 1kg is placed at a distance of 4m from the centre and on the axis of a uniform ring of mass 5kg and radius 3m. The work done to increase the distance of the particle from 4m to $$\sqrt{3}m$$ is.

A
$$\dfrac{G}{3}J$$

B
$$\dfrac{G}{4}J$$

C
$$\dfrac{G}{5}J$$

D
$$\dfrac{G}{6}J$$

Solution

Potential of a ring, $$V=\dfrac{-GM}{y}$$
where, $$y=\sqrt{x^2+r^2}$$
So initial, energy is $$\dfrac{-5G\times 1}{5}$$
Final energy is $$\dfrac{-5G\times 1}{6}$$
So the difference $$=$$ Final $$-$$ Initial $$=\dfrac{G}{6}$$
Question 10
1 / -0

A particle of mass $$10 gm$$ is kept on the surface of a uniform sphere of mass $$100 kg$$ and radius $$10 cm$$. Find the work done against the gravitational force between them, to take the particle far away from the sphere. $$\left ( G= 6.67\times10^{-11}Nm^{2}/Kg^{2} \right )$$

A
$$3.33\times10^{-9}J$$

B
$$3.33\times10^{-10}J$$

C
$$6.67\times10^{-9}J$$

D
$$6.67\times10^{-10}J$$

Solution

At infinity, the gravitational potential energy of the system will be $$0$$.
Initially, the particle is outside sphere and hence sphere will behave as point object at its centre,
Therefore,
$${ E }_{ 0 }=-\dfrac { GMm }{ R } =-\dfrac { (6.67\times { 10 }^{ -11 })\times(100)\times(0.01) }{ 0.1 } =-6.67\times{ 10 }^{ -10 }J$$
We know that, $$W=\Delta U=U_{f}-U_{i} \implies W=-U_{i}$$
Thus, work done is $$6.67\times { 10 }^{ -10 }J$$