Gravitation Test

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Gravitation Test - 40

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Question 1
1 / -0

A body of mass $$m$$ is raised to a height $$10R$$ from the surface of the earth, where $$R$$ is the radius of the earth. The increase in potential energy is ($$G =$$ universal constant of gravitation, $$M =$$ mass of the earth and $$g =$$ acceleration due to gravity)

A
$$\dfrac{GMm}{11R}$$

B
$$\dfrac{GMm}{10R}$$

C
$$\dfrac{mgR}{11G}$$

D
$$\dfrac{10GMm}{11R}$$

Solution

PE at the earth surface $$= \dfrac{-GMm}{R}$$
PE at a height $$h$$ above the earth's surface $$= \dfrac{-GMm}{\left(R+h\right)}$$
$$\therefore$$ Change in PE $$= \dfrac{-GMm}{\left(R+h\right)} - \left(-\dfrac{GMm}{R}\right) = \dfrac{GMmh}{R\left(R+h\right)}$$
Substituting $$h = 10R$$
$$\triangle PE = \dfrac{GMm \times 10R}{R\left(R+10R\right)} = \dfrac{10GMm}{11R}$$
Question 2
1 / -0

Let $$g_h$$ and $$g_d$$ be the acceleration due to gravity at height h above the earth's surface and at depth d below the earth's surface respectively. If $$g_h=g_d$$, then the relation between h and d is

A
$$d = h$$

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

C
$$d = \dfrac{h}{4}$$

D
$$d = 2h$$

Solution

Let the radius of the earth be $$R$$ and acceleration due to gravity at earth surface be $$g$$.
Acceleration due to gravity at height $$h$$ above the earth surface, $$g_h = g \bigg(1 - \dfrac{2h}{R} \bigg)$$
Acceleration due to gravity at depth $$d$$ below the earth surface, $$g_d = g \bigg(1 - \dfrac{d}{R} \bigg)$$
Now for $$g_h = g_d$$ we get, $$ g \bigg(1 - \dfrac{2h}{R} \bigg)$$ $$ = g \bigg(1 - \dfrac{d}{R} \bigg)$$
$$\Rightarrow $$ $$1 - \dfrac{2h}{R} = 1 - \dfrac{d}{R}$$
$$\Rightarrow $$ $$d = 2h$$
Question 3
1 / -0

The value of acceleration due to gravity, at earth surface is $$g$$. Its value at the centre of the earth, which we assume as a.sphere of radius $$R$$ and of uniform mass density, will be:

A
$$10 R m/s^2$$

B
Zero

C
$$5 R m/s$$

D
$$20 R m/s^2$$

Solution

$$Answer:-$$ B
Variation of $$g$$ with depth is given by:-
$$g_d=g(1-\dfrac{d}{R})$$
where d is depth from surface,
As value of $$d$$ approaches $$R$$ value of $$g_d\simeq0$$
Question 4
1 / -0

The motion of planets in the solar system is an exampIe of the conservation of

A
mass

B
linear momentum

C
angular momentum

D
energy

Solution

For any circular motion the angular momentum is conserved as no torque is acting on it because centripetal force acts through the point of axis.
Question 5
1 / -0

Calculate the gravitational force of earth acting on an object weighting 10kg placed 1000km above the surface of earth.
(Mass of earth $$=6\times 10^{24}kg$$ and radius of earth $$=6.4\times 10^6m$$)

A
$$0.84\times 10^3N$$

B
$$0.72\times 10^2N$$

C
$$0.65\times 10^6N$$

D
$$0.42\times 10^4N$$

Solution

Given:
Mass of earth $$=6\times 10^{24}kg$$
Mass of object $$=10kg$$
Distance of the object $$=$$ Radius of earth + distance from earth"s surface
$$= 6.4\times 10^6m+1000 km$$
$$=6.4\times 10^6m+1\times 10^6m$$
  $$= 7.4\times 10^6m$$
$$G = 6.7\times 10^{-11}Nm^2kg^{-2}$$
By using the formula,
$$ F = \displaystyle G\frac{M\times m}{d^2}$$

$$F =  \displaystyle \frac{6.7\times 10^{-11}Nm^2kg^{-2}\times 6\times 10^{24}kg\times 10 kg}{(7.4\times 10^6m)^2}$$

$$F=  \displaystyle \frac{40.2\times 10^{14}Nm^2}{54.76\times 10^{12}m^2}$$

$$F= \displaystyle \frac{4\times 10^{15}N}{5.5\times 10^{13}}$$

$$F = 0.72\times 10^2N$$
Thus, the force exerted by earth on the object is $$0.72\times 10^2N$$
Question 6
1 / -0

A body of mass '$$m$$' is raised to a height '$$10R$$' from the surface of the Earth, where '$$R$$' is the radius of the Earth. The increase in potential energy is ____ . ($$G$$ = universal constant of gravitation, $$M$$ = mass of earth and $$g$$ = acceleration due to gravity).

A
$$\displaystyle\frac{GMm}{11R}$$

B
$$\displaystyle\frac{GMm}{10R}$$

C
$$\displaystyle\frac{mgR}{11G}$$

D
$$\displaystyle\frac{10GMm}{11R}$$

Solution

By conservation of energy, work done = increase in potential energy.
Potential energy at a height $$x$$ above the surface of the Earth is $$-\dfrac{GMm}{R+x}$$.

So, increase in potential energy = work done = $$ \dfrac { GMm }{ R } -\dfrac { GMm }{ 11R } =\dfrac { 10GMm }{ 11R} $$
Question 7
1 / -0

The height at which the acceleration due to gravity is $$25\%$$ of that of the surface of earth is ____________. (R=Radius of the earth)

A
$$h=3R$$

B
$$h=2R$$

C
$$h=R$$

D
$$h=\displaystyle\frac{R}{2}$$

Solution

Acceleration due to gravity at height $$h$$ is $$g' = \dfrac{gR^2}{(R+h)^2}$$
Let the height be $$h$$ where $$g' =25$$ % of $$g = \dfrac{g}{4}$$
$$\Rightarrow$$ $$\dfrac{g}{4} = \dfrac{gR^2}{(R+h)^2}$$
$$\Rightarrow \dfrac{1}{2} = \dfrac{R}{(R+h)}$$
$$\Rightarrow h = R$$
Question 8
1 / -0

The change in the gravitational potential energy when a body of mass m is raised to a height $$nR$$ above the surface of the Earth is (Here R is the radius of the earth)

A
$$\begin{pmatrix}\dfrac{n}{n+1}\end{pmatrix}mgR$$

B
$$\begin{pmatrix}\dfrac{n}{n-1}\end{pmatrix}mgR$$

C
$$nmgR$$

D
$$\dfrac{mgR}{n}$$

Solution

The acceleration due to gravity varies with height as

$$g'=\dfrac{g}{\begin{pmatrix}1+\dfrac{h}{R}\end{pmatrix}}$$

Hence, $$h=nR\Rightarrow g'=\dfrac{g}{\begin{pmatrix}1+\dfrac{nR}{R}\end{pmatrix}}=\dfrac{g}{(1+n)}$$

The change in potential energy is equal to

$$\Delta U=mgh=\dfrac{mg(nR)}{(1+n)}=\begin{pmatrix}\dfrac{n}{n+1}\end{pmatrix}mgR$$
Question 9
1 / -0

If $$g$$ is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass $$m$$ raised from the earth's surface to a height equal to the radius $$R$$ of the earth is

A
$$\cfrac{mgR}{4}$$

B
$$\cfrac{mgR}{2}$$

C
$$mgR$$

D
$$2mgR$$

Solution

The object is at earth surface initially.
Initial potential energy, $$P.E_i =\dfrac{-GMm}{R}$$ where, $$GM = gR^2$$
$$\Rightarrow$$ $$P.E_i = -mgR$$

Final Potential energy, $$P.E_f = \dfrac{-GMm}{2R}= -\dfrac{mgR}{2}$$

$$\therefore$$ Change in potential energy, $$\Delta P.E = -\dfrac{mgR}{2} - (-mgR) = \dfrac{mgR}{2}$$
Question 10
1 / -0

A body is taken to a height of $$nR$$ from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is

A
$$(n\, +\, 1)^2$$

B
$$(n\, +\, 1)^{-2}$$

C
$$(n\, +\, 1)^{-1}$$

D
$$(n\, +\, 1)$$

Solution

Acceleration due to gravity at the surface of the earth=$$g=\dfrac{GM}{R^2}$$
where $$R $$ is the radius of earth.
Acceleration due to gravity at height $$nR=g'=\dfrac{GM}{(R+nR)^2}=\dfrac{GM}{R^2}\dfrac{1}{(n+1)^2}$$
$$\implies \dfrac{g}{g'}=(n+1)^2$$

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Gravitation Test - 40

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Gravitation Test - 40

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SHARING IS CARING

Question 1

$$\dfrac{GMm}{11R}$$

$$\dfrac{GMm}{10R}$$

$$\dfrac{mgR}{11G}$$

$$\dfrac{10GMm}{11R}$$

Solution

Question 2

$$d = h$$

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

$$d = \dfrac{h}{4}$$

$$d = 2h$$

Solution

Question 3

$$10 R m/s^2$$

Zero

$$5 R m/s$$

$$20 R m/s^2$$

Solution

Question 4

mass

linear momentum

angular momentum

energy

Solution

Question 5

$$0.84\times 10^3N$$

$$0.72\times 10^2N$$

$$0.65\times 10^6N$$

$$0.42\times 10^4N$$

Solution

Question 6

$$\displaystyle\frac{GMm}{11R}$$

$$\displaystyle\frac{GMm}{10R}$$

$$\displaystyle\frac{mgR}{11G}$$

$$\displaystyle\frac{10GMm}{11R}$$

Solution

Question 7

$$h=3R$$

$$h=2R$$

$$h=R$$

$$h=\displaystyle\frac{R}{2}$$

Solution

Question 8

$$\begin{pmatrix}\dfrac{n}{n+1}\end{pmatrix}mgR$$

$$\begin{pmatrix}\dfrac{n}{n-1}\end{pmatrix}mgR$$

$$nmgR$$

$$\dfrac{mgR}{n}$$

Solution

Question 9

$$\cfrac{mgR}{4}$$

$$\cfrac{mgR}{2}$$

$$mgR$$

$$2mgR$$

Solution

Question 10

$$(n\, +\, 1)^2$$

$$(n\, +\, 1)^{-2}$$

$$(n\, +\, 1)^{-1}$$

$$(n\, +\, 1)$$

Solution

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