Gravitation Test

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Gravitation Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$g$$ at the surface of earth is $$9.8\ m/s^{2}$$ then the value of $$'g'$$ at a place $$480\ km$$ above the surface of the earth will be nearly (radius of the earth is $$6400\ km)$$.

A
$$9.8\ m/s^{2}$$

B
$$7.2\ m/s^{2}$$

C
$$8.3\ m/s^{2}$$

D
$$4.2\ m/s^{2}$$

Solution

Given that,

$$ g=9.8\,m/{{s}^{2}} $$

$$ h=480\,km $$

$$ R=6400\,km $$

We know that,

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

$$ g'=9.8\left( 1-\dfrac{2\times 480}{6400} \right) $$

$$ g'=8.33\,m/{{s}^{2}} $$

Hence, the value of g’ is $$8.33\,m/{{s}^{2}}$$.
Question 2
1 / -0

Three particles each of mass $$'m'$$ are at the vertices of an equilateral triangle of side length $$'l'$$. Then the work done is moving one particle to infinity is:

A
$$\dfrac{3Gm^2}{l}$$

B
$$\dfrac{Gm^2}{2l}$$

C
$$\dfrac{2Gm^2}{l}$$

D
$$\dfrac{4Gm^2}{l}$$

Solution
Question 3
1 / -0

The value of acceleration due to gravity will be $$1\%$$ of its value at the surface of earth at a height of $$(R_{e} = 6400\ km)$$.

A
$$6400\ km$$

B
$$57600\ km$$

C
$$2560\ km$$

D
$$6100\ km$$

Solution
Question 4
1 / -0

The SI unit of the universal gravitational constant G:

A
$$Nm{{Kg}^{ - 2}}$$

B
$$N{m^2}{{Kg}^{ - 2}}$$

C
$$N{m^2}{{Kg}^{ - 1}}$$

D
$$Nm{{Kg}^{ - 1}}$$

Solution

Force $$=\dfrac{G\ M_{1}\ M_{2}}{R^{2}}$$

$$[G]=\dfrac{[Force][R]^{2}}{[M_{1}][M_{2}]}$$

$$=\dfrac{N\ m^{2}}{(kg)^{2}}$$

$$=N\ m^{2}\ kg^{-2}$$

Option $$B$$.
Question 5
1 / -0

At what height from the surface of the earth, the value of g is reduced by 36% from its value at the surface of the earth of radius R=6400 km?

A
1200 km

B
1600 km

C
2000 km

D
2200 km

Solution
Question 6
1 / -0

A body weights $$63 N$$ on the surface of the Earth. At a height $$h$$ above the surface of Earth, its weight is $$28 N$$ while at a depth $$h$$ below the surface Earth, the weight is $$31.5 N$$. The value of $$h$$ is:

A
$$0.4 \ R$$

B
$$0.5 \ R$$

C
$$0.8 \ R$$

D
$$R$$

Solution
Question 7
1 / -0

At what height above the surface of earth the value of "g" decreases by 2%? [radius of the earth is 6400 km]

A
32 km

B
64 km

C
128 km

D
1600 km

Solution

The effective value of a above height h from eaarth is given by :
$$g_{eff}=\frac{g}{\left ( 1+\frac{h}{R} \right )^{2}}$$( R = radius of Earth )
Now given $$g_{eff}=0.98g$$ (As value of g decreases by 2 %)
$$\Rightarrow \left ( 1+\frac{h}{R} \right )^{2}=\frac{g}{0.98g}=1.020$$
$$\Rightarrow 1+\frac{h}{R}=\sqrt{1.020}=1.010$$
$$\Rightarrow h=0.010\times R=\frac{10}{1000}\times 6400km= 64 km$$
Question 8
1 / -0

The period of a simple pendulum on the surface of earth is $$T$$. At an altitude of half of the radius of earth from the surface, its period will be

A
$$\sqrt{\dfrac{3}{2}}T$$

B
$$\dfrac{3T}{2}$$

C
$$\dfrac{2T}{3}$$

D
$$\sqrt{\dfrac{2}{3}}T$$

Solution

Time Period of pendulum is given by:
$$T=2\pi\sqrt{\dfrac{L}{g}}\rightarrow (1)$$(L$$=$$ length of pendulum, g$$=$$acceleration of gravity)
At $$h=\dfrac{R}{2}$$ the acceleration due to gravity is given by:
$$g_{eff}=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^2}=\dfrac{g}{\left(1+\dfrac{1}{2}\right)^2}=\dfrac{4g}{9}$$
So new time period of pendulum will be
$$T_{new}=2\pi\sqrt{\dfrac{L}{\dfrac{4g}{9}}}=\sqrt{\dfrac{3}{2}}\times 2\pi\sqrt{\dfrac{L}{g}}$$
$$T_{new}=\sqrt{\dfrac{3}{2}}T$$.
Question 9
1 / -0

At a certain height above the earth's surface, the gravitational acceleration is $$4$$% of its value at the surface of the earth. Find the height. ($$R$$ is the surface of the earth)

A
$$2R$$

B
$$4R$$

C
$$R$$

D
$$R/2$$

Solution

$$\textbf{Hint: Use formula of acceleration due to gravity at height h from earth's surface}$$
$$\textbf{Step1:Expression of acceleration due to gravity at a height h}$$
The acceleration due to gravity at a height h above Earth's surface is:
$$g_h=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^2}$$
$$\textbf{Step2: Find height h}$$
As $$g_h$$ is $$4\%$$ of g so $$g_h=0.04$$g
$$\Rightarrow 0.04g=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^2}$$
$$\Rightarrow \left(1+\dfrac{h}{R}\right)^2=\dfrac{1}{0.04}=25$$
$$\Rightarrow 1+\dfrac{h}{R}=5$$
$$\Rightarrow \dfrac{h}{R}=4\Rightarrow h=4R$$.
$$\textbf{Hence option B correct}$$
Question 10
1 / -0

A rocket is fired from the earth to the moon. The distance between the earth and the moon is $$r$$ and the mass of the earth is $$81$$ times the mass of the moon. The gravitational force on the rocket will be zero, when its distance from the moon is then

A
$$\dfrac{r}{20}$$

B
$$\dfrac{r}{15}$$

C
$$\dfrac{r}{10}$$

D
$$\dfrac{r}{5}$$

Solution

Net force on the rocket is zero. i.e. force due to the earth and force due to moon is same but in opposite direction.
$$Fe=Fm$$
$$\dfrac{GMe Mr}{re^2}=\dfrac{GMmMr}{rm^2}$$
$$r$$ is the total distance
$$r= re+ rm$$
$$Me=81 Mm$$
$$\dfrac{a 81 Mm Mr}{re^2}=\dfrac{G M m Mr}{rm^2}$$
$$r_m=\dfrac{r_e}{g}$$
$$g rm +rm =r \Rightarrow 10 r_m=r$$
$$\Rightarrow r_m=\dfrac{r}{10}$$

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Gravitation Test - 58

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Gravitation Test - 58

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SHARING IS CARING

Question 1

$$9.8\ m/s^{2}$$

$$7.2\ m/s^{2}$$

$$8.3\ m/s^{2}$$

$$4.2\ m/s^{2}$$

Solution

Question 2

$$\dfrac{3Gm^2}{l}$$

$$\dfrac{Gm^2}{2l}$$

$$\dfrac{2Gm^2}{l}$$

$$\dfrac{4Gm^2}{l}$$

Solution

Question 3

$$6400\ km$$

$$57600\ km$$

$$2560\ km$$

$$6100\ km$$

Solution

Question 4

$$Nm{{Kg}^{ - 2}}$$

$$N{m^2}{{Kg}^{ - 2}}$$

$$N{m^2}{{Kg}^{ - 1}}$$

$$Nm{{Kg}^{ - 1}}$$

Solution

Question 5

1200 km

1600 km

2000 km

2200 km

Solution

Question 6

$$0.4 \ R$$

$$0.5 \ R$$

$$0.8 \ R$$

$$R$$

Solution

Question 7

32 km

64 km

128 km

1600 km

Solution

Question 8

$$\sqrt{\dfrac{3}{2}}T$$

$$\dfrac{3T}{2}$$

$$\dfrac{2T}{3}$$

$$\sqrt{\dfrac{2}{3}}T$$

Solution

Question 9

$$2R$$

$$4R$$

$$R$$

$$R/2$$

Solution

Question 10

$$\dfrac{r}{20}$$

$$\dfrac{r}{15}$$

$$\dfrac{r}{10}$$

$$\dfrac{r}{5}$$

Solution

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