Gravitation Test

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Gravitation Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

The value of $$'g'$$ at a certain height above the surface of the earth is $$16$$% of its value on the surface. The height is $$(R = 6300 \ km)$$

A
$$10500 \ km$$

B
$$12500 \ km$$

C
$$2688 \ km$$

D
$$9450 \ km$$

Solution
Question 2
1 / -0

The depth at which the value of acceleration due to gravity $$\dfrac {1}{n}$$ times the value at the surface is: ($$R=$$radius of the earth).

A
$$\dfrac {R}{n}$$

B
$$R{\dfrac {(n-1)}{n}}$$

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

D
$$R\left(\dfrac {n}{n+1}\right)$$

Solution
Question 3
1 / -0

A pendulum that beats seconds on the surface of the earth were taken to a depth of (1/4)th the radius of the earth.What will be its time period of oscillation?

A
5.107 sec

B
1.208 sec

C
3.40 sec

D
2.309 sec

Solution
Question 4
1 / -0

A body weight 144 N at the surface of earth. When it is taken to a height of h = 3R,where R is radius of earth it would weight.

A
48 N

B
36 N

C
16 N

D
9 N

Solution

Given that,
Weight of body $$W=144\,N$$
Height $$h=3R$$
Radius of earth $$=R$$
We know that,
Acceleration due to gravity at the surface of the earth $$=g$$
Acceleration due to gravity at height $$g'=3R$$
Now,
$$ \dfrac{g}{g'}={{\left( \dfrac{R}{R+h} \right)}^{2}} $$
$$ \dfrac{g}{g'}={{\left( \dfrac{R}{R+3R} \right)}^{2}} $$
$$ \dfrac{g}{g'}=\dfrac{1}{16} $$
Now, weight at the height
$$ \dfrac{w'}{w}=\dfrac{mg'}{mg} $$
$$ \dfrac{w'}{w}=\dfrac{1}{16} $$
$$ w'=\dfrac{w}{16} $$
$$ w'=\dfrac{144}{16} $$
$$ w'=9\,N $$
Hence, the weight at the height is $$9\ N$$
Question 5
1 / -0

When work done by of gravity is negative, which of the following will definitely occur

A
Potential energy increases

B
kinetic energy decreases

C
Gravitational potential energy increase

D
Gravitational potential energy decreases

Solution
Question 6
1 / -0

Two planets are at mean distance $$d_1$$ and $$d_2$$ from the sun and their periods are $$T_1$$ and $$T_2$$ respectively, Then:
(n is the frequency)

A
$$n_1^2d_1^2=n_2^2d^2_2$$

B
$$n_2^2d_2^3=n_1^2d^3_1$$

C
$$n_1d_1=n_2d_2$$

D
$$n_1^2d_1=n_2^2d_2$$

Solution

According to kepler's 3rd law of planetary motion-
$$T^2 \propto r^3$$ [Here, $$r=d$$]
As we know that,
$$T = \dfrac{1}{n}$$ $$\left(\dfrac{1}{frequency}\right)$$

$$\dfrac{1}{n^2} \propto d^3 \Rightarrow \dfrac{n^2_2}{n_1^2} = \dfrac{d^3_1}{d^3_2}$$

$$\Rightarrow n_2^2 d_2^3 = n_1^2 d_1^3$$

option B is correct.
Question 7
1 / -0

If a tunnel is dug along the diameter of the earth and a body is dropped into it, then the time taken by it to cross the tunnel once is:-

A
$$2 \pi \sqrt{\dfrac{Re}{g}}$$

B
$$\sqrt {2gr}$$

C
$$2 \pi \sqrt{\dfrac{g}{Re}}$$

D
$$ \pi \sqrt{\dfrac{g}{Re}}$$

Solution
Question 8
1 / -0

The value of acceleration due to gravity at a height R from surface of the earth is then(R=radius of the earth and g=acceleration due to gravity on earth surface )

A
$$0$$

B
$$\sqrt{g}$$

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

D
$$\dfrac{g}{2}$$

Solution

$$g'=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^2}$$
$$g'=\dfrac{g}{\left(1+\dfrac{R}{R}\right)^2}=\dfrac{g}{(1+1)^2}$$
$$g'=\dfrac{g}{4}$$.
option $$C$$
Question 9
1 / -0

The value of gravitational acceleration at a height equal to the radius of the earth is

A
Equal to its value at the earth's surface

B
50% of its value at the surface

C
25% of its value at the surface

D
75% of its value at the surface

Solution
Question 10
1 / -0

At what height from the surface of the earth will the value of acceleration due to gravity be reduced by $$36\%$$ from the value at the surface?
(Radius of earth=$$6400\ km$$)

A
$$1500\ km$$

B
$$1200\ km$$

C
$$1000\ km$$

D
$$1600\ km$$

Solution