Gravitation Test

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Gravitation Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

The graph represented the relation between acceleration due to gravity(g) and depth (d) from the surface of the earth. The value of g at a depth of $$640\ km$$ is

A
$$0.98\ {ms}^{2}$$

B
$$7.84\ {ms}^{2}$$

C
$$8.82\ {ms}^{2}$$

D
$$9\ {ms}^{2}$$
Question 2
1 / -0

A planet moves with respect to us that light of $$475 nm$$ is observed at $$475.6 nm$$. The speed of the planet is

A
$$206 kms^{-1}$$

B
$$378 kms^{-1}$$

C
$$108 kms^{-1}$$

D
$$100 kms^{-1}$$

Solution
Question 3
1 / -0

If the distance between two masses is doubled, the gravitational attraction between them Is doubled

A
Is doubled

B
Becomes $$4\ times$$

C
Is reduced to half

D
Is reduced to a quarter

Solution

According to Newton's law of law of gravitational force of gravitational is inversely proportional to the square of distance between the bodies so distance is doubled force of gravitational will reduce by $$\dfrac { 1 }{ 4 } $$.
Question 4
1 / -0

Gravitational pull is maximum

A
Underground

B
Above the earth

C
On earth's surface

D
Underwater

Solution
Question 5
1 / -0

The height at which the value of acceleration due to gravity becomes $$50\% $$ of that at the surface of the earth$$.$$ (radius of the earth =$$6400 km$$) is

A
$$2650$$

B
$$2430$$

C
$$2250$$

D
$$2350$$

Solution

Given,
$$R=6400km$$
$$g'=50g$$%
$$g'=0.5g$$
New gravity at height $$h$$ is given by
$$g'=\dfrac{gR^2}{(R+h)^2}$$
$$0.5g=\dfrac{gR^2}{(R+h)^2}$$
$$\dfrac{(R+h)^2}{R^2}=2$$
$$\dfrac{R+h}{R}=\sqrt{2}=1.4142$$
$$1+\dfrac{h}{R}=1.4142$$
$$\dfrac{h}{R}=0.4142$$
$$h=0.4142R=0.4142\times 6400$$
$$h=2650km$$
The correct option is A.
Question 6
1 / -0

The acceleration due to gravity at a depth of 4600 km inside the earth.

A
2.81 $$m/{s^2}$$

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

C
7 $$m/{s^2}$$

D
7.35 $$m/{s^2}$$

Solution
Question 7
1 / -0

A body weighs W newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be:

A
$$\dfrac{W}{2}$$

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

C
$$\dfrac{4W}{9}$$

D
$$\dfrac{W}{4}$$

Solution

$$\text{At the surface}$$
$$W=\dfrac{GMm}{R^2}......(I)$$
$$W'=\dfrac{GMm}{(\dfrac{3R}{2})^2}.......(II)$$
$$\dfrac{W}{W'}=\dfrac{\dfrac{GMm}{R^2}}{\dfrac{GMm}{9R^2/4}}$$
$$\Rightarrow \dfrac{W}{W'}=\dfrac{9}{4}$$
$$\Rightarrow W'=\dfrac{4W}{9}$$
Question 8
1 / -0

A solid sphere is rotating about an axis passing through its centre with period $$T$$. If its volume shrinks to $$\dfrac {1}{27}$$ the earlier volume mass remaining same, the new period will be

A
$$\dfrac {T}{9}$$

B
$$3T$$

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

D
$$9T$$

Solution
Question 9
1 / -0

On account of the earth rotating about Its axis:-

A
The linear velocity of objects at equator is greater than at other places

B
The angular velocity of objects at equator is more than that of objects at poles

C
The linear velocity of objects at all places at the earth is equal, but angular velocity is different.

D
At all places the angular velocity and liner velocity are uniform.

Solution
Question 10
1 / -0

The weight of an object in the coal mine, sea level, at the top of the mountain are $$W_{1},W_{2}$$ and $$W_{3}$$ respectively, then :

A
$$W_{1}< W_{2}> W_{3}$$

B
$$W_{1}= W_{2}= W_{3}$$

C
$$W_{1}< W_{2}< W_{3}$$

D
$$W_{1}> W_{2}> W_{3}$$

Solution

Lets $$m=$$ mass of an object

The weight of an object depends on the acceleration due to gravity is given by
$$W=mg$$

The weight of an object in the coal mine,

$$W_1=mg(1-\dfrac{d}{R})$$. . . . .(1)

where, $$R=$$ radius of the earth

$$d=$$ depth of coil mine

The weight of an object at the sea level,

$$W_2=mg$$. . . . . . . (2)

The weight of an object at the top of the mountain

$$W_3=mg(1-\dfrac{2h}{R})$$. . . .(3)

From equation (1), (2) and (3), we conclude that

$$W_1<W_2>W_3$$

The correct option is A.