Gravitation Test

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Gravitation Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

Body is projected vertically upward from the surface of the earth with a velocity equal to half the escape velocity. If $$R$$ is radius of the earth, the maximum height attained by the body is :-

A
$$\dfrac{R}{6}$$

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

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

D
$$R$$

Solution
Question 2
1 / -0

How much above the surface of earth does the acceleration due to gravity reduce by 36% of its value on the surface of earth? (radius of earth 6400 km)

A
$$800$$ km

B
$$1600$$ km

C
$$3200$$ km

D
$$4266$$ km

Solution
Question 3
1 / -0

Velocity of water in a river is

A
Same everywhere

B
More in middle and less near its bank

C
Less in middle and more at banks

D
Increases from one bank to other

Solution

The velocity is more in the middle as compare to its bank because while flowing it collide with near its banks which results in lesser velocity while in the middle there is no obstruction so the velocity is more in the middle .
Question 4
1 / -0

Calculate the escape velocity for an atmospheric particle 9600 km above the earth 's surface , given acceleration due to gravity on the surface of earth is 10 $$ms^{ -2 }$$.

A
7.2 km/sec.

B
8.6 km/sec.

C
9.8 km/sec.

D
6.4 km/sec.

Solution
Question 5
1 / -0

The weight of a body at a distance 2R from the centre of earth of radius R is 2.5 N. If the distance is 3R from the centre of earth the weight of the body will be --------

A
$$1.1 N$$

B
$$2.1 N$$

C
$$3.1 N$$

D
$$4.1 N$$

Solution
Question 6
1 / -0

A body weight W newton at the surface of the each, In weight at a height at a 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
Question 7
1 / -0

Let $$g$$ be the acceleration due to gravity at earth's surface and $$K$$ be the rotational kinetic energy of the earth Suppose the earth's radius decreases by $$2$$% keeping all other quantities same, then

A
$$g$$ increases by $$2$$ % and $$K$$ increases by $$2$$ %

B
$$g$$ increases by $$4$$ % and $$K$$ increases by $$4$$ %

C
$$g$$ increases by $$4$$ % and $$K$$ increases by $$2$$ %

D
$$g$$ increases by $$2$$ % and $$K$$ increases by $$4$$ %

Solution
Question 8
1 / -0

The escape velocity for a body of mass 1 kg from the earth surface is $$11.2 kms^{-1}$$. The escape velocity for a body of mass $$100 kg$$ would be

A
$$11.2 \times 10^2 kms^{-1}$$

B
$$112 kms^{-1}$$

C
$$11.2 kms^{-1}$$

D
$$11.2 \times 10^{-2} kms^{-1}$$

Solution

Escape velocity is independent of mass of body$$.$$
So$$,$$ body of mass $$1kg$$ from the earth surface is $$11.2km/s.$$
Hence,
option $$(C)$$ is correct answer.
Question 9
1 / -0

A tunnel is dug along the diameter of the earth. There is a particle of mass m at the centre of the tunnel. Find the minimum velocity given to the particle so that just reaches to the surface of the earth. (R=radius of earth)

A
$$

\sqrt{\dfrac{G M}{R}}

$$

B
$$

\sqrt{\dfrac{G M}{2 R}}

$$

C
$$

\sqrt{\dfrac{2 G M}{R}}

$$

D
None of the above

Solution

By conservation of energy,

Gain in K.E $$=$$ Change in potential energy

$$\dfrac 12 mv^2=m[-\dfrac{GM}{R}-[-\dfrac{3GM}{2R}]]$$

A gravitational potential at centre $$=-\dfrac 32 \dfrac{GM}{R}$$

Gravitational potential at surface $$=-\dfrac{GM}{R}$$

Therefore,

$$\dfrac 12mv^2=-\dfrac{GMm}{R}+\dfrac 32\dfrac{GMm}{R}$$

$$=\dfrac 12 mv^2=\dfrac{GMm}{2R}$$

$$v=\sqrt{\dfrac{GM}{R}}$$
Question 10
1 / -0

An aeroplane is moving with constant horizontal velocity u at height h. The velocity of a packet dropped from aeroplane, when it reaches on the earth's surface will be (g is acceleration due to gravity)

A
$$\sqrt { { u }^{ 2 }+2gh } $$

B
$$\sqrt { 2gh } $$

C
$$2gh$$

D
$$\sqrt { { u }^{ 2 }-2gh } $$

Solution

Here, since there are no external forces and no non conservative forces, we can conserve the total mechanical energy of the system in the initial and final cases.
Assuming the mass of the packet to be $$m$$,
For when the aeroplane is carrying the packet.
$$TME_i = KE_i+PE_i$$
$$TME_i = \dfrac{1}{2}mu^2 + mgh$$

When the packet reaches the ground, all the potential energy is converted into its kinetic energy. ($$PE_f = 0$$)
$$TME_f = KE_f + PE_f$$
$$TME_f = \dfrac{1}{2}mv^2$$

SInce, $$TME_i = TME_f$$
$$\dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 + mgh$$

$$\dfrac{1}{2}v^2 = \dfrac{1}{2}u^2 + gh$$

$$v^2 = u^2+2gh$$

$$v = \sqrt{u^2+2gh}$$