Gravitation Test

Result

Gravitation Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

At what altitude (h) above the earth's surface would the acceleration due to gravity be one fourth of its value at the earths surface?

A
$$h=R$$

B
$$h=4R$$

C
$$h=2R$$

D
$$h=16R$$

Solution
Question 2
1 / -0

A particle weights 120N on the surface of the earth. At what height above the earth's surface will its weight be 30N? Radius of the earth is 6400 km.

A
6000

B
6400

C
5800

D
7000

Solution
Question 3
1 / -0

A particle is taken to a height $$R$$ above the surface, where $$R$$ is the radius of the earth. The acceleration due to gravity there is:

A
$$2.45\ m/s^{2}$$

B
$$4.9\ m/s^{2}$$

C
$$4.8\ m/s^{2}$$

D
$$19.6\ m/s^{2}$$

Solution
Question 4
1 / -0

The altitude at which the weight of a body is only $$64\%$$ of its weight on the surface of the earth is (Radius of the earth $$6,400km$$)

A
$$1600 M$$

B
$$2Km$$

C
$$160 Km$$

D
$$1600 Km$$

Solution

Given,

$$R=6400km$$

So, $$g'=g(1-\dfrac{2h}{R})=g-\dfrac{2gh}{R}\Rightarrow g-g'=\dfrac{2gh}{R}$$

Percentage decreases in weight,

$$=\dfrac{mg-mg'}{mg}\times100=\dfrac{g-g}{g}\times100=\dfrac{2gh}{gR}\times100=\dfrac{2h}{R}\times100\Rightarrow 64=\dfrac{2\times h}{6400}\times100\Rightarrow h=2.048km$$
Question 5
1 / -0

The height at which the weight of a body becomes $$\dfrac{1}{16^th}$$,its weight on the surface of earth (radius R), is :-

A
$$3R$$

B
$$4R$$

C
$$5R$$

D
$$15R$$

Solution
Question 6
1 / -0

If the radius of the earth is $$6400km$$, the height above the surface of the earth ,where the value of acceleration due to gravity will be $$1\%$$ of its value from the surface of the earth is

A
$$6400km$$

B
$$64km$$

C
$$711km$$

D
$$5760km$$

Solution
Question 7
1 / -0

Escape velocity on the earth is

A
$$15.0\ km/s$$

B
$$21.1\ m/s$$

C
$$7.0\ m/s$$

D
$$11.2\ m/s$$

Solution

Escape velocity is the speed that as object need to be traveling to break free of a planet on moon's gravity well and leave it without further propulsion. The escape velocity of Earth is $$11.24m/s$$.
Question 8
1 / -0

A body of mass m is taken from earth's surface to q a height equal to radius of earth . the change in potential will be

A
$$mg$$

B
$$\dfrac{1}{2}mg R$$

C
$$2mgR$$

D
$$\dfrac{1}{4}mgR$$

Solution

Given,
Mass $$=m,$$
Height $$ h=R$$
Potential energy at surface   $$U_s = \dfrac{-GMm}{R} = -mgR$$          $$(gR^2 = GM)$$
Potential energy at height $$h = R$$,    $$U_h = -\dfrac{GMm}{R+R} = \dfrac{-mgR}{2}$$
Change in potential energy   $$\Delta U = \dfrac{-mgR}{2}-(-mgR) = \dfrac{mgR}{2}$$
Question 9
1 / -0

At what height above the surface of earth, acceleration due to gravity is $$\Bigg \lgroup \frac{1}{8} \Bigg \rgroup^{th}$$ of its value at the surface of earth: ($$R_e$$ = Radoius of earth)

A
$$\sqrt{3}R_e$$

B
$$2R_e$$

C
$$2\sqrt{2} R_e$$

D
$$(\sqrt{3} - 1) R_e$$

Solution
Question 10
1 / -0

Earth can be considered as uniform solid sphere of radius $$R. g_{1}$$ and $$g_{2}$$ be acceleration due to gravity at points $$R/2$$ below the surface and $$R/2$$ above the surface of earth respectively. Then $$\dfrac {g_{1}}{g_{2}}$$ equals to

A
$$\dfrac {1}{4}$$

B
$$\dfrac {1}{2}$$

C
$$\sqrt {2}$$

D
$$\dfrac {9}{8}$$

Solution