Gravitation Test

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Gravitation Test - 69

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Weekly Quiz Competition

Question 1
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Consider a satellite moving in a circular orbit around Earth. If K and V denote its kinetic energy and potential energy respectively, then(Choose the convention, where $$V=0$$ as $$r\rightarrow \infty$$)

A
$$K=V$$

B
$$K=2V$$

C
$$V=2K$$

D
$$K=-2V$$

E
$$V=-2K$$

Solution

When a satellite moves in a circular orbit around the earth its
(i) Potential energy,
$$\because V=-\dfrac{GMm}{r}$$
(ii) Kinetic energy,
$$K=\dfrac{1}{2}mv^2=\dfrac{GMm}{2r}\left[\because v=\sqrt{\dfrac{GM}{m}}\right]$$
$$\therefore v=-2K$$.
Question 2
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A body of mass $$m$$ is raised from the earths, surface to an altitude equal to the radius (R) of the Earth. If $$g$$ is the gravitational acceleration on the Earth's surface

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

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

C
$$mgR$$

D
$$2mgR$$
Question 3
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Weight of a body at the equator of a planet is half of that at the poles. If peripheral velocity of a point on the equator of this planet is $$ v_0 $$, what is the escape velocity of a polar particle?

A
$$ v_0 $$

B
$$ 2v_0 $$

C
$$ 3v_0 $$

D
$$ 4v_0 $$

Solution
Question 4
1 / -0

Directions For Questions

The gravitational field in a region is given by $$\vec{E}=(5\ N\ kg^{-1})\vec{i}+(12\ N\ kg^{-1})\vec{j}$$.

...view full instructions

Find the changes is gravitational potential energy if a particle of mass $$2\ kg$$ is taken from the origin to the point $$(12\ m, 5\ m)$$

A
$$-225\ J$$

B
$$-240\ J$$

C
$$-245\ J$$

D
$$-250\ J$$

Solution

$$\Delta v=+\displaystyle \int^{(0,0)}_{(12\ m,5\ m)}Edr =+ \displaystyle \int^{(0,0)}_{(12\ m,5\ m)} (5\hat i +12\hat j)2dr$$
$$=+(10\hat i +24\hat j) |r|_{(12\ m, 5\ m)}^{(0,0)}$$
$$=-(10\hat i +24\hat j) (12\hat i +5\hat j) =-240\ J$$
Question 5
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Directions For Questions

The gravitational field in a region is given by $$\vec{E}=(5\ N\ kg^{-1})\vec{i}+(12\ N\ kg^{-1})\vec{j}$$.

...view full instructions

Find the change in potential energy if the particle is taken from $$(12\ m,0)$$ to $$(0,5\ m)$$

A
$$-10\ j$$

B
$$-50\ J$$

C
zero

D
$$-60\ J$$

Solution

$$\Delta v=+\displaystyle \int_{(0,5\ m)}^{(12\ m, 0)}Edr$$
$$=E \displaystyle \int^{5\hat i}_{5\hat j}dr=(10\hat i +24\hat j) [12\hat i -5\hat j]=0$$
Question 6
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The value of G is

A
$$9.8 \ N m^{2} kg^{-2}$$

B
$$6.7 \times 10^{-11} \ Nm^{2}kg^{-2}$$

C
$$6.7 \times 10^{-11} \ ms^{-2}$$

D
$$6.7 \ Nkg^{-1}$$

Solution

Universal gravitational constant is represented as 'G'. Its value is $$G = 6.7 \times 10^{-11} \ Nm^{2}kg^{-2}$$
Question 7
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Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine $$100 \ km$$ below the earth's surface (Given $$R= 6400 \ km$$)

A
$$9.66 \ m/s^2$$

B
$$7.64 \ m/s^2$$

C
$$5.06 \ m/s$$

D
$$3.10 \ m/s^2$$

Solution

$$g'= g \bigg(1- \dfrac{d}{R} \bigg)$$

On putting above value.
$$g' = 9.8 \bigg( 1- \dfrac{100}{6400} \bigg) = 9.66 \ m/s^2$$
Question 8
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Acceleration due to gravity is '$$g$$' on the surface of the earth. The value of acceleration due to gravity at a height of $$32 \ km$$ above earth's surface is (Radius of the earth $$=6400 \ km$$)

A
$$0.9 \ g$$

B
$$0.99 \ g$$

C
$$0.8 \ g$$

D
$$1.01 \ g$$

Solution

$$h= 32 \ km, R=6400 \ km, $$ so $$h<<R$$

$$g'= g\bigg( 1- \dfrac{2h}{R} \bigg) = g \bigg(1- \dfrac{2 \times 32}{6400} \bigg) $$

$$\implies g'= \dfrac{99}{100} g= 0.99 \ g$$
Question 9
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At the center of the earth:

A
the mass of a body is zero but the weight is not zero

B
the mass of a body is not zero but the weight is zero

C
both mass and weight of a body are zero

D
both mass and weight of a body are not zero

Solution

At the center of earth the weight becomes zero because acceleration due to gravity becomes zero.
Question 10
1 / -0

At what distance from the centre of the earth, the value of acceleration due to gravity g will be half that on the surface ( R = radius of earth)

A
$$2R$$

B
$$R$$

C
$$1.44 \ R$$

D
$$0.414 \ R$$

Solution

$$g'= g\bigg( \dfrac{R}{R+h} \bigg)^2$$

$$ \implies \dfrac{1}{\sqrt{2}}= \dfrac{R}{R+h}$$

$$\implies R+h= \sqrt{2} R$$

$$ \implies h= ( \sqrt{2}-1) R= 0.414 \ R$$

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Re-Attempt Weekly Quiz Competition

Gravitation Test - 69

Result

Gravitation Test - 69

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SHARING IS CARING

Question 1

$$K=V$$

$$K=2V$$

$$V=2K$$

$$K=-2V$$

$$V=-2K$$

Solution

Question 2

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

$$\dfrac{1}{2}mgR$$

$$mgR$$

$$2mgR$$

Question 3

$$ v_0 $$

$$ 2v_0 $$

$$ 3v_0 $$

$$ 4v_0 $$

Solution

Question 4

$$-225\ J$$

$$-240\ J$$

$$-245\ J$$

$$-250\ J$$

Solution

Question 5

$$-10\ j$$

$$-50\ J$$

zero

$$-60\ J$$

Solution

Question 6

$$9.8 \ N m^{2} kg^{-2}$$

$$6.7 \times 10^{-11} \ Nm^{2}kg^{-2}$$

$$6.7 \times 10^{-11} \ ms^{-2}$$

$$6.7 \ Nkg^{-1}$$

Solution

Question 7

$$9.66 \ m/s^2$$

$$7.64 \ m/s^2$$

$$5.06 \ m/s$$

$$3.10 \ m/s^2$$

Solution

Question 8

$$0.9 \ g$$

$$0.99 \ g$$

$$0.8 \ g$$

$$1.01 \ g$$

Solution

Question 9

the mass of a body is zero but the weight is not zero

the mass of a body is not zero but the weight is zero

both mass and weight of a body are zero

both mass and weight of a body are not zero

Solution

Question 10

$$2R$$

$$R$$

$$1.44 \ R$$

$$0.414 \ R$$

Solution

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