Gravitation Test

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Gravitation Test - 52

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Question 1
1 / -0

The value of g is _____ at poles

A
Maximum

B
Minimum

C
same

D
cant say

Solution

Acceleration due to gravity at latitude $$\lambda $$ is $${ g }^{ \prime }=g-{ w }^{ 2 }R{ \cos { ^{ \prime } } }\lambda $$ [W=angular velocity, R=radius of the Earth]
At poles $$\lambda =90°\quad \therefore \cos { ^{ 2 }\lambda } =0\\ \therefore { g }^{ \prime }={ g }_{ max }=g$$
The value of g is $$A$$ maximum at poles.
Question 2
1 / -0

The value of g on the moon is $$\dfrac {1}{6}th$$ that on the earth. A body weighing $$60\ N$$ on the earth weighs ________ on the moon.

A
$$20\ N$$

B
Zero

C
$$360\ N$$

D
$$6\ N$$

Solution

Given,
The value of g on the moon will be $$g_m=\dfrac16$$ of earth, ie $$g_m=(g/6) m/s^2$$
where, $$g=$$ acceleration due to gravity on earth's surface.
$$\because mg=60$$
Thus, the weight of body on the moon $$=mg_m=mg/6=60/10=6 kg$$
Question 3
1 / -0

According to Newton's law of gravitation, the apple and the earth experience equal and opposite forces due to gravitation. But it is the apple that falls towards the earth and not vice-versa. Why?

A
$$a_{apple} > a_{earth}$$

B
$$a_{apple} < a_{earth}$$

C
$$a_{apple} = a_{earth}$$

D
cant say

Solution

According to Newton's third law of motion, the force with which the earth is attracted towards the apple is equal to the force with which earth attracts the apple. However, the mass of the earth is extremely large as compared to that of apple. So acceleration of the earth is very small and is not noticeable
F=$$M{ a }_{ earth }=m{ a }_{ apple }$$ [M=mass of the Earth, m=mass of the apple]
=> $$\cfrac { { a }_{ apple } }{ { a }_{ earth } } =\cfrac { M }{ m } >1$$ [Mass of the Earth is very large than the mass of the apple]
$${ a }_{ apple }>{ a }_{ earth }$$
Question 4
1 / -0

Suppose gravitational force between two masses were to be given by $$F = k\dfrac {\sqrt {m_{1}m_{2}}}{d^{3}}$$ where k is some constant, two equal masses attract each other with a certain force when the distance is d. If each of the masses is doubled, than what value the distance between them must be maintained for the force to remain the same as earlier?

A
$$\sqrt [3]{3d}$$

B
$$\sqrt [13]{3d}$$

C
$$\sqrt [13]{2d}$$

D
$$\sqrt [3]{2d}$$

Solution

$$\qquad F=\cfrac { k\sqrt { { m }_{ 1 }^{ \prime }{ m }_{ 2 }^{ \prime } } }{ { d }^{ 3 } } $$ (given) [k=constant]
Now, $$\qquad { F }^{ \prime }=\cfrac { k\sqrt { { m }_{ 1 }^{ \prime }{ m }_{ 2 }^{ \prime } } }{ { d }_{ 3 }^{ \prime } } $$
$${ d }^{ \prime }$$=? for $${ F }^{ \prime }$$=F, $${ m }_{ 1 }^{ \prime }=2{ m }_{ 1 }$$, $${ m }_{ 2 }^{ \prime }=2{ m }_{ 2 }$$
$$\therefore { F }=\cfrac { k\sqrt { { 2m }_{ 1 }{ m }_{ 2 } } }{ { d }^{ \prime 3 } } \\ =>\cfrac { k\sqrt { { m }_{ 1 }{ m }_{ 2 } } }{ { d }^{ 3 } } =\cfrac { 2k\sqrt { { m }_{ 1 }{ m }_{ 2 } } }{ { d }^{ \prime 3 } } \\ =>{ d }^{ \prime 3 }=2{ d }^{ 3 }\\ =>{ d }^{ \prime }=\sqrt [ 3 ]{ 2d } $$
Question 5
1 / -0

A body weighs $$5\ N$$ in air and $$2\ N$$ when immersed in a liquid. The buoyant force is

A
$$3\ N$$

B
$$5\ N$$

C
$$7\ N$$

D
Zero

Solution

Weight of the body in air $$W = 5\textrm{ N}$$

Weight of the body in liquid $$W' = 2\textrm{ N}$$

Thus, Apparent weight $$=W-B=2\textrm{ N}$$

Buoyant Force $$B = W-W' = 5-2 = 3\textrm{ N}$$
Question 6
1 / -0

If the diameter of earth becomes two times its present value and its mass remains unchanged then how would the weight of an object on the surface of the earth be affected?

A
One fifth of the present value

B
One fourth of the present value

C
One third of the present value

D
One half of the present value

Solution

Acceleration due to gravity is g=$$\cfrac { GM }{ { R }^{ 2 } } =\cfrac { GM }{ \left( \cfrac { D }{ 2 } \right) ^{ 2 } } =\cfrac { 4GM }{ { D }^{ 2 } } $$
Where,G=Universal gravitational constant
M=mass of the Earth
R=radius of the Earth
D=Diameter of the Earth
Now, $${ g }^{ \prime }=\cfrac { GM }{ \left(2\times \cfrac { D }{ 2 } \right) ^{ 2 } } =\cfrac { GM }{ { D }^{ 2 } } =\cfrac { g }{ 4 } $$
Weight of the body=$$m{ g }^{ \prime }=\cfrac { mg }{ 4 } $$

Hence, weight of the object on the surface of the earth will become one fourth of the present value.
Question 7
1 / -0

A rectangular tank of $$6\ m$$ long, $$2\ m$$ broad, and $$2\ m$$ deep is full of water. find the thrust acting on the bottom of the tank. (Density of water, $$\rho_w=1000\ kg/m^3$$)

A
$$23.52\times 10^{4}N$$

B
$$23.52\ N$$

C
$$11.76\times 10^{4}N$$

D
$$3.92\times 10^{4}N$$

Solution

Volume of the water tank, $$V=6\ m \times 2\ m \times 2\ m $$
$$V=24\ m^3$$
Density of water, $$\rho_w=1000\ kg/m^3$$
Mass, $$m=\rho_w \times V$$
$$m= 1000\times 24$$
$$m=24000\ kg$$

We know,
Thrust is the net force acting in a particular direction.
$$F=mg$$
$$F=24000\times 9.8$$
$$F=23.52\times 10^4\ N$$
Question 8
1 / -0

The maximum vertical distance through which a full dressed astronaut can jump on the earth is 0.5 m. Estimate the maximum vertical distance through which he can jump on the moon, which has a mean density 2/3 rd that of the earth and radius one quarter that of the earth.

A
1.5 m

B
3 m

C
6 m

D
7.5 m

Solution

The gravitational acceleration on a planet $$=g=\dfrac{GM}{R^2}$$
$$=\dfrac{G\rho\dfrac{4}{3}\pi R^3}{R^2}$$ $$=G\dfrac{4}{3}\pi \rho R\propto \rho R$$
Hence the value of $$g$$ on moon $$=g_m=\dfrac{g_e}{6}$$
Since the astronaut has energy to jump only upto $$0.5m$$ on earth,
$$Energy=mg_eh_e=mg_mh_m$$
$$\implies h_m=\dfrac{g_e}{g_m}h_e$$ $$=6\times h_e=3m$$
Question 9
1 / -0

The value of universal gravitational constant $$G$$ is-

A
$$6.67\times 10^{-11} \dfrac{Nm^2}{kg}$$

B
$$6.67\times 10^{-11} \dfrac{Nm^2}{kg^2}$$

C
$$66.7\times 10^{-11} \dfrac{Nm^2}{kg^2}$$

D
$$66.7\times 10^{-11} \dfrac{Nm^2}{kg}$$

Solution

Gravitational force between two objects of mass $$m_1$$ and $$m_2$$ separated by a distance $$r$$, $$F = \dfrac{Gm_1 m_2}{r^2}$$
$$\implies$$ $$G = \dfrac{F r^2 }{m_1 m_2}$$
Value of universal gravitational constant $$G$$ is $$6.67\times 10^{-11}$$ $$\dfrac{Nm^2}{kg^2}$$.
Question 10
1 / -0

S.I. Unit of universal gravitational constant $$G$$ is-

A
$$\dfrac{Nm^2}{Kg}$$

B
$$\dfrac{Nm^2}{Kg^2}$$

C
$$\dfrac{Nm}{Kg^2}$$

D
$$\dfrac{Nm}{Kg}$$

Solution

Gravitational force between two objects of mass $$m_1$$ and $$m_2$$ separated by a distance $$r$$, $$F = \dfrac{Gm_1 m_2}{r^2}$$
$$\implies$$ $$G = \dfrac{F r^2 }{m_1 m_2}$$
SI unit of universal gravitational constant $$G$$ is $$\dfrac{Nm^2}{Kg^2}$$.

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Gravitation Test - 52

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Gravitation Test - 52

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SHARING IS CARING

Question 1

Maximum

Minimum

same

cant say

Solution

Question 2

$$20\ N$$

Zero

$$360\ N$$

$$6\ N$$

Solution

Question 3

$$a_{apple} > a_{earth}$$

$$a_{apple} < a_{earth}$$

$$a_{apple} = a_{earth}$$

cant say

Solution

Question 4

$$\sqrt [3]{3d}$$

$$\sqrt [13]{3d}$$

$$\sqrt [13]{2d}$$

$$\sqrt [3]{2d}$$

Solution

Question 5

$$3\ N$$

$$5\ N$$

$$7\ N$$

Zero

Solution

Question 6

One fifth of the present value

One fourth of the present value

One third of the present value

One half of the present value

Solution

Question 7

$$23.52\times 10^{4}N$$

$$23.52\ N$$

$$11.76\times 10^{4}N$$

$$3.92\times 10^{4}N$$

Solution

Question 8

1.5 m

3 m

6 m

7.5 m

Solution

Question 9

$$6.67\times 10^{-11} \dfrac{Nm^2}{kg}$$

$$6.67\times 10^{-11} \dfrac{Nm^2}{kg^2}$$

$$66.7\times 10^{-11} \dfrac{Nm^2}{kg^2}$$

$$66.7\times 10^{-11} \dfrac{Nm^2}{kg}$$

Solution

Question 10

$$\dfrac{Nm^2}{Kg}$$

$$\dfrac{Nm^2}{Kg^2}$$

$$\dfrac{Nm}{Kg^2}$$

$$\dfrac{Nm}{Kg}$$

Solution

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