Gravitation Test

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Gravitation Test - 50

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Question 1
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The centers of two bowling balls are $$2.0\,m$$ apart. The mass of one bowling ball is $$5.0\ Kg$$. The mass of the other is $$6.0\ kg$$. How much gravitational force do they exert on each other?
The universal gravitational constant is: $$6.67 \ \times \,10^{-11}\,\dfrac{N.\,m^2}{Kg^2}$$

A
$$5.0\, \times\,10^{-9}\, N$$

B
$$15.0\, N$$

C
$$5.0\, \times\,10^{-10}\, N$$

D
$$30.0\, N$$

E
$$3.4\, \times\,10^{-9}\, N$$

Solution

Given : $$M_1 =5 kg$$ $$M_2 = 6 kg$$ $$r = 2$$ m $$G = 6.67 \times 10^{-11}$$ $$ Nm^2/kg^2$$

Gravitational force between them $$F = \dfrac{GM_1M_2}{r^2} = \dfrac{6.67 \times 10^{-11} \times 5\times 6}{2^2} = 5 \times 10^{-10}$$ N
Question 2
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The distance between the two point masses $$m_1$$ and $$m_2$$ is d. Now, the distance between them is reduced by two-thirds, Calculate by which factor new gravitational force would be change?

A
It increases by a factor of $$9$$

B
It increases by a factor of $$\dfrac{4}{9}$$

C
It increases by a factor of $$3$$

D
It decreases by factor of $$9$$

E
It decreases by a factor of $$3$$

Solution

Given that the distance reduces by two-thirds, means that the new distance is
$$d_1 = d - (\dfrac{2}{3} \times d) = \dfrac{d}{3}$$
Therefore new force is $$F_1 = G \dfrac{m_1 m_2}{d_1^2} = 9 G \dfrac{m_1 m_2}{d^2} = 9F$$
Hence option A is correct.
Question 3
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Mass differs from weight in which way?

A
weight is a force where as mass is not a force.

B
the mass of an object is always more than its weight.

C
weight is constant but mass is variable.

D
there is no difference

Solution

Mass is the amount of matter in an object but weight is the gravitational force on that object.
Thus, weight is a force but mass is not a force.
Weight $$=$$ mass $$\times $$ gravitational acceleration (g)
Numerically, weight can be equal to, less than or greater than the mass of an object depending upon the magnitude of gravity, but the mass of a body always remains constant.

So the correct option is (A)
Question 4
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Relative density of a body with respect to a medium of density $$\rho_1$$ is $$a$$. What is the relative density of the body with respect to a medium of density $$\rho_2$$?

A
$$\dfrac{a \rho_2}{\rho_1}$$

B
$$\dfrac{a \rho_1}{\rho_2}$$

C
$$\dfrac{\rho_1}{a\rho_2}$$

D
$$\dfrac{\rho_2}{a\rho_1}$$

Solution

Relative density is defined as the ratio of density of substance to the density of any other substance.
$$\therefore$$ Relative density wrt to $$\rho_1$$, $$a = \dfrac{\rho}{\rho_1}$$ where $$\rho$$ is the density of substance.
$$\implies$$ $$\rho = a\rho_1$$
Thus relative density wrt to medium of density $$\rho_2$$, $$a' = \dfrac{\rho}{\rho_2} = \dfrac{a\rho_1}{\rho_2}$$
Question 5
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The pair of physical quantities having the same unit is

A
Thrust and Pressure

B
Thrust and Weight

C
Force and Pressure

D
Weight and Pressure

Solution

Thrust is the force which moves an aircraft through the air. Pressure is the force per unit area. Weight is the gravitational attraction force.
Thus, thrust and weight both represent the force. So they will have same unit.
Question 6
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Four different pairs of people stand in the same park. The tables show the weight of each person, along with the distance between their centers of mass. Which pair has the strongest attraction for each other (gravitationally)?

A

B

C

D

E
Some pairs tie for the greatest attraction

Solution

Force of attraction between the person $$F = \dfrac{Gm_1m_2}{r^2}$$ $$(pounds)^2/(feet)^2$$
(A) :    $$F_A = \dfrac{G(100)(100)}{5^2} = 400 G$$    $$(pounds)^2/(feet)^2$$

(B) :    $$F_B = \dfrac{G(50)(100)}{5^2} =200G$$   $$(pounds)^2/(feet)^2$$

(C) :    $$F_C = \dfrac{G(300)(100)}{(15)^2} = 133.3 G$$    $$(pounds)^2/(feet)^2$$

(D) :    $$F_D = \dfrac{G(300)(200)}{(30)^2} =66.66G$$ $$(pounds)^2/(feet)^2$$   $$\implies F_A>F_B>F_C>F_D$$
Question 7
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A particle is thrown vertically upwards. Its velocity at one-fourth of the maximum height is 20 m$$s^{-1}$$.Then, the maximum height attained by it is
[take g = 10 m$$s^{-2}$$]

A
16.67 m

B
10.67 m

C
26.67 m

D
25 m

Solution

Let maximum height is $$h$$ and the initial velocity is $$u$$.
At maximum height final velocity $$v=0$$,
Using $$v^2=u^2+2as$$ $$\Rightarrow 0= u^2-2gh$$ $$\Rightarrow u^2=2gh$$ ......1
Now at height $$\frac{h}{4}$$ , velocity $$v=20ms^{-1}$$,
Using $$v^2=u^2+2as$$ $$\Rightarrow 20^2= u^2-2g\frac{h}{4}$$
$$\Rightarrow 20^2= 2gh-\dfrac{gh}{2}$$ (Using equation 1)
$$\Rightarrow h=\dfrac{80}{3} m$$
Question 8
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On the earth, a Sumo of 420 kg is checking his reading on a spring balance. To have the same reading of spring balance on the moon, how many Sumos of mass 420 kg each should stand on it. Explain. (g = 30 m s$$^{-2}$$)

A
6

B
12

C
18

D
3

Solution

Acceleration due to gravity of moon $$g_m = \dfrac{g_e}{6}$$
where $$g_e$$ is the acceleration due to gravity on earth.
Mass of a Sumo $$m = 420$$ kg
Weight of Sumo on earth $$W_e = mg_e$$
Weight of Sumo on moon $$W_m = mg_m =\dfrac{mg_e}{6}$$
Thus number of Sumos $$n = \dfrac{W_e}{W_m} = 6$$
Question 9
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Find the gravitational attraction between two atoms in a hydrogen molecule when $$G$$ is $$6.67\times { 10 }^{ -11 }N{ m }^{ 2 }{ kg }^{ -2 }$$, mass of a hydrogen atom is $$1.67\times { 10 }^{ -27 }kg$$ and the distance between two atoms is $$1\mathring { A } $$.

A
$$11.14\times { 10 }^{ -18 }N$$

B
$$2.39\times { 10 }^{ -45 }N$$

C
$$1.86\times { 10 }^{ -44 }N$$

D
$$4.18\times { 10 }^{ -23 }N$$

Solution

We know from Newton's law of gravitation that the gravitational attraction between two masses is $$F=\cfrac { { GM }_{ 1 }{ M }_{ 2 } }{ { R }^{ 2 } }$$, where $$M_1$$, $$M_2$$ are the masses and $$R$$ is the distance between the masses.
Given:
$$ G={ 6.67\times 10 }^{ -11 }N{ m }^{ 2 }{ kg }^{ -2 }\\ { M }_{ 1 }={ M }_{ 2 }=1.67\times { 10 }^{ -27 }kg\\ R=1\overset { \circ }{ A } ={ 10 }^{ -10 }m\\ \therefore F=\cfrac { 6.67\times { 10 }^{ -11 }\times 1.67\times 1.67\times { 10 }^{ -27-27 } }{ { 10 }^{ -10 }\times { 10 }^{ -10 } } =1.86\times { 10 }^{ -44 }N$$
Question 10
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From a place where, g = 9.8 m $$s^{-2}$$, a stone is thrown upwards with a velocity of 4.9 m $$s^{-1}$$. The time taken by the stone to return to the earth is :

A
2 s

B
1 s

C
4 s

D
8 s

Solution

Given : $$g = 9.8 m/s$$
Initial speed of the stone $$u = 4.9 m/s$$
Time taken to return to the earth (or time of flight) $$t = \dfrac{2u}{g}$$
$$\therefore$$ $$t = \dfrac{2\times 4.9}{9.8} = 1$$ s