Gravitation Test

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Gravitation Test - 50

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Question 1
1 / -0

Find the percentage decrease in the weight of the body when taken to a depth of $$32$$ km below the surface of earth. Radius of the earth is $$6400$$km.

A
0.7

B
0.5

C
1.2

D
cant say

Solution

Here, $$d=32$$km; $$R=6400$$km
Weight of body at depth d is $$=mg'=mg\left(1-\dfrac{d}{R}\right)$$
$$\%$$ decrease in weight $$=\dfrac{mg-mg'}{mg}$$
$$=\dfrac{d}{R}\times 100=\dfrac{32}{6400}\times 100=0.5\%$$.
Question 2
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On a planet whose size is the same and mass four times as that of our earth, find the amount of work done to lift $$3$$kg mass vertically upwards through $$3$$m distance on the planet. The value of g on the surface of earth is $$10$$m $$s^{-2}$$

A
$$360$$J.

B
$$160$$J.

C
$$560$$J.

D
$$460$$J.

Solution

Here, $$R_P=R_e; M_P=4M_e$$
$$M=3kg; h=3m; g_e=10m/s^{-2}$$
On the surface of the earth,
$$g_e=\dfrac{GM}{R^2_e}$$
On the surface of the planet,
$$g_P=\dfrac{GM_P}{R^2_P}=\dfrac{G4M_e}{R^2_e}4\dfrac{GM_e}{R^2_e}=4g_e$$
$$g_P=4\times 10=40m/s^2$$
Work done $$=mg_Ph=3\times 40\times 3=360$$J.
Question 3
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Assuming the earth to be a uniform sphere of radius $$6400$$km and density $$5.5$$ g/c.c, find the value of g on its surface. $$G=6.66\times 10^{-11}Nm^2kg^{-2}$$.

A
$$3.82ms^{-2}$$.

B
$$9.82ms^{-2}$$.

C
$$19.82ms^{-2}$$.

D
$$2ms^{-2}$$.

Solution

Here, $$R=6400\times 10^3m=6.4\times 10^6m$$
$$\rho =5.5g/c.c. = 5.5\times 10^3$$ $$kg/m^3$$
Now, $$g=\dfrac{GM}{R^2}=\dfrac{G}{R^2}\times \dfrac{4}{3}\pi R^3\times \rho$$
$$=\dfrac{4}{3}\pi GR\rho$$
$$=\dfrac{4}{3}\times \dfrac{22}{7}\times 6.66\times 10^{-11}\times 6.4\times 10^6\times 5.5\times 10^3$$
$$=9.82ms^{-2}$$.
Question 4
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The radius of a planet is R. A satellite revolves around it in a circle of radius r with angular velocity $$\omega_0$$. The acceleration due to the gravity on planet's surface is?

A
$$\dfrac{r^3\omega_0}{R}$$

B
$$\dfrac{r^3\omega^2_0}{R^2}$$

C
$$\dfrac{r^3\omega^2_0}{R}$$

D
$$\dfrac{r^2\omega^2}{R^2}$$

Solution

We know $$F=Ma$$
$$\dfrac { { GM }_{ m } }{ { r }^{ 2 } } =\dfrac { { mw }^{ 2 } }{ r } $$
$$GM={ w }^{ 2 }{ r }^{ 3 }$$
$$F=Ma$$
$$a=\dfrac { F }{ M } $$
$$a=\dfrac { GM }{ { R }^{ 2 } } $$
$$G=\dfrac { { w }^{ 2 }{ r }^{ 3 } }{ { R }^{ 2 } } $$
Question 5
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The friction of the air causes vertical retardation equal to one-tenth of the acceleration due to gravity (take $$g = 10\, m \,s^{-2}$$). Find the decrease in the time of flight.(in percent)

A
9

B
10

C
11

D
8

Solution

$$T=\dfrac{2u \sin \theta}{g} \therefore \dfrac{T_1}{T_2}=\dfrac{g_2}{g_1}=\dfrac{g+\dfrac{g}{10}}{g}=\dfrac{11}{10}$$
Fractional decrease in time of flight =$$\dfrac{T_1-T_2}{T_1}=\dfrac{1}{11}$$
Percentage decrease = $$9\%$$
Question 6
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A body of mass m rises to a height $$h=R/5$$ from the earth's surface where R is earth's radius. If g is acceleration due to gravity at the earth's surface, the increase in potential energy is?

A
mgh

B
$$\dfrac{4}{5}$$mgh

C
$$\dfrac{5}{6}$$mgh

D
$$\dfrac{6}{7}$$mgh

Solution
Question 7
1 / -0

The orbit of Pluto is much more eccentric than the orbits of the other planets. That is, instead of being nearly circular, the orbit is noticeably elliptical. The point in the orbit nearest to the Sun is called the perihelion and the point farthest from the Sun is called the aphelion. At perihelion, the gravitational potential energy of Pluto in its orbit has :

A
Its maximum value

B
Its minimum value

C
The same value as at every other point in the orbit

D
The value which depends on the sense of rotation

Solution

Gravitational potential energy = $$\dfrac{-GMm}{r}$$

If $$r$$ reduces, it will become more negative , hence decreases.
Question 8
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The value of g at a certain height h above the free surface of the earth is $$x/4$$ where x is the value of g at the surface of the earth. The height h is?

A
R

B
$$2$$R

C
$$3$$R

D
$$4$$R

Solution
Question 9
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Figure shows a method for measuring the acceleration due to gravity. The ball is projected upward by a gun. The ball passes the electronic gets $$1$$ and $$2$$ as it rises and again as it falls. Each get is connected to a separate timer. The passage of the ball through each gate starts the corresponding timer, and the second passage through the same gate stops the timer. The time intervals $$\triangle { t }_{ 1 }$$ and $$\triangle { t }_{ 2 }$$ are thus measured. The vertical distance between the two gates is $$d$$. If $$d=5\ m$$, $$\triangle { t }_{ I }=3\ s$$, $$\triangle { t }_{ 2 }=2\ s$$, then find the measured value of acceleration due to gravity (in $$m/{s}^{2}$$).

A
$$8\ m{ s }^{ -2 }$$

B
$$4\ m{ s }^{ -2 }$$

C
$$2\ m{ s }^{ -2 }$$

D
$$1\ m{ s }^{ -2 }$$

Solution
Question 10
1 / -0

A projectile is fired vertically upwards from the surface of the earth with a velocity $$kv_e$$ where $$v_e$$ is the escape velocity and $$k < 1$$. If R is the radius of the earth, the maximum height to which it will rise measured from the centre of earth will be (neglect air resistance).

A
$$\dfrac{1-k^2}{R}$$

B
$$\dfrac{R}{1-k^2}$$

C
$$R(1-k^2)$$

D
$$\dfrac{R}{1+k^2}$$

Solution