Gravitation Test

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Gravitation Test - 78

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Question 1
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The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is $$16.0\ N$$ when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.

A
19.6 J

B
29 J

C
35 J

D
10 J

Solution

Given that,

Tension T = 16.0 N

Now,

$$ T-2mg+2ma=0....(I) $$

$$ T-mg-ma=0....(II) $$

From equation (I) and (II)

$$ 3ma-mg=0 $$

$$ g=3a $$

$$ a=\dfrac{g}{3} $$

Now, put the value of g in equation (II)

$$ T-3ma-ma=0 $$

$$ T=4ma $$

$$ a=\dfrac{T}{4m} $$

Now, from equation of motion

At $$t= 1$$

$$ s=ut+\dfrac{1}{2}a{{t}^{2}} $$

$$ s=0+\dfrac{1}{2}\times \dfrac{T}{4m}\times {{(1)}^{2}} $$

$$ s=\dfrac{16}{8m} $$

$$ s=\dfrac{2}{m} $$

Thus, change in the height of the block will be;
$$\Delta h=s$$
Net mass

$$2m-m=m$$

Now, decrease potential energy

$$ P.E=mg\Delta h $$

$$ P.E=m\times 9.8\times \dfrac{2}{m} $$

$$ P.E=19.6\,J $$

Hence, the decrease potential energy is $$19.6\ J$$
Question 2
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$$g_e$$ and $$g_p$$ denote the acceleration due to gravity on the surface of earth and another planet whose mass and radius are twice that of the earth, then

A
$$g_p = g_e$$

B
$$g_p = \dfrac{g_e }{ 2}$$

C
$$g_p = 2g_e$$

D
$$g_p = 3g_e$$

Solution
Question 3
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A man weighs $$100\ kg$$ on the surface of the earth of radius $$R$$. At what height above the surface of the earth, he will weigh $$50\ kg$$?

A
$$0.41\ R$$

B
$$0.51\ R$$

C
$$0.31\ R$$

D
$$0.61\ R$$

Solution
Question 4
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If the mass and radius of a planet are doubled, then acceleration due to gravity on its surface will become :

A
One fourth

B
One half

C
Double

D
Four times

Solution

The acceleration due to gravity on the surface is given as

$$\dfrac{{Gm}}{{{r^2}}}$$

If the mass and the radius are doubled then the acceleration due to gravity will become

$$\dfrac{{G2m}}{{4{r^2}}}$$

Hence the acceleration becomes one half.
Question 5
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At what height the acceleration due to gravity will be reduced to $$36\%$$ of its value on the surface of the earth?

A
Use $$g_{h} = g\left [\dfrac {2R}{R + h}\right ]^{2}$$.

B
Use $$g_{h} = g\left [\dfrac 3{R}{R + h}\right ]^{2}$$.

C
Use $$g_{h} = g\left [\dfrac {2R}{2R + h}\right ]^{2}$$.

D
Use $$g_{h} = g\left [\dfrac {R}{R + h}\right ]^{2}$$.

Solution
Question 6
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The decrease in the value of g at height h from earth's surface is

A
$$\dfrac{2h}{R}g$$

B
$$\dfrac{\sqrt2h}{R}g$$

C
$$\dfrac{h}{R}g$$

D
$$\dfrac{R}{2hg}$$

Solution
Question 7
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A body is projected with a velocity double the escape velocity. The velocity of the body after it escape the gravitational field of the earth is:

A
$$0$$

B
$$V_{e}$$

C
$$\sqrt{2} V_{e}$$

D
$$\sqrt{3} V_{e}$$

Solution
Question 8
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The Moon to orbits Jupiter once in $$1.769$$ days. The orbital radius of the Moon to is $$421700$$ km. Calculate the mass of Jupiter?

A
$$3.898\times { 10 }^{ 27 }kg$$

B
$$1.898\times { 10 }^{ 27 }kg$$

C
$$8.898\times { 10 }^{ 27 }kg$$

D
$$5.898\times { 10 }^{ 27 }kg$$

Solution
Question 9
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If the radius of earth's orbit is made $$1/4th$$, the duration of an year will become

A
$$8\ times$$

B
$$4\ times$$

C
$$1/8\ times$$

D
$$1/4\ times$$

Solution

To solve this problem we have to resort to Kepler's Third law of orbital motion and it states that the square if average orbital period of a planet is proportional to the cube of the average orbital radius such that $${ T }^{ 2 }$$ $$\alpha $$ $${ R }^{ 3 }$$
If we reduce radius of earth to form fold.
$$\dfrac { { T }_{ 1 }^{ 2 } }{ { R }_{ 1 }^{ 3 } } =\dfrac { { T }_{ 2 }^{ 2 } }{ { \left( \dfrac { { R }_{ 1 } }{ 4 } \right) }^{ 3 } } $$
$$\dfrac { { T }_{ 1 }^{ 2 } }{ { R }_{ 1 }^{ 3 } } =\dfrac { { 64T }_{ 2 }^{ 2 } }{ { R }_{ 1 }^{ 3 } } $$
$${ T }_{ 2 }^{ 2 }=\dfrac { { T }_{ 1 }^{ 2 } }{ 64 } $$
Taking square root of both sides
$$\sqrt { { T }_{ 2 }^{ 2 } } =\sqrt { \dfrac { { T }_{ 1 }^{ 2 } }{ 64 } } $$
$${ T }_{ 2 }=\dfrac { { T }_{ 1 } }{ 8 } $$
Orbital period of earth duration of gear $$\dfrac { 1 }{ 8 } $$ times.
Question 10
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A narrow tunnel is dug across a planet diametrically and small bod is dropped from a large height, so that it falls the tunnel. The variation of its kinetic energy 'E' with distance r, from the centre is represented by:

A

B

C

D

Solution