Gravitation Test

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Gravitation Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

How much deep inside the earth radius should a an go ,so that his wight become on fourth of that one the earth's surface/

A
$$\dfrac{R}{4}$$

B
$$\dfrac{R}{2}$$

C
$$\dfrac{3R}{4}$$

D
$$\dfrac{2R}{3}$$

Solution

If the weight has to be $$\dfrac{1}{4}$$ of the original, the new weight is$$\dfrac{mg}{4}$$

Therefore, the gravitational force $$=\dfrac{g}{4}=\dfrac{mg}{\dfrac{4}{m}}$$

Now, $$\dfrac{g}{4}=g(1-\dfrac{h}{R})\Rightarrow1-\dfrac{h}{R}=\dfrac{1}{4}\Rightarrow R-\dfrac{h}{R}=\dfrac{1}{4}\Rightarrow 4(R-h)=R\Rightarrow h=\dfrac{3}{4}R$$
Question 2
1 / -0

Assuming the earth as a sphere of uniform density, the acceleration due to gravity half way toward the centre of the earth will be

A
$$0.75g$$

B
$$0.5g$$

C
$$0.25 g$$

D
$$0.125 g$$

Solution
Question 3
1 / -0

An ice cube of size $$a = 10\ cm$$ is floating in a tank $$(base = 50cm\times 50cm)$$ partially filled with water. The change in gravitational potential energy, when ice completely melts is ________ [Density of ice is $$900\ kg\ m^{-3}$$ and $$g = 10\ ms^{-2}]$$.

A
$$-0.0455\ J$$

B
$$-0.016\ J$$

C
$$-0.24\ J$$

D
$$-0.072\ J$$

Solution

Relative density of ice is $$0.9$$ hence $$90$$% of volume of ice is immersed in water when melts completely level of water does not change.
Therefore change in $$CG=$${ GG }^{ 1 }=0.5cm$$
Change in potential energy
$$\Delta V=-mgh$$
$$={ -\left( 0.1 \right) }^{ 3 }\left( 900 \right) \left( 0.5\times { 10 }^{ -2 } \right) 5$$
$$=-0.0455$$
Question 4
1 / -0

$$W$$ is the weight of a man on the surface of the earth. What would be his weight $$(W)$$ if he goes to a height $$(h)$$ equal to radius of the earth, from the surface of the earth?

A
$$W=\dfrac{W}{2}$$

B
$$W=\dfrac{W}{3}$$

C
$$W=\dfrac{W}{4}$$

D
$$W=2W$$

Solution

Given that,
Weight = $$W$$
Height = $$R$$
We know that,
Weight is inversely proportional to square of distance between man and center of earth
Then, initial distance is $$R$$ and initial weight is $$W$$ and when it is at height $$R$$ then distance is $$2R$$
Now,
$$W\propto \dfrac{1}{{{d}^{2}}}$$
Now,
$$ \dfrac{{{W}_{1}}}{{{W}_{2}}}={{\left( \dfrac{{{d}_{2}}}{{{d}_{1}}} \right)}^{2}} $$

$$ \dfrac{{{W}_{1}}}{{{W}_{2}}}={{\left( \dfrac{2R}{R} \right)}^{2}} $$

$$ \dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{4{{R}^{2}}}{{{R}^{2}}} $$

$$ \dfrac{W}{{{W}_{2}}}=4 $$

$$ {{W}_{2}}=\dfrac{W}{4} $$
Hence, the weight at height is $$\dfrac{W}{4}$$
Question 5
1 / -0

When a small mass $$m$$ is suspended at lower end of the elastic wire having upper end fixed with ceiling. There is loss in gravitational potential energy. let it be x, due to extension of wire, mark correct option,

A
The lost energy can be recovered

B
The lost energy can be irrecoverable

C
only $$\dfrac{x}{2}$$ amount of energy recoverable

D
only $$\dfrac{x}{3}$$ amount of energy recoverable

Solution

Work energy theorem
$$\begin{array}{l}{w_{spring}} + {w_{gravity}} + {w_{loss}} = 0\\ - \dfrac{1}{2}k{x^2} + mgx + {w_{loss}} = 0\\ - \dfrac{1}{{2mgx}} + mgx + {w_{loss}} = 0\\{w_{loss}} = - \dfrac{{mgx}}{2}\end{array}$$

Since there is a loss so not recoverable
half of it is lost rest is recoverable
Question 6
1 / -0

Given that mass of earth is $$M$$ and its radius $$R$$. body is dropped from a height equal to the radius of the earth above the surface of the earth. When it reaches the ground velocity of body will be

A
$$\dfrac {GM}{R}$$

B
$$\left(\dfrac {GM}{R}\right)^{1/2}$$

C
$$\dfrac {2GM}{R}$$

D
$$\left(\dfrac {2GM}{R}\right)^{1/2}$$

Solution

This question can be very easily tackled using conservation of energy with a bit care,

Total energy of body at height 2R = $$m_b\dfrac{ GM_e} {4R_e^2}2R$$=Total energy of body at surface of earth=$$\dfrac{m_bv^2} {2} $$

From these equations we get v=$$[\dfrac{GM_e}{R_e}] ^\dfrac{1} {2} $$
Question 7
1 / -0

The mean distance of Earth from the Sun is $$149.6 \times 10^6$$ km and the mean distance of Mercury from the Sun is $$57.9 \times10^6$$ km. The period of Earth’s revolutions is 1 year, what is the period of Mercury’s revolution?

A
Infinite

B
0.24 year on earth

C
3.2 year on earth

D
2.1 year on earth

Solution

$$\cfrac { { T }_{ 1 }^{ 2 } }{ { r }_{ 1 }^{ 3 } } =\cfrac { { T }_{ 2 }^{ 2 } }{ { r }_{ 2 }^{ 3 } } $$
$$1=$$ Earth; $$2=$$ Mercury
$$\cfrac { { 1 }^{ 2 } }{ { \left( 149.6\times { 10 }^{ 6 } \right) }^{ 3 } } =\cfrac { { T }_{ 2 }^{ 2 } }{ { \left( 57.9\times { 10 }^{ 6 } \right) }_{ }^{ 3 } } $$
$$\Rightarrow \cfrac { 1 }{ 3348071.936\times { 10 }^{ 18 } } =\cfrac { { T }_{ 2 }^{ 2 } }{ 194104.539\times { 10 }^{ 18 } } \Rightarrow { T }_{ 2 }^{ 2 }=0.057975\Rightarrow { T }_{ 2 }=\sqrt { 0.057975 } $$
$${T}_{2}=0.24$$ year on Earth
$$1$$ year on Earth $$=365$$ days
Question 8
1 / -0

If radius of earth contracted by 0.1% its mass remaining same,then weight of the body at earth's surface will increase by

A
0.1%

B
0.2%

C
0.3%

D
Remains same

Solution

If radius is decreased by 0.1%
$$r'=r-0.001 r=0.999 r$$
$$W'=\dfrac{GMm}{(0.999r)^2}=\dfrac{GMm}{(0.999)^2r^2}$$
$$=\dfrac{W}{0.999^2}$$
$$=1.002 W$$
$$\Delta W=W'-W=0.002$$
$$0.002 \times 100=0.2$$% increase.
Question 9
1 / -0

At which height above the surface of the earth of radius $$R$$, will the acceleration due to gravity decrease by $$0.1\%$$?

A
$$\dfrac {R}{100}$$

B
$$\dfrac {R}{200}$$

C
$$\dfrac {R}{1000}$$

D
$$\dfrac {R}{2000}$$

Solution
Question 10
1 / -0

An astronaut orbiting in a spaceship round the earth has a centripetal acceleration of $$2.45m/s^{2}$$. The height of spaceship from earth's surface is ($$R=$$radius of earth)

A
$$3R$$

B
$$2R$$

C
$$R$$

D
$$R/2$$

Solution

Given that,

Acceleration $$a=2.45\,m/{{s}^{2}}$$

Mass of earth $$M=6\times {{10}^{24}}\,kg$$

Radius of earth $$R=6\times {{10}^{6}}\,m$$

Gravitational constant $$G=6.67\times {{10}^{-11}}\,N{{m}^{2}}/kg$$

Now centripetal acceleration is

$$ {{a}_{c}}=\dfrac{{{v}^{2}}}{R} $$

$$ {{a}_{c}}=\dfrac{{{\left( \sqrt{\dfrac{GM}{R}} \right)}^{2}}}{R} $$

$$ {{a}_{c}}=\dfrac{GM}{{{R}^{2}}} $$

$$ {{a}_{c}}=\dfrac{GM}{{{\left( R+h \right)}^{2}}} $$

Now, put the values

$$ 2.45=\dfrac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{{{\left( 6\times {{10}^{6}}+h \right)}^{2}}} $$

$$ {{\left( 6\times {{10}^{6}}+h \right)}^{2}}=\dfrac{6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}}{2.45} $$

$$ {{\left( 6\times {{10}^{6}}+h \right)}^{2}}=\dfrac{40.02\times {{10}^{13}}}{2.45} $$

$$ {{\left( 6\times {{10}^{6}}+h \right)}^{2}}=1.6334\times {{10}^{14}} $$

$$ 6\times {{10}^{6}}+h=\sqrt{1.6334\times {{10}^{14}}} $$

$$ h=1.278\times {{10}^{7}}-6\times {{10}^{6}} $$

$$ h=\left( 12.78-6 \right)\times {{10}^{6}} $$

$$ h=6.78\times {{10}^{6}} $$

$$ h=6.8\times {{10}^{6}}\,m $$

So, $$h\cong R$$

Hence, the height of spaceship is $$R$$

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Gravitation Test - 56

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Gravitation Test - 56

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SHARING IS CARING

Question 1

$$\dfrac{R}{4}$$

$$\dfrac{R}{2}$$

$$\dfrac{3R}{4}$$

$$\dfrac{2R}{3}$$

Solution

Question 2

$$0.75g$$

$$0.5g$$

$$0.25 g$$

$$0.125 g$$

Solution

Question 3

$$-0.0455\ J$$

$$-0.016\ J$$

$$-0.24\ J$$

$$-0.072\ J$$

Solution

Question 4

$$W=\dfrac{W}{2}$$

$$W=\dfrac{W}{3}$$

$$W=\dfrac{W}{4}$$

$$W=2W$$

Solution

Question 5

The lost energy can be recovered

The lost energy can be irrecoverable

only $$\dfrac{x}{2}$$ amount of energy recoverable

only $$\dfrac{x}{3}$$ amount of energy recoverable

Solution

Question 6

$$\dfrac {GM}{R}$$

$$\left(\dfrac {GM}{R}\right)^{1/2}$$

$$\dfrac {2GM}{R}$$

$$\left(\dfrac {2GM}{R}\right)^{1/2}$$

Solution

Question 7

Infinite

0.24 year on earth

3.2 year on earth

2.1 year on earth

Solution

Question 8

0.1%

0.2%

0.3%

Remains same

Solution

Question 9

$$\dfrac {R}{100}$$

$$\dfrac {R}{200}$$

$$\dfrac {R}{1000}$$

$$\dfrac {R}{2000}$$

Solution

Question 10

$$3R$$

$$2R$$

$$R$$

$$R/2$$

Solution

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