Gravitation Test

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Gravitation Test - 77

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Question 1
1 / -0

Use the assumptions of the previous question. An object weighed by a spring balance at the equator gives the same reading as a reading taken at a depth $$d$$ below the earth's surface at a pole $$(d < < R)$$. The value of $$d$$ is

A
$$\dfrac {\omega^{2}R^{2}}{g}$$

B
$$\dfrac {\omega^{2}R^{2}}{2g}$$

C
$$\dfrac {2\omega^{2}R^{2}}{g}$$

D
$$\dfrac {\sqrt {Rg}}{\omega}$$

Solution
Question 2
1 / -0

If $$g$$ on the surface of the earth is $$9.8/ m{s}^{-2}$$, its value at a depth of $$3200\ km$$ (Radius of the earth=$$6400\ km$$) is

A
$$9.8/ m{s}^{-2}$$,

B
Zero

C
$$4.9/ m{s}^{-2}$$,

D
$$2.45/ m{s}^{-2}$$,

Solution

$$g_d=g\left(1-\dfrac d{R}\right)$$
$$\implies 9.8\left(1-\dfrac{3200\times 10^{3}}{6400\times 10^{3}}\right)$$
$$4.9m/s^2$$
Question 3
1 / -0

A person on the surface of the moon:

A
Does not feel the effect of earth's gravity because the gravity due to moon is such stronger.

B
Does not feel the effect of earth's gravity.

C
Does not feel the effect of earth's gravity and moon gravity he is freely towards the earth.

D
Feels only the combined effect of earth's and moon gravity which are comparable in magnet.

Solution

The value of $${g_e}$$ on the surface of moon will be,
$$g' = g{\left( {1 + \frac{h}{R}} \right)^{ - 2}}$$
After putting the values of $$g,h,R$$
we get the value of $${g'}$$ as,
$${g'} = .0026\,{\rm{m}}/{{\rm{s}}^2}$$
This is almost negligible and can’t be felt on the person standing on moon.
Thus, a person on the surface of the moon does not feel the effect of earth's gravity because the gravity due to moon is much stronger.
Question 4
1 / -0

Assume that the earth moves around the sun in an elliptic orbit with sun at one of the foci. If the minimum and the maximum distance of the earth from the sun be $$\alpha \,{\text{and}}\,\beta $$ respectively, then the eccentricity of the elliptical orbit is

A
$$
\dfrac{\alpha }
{\beta }
$$

B
$$
\sqrt {\dfrac{\alpha }
{\beta }}
$$

C
$$
\sqrt {\dfrac{{\beta - \alpha }}
{{\beta + \alpha }}}
$$

D
$$
\dfrac{{\beta - \alpha }}
{{\beta + \alpha }}
$$
Question 5
1 / -0

A body is released from a point distance r from the center of earth. If R is the earth r > R, then the velocity of the body at the time of striking the earth will be:

A
$$\sqrt{gR}$$

B
$$\sqrt{2gR}$$

C
$$\sqrt{\frac{2gR}{r - R}}$$

D
$$\sqrt{\dfrac{2gR(r - R)}{r}}$$

Solution

$${ \left( PE \right) }_{ height }={ \left( PE \right) }_{ surface }+{ \left( KE \right) }_{ surface }\\ -\dfrac { GMm }{ r } =-\dfrac { GMm }{ R } +\dfrac { 1 }{ 2 } m{ v }^{ 2 }\\ \therefore \dfrac { 1 }{ 2 } { v }^{ 2 }=GM\left( \dfrac { 1 }{ R } -\dfrac { 1 }{ r } \right) \\ \Rightarrow v=\sqrt { \dfrac { 2GM }{ R } \left( \dfrac { r-R }{ r } \right) } \dfrac { R }{ R } \\ \Rightarrow \sqrt { \dfrac { 2GMR }{ R } \left( \dfrac { r-R }{ r } \right) } \\ \Rightarrow \sqrt { \dfrac { 2gR(r-R) }{ r } } $$
Question 6
1 / -0

The depth at which the value of $$g$$ becomes $$25$$% of that at the surface of the earth is (in Km)

A
4800

B
2400

C
3600

D
1200

Solution

Given that $$g_d=0.25g$$
We have, $$g_d=g\left(1-\dfrac d{R}\right)$$
$$\therefore 0.25g=g\left(1-\dfrac d{R}\right)$$
$$\implies d=0.75R$$
$$\implies 0.75\times 6400$$
$$\implies 4800km$$
Question 7
1 / -0

For escape of oxygen molecules from the earth's surface, its temperature should be:

A
$$4.11\times {10}^{3}K$$

B
$$2.09\times {10}^{4}K$$

C
$$8.3\times {10}^{4}K$$

D
$$3\times {10}^{5}K$$

Solution

rms velocity of gas molecule is given by,
$$V_{rms}=\sqrt{\dfrac{3kT}{M}}=11.2\times 10^3$$
$$\therefore T=\dfrac{(11.2\times 10^3)^2\times}{3k}$$
$$\implies \dfrac{(11.2\times 10^3)^2(2.76\times 10^{-26})}{3\times 1.38\times 10^{-23}}$$
$$\implies 8.3\times 10^{4}k$$
Question 8
1 / -0

A particle hanging from a massless spring stretches it by $$2cm$$ at the eath's surface. How much will the same particle stretch the spring at a height of $$2624Km$$ from the surface of the earth? (Radius of the earth$$=6400Km$$)

A
$$1cm$$

B
$$2cm$$

C
$$3cm$$

D
$$4cm$$

Solution

$$\triangle l\alpha \quad F\\ \Rightarrow \triangle l\alpha \quad mg\\ \Rightarrow \triangle l\alpha \quad g\\ \therefore \dfrac { \triangle { l }_{ 2 } }{ \triangle { l }_{ 1 } } =\dfrac { { g }_{ 2 } }{ { g }_{ 1 } } \\ \Rightarrow \dfrac { \triangle { l }_{ 2 } }{ 2 } =\dfrac { \dfrac { GM }{ { \left( 6400+2624 \right) }^{ 2 } } }{ \dfrac { GM }{ { \left( 6400 \right) }^{ 2 } } } \\ \Rightarrow \triangle { l }_{ 2 }=1$$
$$\therefore \triangle { l }_{ 2 }$$ will be $$1cm$$.
Question 9
1 / -0

The period of rotaion of the earth so as to make any object weightless on its equator is

A
$$84$$ min

B
$$74$$ minutes

C
$$64$$ minutes

D
$$54 $$ minutes

Solution
Question 10
1 / -0

Assuming the earth as a sphere of uniform density, the acceleration due to gravity half way towards the centre of the earth will be if it weighed 250N on the surface?

A
$$0.25\ g$$

B
$$0.50\ g$$

C
$$0.125\ g$$

D
$$0.75\ g$$

Solution

Suppose $${ g }_{ 1 }{ g }_{ a }$$ be the acceleration due to gravity on earth's surface and at a depth $$d$$ from surface respectively. Also suppose $$w$$ and $${ w }_{ d }$$ be weight of body on earth's surface at a depth $$d$$.
$$w=mg=250N\quad \longrightarrow \left( 1 \right) $$
$${ w }_{ d }={ mg }_{ d }\quad \longrightarrow \left( 2 \right) $$
We know that $${ g }_{ d }=g\left( 1-\dfrac { d }{ R } \right) \quad \longrightarrow \left( 3 \right) $$
Here $${ d }_{ 2 }=\dfrac { R }{ 2 } $$ $$R=$$ radius of earth
From equation $$(3)$$ and $$(4)$$
$${ g }_{ d }=g\left( 1-\dfrac { R/2 }{ R } \right) $$
$$=g\left( 1-1/2 \right) =g\times \dfrac { 1 }{ 2 } =\dfrac { g }{ 2 } \quad \longrightarrow \left( 5 \right) $$
$${ w }_{ d }={ mg }_{ d }=\dfrac { mg }{ 2 } =\dfrac { 1 }{ 2 } mg=\dfrac { 1 }{ 2 } w=\dfrac { 1 }{ 2 } \times 250=125N$$.

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Re-Attempt Weekly Quiz Competition

Gravitation Test - 77

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Gravitation Test - 77

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SHARING IS CARING

Question 1

$$\dfrac {\omega^{2}R^{2}}{g}$$

$$\dfrac {\omega^{2}R^{2}}{2g}$$

$$\dfrac {2\omega^{2}R^{2}}{g}$$

$$\dfrac {\sqrt {Rg}}{\omega}$$

Solution

Question 2

$$9.8/ m{s}^{-2}$$,

Zero

$$4.9/ m{s}^{-2}$$,

$$2.45/ m{s}^{-2}$$,

Solution

Question 3

Does not feel the effect of earth's gravity because the gravity due to moon is such stronger.

Does not feel the effect of earth's gravity.

Does not feel the effect of earth's gravity and moon gravity he is freely towards the earth.

Feels only the combined effect of earth's and moon gravity which are comparable in magnet.

Solution

Question 4

$$\dfrac{\alpha }{\beta }$$

$$\sqrt {\dfrac{\alpha }{\beta }} $$

$$\sqrt {\dfrac{{\beta - \alpha }}{{\beta + \alpha }}} $$

$$\dfrac{{\beta - \alpha }}{{\beta + \alpha }}$$

Question 5

$$\sqrt{gR}$$

$$\sqrt{2gR}$$

$$\sqrt{\frac{2gR}{r - R}}$$

$$\sqrt{\dfrac{2gR(r - R)}{r}}$$

Solution

Question 6

4800

2400

3600

1200

Solution

Question 7

$$4.11\times {10}^{3}K$$

$$2.09\times {10}^{4}K$$

$$8.3\times {10}^{4}K$$

$$3\times {10}^{5}K$$

Solution

Question 8

$$1cm$$

$$2cm$$

$$3cm$$

$$4cm$$

Solution

Question 9

$$84$$ min

$$74$$ minutes

$$64$$ minutes

$$54 $$ minutes

Solution

Question 10

$$0.25\ g$$

$$0.50\ g$$

$$0.125\ g$$

$$0.75\ g$$

Solution

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$$
\dfrac{\alpha }
{\beta }
$$

$$
\sqrt {\dfrac{\alpha }
{\beta }}
$$

$$
\sqrt {\dfrac{{\beta - \alpha }}
{{\beta + \alpha }}}
$$

$$
\dfrac{{\beta - \alpha }}
{{\beta + \alpha }}
$$