Gravitation Test

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Gravitation Test - 47

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Question 1
1 / -0

The change in the value of $$g$$ at a height $$h$$ about the surface of the earth is the same as at a depth $$d$$ below the surface of earth. When both $$d$$ and $$h$$ are much smaller than the radius of earth, then which one of the following is correct?

A
$$d = h$$

B
$$d = 2h$$

C
$$d = \dfrac {3h}{2}$$

D
$$d = \dfrac {h}{2}$$

Solution

Acceleration due to gravity at a height $$h$$, $$g'_h = g \left (1 - \dfrac {2h}{R}\right )$$
Acceleration due to gravity at a depth $$d$$, $$g'_d = g \left (1 - \dfrac {d}{R}\right )$$
$$\therefore \ g \left (1 - \dfrac {2h}{R}\right )$$ $$ = g \left (1 - \dfrac {d}{R}\right )$$
or $$\dfrac {2h}{R} = \dfrac {d}{R}$$
or $$d = 2h$$.
Question 2
1 / -0

An object falls a distance $$H$$ in $$50s$$ when dropped on the surface of the earth. How long would it take for the same object to fall through the same distance on the surface of a planet whose mass and radius are twice that of the earth? (Neglect air resistance)

A
$$35.4s$$

B
$$50.0s$$

C
$$70.7s$$

D
$$100.0s$$

Solution

$$g=\cfrac { GM }{ { r }^{ 2 } } $$
$$\quad g'=\cfrac { G2M }{ { (2r) }^{ 2 } } =\cfrac { 2GM }{ { 4r }^{ 2 } } $$
$$\quad g'=\cfrac { g }{ 2 } \quad \cfrac { H }{ H } =\cfrac { \cfrac { 1 }{ 2 } { g }_{ e }{ t }_{ e }^{ 2 } }{ \cfrac { 1 }{ 2 } { g }_{ p }{ t }_{ p }^{ 2 } } $$
$${ g }_{ p }\times { t }_{ p }^{ 2 }={ g }_{ e }\times { t }_{ e }^{ 2 }\quad \quad $$
$$\cfrac { { g }_{ e } }{ 2 } { t }_{ p }^{ 2 }={ g }_{ e }\times 50\times 50$$
$${ t }_{ p }^{ 2 }=2\times 50\times 50\Rightarrow { t }_{ p }=\sqrt { 5000 } =70.7sec$$
Question 3
1 / -0

The escape velocity of $$10g$$ body from the earth is $$11.2km{ s }^{ -1 }$$. Ignoring air resistance, the escape velocity of $$10kg$$ of the iron ball from the earth will be

A
$$0.0112km{ s }^{ -1 }$$

B
$$0.112km{ s }^{ -1 }$$

C
$$11.2km{ s }^{ -1 }$$

D
$$0.56km{ s }^{ -1 }$$

Solution

Escape velocity is independent of the mass of the projected body provided the air resistance is neglected. Hence, the escape velocity of 10kg of the iron ball projected from the earth is same that of escape velocity of 10g projected from the earth i.e., 11.2kms−1
Question 4
1 / -0

If $$g$$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth is

A
$$\cfrac { 1 }{ 2 } mgR$$

B
$$\cfrac { 1 }{ 4 } mgR$$

C
$$mgR$$

D
$$2mgR$$

Solution

Initial potential energy of the system, $${ U }_{ i }=-\cfrac { GMm }{ R } $$
Final potential energy of the system, $${ U }_{ f }=-\cfrac { GMm }{ 2R } $$
$$\therefore \Delta U={ U }_{ f }-{ U }_{ i }=-GMm\left( \cfrac { 1 }{ 2R } -\cfrac { 1 }{ R } \right) $$
$$=\cfrac { GMm }{ 2R } ....(i)$$
But $$\quad g=\cfrac { GM }{ { R }^{ 2 } } \Rightarrow GM=g{ R }^{ 2 }...(ii)$$
From Eqs. (i) and (ii)
$$\Delta U=\cfrac { g{ R }^{ 2 } }{ 2R } =\cfrac { 1 }{ 2 } mgR$$
Question 5
1 / -0

The height vertically above the earth's surface at which the acceleration duo to gravity becomes 1% of its value at the surface is (It is the radius of the earth)

A
8 R

B
9 R

C
10 R

D
20 R

Solution

$$g'=\frac{g}{\left ( 1+\frac{h}{R} \right )^2}\Rightarrow \frac{g}{100}=\frac{g}{\left ( 1+\frac{h}{R} \right )^2}$$
$$\left ( 1+\frac{h}{R} \right )^2=100$$
$$h=9\, R$$
Question 6
1 / -0

The height of a point vertically above the earth's surface at which the acceleration due to gravity becomes $$9\%$$ of its value at the surface is (Given, R$$=$$ radius of earth)

A
$$2R$$

B
$$\displaystyle\frac{7}{3}R$$

C
$$3R$$

D
$$\displaystyle\frac{2}{3}R$$

Solution

Acceleration due to gravity at height h is given as
$$g_h=\left(\displaystyle\frac{R}{R+h}\right)^2g$$
$$\Rightarrow \displaystyle\frac{g_h}{g}=\left(\displaystyle\frac{1}{\displaystyle 1+\frac{h}{R}}\right)^2$$ ..........(i)
$$\because g_h$$ is $$9\%$$ of g
$$\therefore \displaystyle g_h=\frac{9}{100}$$g
$$=0.09$$g
From Eq. (i),
$$\displaystyle\frac{0.09g}{g}=\left(\displaystyle\frac{1}{\displaystyle 1+\frac{h}{r}}\right)^2$$
$$\left(\displaystyle 1+\frac{h}{R}\right)^2=\displaystyle\frac{1}{0.09}$$
$$\Rightarrow \displaystyle 1+\frac{h}{r}=\frac{1}{0.3}$$
$$\Rightarrow \displaystyle \frac{1}{R}=\frac{10}{3}-1=\frac{7}{3}$$
$$\displaystyle h=\frac{7}{3}R$$.
Question 7
1 / -0

The height at which the weight of a body becomes 1/16th its weight on the surface of earth (Radius R ) is:

A
4R

B
5R

C
15R

D
3R

Solution

$$\cfrac { GMm }{ { \left( R+h \right) }^{ 2 } } =\cfrac { 1 }{ 16 } \cfrac { GMm }{ { R }^{ 2 } } \\ \cfrac { 1 }{ { { \left( R+h \right) }^{ 2 } } } =\cfrac { 1 }{ 16{ R }^{ 2 } } \\ or,\quad { \left( \cfrac { R }{ R+h } \right) }^{ 2 }=\cfrac { 1 }{ 16 } \\ \Rightarrow \cfrac { R }{ R+h } =\cfrac { 1 }{ 4 } .\\ \therefore R+h=4R\\ \Rightarrow h=3R$$
Question 8
1 / -0

A planet of radius $$R_p$$ is revolving around a star of radius $$R^{\ast}$$, which is at temperature $$T^{\ast}$$. The distance between the star and the planet is d. If the planet's temperature is $$fT^{\ast}$$, then f is proportional to.

A
$$\sqrt{R^{\ast}/d}$$

B
$$R^{\ast}/d$$

C
$$R^{\ast}R_p/d^2$$

D
$$(R^{\ast}/d)^4$$

Solution

Total energy radiated by star is $$\propto$$ $${T^*}^4{R^*}^2$$
Energy received by planet is $$\propto$$ $$\dfrac{1}{d^2}$$
Combining both energy received by planet $$\propto$$ $$(fT^*)^4$$
$$\Rightarrow$$ $$f^4 \propto \dfrac{ {R^*}^2}{d^2}$$
$$\Rightarrow f \propto (\dfrac{ {R^*}^2}{d^2})^{\dfrac{1}{4}}= \sqrt{\dfrac{R^*}{d}}$$
Question 9
1 / -0

If the acceleration due to gravity inside the earth is to be kept constant, then the relation between the density $$d$$ and the distance $$r$$ from the centre of earth will be

A
$$d\propto r$$

B
$$d\propto r^{1/2}$$

C
$$d\propto 1/r$$

D
$$d\propto \dfrac {1}{r^{2}}$$

Solution

$$g=\cfrac{GM}{m^{2}}$$
Now, $$m=\cfrac{4}{3}\pi r^{3}d\quad $$ [$$d \rightarrow$$ density]
$$\therefore g=\cfrac{4G\pi r^{3}d}{3r^{2}}=\cfrac{4}{3}G\pi rd$$
$$\Rightarrow d=\cfrac{3g}{4G\pi r}$$
or, $$d \propto \cfrac{1}{R}$$
Question 10
1 / -0

The escape velocity of a body projected vertically upwards from the surface of the earth is 'v' if the body is projected at an angle of $$30^o$$ with the horizontal, the escape velocity would be

A
$$\dfrac{v}{2}$$

B
$$\dfrac{sqrt3}{2}v$$

C
$$2v$$

D
$$v$$

Solution

Escape velocity $${ V }_{ e }=\sqrt { 2gR } $$ is independent of the projection angle. So, the escape velocity will be $$v$$.

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Re-Attempt Weekly Quiz Competition

Gravitation Test - 47

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Gravitation Test - 47

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SHARING IS CARING

Question 1

$$d = h$$

$$d = 2h$$

$$d = \dfrac {3h}{2}$$

$$d = \dfrac {h}{2}$$

Solution

Question 2

$$35.4s$$

$$50.0s$$

$$70.7s$$

$$100.0s$$

Solution

Question 3

$$0.0112km{ s }^{ -1 }$$

$$0.112km{ s }^{ -1 }$$

$$11.2km{ s }^{ -1 }$$

$$0.56km{ s }^{ -1 }$$

Solution

Question 4

$$\cfrac { 1 }{ 2 } mgR$$

$$\cfrac { 1 }{ 4 } mgR$$

$$mgR$$

$$2mgR$$

Solution

Question 5

8 R

9 R

10 R

20 R

Solution

Question 6

$$2R$$

$$\displaystyle\frac{7}{3}R$$

$$3R$$

$$\displaystyle\frac{2}{3}R$$

Solution

Question 7

4R

5R

15R

3R

Solution

Question 8

$$\sqrt{R^{\ast}/d}$$

$$R^{\ast}/d$$

$$R^{\ast}R_p/d^2$$

$$(R^{\ast}/d)^4$$

Solution

Question 9

$$d\propto r$$

$$d\propto r^{1/2}$$

$$d\propto 1/r$$

$$d\propto \dfrac {1}{r^{2}}$$

Solution

Question 10

$$\dfrac{v}{2}$$

$$\dfrac{sqrt3}{2}v$$

$$2v$$

$$v$$

Solution

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