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CBSE Questions for Class 11 Engineering Physics Gravitation Quiz 8 - MCQExams.com
CBSE
Class 11 Engineering Physics
Gravitation
Quiz 8
The ratio of the acceleration due to gravity as the bottom of a deep mine and that on the surface of the earth is 978/Find the depth of the mine, if the density of the earth is uniform throughout and the radius of the earth is 6300 km.
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12.88 km
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13.0 km
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25.38 km
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90.9 km
Explanation
Ratio of acceleration due to gravity
$$ \dfrac { g'}{g} = \dfrac {978}{9800} = I - \dfrac {d}{R} $$
$$ \dfrac {d}{R} = I - \dfrac {978}{9800} $$
$$ d = \dfrac {2}{980} R $$
$$ = \dfrac {2 \times 6300}{980} $$
= 12.86 km
If g is the acceleration due to gravity at the Earths, surface, the gain of the potential energy of an object of mass m raised from the surface of the Earth to height equal to the radius R of the Earth is :
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$$\dfrac{mgR}{4}$$
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$$\dfrac{mgR}{2}$$
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$$mgR$$
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$$2 mgR$$
Explanation
Potential energy of an object at the surface of the Earth is
$$U_1=-\dfrac{GMm}{R}$$ ...(i)
Potential energy of the object at a height, h = R from the surface of the Earth.
$$U_2=-\dfrac{GMm}{R+h}=-\dfrac{GMm}{R+R}$$
Gain in potential energy
$$\triangle U=U_2-U_1=-\dfrac{GMm}{2R}+\dfrac{GMm}{R}=\dfrac{1}{2}\dfrac{GMm}{R}$$
Also, $$GM=gR^2$$
$$\triangle U=\dfrac{1}{2}\dfrac{gR^2m}{R}=\dfrac{mgR}{2}$$
You are to examine these two statements carefully and select the answers to these items using the code given below :
Statement I :
The acceleration due to gravity decreases with increase in height from the surface of the Earth.
Statement II :
The acceleration due to gravity is inversely proportional to the square of the distance from the centre of the Earth.
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Both the statements are individually true and Statement II is the correct explanation of Statement I
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Both the statements are individually true but Statement II is not the correct explanation of Statement I
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Statement I is true but Statement II is false
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Statement I is false but Statement II is true
Explanation
The acceleration due to gravity decreases with increases in height from the surface of the earth because it is universally proportional to the square of the distance from the center of the earth.
The depth $$'d'$$ at which the value of acceleration due to gravity becomes $$\dfrac {1}{n}$$ times the value at the earth's surface is $$(R =$$ radius of earth)
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$$d = R \left (\dfrac {n}{n - 1}\right )$$
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$$d = R \left (\dfrac {n - 1}{n}\right )$$
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$$d = R \left (\dfrac {n - 1}{2n}\right )$$
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$$d = R^{2} \left (\dfrac {n - 1}{n}\right )$$
Explanation
Acceleration due to gravity at a depth $$d$$ under the earth surface $$g' = g_s\bigg(1-\dfrac{d}{R}\bigg)$$
Given : $$g' = \dfrac{g_s}{n}$$
$$\therefore$$
$$\dfrac{g_s}{n} = g_s\bigg(1-\dfrac{d}{R}\bigg)$$
Or
$$\dfrac{1}{n} = \bigg(1-\dfrac{d}{R}\bigg)$$
$$\implies$$ $$d =R\bigg(\dfrac{n-1}{n}\bigg)$$
Let '$${ g }_{ h }$$' and '$${ g }_{ d }$$' be the acceleration due to gravity at height '$$h$$' above the earth's surface and at depth '$$d$$' below the earth's surface respectively. IF $${ g }_{ h }={ g }_{ d }$$ then the relation between '$$h$$' and '$$d$$' is
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$$d=h$$
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$$d=\dfrac { h }{ 2 } $$
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$$d=\dfrac { h }{ 4 } $$
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$$d=2h$$
Explanation
Acceleration due to gravity at height $$h$$ above the earth surface $$g_h = g(1-\dfrac{2h}{R})$$
where $$R$$ is the radius of earth.
Acceleration due to gravity at depth $$d$$ below the earth surface $$g_d = g(1-\dfrac{d}{R})$$
But $$g_h = g_d$$ implies
$$ g(1-\dfrac{2h}{R})=$$
$$g(1-\dfrac{d}{R})$$
$$\implies$$ $$d = 2h$$
If the escape speed of a projectile on Earth's surface is $$11.2{ kms }^{ -1 }$$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth :
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$$56.63{ kms }^{ -1 }$$
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$$33{ kms }^{ -1 }$$
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$$39{ kms }^{ -1 }$$
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$$31.7{ kms }^{ -1 }$$
Explanation
According to the principle of conservation of energy, we have
Initial kinetic energy $$+$$ Initial potential energy $$=$$ Final kinetic energy $$+$$ Final potential energy
$$\Rightarrow \dfrac { 1 }{ 2 } m{ v }^{ 2 }-\dfrac { GMm }{ R } =\dfrac { 1 }{ 2 } m{ v }^{ { \prime }^{ 2 } }+0$$
$$\Rightarrow \dfrac { 1 }{ 2 } m{ v }^{ { \prime }^{ 2 } }=\dfrac { 1 }{ 2 } m{ v }^{ 2 }-\dfrac { GMm }{ R } $$ ....(i)
Also consider, $${ v }_{ e }=$$ escape velocity
$$\dfrac { 1 }{ 2 } m{ v }_{ e }^{ 2 }=\dfrac { GMm }{ R } $$ .....(ii)
$$\therefore$$ From equations (i) and (ii), we get
$$\dfrac { 1 }{ 2 } m{ v }^{ { \prime }^{ 2 } }=\dfrac { 1 }{ 2 } m{ v }^{ 2 }-\dfrac { 1 }{ 2 } m{ v }_{ e }^{ 2 }$$ .....(iii)
$$\Rightarrow { v }^{ { \prime }^{ 2 } }={ v }^{ 2 }-{ v }_{ e }^{ 2 }$$
Now, $${ v }_{ e }=11.2{ kms }^{ -1 } $$ and $$v=3{ v }_{ e } $$ .....(iv)
From equations (iii) and (iv), we get
$${ v }^{ { \prime }^{ 2 } }={ \left( 3{ v }_{ e } \right) }^{ 2 }-{ v }_{ e }^{ 2' }$$
$${ v }^{ { \prime }^{ 2 } }=9{ v }_{ e }^{ 2 }-{ v }_{ e }^{ 2 }=8{ v }_{ e }^{ 2 }=8\times { \left( 11.2 \right) }^{ 2 }$$
$$v'^2=8\times { \left( 11.2 \right) }^{ 2 }$$
$$\Rightarrow { v }^{ \prime }=\sqrt { 8\times { \left( 11.2 \right) }^{ 2 } } =\sqrt { 8 } \times 11.2$$
$${ v }^{ \prime }=2\times 1.414\times 11.2=31.68{ kms }^{ -1 }$$
$$\therefore $$ Speed of the body far away from the Earth $${ v }^{ \prime }=31.68{ kms }^{ -1 }=31.7{ kms }^{ -1 }$$
A man weighs $$60kg$$ at earth surface. At what height above the earth's weight become $$30kg$$, Given radius of earth is $$6400km$$:
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$$2624km$$
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$$3000km$$
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$$2020km$$
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None of these
Explanation
Let acceleration due to gravity at a height $$h$$ above the surface of earth be $$g'$$
So, $$mg'=m\dfrac{gR^2}{(R+h)^2}$$
or, $$30=60\times\dfrac{(6400)^2}{(6400+h)^2}$$
So, $$h=6400\sqrt{2}-6400=2624km$$
The earth's radius is R and acceleration due to gravity at its surface is g. If a body of mass m is sent to a height $$\displaystyle h=\frac { R }{ 5 } $$ from the earth's surface, the potential energy increases by
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$$\displaystyle mgh$$
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$$\displaystyle \frac { 4 }{ 5 } mgh$$
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$$\displaystyle \frac { 5 }{ 6 } mgh$$
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$$\displaystyle \frac { 6 }{ 7 } mgh$$
Explanation
Increase in potential energy
$$\displaystyle \Delta U={ U }_{ 2 }-{ U }_{ 1 }$$
$$\displaystyle =\left[ -\frac { GMm }{ \left( R+\frac { R }{ 5 } \right) } \right] -\left[ -\frac { GMm }{ R } \right] $$
$$\displaystyle =\frac { GMm }{ R } -\frac { 5GMm }{ 6R } $$
$$\displaystyle =\frac { GMm }{ 6R } =\frac { g{ R }^{ 2 }m }{ 6R } =\frac { mgR }{ 6 } $$
$$\displaystyle =mg.\frac { 5h }{ 6 } =\frac { 5 }{ 6 } mgh$$
Weight of a body of mass m decreases by $$1\%$$ when it is raised to height h above the Earth's surface. If the body is taken to a depth h in a mine, then its weight will:
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Decreases by $$0.5\%$$
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Decreases by $$2\%$$
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Increases by $$0.5\%$$
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Increase by $$1\%$$
Explanation
Hint :- Gravitational acceleration g varies decreasingly with height as well as depth.
Step 1: Calculate height h
g = $$\dfrac{GM}{r^{2}}$$
Then g$$_{h}$$ = $$\dfrac{GM}{(r + h)^{2}}$$
$$\dfrac{g}{g_{h}}$$ = $$\dfrac{100}{99}$$ = ($$\dfrac{r + h}{r}$$)$$^{2}$$
$$\dfrac{r + h}{r}$$ = 1.005
r + h = 1.005r
h= 0.005r
Step 2: Calculate % variation in g with given depth
g$$_{d}$$ = g(1 - $$\dfrac{d}{r}$$)
d =h= 0.005r
g$$_{d}$$ = 1 - 0.005 = 0.995
Thus the weight will be decreased by 0.5%.
Which one of the following statements is correct?
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Acceleration due to gravity decreases with the increase of altitude
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Acceleration due to gravity increase with the increase of depth (assuming earth to be a sphere of uniform density)
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Acceleration due to gravity decreases with the increase of latitude
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Acceleration due to gravity is independent of the mass of the earth
The ratio of acceleration due to gravity at a height h above the surface of the earth and at a depth h below the surface of the earth h < radius of earth
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Is constant
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Increases linearly with h
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Decreases linearly with h
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Decreases parabolically with h
Explanation
Acceleration due to gravity height h,
$$\displaystyle g_1= g \left( 1-\frac{2h}{R} \right)$$
Acceleration due to gravity at depth h,
$$\displaystyle g_2= g \left( 1-\frac{h}{R} \right)$$
$$\therefore$$ $$\displaystyle \frac{g_1}{g_2} =\frac{1-2h/R}{1-h/R}$$
$$\displaystyle = \left( 1-\frac{2h}{R} \right) \left( 1-\frac{h}{R} \right)^{-1}$$
$$\displaystyle =\left( 1-\frac{h}{R} \right)$$
[neglecting higher power of $$\displaystyle \frac{h}{R}$$]
$$\displaystyle \therefore \frac{g_1}{g_2}$$ decreases linearly with h.
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?
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$$d = h$$
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$$d = 2h$$
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$$d = \dfrac {3h}{2}$$
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$$d = \dfrac {h}{2}$$
Explanation
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 $$d = 2h$$.
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)
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$$35.4s$$
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$$50.0s$$
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$$70.7s$$
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$$100.0s$$
Explanation
$$g=\cfrac { GM }{ { r }^{ 2 } } $$
$$\quad g'=\cfrac { G2M }{ { (2r) }^{ 2 } } =\cfrac { 2GM }{ { 4r }^{ 2 } } $$
$$\quad g'=\cfrac { g }{ 2 } \quad \cfrac { H }{ H } =\cfrac { \cfrac { 1 }{ 2 } { g }_{ e }{ t }_{ e }^{ 2 } }{ \cfrac { 1 }{ 2 } { g }_{ p }{ t }_{ p }^{ 2 } } $$
$${ g }_{ p }\times { t }_{ p }^{ 2 }={ g }_{ e }\times { t }_{ e }^{ 2 }\quad \quad $$
$$\cfrac { { g }_{ e } }{ 2 } { t }_{ p }^{ 2 }={ g }_{ e }\times 50\times 50$$
$${ t }_{ p }^{ 2 }=2\times 50\times 50\Rightarrow { t }_{ p }=\sqrt { 5000 } =70.7sec$$
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
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$$0.0112km{ s }^{ -1 }$$
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$$0.112km{ s }^{ -1 }$$
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$$11.2km{ s }^{ -1 }$$
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$$0.56km{ s }^{ -1 }$$
Explanation
Escape velocity is independent of the mass of the projected body provided the air resistance is neglected. Hence, the escape velocity of
10
k
g
10kg
of the iron ball projected from the earth is same that of escape velocity of
10
g
10g
projected from the earth i.e.,
11.2
k
m
s
−
1
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
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$$\cfrac { 1 }{ 2 } mgR$$
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$$\cfrac { 1 }{ 4 } mgR$$
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$$mgR$$
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$$2mgR$$
Explanation
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)
$$\Delta U=\cfrac { g{ R }^{ 2 } }{ 2R } =\cfrac { 1 }{ 2 } mgR$$
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)
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8 R
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9 R
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10 R
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20 R
Explanation
$$g'=\frac{g}{\left ( 1+\frac{h}{R} \right )^2}\Rightarrow \frac{g}{100}=\frac{g}{\left ( 1+\frac{h}{R} \right )^2}$$
$$\left ( 1+\frac{h}{R} \right )^2=100$$
$$h=9\, R$$
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)
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$$2R$$
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$$\displaystyle\frac{7}{3}R$$
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$$3R$$
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$$\displaystyle\frac{2}{3}R$$
Explanation
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),
$$\displaystyle\frac{0.09g}{g}=\left(\displaystyle\frac{1}{\displaystyle 1+\frac{h}{r}}\right)^2$$
$$\left(\displaystyle 1+\frac{h}{R}\right)^2=\displaystyle\frac{1}{0.09}$$
$$\Rightarrow \displaystyle 1+\frac{h}{r}=\frac{1}{0.3}$$
$$\Rightarrow \displaystyle \frac{1}{R}=\frac{10}{3}-1=\frac{7}{3}$$
$$\displaystyle h=\frac{7}{3}R$$.
The height at which the weight of a body becomes 1/16th its weight on the surface of earth (Radius R ) is:
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4R
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5R
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15R
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3R
Explanation
$$\cfrac { GMm }{ { \left( R+h \right) }^{ 2 } } =\cfrac { 1 }{ 16 } \cfrac { GMm }{ { R }^{ 2 } } \\ \cfrac { 1 }{ { { \left( R+h \right) }^{ 2 } } } =\cfrac { 1 }{ 16{ R }^{ 2 } } \\ or,\quad { \left( \cfrac { R }{ R+h } \right) }^{ 2 }=\cfrac { 1 }{ 16 } \\ \Rightarrow \cfrac { R }{ R+h } =\cfrac { 1 }{ 4 } .\\ \therefore R+h=4R\\ \Rightarrow h=3R$$
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.
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$$\sqrt{R^{\ast}/d}$$
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$$R^{\ast}/d$$
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$$R^{\ast}R_p/d^2$$
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$$(R^{\ast}/d)^4$$
Explanation
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 $$\propto$$ $$(fT^*)^4$$
$$\Rightarrow$$ $$f^4 \propto \dfrac{ {R^*}^2}{d^2}$$
$$\Rightarrow f \propto (\dfrac{ {R^*}^2}{d^2})^{\dfrac{1}{4}}= \sqrt{\dfrac{R^*}{d}}$$
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
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$$d\propto r$$
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$$d\propto r^{1/2}$$
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$$d\propto 1/r$$
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$$d\propto \dfrac {1}{r^{2}}$$
Explanation
$$g=\cfrac{GM}{m^{2}}$$
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, $$d \propto \cfrac{1}{R}$$
On a planet where $${ g }_{ planet }=0.2{ g }_{ earth }$$. What will be the difference in the height of column filled with mercury in a closed end manometer when the gas is filled withe pressure of $$2atm$$ on earth (Assuming:outside pressure to be $$1atm$$ on both planet; Volume of gas remain constant)
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$$30.4cm$$
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$$760cm$$
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$$380cm$$
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$$152cm$$
Explanation
$$P=egh$$
$$2atm=760\times2$$ mm of hg
$$\Rightarrow\cfrac{P_1}{P_2}=\cfrac{e_1g_1h_1}{e_2g_2h_2}\\ \because P_1=P_2\& e_1=e_2\Rightarrow\cfrac{h_2}{h_1}=\cfrac{g_1}{g_2}\\ \quad\cfrac{h_2}{760\times2}=\cfrac{g}{0.2g}\\ \quad\Rightarrow h_2=\cfrac{760\times2}{0.2}=7600mm\\ \quad\quad=760cm$$
At what altitude will the acceleration due to gravity be $$25$$% of that at the earth's surface (given radius of earth is $$R$$)?
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$$R/ 4$$
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$$R$$
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$$3R/ 8$$
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$$R/ 2$$
A particle of mass M is situated at the centre of spherical shell of same mass and radius a. The gravitational potential at a point situated at a/2 distance from the centre will be
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$$-\dfrac { 3\ GM }{ a } $$
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$$-\dfrac { 2\ GM }{ a } $$
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$$-\dfrac { GM }{ a } $$
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$$-\dfrac { 4GM }{ a } $$
Explanation
Let point $$A$$ is situated at $$a/2$$ distance from the center.
Gravitational Potential at point $$A$$ due to spherical shell, $$V_1=-\dfrac{GM}{a}$$,
Gravitational Potential at point $$A$$ due to mass $$M$$ at point $$O$$ , $$V_2=-\dfrac{GM}{(a/2)}=-\dfrac{2GM}{a}$$,
$$\therefore $$ Total Gravitational Potential $$=V_1+V_2= -\dfrac{GM}{a} -\dfrac{2GM}{a}=-\dfrac{3GM}{a}$$
Suppose(God forbid) due to some reason, the earth expands to make its volume eight-fold. What you expect your weight to be?
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Two-fold
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One-half
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One-fourth
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Unaffected
Explanation
we know , gravity is
g= $$\dfrac{GM}{r^2}$$
the radius of new one will become r'=2r
then, g'=$$\dfrac{GM}{4r^2}$$
The height from earth's surface at which acceleration due to gravity becomes $$\dfrac {g}{4}$$ is (where $$g$$ is acceleration due to gravity on the surface of earth and $$R$$ is radius of earth).
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$$\sqrt {2}R$$
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$$R$$
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$$\dfrac {R}{\sqrt {2}}$$
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$$2R$$
Explanation
We know that $$g=\dfrac{GM_e}{R^2}$$
And at height h from the earth surface $$ g_1=\dfrac{GM_e}{(R+h)^2}$$
And here $$g_1=\dfrac{g}{4}$$
So, $$ \dfrac{g}{4}=\dfrac{GM_e}{(R+h)^2}$$
so, $$\dfrac{GM_e}{4R^2}=\dfrac{GM_e}{(R+h)^2}$$
Hence, $$\dfrac{1}{4R^2}=\dfrac{1}{(R+h)^2}$$
$$ \implies\dfrac{(R+h)^2}{R^2}=4$$
$$\implies\dfrac{R+h}{R}=2$$
$$\implies R+h=2R$$
$$\implies R=h$$
Hence, at height equal to R from earth surface gravity will become $$\dfrac{g}{4}$$.
Answer-(B)
'G' represents
I. Acceleration due to gravity.
II. Weight
III. Gravitational constant
Which combination is correct?
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II and III only
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I and III only
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III only
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I, II and III
Explanation
$$G$$ denotes gravitational constant where as $$g$$ denotes acceleration due to gravity and weight is given by $$W=mg$$
Architects define a physical structure as a set of materials arranged in such a way that these materials can the downward pull of gravity.
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Mimic
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Resist
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Amplify
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Dislodge
A smooth and bottomless tunnel is dug through the centre of earth. A particle is released from the surface of earth into the tunnel. Time to reach centre of tunnel is (approximately) equal to (where $$R =$$ Radius of earth).
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$$56.4\ minute$$
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$$84.35\ minute$$
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$$42.3\ minute$$
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$$28.2\ minute$$
Explanation
Force on particle at any $$x$$ is $$\dfrac{GM'm}{x^2}$$
But $$\dfrac{M'}{M}=\dfrac{x^3}{R^3}$$
$$\dfrac{GM x^3 m}{R^3 x^2}=(\dfrac{GMm}{R^3})x$$
Acceleration of particle, $$a=\dfrac{GM}{R^3}x=\dfrac{g}{R}x$$
It is an SHM motion, $$T=2 \pi \sqrt{\dfrac{R}{g}}$$
Time to reach center of earth
$$=\pi \sqrt{\dfrac{R}{g}}$$
$$=3.14 \times \sqrt{\dfrac{6400 \times 10^3}{9.81}}=42.283$$ minutes
The areal velocity and the angular momentum of the planet are related by which of the following relations?
(where $${m}_{p}$$ is the mass of the planet)
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$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{\overset{-}{L}}{{2m}_{p}}$$
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$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{\overset{-}{L}}{{m}_{p}}$$
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$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{2\overset{-}{L}}{{m}_{p}}$$
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$$\frac{\Delta\overset{-}{A}}{\Delta{t}}=\frac{\overset{-}{L}}{\sqrt{2m}_{p}}$$
Explanation
Lets consider,
$$m_p=$$ mass of the planet.
The angular momentum, $$L=m_pvr$$
$$vr=\dfrac{ L}{m_p}$$. . .. . .(1)
Time period, $$t=\dfrac{2\pi r}{v}$$
Total area swapped in time $$t$$ , $$A=\pi r^2$$
Areal velocity, $$\dfrac{A}{t}=\dfrac{\pi r^2}{2\pi r/ v}=\dfrac{vr}{2}$$
$$\dfrac{A}{t}=\dfrac{L}{2m_p}$$ (from equation 1)
In vector form, areal velocity is given by
$$\dfrac{\vec A}{t}=\dfrac{\vec L}{2m_p}$$
The correct option is A.
A hole is drilled along the earth's diameter and a stone is dropped into it. When the stone is at the centre of the earth, it has.
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Acceleration
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Weight
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Mass
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Potential energy
Explanation
At the center of the earth everything is weightless as acceleration and potential energy vanish. So it has only mass.
From a solid sphere of mass $$M$$ and radius $$R$$ a spherical portion of radius $$\dfrac {R}{2}$$ is removed, as shown in the figure. Taking gravitational potential $$V = 0$$ at $$r = \infty$$, the potential at the centre of the cavity thus formed is : $$(G =$$ gravitational constant).
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$$\dfrac {-2GM}{3R}$$
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$$\dfrac {-2GM}{R}$$
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$$\dfrac {-GM}{2R}$$
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$$\dfrac {-GM}{R}$$
Explanation
By superposition principle, $$v_1=\dfrac{-GM}{2R^3}[3R^2-(\dfrac{R}{2})^2]$$
$$=-\dfrac{11GM}{8R^3}$$
Also, $$v_2=-\dfrac{3}{2} \dfrac{G (M/8)}{(R/2)}=\dfrac{-3GM}{8R}$$
The required potential is, $$v=v_1-v_2$$
$$=-\dfrac{11GM}{8R}-(-\dfrac{3GM}{8R})$$
$$V=-\dfrac{GM}{R}$$
ACCELERATION DUE TO GRAVITY BELOW AND ABOVE THE SURFACE OF EARTH
Which of the following statements is correct?
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Acceleration due to gravity increases with increasing altitude.
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Acceleration due to gravity increases with increasing depth.
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Acceleration due to gravity increases with increasing latitude.
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Acceleration due to gravity is independent of the mass of the earth.
Explanation
Acceleration due to gravity at a altitude h above the earth's surface is $$g_{h}=\dfrac{gR_{E}^{2}}{\left(R_{E}+{h}\right)^{2}}$$ ...(i)
where g is the acceleration due to gravity on the earth's surface and R_{E} is the radius of the earth.
Eq. (i) shows that acceleration due to gravity decreases with increasing altitude.
Acceleration due to gravity at a depth d below the earth's surface is $$g_{d}=g\left(1-\dfrac{d}{R_{E}}\right)$$ ....(ii)
Eq. (ii) shows that acceleration due to gravity decreases with increasing depth.
Acceleration due to gravity at latitude $$\lambda$$
$$g\lambda=g-R_{E}\omega^{2}\cos^{2}\lambda$$ .....(iii)
where $$\omega$$ is the angular speed of rotation of the earth.
Eq.(iii) shows that acceleration due to gravity increases with increasing latitude.
Acceleration due to gravity of body of mass m is placed n the earth's surface is $$g=\dfrac{GM_{E}}{R_{E}^{2}}$$ ....(iv)
Eq.(iv) shows that acceleration due to gravity is independent of the mass of the body but it depends upon the mass of the earth.
ESCAPE SPEED
If $$v_{e}$$ is escape velocity and $$v_{o}$$ is orbital velocity of a satellite for orbit close to the earth's surface. Then these are related by:
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$$v_{o}=\sqrt{2}v_{e}$$
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$$v_{o}=v_{e}$$
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$$v_{e}=\frac{v_{o}}{2}$$
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$$v_{e}=\sqrt{2}v_{o}$$
Explanation
Escape velocity, $$v_{e}=\sqrt{\dfrac{2GM_{E}}{R_{E}}}$$ ...(i)
where $$M_{E}$$ and $$R_{E}$$ be the mass and radius of the earth respectively.
The orbital velocity of a satellite close to the earth's surface is $$v_{o}=\sqrt{\dfrac{GM_{E}}{R}}$$ ....(ii)
From (i) and (ii), we get $$v_{e}=\sqrt{2}v_{o}$$
ESCAPE SPEED
The escape speed velocity of a body from the earth depends on
(i) the mass of the body
(ii) the location from where it is projected.
(iii) the direction of projection
(iv) the height of the location from where the body is launched.
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(i) and (ii)
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(ii) and (iv)
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(i) and (iii)
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(iii) and (iv)
Explanation
The escape velocity is independent of mass of the body and the direction of projection. It depends upon the gravitational potential at the point from where the body is launched. Since this potential depends slightly on the latitude and height of the point, the escape velocity depends slightly on these factors.
Which of the following statements is correct regarding the universal gravitational constant G?
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G has same value in all systems of units.
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The value of G is same everywhere in the universe.
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The value of G was first experimentally determined by Johannes Kepler.
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G is a vector quantity
Explanation
G has different value in different system of units.
In SI system the value of G is $$6.67\times 10^{-11}N m^{2} kg^{-2}$$ whereas in CGS its value is $$6.67\times 10^{-8}dyne cm^{2} g^{-2}$$.
The value of G is same throughout the universe.
The value of G was first experimentally determined by English scientist Henry Cavendish.
G is a scalar quantity.
The escape velocity from the surface of the earth is (where $$R_{E}$$ is the radius of the earth)
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$$\sqrt{2gR_{E}}$$
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$$\sqrt{gR_{E}}$$
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$$2\sqrt{gR_{E}}$$
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$$\sqrt{3gR_{E}}$$
Explanation
The escape velocity from the surface of the earth is
$$v_e=\sqrt{\dfrac{2GM}{R_E}}$$. . . . .(1)
Acceleration due to gravity,
$$g=\dfrac{GM}{R_E^2}$$
$$\dfrac{GM}{R_E}=gR_E$$. . . . . . . . . .(2)
Substitute equation (2) in equation (1), we get
$$v_e=\sqrt{2gR_E}$$
The correct option is A.
The dependence of acceleration due to gravity g on the distance r from the centers of the earth assumed to be a sphere of radius R of uniform density is as shown figure below.
The correct figure is
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(i)
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(ii)
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(iii)
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(iv)
Explanation
The acceleration due to gravity at a depth d below the surface of earth is $${g}^{\prime}=g\left(1-\dfrac{d}{R}\right)=g\left(\dfrac{R-d}{R}\right)=g\dfrac{r}{R}$$ .....(i)
where R-d=r=distance of location from the centre of the earth. When r=0, $$g^{\prime}=0$$
From (i), $$g\propto{r}$$ till R=r, for which $$g^{\prime}=g$$
For $$r>R, g^{\prime}=\dfrac{gR^{2}}{\left(R+h\right)^{2}}=\dfrac{gR^{2}}{r^{2}}$$ or $$g^{\prime}\propto\dfrac{1}{r^{2}}$$
Here, $$ R + h = r $$
Therefore, the variation of g with distance r from centre of earth will be as shown in figure (4). Thus, option (d) is correct.
The acceleration due to gravity $$g$$ and density of the earth $$\rho$$ are related by which of the following relations.? (where G is the gravitational constant and $${R}_{E}$$ is the radius of the earth)
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$$\rho=\dfrac{4\pi GR_{E}}{3g}$$
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$$\rho=\dfrac{3g}{4\pi GR_{E}}$$
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$$\rho=\dfrac{3G}{4\pi gR_{E}}$$
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$$\rho=\dfrac{4\pi gR_{E}}{3G}$$
Explanation
Acceleration due to gravity on earth is $$g=\dfrac{GM_{E}}{R_{E}^{2}}$$ ....(i)
As
$$\rho=\dfrac{M_{E}}{\dfrac{4}{3}\pi R_{E}^{3}} \Rightarrow {M}_{E}=\rho\dfrac{4}{3}\pi {R_{E}^{3}}$$
Substituting this value in Eq. (i), we get
$$g=\dfrac{G\left(\rho\dfrac{4}{3}\pi {R_{E}^{3}}\right)}{{R}_{E}^{2}}=\dfrac{4}{3}\pi\rho GR_{E}$$
or $$\rho=\dfrac{3g}{4\pi GR_{E}}$$
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)
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9
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10
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11
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8
Explanation
$$T=\dfrac{2u \sin \theta}{g} \therefore \dfrac{T_1}{T_2}=\dfrac{g_2}{g_1}=\dfrac{g+\dfrac{g}{10}}{g}=\dfrac{11}{10}$$
Fractional decrease in time of flight =$$\dfrac{T_1-T_2}{T_1}=\dfrac{1}{11}$$
Percentage decrease = $$9\%$$
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?
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R
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$$2$$R
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$$3$$R
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$$4$$R
The escape velocity for a body projected vertically upwards from the surface of the earth is $$11.2$$ km $$s^{-1}$$. If the body is projected in a direction making an angle $$45^o$$ with the vertical, the escape velocity will be:
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$$\dfrac{11.2}{\sqrt{2}}$$km $$s^{-1}$$
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$$11.2\times \sqrt{2}$$km $$s^{-1}$$
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$$11.2\times 2$$km $$s^{-1}$$
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$$11.2$$km $$s^{-1}$$
Explanation
$$V_e=\sqrt{2gr}$$
It doesn't depend upon direction of projection.
In order to shift a body of mass m from a circular orbit of radius $$3R$$ to a higher orbit of radius $$5R$$ around the earth, the work done is?
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$$\dfrac{3GMm}{5R}$$
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$$\dfrac{GMm}{2R}$$
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$$\dfrac{2}{15}\dfrac{GMm}{R}$$
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$$\dfrac{GMm}{5R}$$
Explanation
Work done is shifting a body from an orbit of radius $$3R$$ to $$5R$$ is equal to change in potential energy of the body.
Potential energy $$=\dfrac { -{ GM }{ m } }{ r } $$
where $$M$$ is mass of the earth $$m$$ is mass of satellite $$r$$ is radius of orbit
$$W=$$ Final potential energy $$-$$ Initial potential energy
$$W=\dfrac { { -GM }{ m } }{ 5R } -\left( \dfrac { { -GM }{ m } }{ 3R } \right) $$
$$=\dfrac { { GM }{ m } }{ R } \left( \dfrac { 1 }{ 3 } -\dfrac { 1 }{ 5 } \right) =\dfrac { { 2GM }{ m } }{ 15R } $$
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.
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0.7
0%
0.5
0%
1.2
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cant say
Explanation
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 $$=\dfrac{mg-mg'}{mg}$$
$$=\dfrac{d}{R}\times 100=\dfrac{32}{6400}\times 100=0.5\%$$.
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 $$s^{-2}$$
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$$360$$J.
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$$160$$J.
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$$560$$J.
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$$460$$J.
Explanation
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.
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. $$G=6.66\times 10^{-11}Nm^2kg^{-2}$$.
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$$3.82ms^{-2}$$.
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$$9.82ms^{-2}$$.
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$$19.82ms^{-2}$$.
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$$2ms^{-2}$$.
Explanation
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, $$g=\dfrac{GM}{R^2}=\dfrac{G}{R^2}\times \dfrac{4}{3}\pi R^3\times \rho$$
$$=\dfrac{4}{3}\pi GR\rho$$
$$=\dfrac{4}{3}\times \dfrac{22}{7}\times 6.66\times 10^{-11}\times 6.4\times 10^6\times 5.5\times 10^3$$
$$=9.82ms^{-2}$$.
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?
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$$\dfrac{r^3\omega_0}{R}$$
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$$\dfrac{r^3\omega^2_0}{R^2}$$
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$$\dfrac{r^3\omega^2_0}{R}$$
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$$\dfrac{r^2\omega^2}{R^2}$$
Explanation
We know $$F=Ma$$
$$\dfrac { { GM }_{ m } }{ { r }^{ 2 } } =\dfrac { { mw }^{ 2 } }{ r } $$
$$GM={ w }^{ 2 }{ r }^{ 3 }$$
$$F=Ma$$
$$a=\dfrac { F }{ M } $$
$$a=\dfrac { GM }{ { R }^{ 2 } } $$
$$G=\dfrac { { w }^{ 2 }{ r }^{ 3 } }{ { R }^{ 2 } } $$
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?
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mgh
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$$\dfrac{4}{5}$$mgh
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$$\dfrac{5}{6}$$mgh
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$$\dfrac{6}{7}$$mgh
Which of the following are correct?
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Out of electrostatic, electromagnetic, nuclear and gravitational interactions, the gravitational interaction is the weaker
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If the earth were to rotate faster than its present speed, the weight of an object would decrease at the equator but remain unchanged at the poles
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The mass of the earth in terms of g, R and G is $$(gR^2/G)$$
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If the earth stops rotating in its orbit around the Sun there will be no variation in the weight of a body on the surface of earth
Velocity of the planet is minimum at
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$$C$$
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$$D$$
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$$A$$
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$$B$$
Explanation
As, here gravitational force is the only acting force which always passes through centre i.e $$C$$ about centre of sun $$=0$$.
So, angular momentum about centre of sun $$=$$ conserved
i.e $$MVr=$$ constant
$$\Rightarrow$$ Velocity will be minimum when $$'r'$$ will be maximum
$$'r'$$ is maximum at $$D$$.
$$\Rightarrow$$ $$'V'$$ will be minimum at $$D$$.
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 $$m/{s}^{2}$$).
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$$8\ m{ s }^{ -2 }$$
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$$4\ m{ s }^{ -2 }$$
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$$2\ m{ s }^{ -2 }$$
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$$1\ m{ s }^{ -2 }$$
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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