Q41JEE Main 2024MCQ

At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R.) [29-Jan-2024 Shift 1]

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📖 Explanation

$\;g_p=\;\frac{g^2}{(R+h)^2}$ $\;g_q=g (1-\;\frac{h}{R})$ $\;g_p=g_q$ $\;\;\frac{g}{ (1+\;\frac{h}{R}) ^2}=g (1-\;\frac{h}{R})$ $\; (1-\;\frac{h^2}{R^2}) (1+\;\frac{h}{R}) =1$ $\;\;\text{ Take }\; \;\frac{h}{R}=x$ So $\;x^{3-}x+x^2=0$ $\;x=\;{{\sqrt{5}}-1}/{2}$ $\;h=\;\frac{R}{2}({\sqrt{5}}-1)$

Q42JEE Main 2024MCQ

If ${\text{G}}$ be the gravitational constant and ${\text{u}}$ be the energy density then which of the following quantity have the dimension as that the $\sqrt{ {\text{uG}}}$ :

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📖 Explanation

$\;{[ {\text{uG}}]= \$ [ ( {\text{M}}^1 {\text{L}}^{-1} {\text{T}}^{-2}) ( {\text{M}}^{-1} {\text{L}}^3 {\text{T}}^{-2}) ]$ };{[ {\text{uG}}]= $[ {\text{M}}^0 {\text{L}}^2 {\text{T}}^{-4}]$ };{[\sqrt{ {\text{uG}}}]= $[ {\text{L}}^1 {\text{T}}^{-2}]$ }$Option (2) is correct

Q43JEE Main 2023MCQ

The weight of a body on the surface of the earth is $100 {\text{N}}$ . The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is : [6-Apr-2023 shift 2]

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📖 Explanation

using newton's formula $F=\;\frac{G M m}{r^2}$ at surface of earth, $100=\;{ {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{Re}}^2}$ . . . (1) $\;\;\text{ at }\; {\text{r}}= {\text{R}}_{ {\text{e}}}+\;{ {\text{R}}_{ {\text{e}}}}/{4}=\;\frac{5}{4} {\text{R}}_{ {\text{e}}}$ $\; {\text{F}}^{'}=\;{ {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ (\;\frac{5}{4} {\text{R}}_{ {\text{e}}}) ^2}=\;\frac{16}{25} \times \;{ {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{R}}_{ {\text{e}}}{ }^2}$ $\; {\text{F}}^{'}=\;\frac{16}{25} \times 100=64 {\text{N}}$

Q44JEE Main 2023MCQ

If the distance of the earth from Sun is $1.5 \times 10^6$ ${\text{km}}$ . Then the distance of an imaginary planet from Sun, if its period of revolution is $2.83$ years is: [24-Jan-2023 Shift 2]

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📖 Explanation

$\; {\text{T}}^2 \propto {\text{R}}^3 \Rightarrow (\;{ {\text{T}}_1}/{ {\text{T}}_2}) ^2= (\;{ {\text{R}}_1}/{ {\text{R}}_2}) ^3$ $\;\Rightarrow (\;\frac{1}{2.83}) ^2= (\;{1.5 \times 10^6}/{ {\text{R}}_2}) ^3$ $\;\Rightarrow {\text{R}}_2= \$ [(2.83)^2 × (1.5 × 10^6) ^3]$ ^{1 / 3};=8^{1 / 3} \times 1.5 \times 10^6=3 \times 10^6 {\text{km}}$

Q45JEE Main 2023MCQ

A body is released from a height equal to the radius (r) of the earth. The velocity of the body when it strikes the surface of the earth will be :(Given ${\text{g}}=$ acceleration due to gravity on the earth.) [15-Apr-2023 shift 1]

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📖 Explanation

By energy conservation, $\; {\text{K}}_1+ {\text{U}}_1= {\text{K}}_2+ {\text{U}}_2$ $\; {\text{O}}-\;{ {\text{GMm}}}/{2 {\text{R}}}=\;\frac{1}{2} {\ mv}^{2-}\;{ {\text{GMm}}}/{ {\text{R}}}$ $\; {\text{V}}=\sqrt{\;{ {\text{GM}}}/{ {\text{R}}} \times \;{ {\text{R}}}/{ {\text{R}}}}$ $\; {\text{V}}=\sqrt{ {\text{gR}}}$

Q46JEE Main 2023MCQ

The weight of a body on the earth is $400 {\text{N}}$ . Then weight of the body when taken to a depth half of the radius of the earth will be: [8-Apr-2023 shift 1]

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📖 Explanation

Weight on the earth surface $=m g$ ${\text{mg}}=400 {\text{N}}$ (given)Weight at a depth ${\text{d}} {\text{w}}= {\text{m}} (\;{ {\text{GM}}( {\text{R}}- {\text{d}})}/{ {\text{R}}^3})$ $\; {\text{W}}= {\text{mg}} (1-\;{ {\text{d}}}/{ {\text{R}}})$ $\; {\text{d}}=\;{ {\text{R}}}/{2} \Rightarrow {\text{w}}= {\text{mg}} (1-\;\frac{1}{2}) \Rightarrow {\text{w}}=\;{ {\text{mg}}}/{2}$ $\; {\text{w}}=200 {\text{N}}$

Q47JEE Main 2023MCQ

The orbital angular momentum of a satellite is ${\text{L}}$ , when it is revolving in a circular orbit at height ${\text{h}}$ from earth surface. If the distance of satellite from the earth center is increased by eight times to its initial value, then the new angular momentum will be- [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{L}}= {\ mv} {\text{v}}_0 {\text{r}} ( {\text{v}}_0=\sqrt{\;{ {\text{GM}}}/{ {\text{h}}}})$ $\; {\text{L}}= {\text{m}} \sqrt{ {\text{GMh}}}$ $\; {\text{h}}^{'} \to {\text{h}}+8 {\text{h}}=9 {\text{h}}$ . . . (1) $\; {\text{L}}^{'}= {\text{m}} \sqrt{ {\text{GM}9h}}$ . . . (2) $\;\;{ {\text{L}}^{'}}/{ {\text{L}}^{'}}=3$ $\; {\text{L}}^{'}=3 {\text{L}}$

Q48JEE Main 2023MCQ

If the gravitational field in the space is given as $(-\;{ {\text{K}}}/{ {\text{r}}^2})$ . Taking the reference point to be at ${\text{r}}=2 {\text{cm}}$ with gravitational potential ${\text{V}}=10 {\text{J}} / {\text{kg}}$ . Find the gravitational potential at ${\text{r}}=3 {\text{cm}}$ in ${\text{SI}}$ unit (Given, that ${\text{K}}=6 {\text{J}} {\text{cm}} / {\text{kg}}$ ) [30-Jan-2023 Shift 1]

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📖 Explanation

$\;-\;{ {\text{dV}}}/{ {\text{dr}}}=-\;{ {\text{k}}}/{ {\text{r}}^2} \Rightarrow ∫_{10}^{ {\text{V}}} {\text{dV}}=∫_{2}^{3} \;{ {\text{k}}}/{ {\text{r}}^2} {\text{dr}}$ $\; {\text{V}}-10= {\text{k}} [\;\frac{1}{2}-\;\frac{1}{3}]$ $\; {\text{V}}-10=\;{ {\text{k}}}/{6} \Rightarrow {\text{V}}=11 \;\text{ volts }\;$

Q49JEE Main 2023MCQ

For a body projected at an angle with the horizontal from the ground, choose the correct statement. [1-Feb-2023 Shift 2]

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📖 Explanation

At highest point $\; {\text{V}}_{ {\text{y}}}=0$ $\; {\text{V}}_{ {\text{x}}}= {\text{u}}_{ {\text{x}}}= {\text{u}} \cos \theta$ $\; {\text{U}}_{ {\text{g}}}= {\text{mgh}}, \;\text{ it is maximum at }\; {\text{H}}_{\max } .$

Q50JEE Main 2023NAT

If the earth suddenly shrinks to $\;\frac{1}{64}$ th of its original volume with its mass remaining the same, the period of rotation of earth becomes $\;\frac{24}{x} h$ . The value of ${\text{x}}$ is _______. [10-Apr-2023 shift 1]

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📖 Explanation

By AMC $\;\;\frac{2}{5} {\text{MR}}^2 \omega _1^2=\;\frac{2}{5} {\text{M}} (\;{ {\text{R}}}/{4}) ^2 \omega _2$ $\;\;\frac{\omega _1}{\omega _2}=\;\frac{1}{16}=\;\frac{T_2}{T_1}=\;\frac{T_2}{24}$ $\;T_2=\;\frac{24}{16} \;\; \therefore {\text{x}}=16 \;\text{ Ans. }\;$

Q51JEE Main 2023MCQ

A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $( {\text{R}}_{ {\text{e}}})$ , the increase in potential energy will be : ( $g=$ acceleration due to gravity on the surface of Earth) [25-Jan-2023 Shift 2]

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📖 Explanation

${\text{U}}=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{r}}}$ ${\text{U}}_{ {\text{i}}}=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{R}}_{ {\text{e}}}}$ ${\text{U}}_{ {\text{f}}}=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ ( {\text{R}}_{ {\text{e}}}+ {\text{h}}) }=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{R}}_{ {\text{e}}}+2 {\text{R}}_{ {\text{e}}}}$ $\;{- {\text{GM}}_e {\text{m}}}/{3 {\text{R}}_{ {\text{e}}}}$ Increase in internal energy $\Delta {\text{U}}= {\text{U}}_{ {\text{f}}}- {\text{U}}_{ {\text{i}}}$ $\;=\;\frac{2}{3} \;{ {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{R}}_{ {\text{e}}}}$ $\;\;\frac{2}{3} \;{ {\text{GM}}_{ {\text{e}}}}/{ {\text{R}}_{ {\text{e}}}^2} {\text{mR}}_{ {\text{e}}}$ $\;=\;\frac{2}{3} {\text{mgR}}_{ {\text{e}}}$

Q52JEE Main 2023MCQ

Given below are two statements:Statement I : If $E$ be the total energy of a satellite moving around the earth, then its potential energy will be $\;\frac{E}{2}$ Statement II : The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy E.In the light of the above statements, choose the most appropriate answer from the options given below [8-Apr-2023 shift 1]

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📖 Explanation

For satellite K.E. $=\;\frac{1}{2} {\ mv}^2=\;\frac{1}{2} {\text{m}} (\sqrt{\;{ {\text{GM}}}/{ {\text{r}}}}) ^2$ ${\text{K}} . {\text{E}} .=\;{ {\text{GMm}}}/{2 {\text{r}}}$ Potential energy $U=-\;{ {\text{GMm}}}/{ {\text{r}}}$ Total energy $= {\text{K}} . {\text{E}}+ {\text{U}}$ $\;E=-\;\frac{G M m}{2 r}$ $\;U=2 E \;\; \;\text{ St I }\;- \;\text{ incorrect }\;$ $\;\;\text{ K.E. }\;=| {\text{E}}| {\text{St}} \;\text{ II - incorrect }\;$

Q53JEE Main 2023MCQ

The time period of a satellite, revolving above earth's surface at a height equal to ${\text{R}}$ will be (Given ${\text{g}}=\pi ^2 {\text{m}} / {\ s}^2, {\text{R}}=$ radius of earth) [10-Apr-2023 shift 2]

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📖 Explanation

$\;T^2=\;\frac{4 \pi ^2 r^3}{G M}$ $\;T^2=\;{4 \pi ^2(2 {\text{R}})^3}/{G M}$ $\;=4 \times 8 \times \;\frac{\pi 2 \times R^3}{G M}$ $\;=4 \times 8 \times \;{(g) (R^3) }/{ {\text{GR}}^2}$ At surface of earth $\Rightarrow {\text{g}}=\;{ {\text{GM}}}/{ {\text{R}}^2}$ So, ${\text{GM}}= {\text{g}} \cdot {\text{R}}^2$ Also $\pi ^2=g$ $\; {\text{T}}^2=32 {\text{R}}$ $\;\Rightarrow {\text{T}}=\sqrt{32 {\text{R}}}$

Q54JEE Main 2023MCQ

The acceleration due to gravity at height ${\text{h}}$ above the earth if ${\text{h}}< < {\text{R}}$ (Radius of earth) is given by [8-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{g}}^{'}=\;{ {\text{GM}}}/{( {\text{R}}+ {\text{h}})^2}$ $\; {\text{g}}^{'}=\;{ {\text{GM}}}/{ {\text{R}}^2 (1+\;{ {\text{h}}}/{ {\text{R}}}) ^2}$ using binomial expansion & neglect higher order term $\Rightarrow {\text{g}}= {\text{g}} (1-\;{2 {\text{h}}}/{ {\text{R}}})$

Q55JEE Main 2023MCQ

A planet having mass $9 {\text{Me}}$ and radius $4 {\text{Re}}$ , where ${\text{Me}}$ and ${\text{Re}}$ are mass and radius of earth respectively, has escape velocity in ${\text{km}} / {\ s}$ given by:(Given escape velocity on earth ${\text{V}}_{ {\text{e}}}=11.2 \times 10^3 {\text{m}} / {\ s}$ ) [13-Apr-2023 shift 1]

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📖 Explanation

$\;\;\text{ Escape velocity, }\; {\text{v}}_{ {\text{e}}}=\sqrt{\;{2 {\text{GM}}}/{ {\text{R}}}}$ $\; {\text{V}}_{ {\text{p}}}=\sqrt{\;{2 {\text{G}} (9 {\text{m}}_{ {\text{e}}}) }/{4 {\text{R}}_{ {\text{e}}}}}=\;\frac{3}{2} ( {\text{V}}_{ {\text{e}}}) _{\;\text{earth }\;}$ $\; {\text{v}}_{ {\text{p}}}=\;\frac{3}{2} \times 11.2 {\text{km}} / {\ s}$ $\; {\text{v}}_{ {\text{p}}}=16.8 {\text{km}} / {\ s}$

Q56JEE Main 2023MCQ

Given below are two statements : one is labelled as Assertion $\text{bo} {A}$ and the other is labelled as Reason R.Assertion A : Earth has atmosphere whereas moon doesn't have any atmosphere.Reason R : The escape velocity on moon is very small as compared to that on earth.In the light of the above statements. choose the correct answer from the options given below: [6-Apr-2023 shift 1]

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📖 Explanation

$V_{e s c}=\sqrt{\;\frac{2 G M}{r}}={\sqrt{2 g r}}$ Radius of moon is less than that of earth and acceleration due to gravity is also less on moon as compared to that on earth.Thus, $V_{\;\text{esc }\;}$ of Moon $< V_{\;\text{esc }\;}$ of EarthThis is also the reason behind escape of atmosphere from moon.

Q57JEE Main 2023MCQ

If ${\text{V}}$ is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:- [11-Apr-2023 shift 2]

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📖 Explanation

$\; {\text{V}}=-\;{ {\text{GM}}}/{ {\text{R}}^3} (1.5 {\text{R}}^{2-}0.5 {\text{r}}^2)$ $\; {\text{V}}=-\;{ {\text{GM}}}/{ {\text{R}}}[\;\text{ At the surface] }\;$ $\; {\text{V}}_{\;\text{centre }\;}=-\;{3 {\text{GM}}}/{2 {\text{R}}}=\;\frac{3}{2} {\text{V}}$

Q58JEE Main 2023MCQ

The weight of a body at the surface of earth is 18 ${\text{N}}$ . The weight of the body at an altitude of 3200 ${\text{km}}$ above the earth's surface is (given, radius of earth ${\text{R}}_{ {\text{e}}}=6400 {\text{km}}$ ) [24-Jan-2023 Shift 1]

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📖 Explanation

Acceleration due to gravity at height ${\text{h}}$ ${\text{g}}^{'}=\;{ {\text{g}}}/{ [1+\;{ {\text{h}}}/{ {\text{R}}}] ^2}$ So weight at given height ${\text{mg}}^{'}=\;{ {\text{mg}}}/{ [1+\;{ {\text{h}}}/{ {\text{R}}}] ^2}=\;\frac{18}{ [1+\;\frac{1}{2}] ^2}=8 {\text{N}}$

Q59JEE Main 2023MCQ

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$ . Assertion A: A pendulum clock when taken to Mount Everest becomes fast. Reason R: The value of g (acceleration due to gravity) is less at MountEverest than its value on the surface of earth. In the light of the above statements, choose the most appropriate answer from the options given below [24-Jan-2023 Shift 2]

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📖 Explanation

${\text{T}} \propto \;{1}/{\sqrt{ {\text{g}}}}$

Q60JEE Main 2023MCQ

If earth has a mass nine times and radius twice to the of a planet ${\text{P}}$ . Then $\;\frac{v_e}{3} {\sqrt{x}} {\text{ms}}^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of ${\text{P}}$ , where $v_e$ is escape velocity on earth. The value of $x$ is [1-Feb-2023 Shift 1]

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📖 Explanation

$\;v_{\;\text{(escape) plant }\;}=\sqrt{\;\frac{2 G M_P}{R_P}}$ $\;=\sqrt{\;\frac{2 G (\frac{M_e}{9}) }{ (\;\frac{R_e}{2}) }}=\;{v_e {\sqrt{2}}}/{3} \therefore x=2$