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jaoe38c1lsc2wpgx | physics | geometrical-optics | refraction,-tir-and-prism | <p>If the refractive index of the material of a prism is $$\cot \left(\frac{A}{2}\right)$$, where $$A$$ is the angle of prism then the angle of minimum deviation will be</p> | [{"identifier": "A", "content": "$$\\pi-2 \\mathrm{~A}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{2}-2 \\mathrm{~A}$$\n"}, {"identifier": "C", "content": "$$\\pi-\\mathrm{A}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{2}-\\mathrm{A}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \cot \frac{\mathrm{A}}{2}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right)}{\sin \frac{\mathrm{A}}{2}} \\
& \Rightarrow \cos \frac{\mathrm{A}}{2}=\sin \left(\frac{\mathrm{A}+\delta_{\min }}{2}\right) \\
& \frac{\mathrm{A}+\delta_{\min }}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2} \\
& \delt... | mcq | jee-main-2024-online-27th-january-morning-shift | 11,052 |
jaoe38c1lsd8jsi1 | physics | geometrical-optics | refraction,-tir-and-prism | <p>Light from a point source in air falls on a convex curved surface of radius $$20 \mathrm{~cm}$$ and refractive index 1.5. If the source is located at $$100 \mathrm{~cm}$$ from the convex surface, the image will be formed at ________ $$\mathrm{cm}$$ from the object.</p> | [] | null | 200 | <p>In the problem, you're dealing with refraction at a convex surface. The light is coming from a point source located in air (with a refractive index, $$ \mu_1 = 1.0 $$) and entering a medium with a refractive index of $$ \mu_2 = 1.5 $$. The convex surface has a radius of curvature $$ R = 20 \, \text{cm} $$, and t... | integer | jee-main-2024-online-31st-january-evening-shift | 11,054 |
jaoe38c1lse6cweq | physics | geometrical-optics | refraction,-tir-and-prism | <p>The refractive index of a prism with apex angle $$A$$ is $$\cot A / 2$$. The angle of minimum deviation is :</p> | [{"identifier": "A", "content": "$$\\delta_m=180^{\\circ}-3 \\mathrm{~A}$$\n"}, {"identifier": "B", "content": "$$\\delta_m=180^{\\circ}-4 A$$\n"}, {"identifier": "C", "content": "$$\\delta_m=180^{\\circ}-2 A$$\n"}, {"identifier": "D", "content": "$$\\delta_m=180^{\\circ}-A$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \mu=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\
& \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta m}{2}\right)}{\sin \frac{A}{2}} \\
& \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right) \\
& \frac{\pi}{2}-\frac{A}{2}... | mcq | jee-main-2024-online-31st-january-morning-shift | 11,055 |
lv2erzev | physics | geometrical-optics | refraction,-tir-and-prism | <p>A light ray is incident on a glass slab of thickness $$4 \sqrt{3} \mathrm{~cm}$$ and refractive index $$\sqrt{2}$$ The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ______ $$\mathrm{cm}$$.</p>
<p>(Given $$\sin 15^{\c... | [] | null | 2 | <p>$$\begin{aligned}
& \mu=\sqrt{2} \\
& \sin \theta_C=\frac{1}{\sqrt{2}} \\
& Q_C=45^{\circ} \\
& i=Q_C=45^{\circ}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { ( } \phi \text { ) lateral displacement }=\frac{t \sin (i-r)}{\cos r} \\
& \sin 45^{\circ}=\sqrt{2} \sin r \\
& \Rightarrow \quad r=30^{\circ} \\
& \ther... | integer | jee-main-2024-online-4th-april-evening-shift | 11,056 |
lv5gswap | physics | geometrical-optics | refraction,-tir-and-prism | <p>Critical angle of incidence for a pair of optical media is $$45^{\circ}$$. The refractive indices of first and second media are in the ratio:</p> | [{"identifier": "A", "content": "$$1: 2$$\n"}, {"identifier": "B", "content": "$$1: \\sqrt{2}$$\n"}, {"identifier": "C", "content": "$$2: 1$$\n"}, {"identifier": "D", "content": "$$\\sqrt{2}: 1$$"}] | ["D"] | null | <p>The critical angle is the angle of incidence at which light is refracted along the boundary, meaning the angle of refraction is $$90^{\circ}$$. The relationship between the critical angle and the refractive indices of two media can be understood using Snell's Law, given as:</p>
<p>$$ n_1 \sin(\theta_c) = n_2 \sin(9... | mcq | jee-main-2024-online-8th-april-morning-shift | 11,057 |
lvc57awz | physics | geometrical-optics | refraction,-tir-and-prism | <p>The refractive index of prism is $$\mu=\sqrt{3}$$ and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is _________$$^\circ$$.</p> | [] | null | 60 | <p>To find the value of the angle of the prism given the refractive index of the prism $$\mu = \sqrt{3}$$ and the ratio of the angle of minimum deviation ($$\delta_m$$) to the angle of the prism ($$A$$) is 1 (i.e., $$\frac{\delta_m}{A} = 1 \Rightarrow \delta_m = A$$), we can use the prism formula relating these variabl... | integer | jee-main-2024-online-6th-april-morning-shift | 11,058 |
u489dvI5VYscDU0X | physics | gravitation | acceleration-due-to-gravity-and-its-variation | If $${g_E}$$ and $${g_M}$$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio
<br/>$${{electro\,\,ch\arg e\,\,on\,\,the\,\,moon} \over {electronic\,\,ch\arg e\,\,on\,\,the\,\,... | [{"identifier": "A", "content": "$${g_M}/{g_E}$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$0$$ "}, {"identifier": "D", "content": "$${g_E}/{g_M}$$ "}] | ["B"] | null | electronic charge does does not depend on acceleration due to gravity as it is a universal constant.
<br><br>So, electronic charge on earth
<br><br>$$=$$ electronic charge on moon
<br><br>$$\therefore$$ Required ratio $$=1.$$ | mcq | aieee-2007 | 11,060 |
DJA6gakv8s78Q6bP | physics | gravitation | acceleration-due-to-gravity-and-its-variation | The height at which the acceleration due to gravity becomes $${g \over 9}$$ (where $$g=$$ the acceleration due to gravity on the surface of the earth) in terms of $$R,$$ the radius of the earth, is: | [{"identifier": "A", "content": "$${R \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$$R/2$$ "}, {"identifier": "C", "content": "$$\\sqrt 2 \\,\\,R$$ "}, {"identifier": "D", "content": "$$2\\,R$$ "}] | ["D"] | null | Given that, at height h from ground the acceleration due to gravity becomes $${g \over 9}$$.
<br><br>We know acceleration at earth surface due to gravity g = $${{GM} \over {{R^2}}}$$
<br><br>and acceleration at height h due to gravity g' = $${{GM} \over {{{\left( {R + h} \right)}^2}}}$$
<br><br>So $${{g} \over 9} $$ = ... | mcq | aieee-2009 | 11,061 |
cHKCIlto0NLtWaegyU28r | physics | gravitation | acceleration-due-to-gravity-and-its-variation | If the Earth has no rotational motion, the weight of a person on the equator is W. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weigh $${3 \over 4}$$ W. Radius of the Earth is 6400 km and g=10 m/s<sup>2</sup>. | [{"identifier": "A", "content": "1.1 $$ \\times $$ 10<sup>\u22123</sup> rad/s "}, {"identifier": "B", "content": "0.83 $$ \\times $$ 10<sup>\u22123</sup> rad/s "}, {"identifier": "C", "content": "0.63 $$ \\times $$ 10<sup>\u22123</sup> rad/s "}, {"identifier": "D", "content": "0.28 $$ \\times $$ 10<sup>\u22123</sup>... | ["C"] | null | Initially when earth is not rotating then weight of the person is $$w$$.
<br><br>When earth rotares about it's axis then weight = $${{3\omega } \over 4}$$
<br><br>Then,
<br><br>g' = g $$-$$ $$\omega $$<sup>2</sup>R cos<sup>2</sup>$$\theta $$
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ $${{3g} \over 4}$$ = g $$-$$ $$\omega $$... | mcq | jee-main-2017-online-8th-april-morning-slot | 11,062 |
z2TAdzPPQdmcvSpt | physics | gravitation | acceleration-due-to-gravity-and-its-variation | The variation of acceleration due to gravity $$g$$ with distance d from centre of the earth is best represented by
(R = Earth’s radius): | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l7ymblqr/1f11bc32-42db-470d-aecc-bce817906456/133a7230-3285-11ed-8893-19b23ee4c66d/file-1l7ymblqs.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l7ymblqr/1f11bc32-42db-470d-aecc-bce817906456/133... | ["A"] | null | When d < R means distance of a point is d from the center of the circle where d is inside the earth surface. In this case the value of acceleration, $$g = -{{GMd} \over {{R^3}}}$$
<br><br>$$\therefore$$ $$g \propto d$$
<br><br>So inside of the earth surface g - d graph is straight line.
<br><br>When d > R means d... | mcq | jee-main-2017-offline | 11,063 |
OSuATH4nBcRDsemOQm5yS | physics | gravitation | acceleration-due-to-gravity-and-its-variation | Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence : | [{"identifier": "A", "content": "Weight of the object, everywhere on the earth, will increase. "}, {"identifier": "B", "content": "Weight of the object, everywhere on the earth, will decrease. "}, {"identifier": "C", "content": "There will be no change in weight anywhere on the earth."}, {"identifier": "D", "content": ... | ["D"] | null | With rotation of earth the effect on acceleration due to gravity vary as
<br><br>g' = g $$-$$ $$\omega $$<sup>2</sup> R cos<sup>2</sup> $$\theta $$
<br><br>Here $$\theta $$ is lattitude.
<br><br>At poles $$\theta $$ = 90<sup>o</sup> so those will be no change in gravity.
<br><br>At all other point $$\omega $$ wil... | mcq | jee-main-2018-online-16th-april-morning-slot | 11,064 |
HLuqiuPqmYocjJR4xL18hoxe66ijvztpv00 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | The value of acceleration due to gravity at
Earth's surface is 9.8 ms<sup>–2</sup>. The altitude above
its surface at which the acceleration due to
gravity decreases to 4.9 ms<sup>–2</sup>, is close to :
(Radius of earth = 6.4 × 10<sup>6</sup> m) | [{"identifier": "A", "content": "1.6 \u00d7 10<sup>6</sup> m"}, {"identifier": "B", "content": "9.0 \u00d7 10<sup>6</sup> m"}, {"identifier": "C", "content": "6.4 \u00d7 10<sup>6</sup> m"}, {"identifier": "D", "content": "2.6 \u00d7 10<sup>6</sup> m"}] | ["D"] | null | $${{GM} \over {{{\left( {R + h} \right)}^2}}} = {{GM} \over {2{R^2}}}$$<br><br>
$$R + h = \sqrt 2 R$$<br><br>
$$h = \left( {\sqrt 2 - 1} \right)R \simeq 2.6 \times {10^6}m$$ | mcq | jee-main-2019-online-10th-april-morning-slot | 11,065 |
Pf4Deh0X6siHTFmGIp7k9k2k5gvxnv3 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | Consider two solid spheres of radii R<sub>1</sub> = 1m,
R<sub>2</sub> = 2m and masses M<sub>1</sub> and M<sub>2</sub>, respectively.
The gravitational field due to sphere (1) and (2) are shown. The value of $${{{M_1}} \over {{M_2}}}$$ is :
<img src="data:image/png;base64,UklGRkQQAABXRUJQVlA4IDgQAACwUwCdASpbAdUAPm02l0gk... | [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over 6}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["B"] | null | Gravitational field on the surface of a solid
sphere
<br><br>E = $${{GM} \over {{R^2}}}$$
<br><br>By the graph
<br><br>E<sub>1</sub> = $${{G{M_1}} \over {R_1^2}}$$
<br><br>and E<sub>2</sub> = $${{G{M_2}} \over {R_2^2}}$$
<br><br>$$ \therefore $$ $${{{E_1}} \over {{E_2}}} = {\left( {{{{R_2}} \over {{R_1}}}} \right)^2}\... | mcq | jee-main-2020-online-8th-january-morning-slot | 11,067 |
ugAf4Kk4E3mVZ6bJGt7k9k2k5hi2w69 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | A ball is dropped from the top of a 100 m high
tower on a planet. In the last $${1 \over 2}s$$ before hitting
the ground, it covers a distance of 19 m.
Acceleration due to gravity (in ms<sup>–2</sup>) near the
surface on that planet is _____. | [] | null | 8 | Let time to travel 81 m is t sec.
<br><br>Time to travel 100 m is t + $${1 \over 2}$$ sec.
<br><br>$$ \therefore $$ 81 = $${1 \over 2}$$ $$ \times $$ a $$ \times $$ t<sup>2</sup>
<br><br>$$ \Rightarrow $$ t = $$9\sqrt {{2 \over a}} $$
<br><br>And 100 = $${1 \over 2}$$ $$ \times $$ a $$ \times $$ $${\left( {{1 \over 2} ... | integer | jee-main-2020-online-8th-january-evening-slot | 11,068 |
v73vAqbrjACDGVn0Xijgy2xukexxbbjs | physics | gravitation | acceleration-due-to-gravity-and-its-variation | The height ‘h’ at which the weight of a body will
be the same as that at the same depth ‘h’ from
the surface of the earth is (Radius of the earth
is R and effect of the rotation of the earth is
neglected) | [{"identifier": "A", "content": "$${R \\over 2}$$"}, {"identifier": "B", "content": "$${{\\sqrt 5 R - R} \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt 3 R - R} \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt 5 } \\over 2}R - R$$"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263576/exam_images/be2g19uggo9uo2zyreui.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266583/exam_images/u8ivbxs9z9n7ugxruur1.webp"><source media="(max-wid... | mcq | jee-main-2020-online-2nd-september-evening-slot | 11,069 |
tVvLTC6iX4v2OZy4rjjgy2xukf3qy45t | physics | gravitation | acceleration-due-to-gravity-and-its-variation | The mass density of a planet of radius R varies with the distance r from its centre as
<br/>$$\rho $$(r) = $${\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$$.
<br/>Then the gravitational field is maximum at : | [{"identifier": "A", "content": "$$r = {1 \\over {\\sqrt 3 }}R$$"}, {"identifier": "B", "content": "r = R"}, {"identifier": "C", "content": "$$r = \\sqrt {{3 \\over 4}} R$$"}, {"identifier": "D", "content": "$$r = \\sqrt {{5 \\over 9}} R$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266273/exam_images/oeq8e819cefrue1iqjgh.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Evening Slot Physics - Gravitation Question 130 English Explanation">
<br><br>dm = $... | mcq | jee-main-2020-online-3rd-september-evening-slot | 11,070 |
JZM1W6Nb01kbMNm2T7jgy2xukf6hcdq6 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is
given by $${{Ax} \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$$ in the x-direction. The magnitude of gravitational potential on the x-axis at a
distance x, taking its value to be zero at infinity, is: | [{"identifier": "A", "content": "$${A{{\\left( {{x^2} + {a^2}} \\right)}^{3/2}}}$$"}, {"identifier": "B", "content": "$${A{{\\left( {{x^2} + {a^2}} \\right)}^{1/2}}}$$"}, {"identifier": "C", "content": "$${A \\over {{{\\left( {{x^2} + {a^2}} \\right)}^{1/2}}}}$$"}, {"identifier": "D", "content": "$${A \\over {{{\\left(... | ["C"] | null | Given $${E_x} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$$<br><br>$$ \therefore $$ $${{ - dV} \over {dx}} = {{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}$$<br><br>$$ \Rightarrow $$ $$\int\limits_0^V {dV} = - \int\limits_\infty ^x {{{Ax} \over {{{({x^2} + {a^2})}^{3/2}}}}dx} $$<br><br>$$ \Rightarrow $$ $$V = {A \over {{{({... | mcq | jee-main-2020-online-4th-september-morning-slot | 11,071 |
uieo9M9hxL3WUNSGk1jgy2xukfg7aaur | physics | gravitation | acceleration-due-to-gravity-and-its-variation | The value of the acceleration due to gravity is
g<sub>1</sub> at a height h = $${R \over 2}$$ (R = radius of the earth) from the surface of the earth. It is again equal
to g<sub>1</sub> at a depth d below the surface of the
earth. The ratio $$\left( {{d \over R}} \right)$$ equals : | [{"identifier": "A", "content": "$${5 \\over 9}$$"}, {"identifier": "B", "content": "$${1 \\over 9}$$"}, {"identifier": "C", "content": "$${7 \\over 9}$$"}, {"identifier": "D", "content": "$${4 \\over 9}$$"}] | ["A"] | null | Given, $${g_{at\,high}} = {g_{at\,depth}}$$<br><br>We know, $${g_{depth}} = $$$$g\left( {1 - {d \over R}} \right)$$<br><br>$$ \therefore $$ $$g\left( {1 - {d \over R}} \right) = {{GM_e} \over {{{(R + h)}^2}}}$$<br><br>$$ \Rightarrow $$ $$g\left( {1 - {d \over R}} \right) =$$$${{G{M_e}} \over {{{\left( {R + {R \over 2}}... | mcq | jee-main-2020-online-5th-september-morning-slot | 11,072 |
LIfiREm7HD14hnjG9cjgy2xukfl43bnp | physics | gravitation | acceleration-due-to-gravity-and-its-variation | The acceleration due to gravity on the earth’s
surface at the poles is g and angular velocity of
the earth about the axis passing through the
pole is $$\omega $$. An object is weighed at the equator
and at a height h above the poles by using a
spring balance. If the weights are found to be
same, then h is (h << R... | [{"identifier": "A", "content": "$${{{R^2}{\\omega ^2}} \\over {2g}}$$"}, {"identifier": "B", "content": "$${{{R^2}{\\omega ^2}} \\over g}$$"}, {"identifier": "C", "content": "$${{{R^2}{\\omega ^2}} \\over {8g}}$$"}, {"identifier": "D", "content": "$${{{R^2}{\\omega ^2}} \\over {4g}}$$"}] | ["A"] | null | At equator, g<sub>1</sub> = g - R$${\omega ^2}$$
<br><br>At height h, g<sub>2</sub> = $$g\left( {1 - {{2h} \over R}} \right)$$ [as given h << R]
<br><br>$$ \because $$ Weight same at poles and at h (so g<sub>1</sub>
= g<sub>2</sub>)
<br><br>$$ \therefore $$ g - R$${\omega ^2}$$ = $$g\left( {1 - {{2h} \over R}} \... | mcq | jee-main-2020-online-5th-september-evening-slot | 11,073 |
DTg56QR7gdFJEwDrmI1klrnrxks | physics | gravitation | acceleration-due-to-gravity-and-its-variation | A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?<br/><br/>[Use $$g = {{GM} \over {{R^2}}}$$ = 9.8 ms<sup>$$-$$2</sup> and radius of earth, R = 6400 km.] | [{"identifier": "A", "content": "49 N"}, {"identifier": "B", "content": "49.83 N"}, {"identifier": "C", "content": "48.83 N"}, {"identifier": "D", "content": "49.17 N"}] | ["C"] | null | Given, weight of body at North pole,<br/><br/>w<sub>p</sub> = mg = 49 N<br/><br/>Radius of Earth, R = 6400 km<br/><br/>Let weight of body at equator be w<sub>e</sub>.<br/><br/>At equator, g<sub>e</sub> = g $$-$$ R$$\omega$$<sup>2</sup><br/><br/>$$\therefore$$ w<sub>e</sub> = mg<sub>e</sub> = m(g $$-$$ R$$\omega$$<sup>2... | mcq | jee-main-2021-online-24th-february-evening-slot | 11,074 |
xHpLB18sh0pDmj82Tq1klunvds2 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA : AB will be x : y. The value of x is ________.<br/><br/><img src="data:image/png;base64,UklGRkIPAABXRUJQVlA4IDYPAACQSgCdASq+AOwA... | [] | null | 4 | $${g_A} = {{GM(r)} \over {{R^3}}}$$<br><br>$${g_C} = {{GM} \over {{{\left( {R + {R \over 2}} \right)}^2}}}$$<br><br>Given, g<sub>A</sub> = g<sub>C</sub><br><br>$$ \Rightarrow {{GM(r)} \over {{R^3}}} = {{GM} \over {{{\left( {R + {R \over 2}} \right)}^2}}}$$<br><br>$$ \Rightarrow {r \over {{R^3}}} = {4 \over {9{R^2}}}$$<... | integer | jee-main-2021-online-26th-february-evening-slot | 11,075 |
UKrdZK6HTkDJS4Z1Kz1kmlvfe2f | physics | gravitation | acceleration-due-to-gravity-and-its-variation | If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately : [Take g = 10 ms<sup>$$-$$2</sup>, the radius of earth, R = 6400 $$\times$$ 10<sup>3</sup> m, Take $$\pi$$ = 3.14] | [{"identifier": "A", "content": "84 minutes"}, {"identifier": "B", "content": "1200 minutes"}, {"identifier": "C", "content": "60 minutes"}, {"identifier": "D", "content": "does not change"}] | ["A"] | null | For objects to float<br><br>mg = 2$$\omega$$<sup>2</sup>R<br><br>$$\omega$$ = angular velocity of earth.<br><br>R = Radius of earth<br><br>$$\omega = \sqrt {{g \over R}} $$ ..... (1)<br><br>Duration of day = T<br><br>$$T = {{2\pi } \over \omega }$$ ..... (2)<br><br>$$ \Rightarrow T = 2\pi \sqrt {{R \over g}} $$<br><br... | mcq | jee-main-2021-online-18th-march-evening-shift | 11,076 |
1krpo0kkb | physics | gravitation | acceleration-due-to-gravity-and-its-variation | A person whose mass is 100 kg travels from Earth to Mars in a spaceship. Neglect all other objects in sky and take acceleration due to gravity on the surface of the Earth and Mars as 10 m/s<sup>2</sup> and 4 m/s<sup>2</sup> respectively. Identify from the below figures, the curve that fits best for the weight of the p... | [{"identifier": "A", "content": "(b)"}, {"identifier": "B", "content": "(c)"}, {"identifier": "C", "content": "(d)"}, {"identifier": "D", "content": "(a)"}] | ["B"] | null | Given,<br/><br/>Acceleration due to gravity on Earth, g<sub>E</sub> = 10 m/s<sup>2</sup><br/><br/>and acceleration due to gravity on Mars, g<sub>M</sub> = 4 m/s<sup>2</sup><br/><br/>We know that, g<sub>E</sub> (at height h) = $${g_{Earth}}\left( {1 - {{2h} \over R}} \right)$$<br/><br/>$$\therefore$$ Weight at Earth, mg... | mcq | jee-main-2021-online-20th-july-morning-shift | 11,077 |
1krw97dzk | physics | gravitation | acceleration-due-to-gravity-and-its-variation | Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is W, the weight of the same object on that planet will be : | [{"identifier": "A", "content": "2W"}, {"identifier": "B", "content": "W"}, {"identifier": "C", "content": "$${2^{{1 \\over 3}}}$$W"}, {"identifier": "D", "content": "$$\\sqrt 2 $$W"}] | ["C"] | null | Density is same<br><br>$$M = {4 \over 3}\pi {R^3}\rho ,2m = {4 \over 3}\pi R{'^3}\rho $$<br><br>$$R' = {2^{1/3}}R$$<br><br>$$\omega = {{GMm} \over {{R^2}}}$$<br><br>$${\omega _2} = {{G2Mm} \over {R{'^2}}}$$<br><br>$${\omega _2} = {2^{1/3}}\omega $$ | mcq | jee-main-2021-online-25th-july-evening-shift | 11,078 |
1ktjq4l8a | physics | gravitation | acceleration-due-to-gravity-and-its-variation | If R<sub>E</sub> be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : r < R<sub>E</sub>) | [{"identifier": "A", "content": "$$1 - {r \\over {{R_E}}} - {{{r^2}} \\over {R_E^2}} - {{{r^3}} \\over {R_E^3}}$$"}, {"identifier": "B", "content": "$$1 + {r \\over {{R_E}}} + {{{r^2}} \\over {R_E^2}} + {{{r^3}} \\over {R_E^3}}$$"}, {"identifier": "C", "content": "$$1 + {r \\over {{R_E}}} - {{{r^2}} \\over {R_E^2}} + {... | ["D"] | null | $${g_{up}} = {g \over {{{\left( {1 + {r \over R}} \right)}^2}}}$$<br><br>$${g_{down}} = g\left( {1 - {r \over R}} \right)$$<br><br>$${{{g_{down}}} \over {{g_{up}}}} = \left( {1 - {r \over R}} \right){\left( {1 + {r \over R}} \right)^2}$$<br><br>$$ = \left( {1 - {r \over R}} \right)\left( {1 + {{2r} \over R} + {{{r^2}} ... | mcq | jee-main-2021-online-31st-august-evening-shift | 11,079 |
1l57ptkmv | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>Given below are two statements :</p>
<p>Statement I : The law of gravitation holds good for any pair of bodies in the universe.</p>
<p>Statement II : The weight of any person becomes zero when the person is at the centre of the earth.</p>
<p>In the light of the above statements, choose the correct answer from the op... | [{"identifier": "A", "content": "Both Statement I and Statement II are true"}, {"identifier": "B", "content": "Both Statement I and Statement II are false"}, {"identifier": "C", "content": "Statement I is true but Statement II is false"}, {"identifier": "D", "content": "Statement I is false but Statement II is true"}] | ["A"] | null | <p><b>Statement I</b> is true. Newton's law of universal gravitation states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between their cent... | mcq | jee-main-2022-online-27th-june-morning-shift | 11,080 |
1l58bhpq3 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The variation of acceleration due to gravity (g) with distance (r) from the center of the earth is correctly represented by :</p>
<p>(Given R = radius of earth)</p> | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l5bpb45a/1d5909a9-3e5b-44d0-ba26-53bca00e7607/4b541de0-fe52-11ec-b169-b5046c590266/file-1l5bpb45b.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l5bpb45a/1d5909a9-3e5b-44d0-ba26-53bca00e7607/4b5... | ["A"] | null | For $r < R g=\frac{G m r}{R^3}=\operatorname{Cr}(C=$ Constant $)$
<br/><br/>For $r > R g=\frac{G m}{r^2}=\frac{C^{\prime}}{r^2}\left(C^{\prime}=\right.$ Constant $)$
<br/><br/>For the above equations the best suited graph is as given in option $(A)$ | mcq | jee-main-2022-online-26th-june-morning-shift | 11,081 |
1l58hk8w7 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.</p>
<p>Assertion A : If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude.</p>
<p>Reason R : At eq... | [{"identifier": "A", "content": "Both A and R are true and R is the correct explanation of A."}, {"identifier": "B", "content": "Both A and R are true but R is NOT the correct explanation of A."}, {"identifier": "C", "content": "A is true but R is false."}, {"identifier": "D", "content": "A is false but R is true."}] | ["D"] | null | <p><b>Assertion A</b> is false. While the direction of acceleration due to gravity does always point towards the center of the Earth, its magnitude actually varies from the poles to the equator. This is due to the Earth's rotation and the fact that Earth is not a perfect sphere but an oblate spheroid, meaning it... | mcq | jee-main-2022-online-26th-june-evening-shift | 11,082 |
1l5ak9ev3 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given g = acceleration due to gravity at the surface of earth).</p> | [{"identifier": "A", "content": "g/2"}, {"identifier": "B", "content": "g/4"}, {"identifier": "C", "content": "g/3"}, {"identifier": "D", "content": "g/9"}] | ["D"] | null | <p>The acceleration due to gravity (g) at a distance (r) from the center of a planet is given by:</p>
<p>$$g = \frac{G M}{r^2}$$</p>
<p>where:</p>
<ul>
<li>G is the gravitational constant,</li>
<li>M is the mass of the planet,</li>
<li>r is the distance from the center of the planet.</li>
</ul>
<p>If the height h of a ... | mcq | jee-main-2022-online-25th-june-morning-shift | 11,083 |
1l5c3sj2c | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The approximate height from the surface of earth at which the weight of the body becomes $${1 \over 3}$$ of its weight on the surface of earth is :</p>
<p>[Radius of earth R = 6400 km and $$\sqrt 3 $$ = 1.732]</p> | [{"identifier": "A", "content": "3840 km"}, {"identifier": "B", "content": "4685 km"}, {"identifier": "C", "content": "2133 km"}, {"identifier": "D", "content": "4267 km"}] | ["B"] | null | <p>According to the given information</p>
<p>$${{GM} \over {{{(R + h)}^2}}} = {1 \over 3} \times {{GM} \over {{R^2}}}$$</p>
<p>$$ \Rightarrow R + h = \sqrt 3 R$$</p>
<p>$$ \Rightarrow h = (\sqrt 3 - 1)R \simeq 4685$$ km</p> | mcq | jee-main-2022-online-24th-june-morning-shift | 11,084 |
1l5w1w33h | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The radii of two planets A and B are in the ratio 2 : 3. Their densities are 3$$\rho$$ and 5$$\rho$$ respectively. The ratio of their acceleration due to gravity is :</p> | [{"identifier": "A", "content": "9 : 4"}, {"identifier": "B", "content": "9 : 8"}, {"identifier": "C", "content": "9 : 10"}, {"identifier": "D", "content": "2 : 5"}] | ["D"] | null | <p>Given,</p>
<p>$${{{r_A}} \over {{r_B}}} = {2 \over 3}$$</p>
<p>$${{{\rho _A}} \over {{\rho _B}}} = {3 \over 5}$$</p>
<p>We know,</p>
<p>Acceleration due to gravity</p>
<p>$$g = {{GM} \over {{r^2}}} = {G \over {{r^2}}} \times {4 \over 3}\pi {r^3} \times \rho $$</p>
<p>$$ = {{4\pi Gr\rho } \over 3}$$</p>
<p>$$\therefo... | mcq | jee-main-2022-online-30th-june-morning-shift | 11,085 |
1l6f509ec | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The length of a seconds pendulum at a height h = 2R from earth surface will be:</p>
<p>(Given R = Radius of earth and acceleration due to gravity at the surface of earth, g = $$\pi$$<sup>2</sup> ms<sup>$$-$$2</sup>)</p> | [{"identifier": "A", "content": "$${2 \\over 9}$$ m"}, {"identifier": "B", "content": "$${4 \\over 9}$$ m"}, {"identifier": "C", "content": "$${8 \\over 9}$$ m"}, {"identifier": "D", "content": "$${1 \\over 9}$$ m"}] | ["D"] | null | <p>$$g = {{GM} \over {{{(R + h)}^2}}} = {{GM} \over {9{R^2}}} = {{{g_0}} \over 9}$$</p>
<p>$$ \Rightarrow T = 2\pi \sqrt {{l \over g}} = 2\pi \sqrt {{l \over {{{{g_0}} \over 9}}}} $$</p>
<p>$$ \Rightarrow 2 = 2\pi \sqrt {{{9l} \over {{g_0}}}} $$</p>
<p>$$ \Rightarrow l = {{{g_0}} \over {9{\pi ^2}}} = {1 \over 9}\,m$$<... | mcq | jee-main-2022-online-25th-july-evening-shift | 11,086 |
1l6gmz34p | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The percentage decrease in the weight of a rocket, when taken to a height of $$32 \mathrm{~km}$$ above the surface of earth will, be :</p>
<p>$$($$ Radius of earth $$=6400 \mathrm{~km})$$</p> | [{"identifier": "A", "content": "1%"}, {"identifier": "B", "content": "3%"}, {"identifier": "C", "content": "4%"}, {"identifier": "D", "content": "0.5%"}] | ["A"] | null | <p>$$\because$$ $$g = {{GM} \over {{r^2}}}$$</p>
<p>$$ \Rightarrow {{\Delta g} \over g} = 2{{\Delta r} \over r}$$</p>
<p>$$ \Rightarrow {{\Delta g} \over g} \times 100 = 2 \times {{32} \over {6400}} \times 100\% = 1\% $$</p>
<p>$$\Rightarrow$$ % decrease in weight = 1%</p> | mcq | jee-main-2022-online-26th-july-morning-shift | 11,088 |
1l6m9tn6b | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>If the radius of earth shrinks by $$2 \%$$ while its mass remains same. The acceleration due to gravity on the earth's surface will approximately :</p> | [{"identifier": "A", "content": "decrease by $$2 \\%$$"}, {"identifier": "B", "content": "decrease by $$4 \\%$$"}, {"identifier": "C", "content": "increase by $$2 \\%$$"}, {"identifier": "D", "content": "increase by $$4 \\%$$"}] | ["D"] | null | <p>The acceleration due to gravity (g) on the surface of a planet is given by the formula:</p>
<p>$$g = \frac{G M}{R^2}$$</p>
<p>where:</p>
<ul>
<li>G is the gravitational constant,</li>
<li>M is the mass of the planet,</li>
<li>R is the radius of the planet.</li>
</ul>
<p>If the radius of the Earth shrinks by 2% but i... | mcq | jee-main-2022-online-28th-july-morning-shift | 11,089 |
1l6p5yz0a | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth $$\alpha \mathrm{h}\left(\mathrm{h}<<\mathrm{R}_{\mathrm{e}}\right)$$ from the earth surface. The value of $$\alpha$$ will be _________.</p>
<p>(... | [] | null | 2 | <p>$$g\left( {1 - {{2h} \over R}} \right) = g\left( {1 - {d \over R}} \right)$$</p>
<p>$$ \Rightarrow 2h = d$$</p>
<p>$$ \Rightarrow \alpha = 2$$</p> | integer | jee-main-2022-online-29th-july-morning-shift | 11,090 |
1ldnx7uwh | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>For a body projected at an angle with the horizontal from the ground, choose the correct statement.</p> | [{"identifier": "A", "content": "Gravitational potential energy is maximum at the highest point."}, {"identifier": "B", "content": "The vertical component of momentum is maximum at the highest point."}, {"identifier": "C", "content": "The horizontal component of velocity is zero at the highest point."}, {"identifier": ... | ["A"] | null | At highest point height is maximum and vertical component of velocity is zero.
<br/><br/>So momentum is zero.
<br/><br/>At highest point horizontal component of velocity will not be zero but vertical component of velocity is equal to zero and because of this K.E. will not be equal to zero.
<br/><br/>Gravitational poten... | mcq | jee-main-2023-online-1st-february-evening-shift | 11,091 |
1ldofuwj7 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>Given below are two statements:</p>
<p>Statement I: Acceleration due to gravity is different at different places on the surface of earth.</p>
<p>Statement II: Acceleration due to gravity increases as we go down below the earth's surface.</p>
<p>In the light of the above statements, choose the correct answer from the... | [{"identifier": "A", "content": "Both Statement I and Statement II are true"}, {"identifier": "B", "content": "Both Statement I and Statement II are false"}, {"identifier": "C", "content": "Statement I is false but Statement II is true"}, {"identifier": "D", "content": "Statement I is true but Statement II is false"}] | ["D"] | null | <p><b>Statement I</b> is true. The acceleration due to gravity varies slightly over the surface of the Earth, being affected by factors such as latitude (because of the Earth's oblate shape or its equatorial bulge), altitude (it decreases with height above the Earth's surface), and local geological variations i... | mcq | jee-main-2023-online-1st-february-morning-shift | 11,093 |
1ldpkdljv | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>At a certain depth "d " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $$\mathrm{3 R}$$ above earth surface. Where $$\mathrm{R}$$ is Radius of earth (Take $$\mathrm{R}=6400 \mathrm{~km}$$ ). The depth $$\mathrm{d}$$ is equal to</p> | [{"identifier": "A", "content": "5260 km"}, {"identifier": "B", "content": "2560 km"}, {"identifier": "C", "content": "640 km"}, {"identifier": "D", "content": "4800 km"}] | ["D"] | null | The acceleration due to gravity $$g$$ at a distance $$d$$ below the surface of the earth is given by :
<br/><br/>$g_{d}=\frac{G M}{R^{3}}(R-d)$
(depth variation)
<br/><br/>where $$G$$ is the gravitational constant and $$M$$ is the mass of the Earth.
<br/><br/>At a height $$3R$$ above the surface of the Earth, the ac... | mcq | jee-main-2023-online-31st-january-morning-shift | 11,094 |
1lduhejk2 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approxim... | [{"identifier": "A", "content": "12 hours"}, {"identifier": "B", "content": "1 hour 24 minutes"}, {"identifier": "C", "content": "24 hours"}, {"identifier": "D", "content": "1 hour 40 minutes"}] | ["B"] | null | Gravitational acceleration at a distance of $r$ from centre of earth is given by
<br/><br/>
$$
g^{\prime}=\frac{g}{R} r
$$
<br/><br/>
Where $R$ is the radius of earth
<br/><br/>
So, $\frac{d^{2} r}{d t^{2}}=-\frac{g}{R} r$
<br/><br/>
$\Rightarrow \quad T=2 \pi \sqrt{\frac{R}{g}}=2 \pi \sqrt{\frac{6400000}{10}}$
<br/><b... | mcq | jee-main-2023-online-25th-january-morning-shift | 11,095 |
1ldwqz3lt | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.</p>
<p>Assertion A : A pendulum clock when taken to Mount Everest becomes fast.</p>
<p>Reason R : The value of g (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth.</p>
<p>... | [{"identifier": "A", "content": "Both A and R are correct but R is NOT the correct explanation of A"}, {"identifier": "B", "content": "A is correct but R is not correct"}, {"identifier": "C", "content": "Both A and R are correct and R is the correct explanation of A"}, {"identifier": "D", "content": "A is not correct b... | ["D"] | null | When we go on the Mount Everest the value of gravitational acceleration decreases $\left(g=\frac{g_{0}}{\left(1+\frac{h}{R_{e}}\right)^{2}}\right)$. Therefore, the time period of oscillation $\left(T=2 \pi \sqrt{\frac{I}{g}}\right)$ increases and the pendulum clock becomes slow thus the assertion is wrong but reason is... | mcq | jee-main-2023-online-24th-january-evening-shift | 11,096 |
1ldwrlc96 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>Given below are two statements:</p>
<p>Statement I : Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface.</p>
<p>Statement II : Acceleration due to earth's gravity is same at a height 'h' and depth 'd' from earth's surface, if h = d.</p>
<p>In the light of above statements, ch... | [{"identifier": "A", "content": "Statement I is correct but statement II is incorect"}, {"identifier": "B", "content": "Both Statement I and II are correct"}, {"identifier": "C", "content": "Statement I is incorrect but statement II is correct"}, {"identifier": "D", "content": "Both Statement I and Statement II are inc... | ["A"] | null | <p>The most appropriate answer is Statement I is correct but statement II is incorrect.</p>
<p><b>
Statement I</b> is correct as acceleration due to Earth's gravity decreases as you move away from its surface either upward or downward. This is because the gravity of the Earth follows an inverse square law, which means ... | mcq | jee-main-2023-online-24th-january-evening-shift | 11,097 |
1ldyd2c7x | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth's surface is (given, radius of earth $$\mathrm{R_e=6400~km}$$) :</p> | [{"identifier": "A", "content": "9.8 N"}, {"identifier": "B", "content": "4.9 N"}, {"identifier": "C", "content": "19.6 N"}, {"identifier": "D", "content": "8 N"}] | ["D"] | null | <p>The weight of an object at a height h from the Earth's surface is given by:</p>
<p>$$W' = W \left(\frac{R}{{R + h}}\right)^2$$</p>
<p>where:</p>
<ul>
<li>W' is the weight at height h,</li>
<li>W is the weight at the Earth's surface,</li>
<li>R is the radius of the Earth,</li>
<li>h is the height abov... | mcq | jee-main-2023-online-24th-january-morning-shift | 11,098 |
1lgvrxzo1 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>Given below are two statements:</p>
<p>Statement I : Rotation of the earth shows effect on the value of acceleration due to gravity (g)</p>
<p>Statement II : The effect of rotation of the earth on the value of 'g' at the equator is minimum and that at the pole is maximum.</p>
<p>In the light of the above statements,... | [{"identifier": "A", "content": "Statement I is false but statement II is true"}, {"identifier": "B", "content": "Statement I is true but statement II is false"}, {"identifier": "C", "content": "Both Statement I and Statement II are true"}, {"identifier": "D", "content": "Both Statement I and Statement II are false"}] | ["B"] | null | <p>Let's analyze both statements:</p>
<p><strong>Statement I:</strong> Rotation of the earth shows an effect on the value of acceleration due to gravity (g).</p>
<p>This is <strong>true</strong>. The value of $ g $ is affected by the rotation of the Earth. The centripetal force due to the Earth's rotation causes a red... | mcq | jee-main-2023-online-10th-april-evening-shift | 11,101 |
1lgyqka47 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The acceleration due to gravity at height $$h$$ above the earth if $$h << \mathrm{R}$$ (Radius of earth) is given by</p> | [{"identifier": "A", "content": "$$g^{\\prime}=g\\left(1-\\frac{2 h}{R}\\right)$$"}, {"identifier": "B", "content": "$$g^{\\prime}=g\\left(1-\\frac{2 h^{2}}{R^{2}}\\right)$$"}, {"identifier": "C", "content": "$$g^{\\prime}=g\\left(1-\\frac{h^{2}}{2 R^{2}}\\right)$$"}, {"identifier": "D", "content": "$$g^{\\prime}=g\\le... | ["A"] | null | <p>The acceleration due to gravity ($g$) at the surface of Earth is approximately $9.81 \, \text{m/s}^2$. This acceleration decreases as we move away from the Earth's surface because gravity is a force that attracts objects towards the center of the Earth. This force decreases with the square of the distance from t... | mcq | jee-main-2023-online-8th-april-evening-shift | 11,103 |
1lh30sxby | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The weight of a body on the surface of the earth is $$100 \mathrm{~N}$$. 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:</p> | [{"identifier": "A", "content": "50 N"}, {"identifier": "B", "content": "64 N"}, {"identifier": "C", "content": "25 N"}, {"identifier": "D", "content": "100 N"}] | ["B"] | null | <p>To find the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth, we can use the formula for gravitational force:</p>
<p>$$F = G \frac{m_1 m_2}{r^2}$$</p>
<p>where $$F$$ is the gravitational force, $$G$$ is the gravitational constant, $$m_1$$ and $$m_2$$ are the masses o... | mcq | jee-main-2023-online-6th-april-evening-shift | 11,105 |
lsbkkgb5 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | If $\mathrm{R}$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$, then the length of the second's pendulum at a height $\mathrm{h}=2 R$ from the surface of earth will be, : | [{"identifier": "A", "content": "$\\frac{1}{9} \\mathrm{~m}$"}, {"identifier": "B", "content": "$\\frac{8}{9} \\mathrm{~m}$"}, {"identifier": "C", "content": "$\\frac{2}{9} \\mathrm{~m}$"}, {"identifier": "D", "content": "$\\frac{4}{9} \\mathrm{~m}$"}] | ["A"] | null | <p>To find the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth, we must first understand that the length of a second's pendulum, $L$, is related to the gravitational acceleration, $g$, and the period, $T$, by the formula: $$ T = 2\pi\sqrt{\frac{L}{g}} $$ Since we are talking about a s... | mcq | jee-main-2024-online-1st-february-morning-shift | 11,106 |
jaoe38c1lsc33oqz | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>The acceleration due to gravity on the surface of earth is $$\mathrm{g}$$. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be :</p> | [{"identifier": "A", "content": "g/4"}, {"identifier": "B", "content": "2g"}, {"identifier": "C", "content": "g/2"}, {"identifier": "D", "content": "4g"}] | ["D"] | null | <p>$$\begin{aligned}
& g=\frac{G M}{R^2} \Rightarrow g \propto \frac{1}{R^2} \\
& \frac{g_2}{g_1}=\frac{R_1^2}{R_2^2} \\
& g_2=4 g_1\left(R_2=\frac{R_1}{2}\right)
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift | 11,107 |
lv2erys9 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>A $$90 \mathrm{~kg}$$ body placed at $$2 \mathrm{R}$$ distance from surface of earth experiences gravitational pull of :</p>
<p>($$\mathrm{R}=$$ Radius of earth, $$\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2}$$)</p> | [{"identifier": "A", "content": "300 N"}, {"identifier": "B", "content": "225 N"}, {"identifier": "C", "content": "100 N"}, {"identifier": "D", "content": "120 N"}] | ["C"] | null | <p>The gravitational force $$F$$ that an object of mass $$m$$ placed at a distance $$r$$ (from the center of Earth) experiences can be calculated using the universal law of gravitation, given by:</p>
<p>$$F = \frac{G M m}{r^2}$$</p>
<p>Where:</p>
<ul>
<li>$$G$$ is the gravitational constant ($$6.674 \times 10^{-11}... | mcq | jee-main-2024-online-4th-april-evening-shift | 11,110 |
lvb29ej7 | physics | gravitation | acceleration-due-to-gravity-and-its-variation | <p>Assuming the earth to be a sphere of uniform mass density, a body weighed $$300 \mathrm{~N}$$ on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?</p> | [{"identifier": "A", "content": "75 N"}, {"identifier": "B", "content": "375 N"}, {"identifier": "C", "content": "300 N"}, {"identifier": "D", "content": "225 N"}] | ["D"] | null | <p>To solve this question, we first need to understand how gravitational force (and hence weight) changes with depth under the surface of the Earth. The gravitational force at a depth $d$ is given by the formula:</p>
<p>$$ F = F_0 \left(1 - \frac{d}{R}\right) $$</p>
<p>where $F_0$ is the gravitational force (or the w... | mcq | jee-main-2024-online-6th-april-evening-shift | 11,111 |
FDLqTQutEYbfE6Rf | physics | gravitation | escape-speed-and-motion-of-satellites | The escape velocity of a body depends upon mass as | [{"identifier": "A", "content": "$${m^0}$$ "}, {"identifier": "B", "content": "$${m^1}$$ "}, {"identifier": "C", "content": "$${m^2}$$ "}, {"identifier": "D", "content": "$${m^3}$$ "}] | ["A"] | null | Escape velocity,
<br><br>$${v_e} = \sqrt {2gR} = \sqrt {{{2GM} \over R}} \Rightarrow {V_e}\, \propto \,{m^0}$$
<p>Where $$M,R$$ are the mass and radius of the planet respectively. In this expression the mass of the body $$(m)$$ is not present. The escape velocity is independent of the mass m or it depends on m<sup>0... | mcq | aieee-2002 | 11,113 |
w3I1CXletGh0mbjL | physics | gravitation | escape-speed-and-motion-of-satellites | The kinetic energy needed to project a body of mass $$m$$ from the earth surface (radius $$R$$) to infinity is | [{"identifier": "A", "content": "$$mgR/2$$ "}, {"identifier": "B", "content": "$$2mgR$$ "}, {"identifier": "C", "content": "$$mgR$$ "}, {"identifier": "D", "content": "$$mgR/4$$"}] | ["C"] | null | The required velocity is called escape velocity ($${v_e}$$) to leave the earth surface of a body.
<br><br>$${v_e} = $$ escape velocity $$ = \sqrt {2gR} $$
<br><br>Kinetic Energy $$K.E = {1 \over 2}mv_e^2$$
<br><br>$$\therefore$$ $$K.E = {1 \over 2}m \times 2gR = mgR$$ | mcq | aieee-2002 | 11,114 |
ZgE0IquZtHg5OwIe | physics | gravitation | escape-speed-and-motion-of-satellites | The escape velocity for a body projected vertically upwards from the surface of earth is $$11$$ $$km/s.$$ If the body is projected at an angle of $${45^ \circ }$$ with the vertical, the escape velocity will be | [{"identifier": "A", "content": "$$11\\sqrt 2 \\,\\,km/s$$ "}, {"identifier": "B", "content": "$$22$$ $$km/s$$ "}, {"identifier": "C", "content": "$$11$$ $$km/s$$ "}, {"identifier": "D", "content": "$${{11} \\over {\\sqrt 2 }}km/s$$ "}] | ["C"] | null | We know, Escape velocity, $${v_e} = \sqrt {2gR} $$
<br><br>So the escape velocity is independent of the angle at which the body is projected, hence it will remain same as 11 km/s. | mcq | aieee-2003 | 11,115 |
RjMWoBxj4sK6AiG6 | physics | gravitation | escape-speed-and-motion-of-satellites | A planet in a distant solar system is $$10$$ times more massive than the earth and its radius is $$10$$ times smaller. Given that the escape velocity from the earth is $$11\,\,km\,{s^{ - 1}},$$ the escape velocity from the surface of the planet would be | [{"identifier": "A", "content": "$$1.1\\,\\,km\\,{s^{ - 1}}$$ "}, {"identifier": "B", "content": "$$100\\,\\,km\\,{s^{ - 1}}$$ "}, {"identifier": "C", "content": "$$110\\,\\,km\\,{s^{ - 1}}$$ "}, {"identifier": "D", "content": "$$0.11\\,\\,km\\,{s^{ - 1}}$$ "}] | ["C"] | null | Let M<sub>e</sub> is mass of earth then mass of planet M<sub>p</sub> = 10M<sub>e</sub>.
<br><br>And let R<sub>e</sub> is radius of earth then radius of planet R<sub>p</sub> = $${{{R_e}} \over {10}}$$
<br><br>Escape velocity of earth, $${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}} $$
<br><br>Escape velocity of planet, $${v... | mcq | aieee-2008 | 11,117 |
HRdYbt32HlVxvYeW | physics | gravitation | escape-speed-and-motion-of-satellites | The mass of a spaceship is $$1000$$ $$kg.$$ It is to be launched from the earth's surface out into free space. The value of $$g$$ and $$R$$ (radius of earth ) are $$10\,m/{s^2}$$ and $$6400$$ $$km$$ respectively. The required energy for this work will be: | [{"identifier": "A", "content": "$$6.4 \\times {10^{11}}\\,$$ Joules"}, {"identifier": "B", "content": "$$6.4 \\times {10^8}\\,$$ Joules "}, {"identifier": "C", "content": "$$6.4 \\times {10^9}\\,$$ Joules "}, {"identifier": "D", "content": "$$6.4 \\times {10^{10}}\\,$$ Joules"}] | ["D"] | null | Potential energy at earth surface = $$ - {{GMm} \over R}$$
<br><br>and at free space potential energy = 0
<br><br>Work done for this = $$0 - \left( { - {{GMm} \over R}} \right)$$ = $${{{GMm} \over R}}$$
<br><br>So the required energy for this work is
<br><br>= $${{GMm} \over R}$$
<br><br>=$${{{g{R^2}m} \over R}}$$ [ a... | mcq | aieee-2012 | 11,118 |
Dy8gMv06swuzNAoE | physics | gravitation | escape-speed-and-motion-of-satellites | A satellite is revolving in a circular orbit at a height $$'h'$$ from the earth's surface (radius of earth $$R;h < < R$$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to : (Neglect the effect of atmosphere.) | [{"identifier": "A", "content": "$$\\sqrt{2 g R}$$"}, {"identifier": "B", "content": "$$\\sqrt{g R}$$"}, {"identifier": "C", "content": "$$\\sqrt{g R / 2}$$"}, {"identifier": "D", "content": "$$\\sqrt{g R}(\\sqrt{2}-1)$$"}] | ["D"] | null | Orbital velocity of satellite,
<br><br>$${v_0} = \sqrt {{{GM} \over {R + h}}} $$
<br><br>= $$\sqrt {{{GM} \over R}} $$ [ As $$h < < R$$ then R + h = R ]
<br><br>= $$\sqrt {gR} $$ [ As $$g = {{GM} \over {{R^2}}}$$ ]
<br><br>Escape velocity
<br><br>$${v_e} = \sqrt {{{2GM} \over R}} $$
<br><br>= $$\sqrt {2gR} $$ ... | mcq | jee-main-2016-offline | 11,120 |
i78V2HeDkRD8DPxXHN4Cx | physics | gravitation | escape-speed-and-motion-of-satellites | A rocket has to be launched from earth in such
a way that it never returns. If E is the minimum
energy delivered by the rocket launcher, what
should be the minimum energy that the
launcher should have if the same rocket is to
be launched from the surface of the moon ?
Assume that the density of the earth and the
moon a... | [{"identifier": "A", "content": "E/32"}, {"identifier": "B", "content": "E/16"}, {"identifier": "C", "content": "E/4"}, {"identifier": "D", "content": "E/64"}] | ["B"] | null | Minimum energy required (E) = – (Potential energy of
object at surface of earth)<br><br>
$$ = \left( { - {{GMm} \over R}} \right) = {{GMm} \over R}$$<br><br>
Now M<sub>earth</sub> = 64 M<sub>moon</sub><br><br>
$$\rho .{4 \over 3}\pi R_e^3 = 64.{4 \over 3}\pi R_m^3$$<br><br>
$$ \Rightarrow $$ R<sub>e</sub> = 4R<sub>m</s... | mcq | jee-main-2019-online-8th-april-evening-slot | 11,122 |
ryG1T8AvwxBtSwbdbFL25 | physics | gravitation | escape-speed-and-motion-of-satellites | A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combi... | [{"identifier": "A", "content": "in the same circular orbit of radius R"}, {"identifier": "B", "content": "such that it escapes to infinity"}, {"identifier": "C", "content": "in a circular orbit of a different radius "}, {"identifier": "D", "content": "in an elliptical orbit"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266684/exam_images/ic8lr1ryal9qbn2uyhyy.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Physics - Gravitation Question 150 English Explanation">
<br>mv$$\wide... | mcq | jee-main-2019-online-12th-january-morning-slot | 11,124 |
G15zlAfrVxIHPFMW1E6DQ | physics | gravitation | escape-speed-and-motion-of-satellites | A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -
| [{"identifier": "A", "content": "mv<sup>2</sup>"}, {"identifier": "B", "content": "$${1 \\over 2}$$ mv<sup>2</sup>"}, {"identifier": "C", "content": "$${3 \\over 2}$$ mv<sup>2</sup>"}, {"identifier": "D", "content": "2 mv<sup>2</sup>"}] | ["A"] | null | At height r from center of earth. orbital velocity
<br><br>= $$\sqrt {{{GM} \over r}} $$
<br><br>$$ \therefore $$ By energy conservation
<br><br>KE of 'm' + $$\left( { - {{GMm} \over r}} \right)$$ = 0 + 0
<br><br>(At infinity, PE = KE = 0)
<br><br>$$ \Rightarrow $$ KE of 'm' = $${{{GMm} \over r}}$... | mcq | jee-main-2019-online-10th-january-morning-slot | 11,126 |
joGNcYCzUBeyp3MYtJaGG | physics | gravitation | escape-speed-and-motion-of-satellites | The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 $$ \times $$ 10<sup>3</sup> km) is E<sub>1</sub> and kinetic energy required for the satellite to be in a circular orbit at this height is E<sub>2</sub>. The value of h for which E<sub>1</sub> and E<sub>2</sub> are equal,... | [{"identifier": "A", "content": "1.6 $$ \\times $$ 10<sup>3</sup> km"}, {"identifier": "B", "content": "3.2 $$ \\times $$ 10<sup>3</sup> km"}, {"identifier": "C", "content": "6.4 $$ \\times $$ 10<sup>3</sup> km"}, {"identifier": "D", "content": "1.28 $$ \\times $$ 10<sup>4</sup> km"}] | ["B"] | null | Energy required to move a satellite from earth surface to height h is,
<br><br>E<sub>1</sub> = U<sub>h</sub> $$-$$ U<sub>surface</sub>
<br><br>= $$-$$ $${{GMm} \over {R + h}} - \left( { - {{GMm} \over R}} \right)$$
<br><br>= GMm $$\left( {{1 \over R} - {1 \over {R + h}}} \right)$$
<br><br>= $${{GMm} \over {R(R + h)}} ... | mcq | jee-main-2019-online-9th-january-evening-slot | 11,127 |
zgvITvPSQFNCof3MzzwaR | physics | gravitation | escape-speed-and-motion-of-satellites | A satellite is revolving in a circular orbit at a height h form the earth surface, such that h < < R where R is the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is : | [{"identifier": "A", "content": "$$\\sqrt {gR} \\left( {\\sqrt 2 - 1} \\right)$$"}, {"identifier": "B", "content": "$$\\sqrt {2gR} $$"}, {"identifier": "C", "content": "$$\\sqrt {gR} $$"}, {"identifier": "D", "content": "$${{\\sqrt {gR} } \\over 2}$$"}] | ["A"] | null | v<sub>0</sub> = $$\sqrt {g(R + h)} \approx \sqrt {gR} $$
<br><br>v<sub>e</sub> = $$\sqrt {2g(R + h)} \approx \sqrt {2gR} $$
<br><br>$$\Delta $$v=v<sub>e</sub> $$-$$ v<sub>0</sub> = $$\left( {\sqrt 2 - 1} \right)\sqrt {gR} $$ | mcq | jee-main-2019-online-11th-january-morning-slot | 11,128 |
jMy0gsVS8L9Ul6e15Zjgy2xukf16efbg | physics | gravitation | escape-speed-and-motion-of-satellites | A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to
that of the earth’s radius R<sub>e</sub>
. By firing rockets attached to it, its speed is instantaneously increased
in the direction of its motion so that it become $$\sqrt {{3 \over 2}} $$
times larger. Due to this t... | [{"identifier": "A", "content": "2R<sub>e</sub>"}, {"identifier": "B", "content": "3R<sub>e</sub>"}, {"identifier": "C", "content": "4R<sub>e</sub>"}, {"identifier": "D", "content": "2.5R<sub>e</sub>"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265123/exam_images/hszefzg6upwkvwdi4fej.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Physics - Gravitation Question 131 English Explanation">
<br><br>V<sub>... | mcq | jee-main-2020-online-3rd-september-morning-slot | 11,129 |
VHcy9CkWpZG6nMnj4Ajgy2xukfak8dx2 | physics | gravitation | escape-speed-and-motion-of-satellites | A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the
orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape
velocity from the planet is:
| [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$\\sqrt 2 $$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 2 }}$$"}] | ["D"] | null | $${V_0} = \sqrt {{{GM} \over r}} $$
<br><br>$${V_e} = \sqrt {{{2GM} \over r}} $$
<br><br>$$ \therefore $$ $${{{V_0}} \over {{V_e}}} = \sqrt {{{GM} \over r} \times {{2GM} \over r}} = {1 \over {\sqrt 2 }}$$ | mcq | jee-main-2020-online-4th-september-evening-slot | 11,130 |
1o6C7Qu4KDyhQmZQQ6jgy2xukev1eaty | physics | gravitation | escape-speed-and-motion-of-satellites | The mass density of a spherical galaxy varies
as
$${K \over r}$$ over a large distance ‘r’ from its centre.
In that region, a small star is in a circular orbit
of radius R. Then the period of revolution, T
depends on R as : | [{"identifier": "A", "content": "T<sup>2</sup> $$ \\propto $$ R"}, {"identifier": "B", "content": "T<sup>2</sup> $$ \\propto $$ R<sup>3</sup>"}, {"identifier": "C", "content": "T $$ \\propto $$ R"}, {"identifier": "D", "content": "T<sup>2</sup> $$ \\propto $$ $${1 \\over {{R^3}}}$$"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265575/exam_images/tghangpmtivh6aceem4q.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267513/exam_images/fmzloboalfscgtf06bd2.webp" style="max-width: 100%;height: auto;display: block;margi... | mcq | jee-main-2020-online-2nd-september-morning-slot | 11,131 |
97LU0e1lqmdn5gpKEK7k9k2k5lc1lci | physics | gravitation | escape-speed-and-motion-of-satellites | Planet A has mass M and radius R. Planet B has
half the mass and half the radius of Planet A.
If the escape velocities from the Planets A and
B are v<sub>A</sub> and v<sub>B</sub>, respectively, then $${{{v_A}} \over {{v_B}}} = {n \over 4}$$.
The value of n is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["C"] | null | Escape velocity, V<sub>e</sub> = $$\sqrt {{{2GM} \over R}} $$
<br><br>V<sub>A</sub> = $$\sqrt {{{2GM} \over R}} $$
<br><br>V<sub>B</sub> = $$\sqrt {{{2G{M \over 2}} \over {{R \over 2}}}} $$ = $$\sqrt {{{2GM} \over R}} $$
<br><br>$${{{V_A}} \over {{V_B}}}$$ = 1
<br><br>Given $${{{V_A}} \over {{V_B}}} = {n \over 4}$$
<br... | mcq | jee-main-2020-online-9th-january-evening-slot | 11,132 |
VnTbssCTwNkSY0QC9M7k9k2k5hi3kw8 | physics | gravitation | escape-speed-and-motion-of-satellites | An asteroid is moving directly towards the
centre of the earth. When at a distance of
10R (R is the radius of the earth) from the earths
centre, it has a speed of 12 km/s. Neglecting
the effect of earths atmosphere, what will be the
speed of the asteroid when it hits the surface
of the earth (escape velocity from the e... | [] | null | 16 | U<sub>1</sub> + K<sub>1</sub> = U<sub>2</sub> + K<sub>2</sub>
<br><br>$$ \Rightarrow $$ $$ - {{GMm} \over {10R}} + {1 \over 2}mV_1^2$$ = $$ - {{GMm} \over R} + {1 \over 2}mV_2^2$$
<br><br>$$ \Rightarrow $$ $${1 \over 2}V_2^2 = {1 \over 2}V_1^2 + {{GM} \over R} - {{GM} \over {10R}}$$
<br><br>$$ \Rightarrow $$ $$V_2^2 = ... | integer | jee-main-2020-online-8th-january-evening-slot | 11,133 |
xmTRopIk2GtOx1TuFB7k9k2k5dlt5p2 | physics | gravitation | escape-speed-and-motion-of-satellites | A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket
of mass $${m \over {10}}$$
so that subsequently the
satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravi... | [{"identifier": "A", "content": "$${{3m} \\over 8}{\\left( {u + \\sqrt {{{5GM} \\over {6R}}} } \\right)^2}$$"}, {"identifier": "B", "content": "$${m \\over {20}}\\left( {{u^2} + {{113} \\over {100}}{{GM} \\over R}} \\right)$$"}, {"identifier": "C", "content": "$$5m\\left( {{u^2} - {{119} \\over {100}}{{GM} \\over R}} \... | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267353/exam_images/mly9ped4ovwxylglilgi.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Morning Slot Physics - Gravitation Question 140 English Explanation 1">
<br>Using ener... | mcq | jee-main-2020-online-7th-january-morning-slot | 11,134 |
oEA4uDDqNV6VL2pMk91klrxdz1p | physics | gravitation | escape-speed-and-motion-of-satellites | Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively.<br/><br/>If T<sub>A</sub> and T<sub>B</sub> are the time periods of A and B respectively then the value of T<sub>B</sub> $$-$$ T<sub>A</sub> :<br/><br/><img src="data:image/png;base64,UklGRkYJA... | [{"identifier": "A", "content": "1.33 $$\\times$$ 10<sup>3</sup> s"}, {"identifier": "B", "content": "4.24 $$\\times$$ 10<sup>2</sup> s"}, {"identifier": "C", "content": "3.33 $$\\times$$ 10<sup>2</sup> s"}, {"identifier": "D", "content": "4.24 $$\\times$$ 10<sup>3</sup> s"}] | ["A"] | null | $$T = 2\pi \sqrt {{{{r^3}} \over {GM}}} $$<br><br>$${T_A} = 2\pi \sqrt {{{{{\left( {6400 + 600} \right)}^3}} \over {GM}}} $$<br><br>$${T_A} = 2\pi \times {10^9}\sqrt {{{{7^3}} \over {GM}}} $$<br><br>$${T_B} = 2\pi \times {10^9}\sqrt {{{{8^3}} \over {GM}}} $$<br><br>$${T_B} - {T_A} = {{2\pi {{10}^9}} \over {\sqrt {GM}... | mcq | jee-main-2021-online-25th-february-morning-slot | 11,136 |
srbuItQplQGedf4awm1klrypxi3 | physics | gravitation | escape-speed-and-motion-of-satellites | Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.<br/><br/>Assertion A : The escape velocities of planet A and B are same. But A and B are of unequal mass.<br/><br/>Reason R : The product of their mass and radius must be same. M<sub>1</sub>R<sub>1</sub> = M<sub>2</su... | [{"identifier": "A", "content": "Both A and R are correct and R is the correct explanation of A"}, {"identifier": "B", "content": "Both A and R are correct but R is NOT the correct explanation of A"}, {"identifier": "C", "content": "A is correct but R is not correct"}, {"identifier": "D", "content": "A is not correct b... | ["C"] | null | $${v_e}$$ = escape velocity<br><br>$${v_e} = \sqrt {{{2GM} \over R}} $$<br><br>so for same $${v_e},{{{M_1}} \over {{R_1}}} = {{{M_2}} \over {{R_2}}}$$<br><br>A is true but R is false | mcq | jee-main-2021-online-25th-february-morning-slot | 11,137 |
ocXJ2ZZEAXgwGZ0Unp1klt3inpo | physics | gravitation | escape-speed-and-motion-of-satellites | The initial velocity v<sub>i</sub> required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity v<sub>e</sub> such that $${v_i} = \sqrt {{x \over y}} \times {v_e}$$. The value of x will be _________... | [] | null | 10 | Here R = radius of the earth<br><br>From energy conservation<br><br>$${{ - G{m_e}m} \over R} + {1 \over 2}m{v_i}^2 = {{ - G{m_e}m} \over {11R}} + 0$$<br><br>$${1 \over 2}m{v_i}^2 = {{10} \over {11}}{{G{m_e}m} \over R}$$<br><br>$${v_i} = \sqrt {{{20} \over {11}}{{G{m_e}} \over R}} $$<br><br>$${v_i} = \sqrt {{{10} \over ... | integer | jee-main-2021-online-25th-february-evening-slot | 11,138 |
pCTHnMPcSwrLyRajNb1kmj3xssq | physics | gravitation | escape-speed-and-motion-of-satellites | The radius in kilometer to which the present radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 times is ___________. | [] | null | 64 | $${V_{es}} = \sqrt {{{2GM} \over R}} $$<br><br>$${V_{es}}\sqrt R $$ = const<br><br>$$ \therefore $$ $${V_{es}}.\sqrt R = 10{V_{es}}\sqrt {R'} $$<br><br>$$ \Rightarrow $$ $$R' = {R \over {100}}$$ = 64 km | integer | jee-main-2021-online-17th-march-morning-shift | 11,139 |
56WBuMrdtHRvDxZhIW1kmkbsz5z | physics | gravitation | escape-speed-and-motion-of-satellites | A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is _________. 'P' has the time period of 24 hours. | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$$6\\sqrt 2 $$"}, {"identifier": "D", "content": "$${6 \\over {\\sqrt 2 }}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266402/exam_images/iqkrykzyb4jm0zvdos04.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Physics - Gravitation Question 109 English Explanation">
<br>From Kepler'... | mcq | jee-main-2021-online-17th-march-evening-shift | 11,140 |
1krqauf83 | physics | gravitation | escape-speed-and-motion-of-satellites | A satellite is launched into a circular orbit of radius R around earth, while a second satellite is launched into a circular orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is : | [{"identifier": "A", "content": "1.5"}, {"identifier": "B", "content": "2.0"}, {"identifier": "C", "content": "0.7"}, {"identifier": "D", "content": "3.0"}] | ["D"] | null | $${T^2} \propto {R^3}$$<br><br>$$T = k{R^{3/2}}$$<br><br>$${{dT} \over T} = {3 \over 2}{{dR} \over R}$$<br><br>$$ = {3 \over 2} \times 0.02 = 0.03$$<br><br>% Change = 3% | mcq | jee-main-2021-online-20th-july-evening-shift | 11,141 |
1ks198ixv | physics | gravitation | escape-speed-and-motion-of-satellites | The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of 9.0 $$\times$$ 10<sup>3</sup> km. Find the mass of Mars.<br/><br/>$$\left\{ {Given\,{{4{\pi ^2}} \over G} = 6 \times {{10}^{11}}{N^{ - 1}}{m^{ - 2}}k{g^2}} \right\}$$ | [{"identifier": "A", "content": "5.96 $$\\times$$ 10<sup>19</sup> kg"}, {"identifier": "B", "content": "3.25 $$\\times$$ 10<sup>21</sup> kg"}, {"identifier": "C", "content": "7.02 $$\\times$$ 10<sup>25</sup> kg"}, {"identifier": "D", "content": "6.00 $$\\times$$ 10<sup>23</sup> kg"}] | ["D"] | null | Option D is correct.<br><br>$${T^2} = {{4{\pi ^2}} \over {GM}}.{r^3}$$<br><br>$$M = {{4{\pi ^2}} \over G}.{{{r^3}} \over {{T^2}}}$$<br><br>by putting values<br><br>$$M = 6 \times {10^{23}}$$ | mcq | jee-main-2021-online-27th-july-evening-shift | 11,142 |
1l6i0smpt | physics | gravitation | escape-speed-and-motion-of-satellites | <p>A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be :</p>
<p>(Take radius of earth $$=6400 \mathrm{~km}$$ and $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ )</p> | [{"identifier": "A", "content": "800 km"}, {"identifier": "B", "content": "1600 km"}, {"identifier": "C", "content": "2133 km"}, {"identifier": "D", "content": "4800 km"}] | ["A"] | null | <p>Applying conservation of energy</p>
<p>$$ - {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {{1 \over 3}\sqrt {{{2G{m_e}} \over {{R_e}}}} } \right)^2} = - {{G{M_e}m} \over {{R_e} + h}}$$</p>
<p>$$ - {{G{M_e}m} \over {{R_e}}} + {{G{M_e}m} \over {9{R_e}}} = - {{G{M_e}m} \over {{R_e} + h}}$$</p>
<p>$${8 \over {9{R_e}... | mcq | jee-main-2022-online-26th-july-evening-shift | 11,146 |
1l6jh7ku0 | physics | gravitation | escape-speed-and-motion-of-satellites | <p>Two satellites $$\mathrm{A}$$ and $$\mathrm{B}$$, having masses in the ratio $$4: 3$$, are revolving in circular orbits of radii $$3 \mathrm{r}$$ and $$4 \mathrm{r}$$ respectively around the earth. The ratio of total mechanical energy of $$\mathrm{A}$$ to $$\mathrm{B}$$ is :</p> | [{"identifier": "A", "content": "9 : 16"}, {"identifier": "B", "content": "16 : 9"}, {"identifier": "C", "content": "1 : 1"}, {"identifier": "D", "content": "4 : 3"}] | ["B"] | null | <p>$$U = - {{G{M_e}m} \over {2r}}$$</p>
<p>So, $${{{U_A}} \over {{U_B}}} = {{{m_A}} \over {{m_B}}} \times {{{r_B}} \over {{r_A}}}$$</p>
<p>$$ = {4 \over 3} \times {4 \over 3} = {{16} \over 9}$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift | 11,147 |
1l6kmozat | physics | gravitation | escape-speed-and-motion-of-satellites | <p>A body of mass $$\mathrm{m}$$ is projected with velocity $$\lambda \,v_{\mathrm{e}}$$ in vertically upward direction from the surface of the earth into space. It is given that $$v_{\mathrm{e}}$$ is escape velocity and $$\lambda<1$$. If air resistance is considered to be negligible, then the maximum height from th... | [{"identifier": "A", "content": "$$\\frac{\\mathrm{R}}{1+\\lambda^{2}}$$"}, {"identifier": "B", "content": "$$\\frac{R}{1-\\lambda^{2}}$$"}, {"identifier": "C", "content": "$$\\frac{R}{1-\\lambda}$$"}, {"identifier": "D", "content": "$$\\frac{\\lambda^{2} \\mathrm{R}}{1-\\lambda^{2}}$$"}] | ["B"] | null | <p>Using energy conservation</p>
<p>$$ - {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {\lambda \sqrt {{{2G{M_e}} \over {{R_e}}}} } \right)^2} = - {{G{M_e}m} \over r}$$</p>
<p>$${{G{M_e}m} \over r} = {{G{M_e}m} \over {{R_e}}} - {{G{M_e}m} \over {{R_e}}}{\lambda ^2}$$</p>
<p>$$r = {{{R_e}} \over {1 - {\lambda ^2}}}$$... | mcq | jee-main-2022-online-27th-july-evening-shift | 11,148 |
1ldnw3vyn | physics | gravitation | escape-speed-and-motion-of-satellites | <p>The escape velocities of two planets $$\mathrm{A}$$ and $$\mathrm{B}$$ are in the ratio $$1: 2$$. If the ratio of their radii respectively is $$1: 3$$, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be :</p> | [{"identifier": "A", "content": "$$\\frac{4}{3}$$"}, {"identifier": "B", "content": "$$\\frac{2}{3}$$"}, {"identifier": "C", "content": "$$\\frac{3}{4}$$"}, {"identifier": "D", "content": "$$\\frac{3}{2}$$"}] | ["C"] | null | The escape velocity of a planet is given by the formula:
<br/><br/>$$\mathrm{v}_{\text{escape}} = \sqrt{\frac{2GM}{R}}$$
<br/><br/>where G is the gravitational constant, M is the mass of the planet, and R is its radius.
<br/><br/>If the escape velocity of planet A is v<sub>A</sub> and the escape velocity of planet B... | mcq | jee-main-2023-online-1st-february-evening-shift | 11,149 |
1ldoff4pr | physics | gravitation | escape-speed-and-motion-of-satellites | <p>If earth has a mass nine times and radius twice to that of a planet P. Then $$\frac{v_{e}}{3} \sqrt{x} \mathrm{~ms}^{-1}$$ will be the minimum velocity required by a rocket to pull out of gravitational force of $$\mathrm{P}$$, where $$v_{e}$$ is escape velocity on earth. The value of $$x$$ is</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "18"}] | ["C"] | null | $$
\begin{aligned}
& M_E=9 M_P \\\\
& R_E=2 R_P
\end{aligned}
$$
<br/><br/>Escape velocity $=\sqrt{\frac{2GM}{R}}$
<br/><br/>For earth $v_e=\sqrt{\frac{2 G M_E}{R_E}}$
<br/><br/>$$
\begin{aligned}
& \text { For } P, v_e=\sqrt{\frac{\frac{2 G M_E}{9}}{\frac{R_E}{2}}}=\sqrt{\frac{2 G M_E}{R_E} \times \frac{2}{9}} \\\\
& ... | mcq | jee-main-2023-online-1st-february-morning-shift | 11,150 |
1lgoyfzdm | physics | gravitation | escape-speed-and-motion-of-satellites | <p>Given below are two statements:</p>
<p>Statement I : For a planet, if the ratio of mass of the planet to its radius increases, the escape velocity from the planet also increases.</p>
<p>Statement II : Escape velocity is independent of the radius of the planet.</p>
<p>In the light of above statements, choose the most... | [{"identifier": "A", "content": "Both Statement I and Statement II are correct"}, {"identifier": "B", "content": "Statement I is correct but statement II is incorrect"}, {"identifier": "C", "content": "Both Statement I and Statement II are incorrect"}, {"identifier": "D", "content": "Statement I is incorrect but statem... | ["B"] | null | Statement I suggests that the escape velocity of a planet increases with an increase in the ratio of its mass to its radius. This statement is consistent with the formula for escape velocity, which shows that the escape velocity of a planet is directly proportional to the square root of its mass and inversely proportio... | mcq | jee-main-2023-online-13th-april-evening-shift | 11,151 |
1lgrhl0d6 | physics | gravitation | escape-speed-and-motion-of-satellites | <p>Two satellites $$\mathrm{A}$$ and $$\mathrm{B}$$ move round the earth in the same orbit. The mass of $$\mathrm{A}$$ is twice the mass of $$\mathrm{B}$$. The quantity which is same for the two satellites will be</p> | [{"identifier": "A", "content": "Potential energy"}, {"identifier": "B", "content": " Kinetic energy"}, {"identifier": "C", "content": "Total energy"}, {"identifier": "D", "content": "Speed"}] | ["D"] | null | The quantity which is same for the two satellites will be speed.
<br/><br/>
The orbital speed of a satellite is independent of the mass of the satellite, but it depends on the radius of the orbit. Potential energy, kinetic energy and total energy depend on the mass of the the satellite.
<br/><br/>
Therefore, the only q... | mcq | jee-main-2023-online-12th-april-morning-shift | 11,154 |
1lgsx75tb | physics | gravitation | escape-speed-and-motion-of-satellites | <p>A space ship of mass $$2 \times 10^{4} \mathrm{~kg}$$ is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the space ship in the orbit to overcome the gravitational pull will be (if $$g=10 \mathrm{~m} / \mathrm{s}^{2}$$ and radius of earth $$=6400 \mathrm{~km}$$ ):<... | [{"identifier": "A", "content": "$$7.9(\\sqrt{2}-1) \\mathrm{km} / \\mathrm{s}$$"}, {"identifier": "B", "content": "$$11.2(\\sqrt{2}-1) \\mathrm{km} / \\mathrm{s}$$"}, {"identifier": "C", "content": "$$7.4(\\sqrt{2}-1) \\mathrm{km} / \\mathrm{s}$$"}, {"identifier": "D", "content": "$$8(\\sqrt{2}-1) \\mathrm{km} / \\mat... | ["D"] | null | <p>To find the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit, we first need to find the orbital velocity. The formula for orbital velocity (v) is given by:</p>
<p>$$v = \sqrt{\frac{GM}{r}}$$</p>
<p>where G is the gravitational constant, M is the mass of Earth, and r... | mcq | jee-main-2023-online-11th-april-evening-shift | 11,155 |
1lgvrp2gw | physics | gravitation | escape-speed-and-motion-of-satellites | <p>The time period of a satellite, revolving above earth's surface at a height equal to $$\mathrm{R}$$ will be</p>
<p>(Given $$g=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}, \mathrm{R}=$$ radius of earth)</p> | [{"identifier": "A", "content": "$$\\sqrt{32 R}$$"}, {"identifier": "B", "content": "$$\\sqrt{4 \\mathrm{R}}$$"}, {"identifier": "C", "content": "$$\\sqrt{8 R}$$"}, {"identifier": "D", "content": "$$\\sqrt{2 R}$$"}] | ["A"] | null | <p>For a satellite orbiting the Earth at a height equal to Earth's radius, its distance from the center of the Earth will be 2R, where R is the radius of the Earth.</p>
<p>Using the formula for the gravitational force:</p>
<p>$$F = G\frac{Mm}{r^2}$$</p>
<p>where F is the gravitational force, G is the gravitational ... | mcq | jee-main-2023-online-10th-april-evening-shift | 11,156 |
1lgyqs0ch | physics | gravitation | escape-speed-and-motion-of-satellites | <p>The orbital angular momentum of a satellite is L, when it is revolving in a circular orbit at height h from earth surface. If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be -</p> | [{"identifier": "A", "content": "9L"}, {"identifier": "B", "content": "8L"}, {"identifier": "C", "content": "4L"}, {"identifier": "D", "content": "3L"}] | ["D"] | null | <p>If we take the velocity (v) of the satellite to be $v = \sqrt{GM/r}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance from the center of the Earth to the satellite, then the orbital angular momentum (L) is:</p>
<p>$ L = mvr = m \sqrt{GMr} $</p>
<p>If the distance (r) fr... | mcq | jee-main-2023-online-8th-april-evening-shift | 11,158 |
1lh24ku0k | physics | gravitation | escape-speed-and-motion-of-satellites | <p>Given below are two statements : one is labelled as Assertion $$\mathbf{A}$$ and the other is labelled as Reason $$\mathbf{R}$$.</p>
<p>Assertion A : Earth has atmosphere whereas moon doesn't have any atmosphere.</p>
<p>Reason R : The escape velocity on moon is very small as compared to that on earth.</p>
<p>In the ... | [{"identifier": "A", "content": "$$\\mathbf{A}$$ is false but $$\\mathbf{R}$$ is true"}, {"identifier": "B", "content": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct but $$\\mathbf{R}$$ is NOT the correct explanation of $$\\mathbf{A}$$"}, {"identifier": "C", "content": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ ... | ["C"] | null | <p>The assertion (A) is true: Earth does have an atmosphere, while the Moon does not have a significant atmosphere.</p>
<p>The reason (R) is also true: The escape velocity on the Moon is indeed smaller than that on Earth. The escape velocity is the minimum velocity an object must have to escape the gravitational pull o... | mcq | jee-main-2023-online-6th-april-morning-shift | 11,160 |
jaoe38c1lscp51wv | physics | gravitation | escape-speed-and-motion-of-satellites | <p>Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).</p>
<p>Assertion (A) : The angular speed of the moon in its orbit about the earth is more than the angular speed of the earth in its orbit about the sun.</p>
<p>Reason (R) : The moon takes less time to move aro... | [{"identifier": "A", "content": "Both (A) and (R) are correct but (R) is not the correct explanation of (A)"}, {"identifier": "B", "content": "(A) is correct but (R) is not correct"}, {"identifier": "C", "content": "Both (A) and (R) are correct and (R) is the correct explanation of (A)"}, {"identifier": "D", "content":... | ["C"] | null | <p>The angular speed $$ \omega $$ of an object in circular motion is given by the equation</p>
<p>$ \omega = \frac{2\pi}{T} $</p>
<p>where $$ T $$ is the period of the motion - the time it takes to make one complete revolution.</p>
<p>Assertion (A) states that the angular speed of the Moon in its orbit around the Earth... | mcq | jee-main-2024-online-27th-january-evening-shift | 11,162 |
jaoe38c1lsd83bit | physics | gravitation | escape-speed-and-motion-of-satellites | <p>The mass of the moon is $$\frac{1}{144}$$ times the mass of a planet and its diameter is $$\frac{1}{16}$$ times the diameter of a planet. If the escape velocity on the planet is $$v$$, the escape velocity on the moon will be :</p> | [{"identifier": "A", "content": "$$\\frac{\\mathrm{v}}{4}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\mathrm{v}}{6}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\mathrm{V}}{12}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\mathrm{v}}{3}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \mathrm{V}_{\text {escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\
& \mathrm{V}_{\text {planet }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\mathrm{V} \\
& \mathrm{V}_{\text {Moon }}=\sqrt{\frac{2 \mathrm{GM} \times 16}{144 \mathrm{R}}}=\frac{1}{3} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\
& ... | mcq | jee-main-2024-online-31st-january-evening-shift | 11,163 |
1lsg6odx4 | physics | gravitation | escape-speed-and-motion-of-satellites | <p>Escape velocity of a body from earth is $$11.2 \mathrm{~km} / \mathrm{s}$$. If the radius of a planet be onethird the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is :</p> | [{"identifier": "A", "content": "7.9 km/s"}, {"identifier": "B", "content": "8.4 km/s"}, {"identifier": "C", "content": "4.2 km/s"}, {"identifier": "D", "content": "11.2 km/s"}] | ["A"] | null | <p>$$\begin{aligned}
& \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{E}}}{3}, \mathrm{M}_{\mathrm{P}}=\frac{\mathrm{M}_{\mathrm{E}}}{6} \\
& \mathrm{~V}_{\mathrm{c}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}} \quad \text{.... (i)}\\
& \mathrm{V}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm... | mcq | jee-main-2024-online-30th-january-evening-shift | 11,164 |
luxweb2z | physics | gravitation | escape-speed-and-motion-of-satellites | <p>A satellite of $$10^3 \mathrm{~kg}$$ mass is revolving in circular orbit of radius $$2 R$$. If $$\frac{10^4 R}{6} \mathrm{~J}$$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius</p>
<p>(use $$g=10 \mathrm{~m} / \mathrm{s}^2, R=$$ radius of earth)</p> | [{"identifier": "A", "content": "4 R"}, {"identifier": "B", "content": "6 R"}, {"identifier": "C", "content": "2.5 R"}, {"identifier": "D", "content": "3 R"}] | ["B"] | null | <p>To determine the new radius of the orbit after the energy is supplied to the satellite, we need to compare the initial and final energies of the satellite in its orbit around the Earth. We will use the formula for total energy (the sum of kinetic and potential energy) of a satellite in a circular orbit:</p>
<p>The ... | mcq | jee-main-2024-online-9th-april-evening-shift | 11,165 |
lv9s213g | physics | gravitation | escape-speed-and-motion-of-satellites | <p>A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is :</p>
<p>(Given $$=$$ Radius of geo-stationary orbit for earth is $$4.2 \times 10^4 \mathrm{~km}$$)</p> | [{"identifier": "A", "content": "$$1.68 \\times 10^5 \\mathrm{~km}$$\n"}, {"identifier": "B", "content": "$$1.4 \\times 10^4 \\mathrm{~km}$$\n"}, {"identifier": "C", "content": "$$8.4 \\times 10^4 \\mathrm{~km}$$\n"}, {"identifier": "D", "content": "$$1.05 \\times 10^4 \\mathrm{~km}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \frac{T_1}{T_2}=\left(\frac{r_1}{r_2}\right)^{3 / 2} \sqrt{\frac{m_2}{m_1}} \\
& \Rightarrow \quad \frac{24}{6}=\left(\frac{4.2 \times 10^4}{r_2}\right)^{3 / 2} \sqrt{\frac{m / 4}{m}} \\
& \Rightarrow r_2=1.05 \times 10^4 \mathrm{~km}
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift | 11,168 |
BRlFiH6S5JAIOpYZ | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | Energy required to move a body of mass $$m$$ from an orbit of radius $$2R$$ to $$3R$$ is | [{"identifier": "A", "content": "$${{GMm} \\over {12{R^2}}}$$"}, {"identifier": "B", "content": "$${{GMm} \\over {3{R^2}}}$$"}, {"identifier": "C", "content": "$${{GMm} \\over {8R}}$$ "}, {"identifier": "D", "content": "$${{GMm} \\over {6R}}$$ "}] | ["D"] | null | Gravitational potential energy E = $$ - {{GMm} \over r}$$
<br><br> where M = mass of earth
<br><br>m = mass of body
<br><br>r = radius of earth
<br><br>Energy required to move a body of mass $$m$$ from an orbit of radius $$2R$$ to $$3R$$
<br><br>$$=$$ (Potential energy of the Earth-mass system when mass is at distanc... | mcq | aieee-2002 | 11,169 |
3doGoWyeohUWtqgf | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | 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 | [{"identifier": "A", "content": "$${1 \\over 4}mgR$$ "}, {"identifier": "B", "content": "$$2mgR$$ "}, {"identifier": "C", "content": "$${1 \\over 2}mgR$$ "}, {"identifier": "D", "content": "$$mgR$$ "}] | ["C"] | null | Gravitational potential energy on the earth surface of a body
<br><br>U = $$-{{GmM} \over R}$$
<br><br>And at the height h from the earth surface the potential energy
<br><br> $${U_h} = - {{GmM} \over {R + h}}$$ = $$ - {{GmM} \over {2R}}$$ [ as h = R ]
<br><br>So the gain in the potential energy
<br><br>$$\Delta U =... | mcq | aieee-2004 | 11,170 |
gH901l7W9tkB4quo | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | From a solid sphere of mass $$M$$ and radius $$R,$$ a spherical portion of radius $$R/2$$ is removed, as shown in the figure. Taking gravitational potential $$V=0$$ at $$r = \infty ,$$ the potential at the center of the cavity thus formed is: <br/>($$G=gravitational $$ $$constant$$)<img src="data:image/png;base64,UklGR... | [{"identifier": "A", "content": "$${{ - 2GM} \\over {3R}}$$ "}, {"identifier": "B", "content": "$${{ - 2GM} \\over R}$$ "}, {"identifier": "C", "content": "$${{ - GM} \\over {2R}}$$ "}, {"identifier": "D", "content": "$${{ - GM} \\over R}$$ "}] | ["D"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/a49Jv8rcFlHMGgTxP/PGzAOvXknGnmshjXoIKxO5OwI36wd/Ma8OvyTk33ymmGKc9umGOE/image.svg" loading="lazy" alt="JEE Main 2015 (Offline) Physics - Gravitation Question 165 English Explanation">
<br><br>Before removing the spherical portion, potential at point $$P$$... | mcq | jee-main-2015-offline | 11,173 |
NLt4siTZCUeim2398ojgy2xukg0hqiet | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | Two planets have masses M and 16 M and their radii are $$a$$ and 2$$a$$, respectively. The separation
between the centres of the planets is 10$$a$$. A body of mass m is fired from the surface of the larger
planet towards the smaller planet along the line joining their centres. For the body to be able to
reach at the su... | [{"identifier": "A", "content": "$$2\\sqrt {{{GM} \\over a}} $$"}, {"identifier": "B", "content": "$$\\sqrt {{{G{M^2}} \\over {ma}}} $$"}, {"identifier": "C", "content": "$${3 \\over 2}\\sqrt {{{5GM} \\over a}} $$"}, {"identifier": "D", "content": "$$4\\sqrt {{{GM} \\over a}} $$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265795/exam_images/s4q8yayuwtszongry8fp.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Evening Slot Physics - Gravitation Question 124 English Explanation">
<br><br>Let at... | mcq | jee-main-2020-online-6th-september-evening-slot | 11,174 |
R3koN9GdIiZHtOapbr1kmipx3uy | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | If one wants to remove all the mass of the earth to infinity in order to break it up completely.<br/><br/>The amount of energy that needs to be supplied will be $${x \over 5}{{G{M^2}} \over R}$$ where x is __________ (Round off to the Nearest Integer) (M is the mass of earth, R is the radius of earth, G is the gravitat... | [] | null | 3 | <p>We know that binding energy of earth,</p>
<p>$$BE = - {3 \over 5}{{G{M^2}} \over R}$$</p>
<p>$$\therefore$$ Energy required to break the earth into pieces</p>
<p>$$ = - BE = {3 \over 5}{{G{M^2}} \over R}$$ ...... (i)</p>
<p>According to question, the amount of energy that needs to be supplied is $${x \over 5}{{G{M... | integer | jee-main-2021-online-16th-march-evening-shift | 11,175 |
1krsuid2y | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height h is ___________ s. | [{"identifier": "A", "content": "$$\\sqrt {{{2{R_e}} \\over g}} \\left[ {{{\\left( {1 + {h \\over {{R_e}}}} \\right)}^{{3 \\over 2}}} - 1} \\right]$$"}, {"identifier": "B", "content": "$${1 \\over 3}\\sqrt {{{{R_e}} \\over {2g}}} \\left[ {{{\\left( {1 + {h \\over {{R_e}}}} \\right)}^{{3 \\over 2}}} - 1} \\right]$$"}, {... | ["D"] | null | $${1 \over 2}m{v^2} - {{GMm} \over r} = 0 \Rightarrow v = \sqrt {{{2GM} \over r}} $$<br><br>$${{dr} \over {dt}} = \sqrt {{{2GM} \over r}} $$<br><br>$$ \Rightarrow \int\limits_{{R_e}}^{({R_e} + h)} {\sqrt r dr = \int\limits_0^t {\sqrt {2GM} dt} } $$<br><br>$$ \Rightarrow {2 \over 3}\left[ {{{({R_e} + h)}^{3/2}} - R_e^{3... | mcq | jee-main-2021-online-22th-july-evening-shift | 11,176 |
1kryy4goq | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | Suppose two planets (spherical in shape) in radii R and 2R, but mass M and 9M respectively have a centre to centre separation 8 R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for t... | [] | null | 4 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263875/exam_images/td73mxvwswfr3btpydgj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Physics - Gravitation Question 99 English Explanation"><br>Acceleration du... | integer | jee-main-2021-online-27th-july-morning-shift | 11,177 |
1ktfm4sii | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100 kg and radius 50 m. If the gravitational potential at a point, 25 m from the centre is V kg/m. The value of V is : | [{"identifier": "A", "content": "$$-$$60 G"}, {"identifier": "B", "content": "+2 G"}, {"identifier": "C", "content": "$$-$$20 G"}, {"identifier": "D", "content": "$$-$$4 G"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267842/exam_images/liy4vlvgeuij7zkczva8.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Physics - Gravitation Question 94 English Explanation"><br>$${V_A} = \le... | mcq | jee-main-2021-online-27th-august-evening-shift | 11,180 |
1l55juo36 | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>Water falls from a 40 m high dam at the rate of 9 $$\times$$ 10<sup>4</sup> kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydroelectric energy number of 100 W lamps, that can be lit, is :</p>
<p>(Take g = 10 ms<sup>$$-$$2</sup>)</p> | [{"identifier": "A", "content": "25"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "18"}] | ["B"] | null | <p>Total gravitational PE of water per second $$ = {{mgh} \over T}$$</p>
<p>$$ = {{9 \times {{10}^4} \times 10 \times 40} \over {3600}} = {10^4}$$ J/sec</p>
<p>50% of this energy can be converted into electrical energy so total electrical energy $$ = {{{{10}^4}} \over 2} = 5000$$ W</p>
<p>So total bulbs lit can be $$ =... | mcq | jee-main-2022-online-28th-june-evening-shift | 11,181 |
1l56un577 | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>Four spheres each of mass m from a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is :</p>
<p> <img src="data:image/png;base64,UklGRogLAABXRUJQVlA4IHwLAADQ3gCdASq5AgADP4HA3WU2Ma4nITUJOsAwCWlu8p8d/K6iN+fx9P30d... | [{"identifier": "A", "content": "$$ - {{Gm} \\over d}\\left[ {(4 + \\sqrt 2 )m + 4\\sqrt 2 M} \\right]$$"}, {"identifier": "B", "content": "$$ - {{Gm} \\over d}\\left[ {(4 + \\sqrt 2 )M + 4\\sqrt 2 m} \\right]$$"}, {"identifier": "C", "content": "$$ - {{Gm} \\over d}\\left[ {3{m^2} + 4\\sqrt 2 M} \\right]$$"}, {"identi... | ["A"] | null | <p>Total gravitational potential energy</p>
<p>$$ = - \left\{ {{{4GMm} \over {d/\sqrt 2 }} + {{4G{m^2}} \over d} + {{2G{m^2}} \over {\sqrt 2 d}}} \right\}$$</p>
<p>$$ = - {{Gm} \over d}\left\{ {M4\sqrt 2 + (4 + \sqrt 2 )m} \right\}$$</p>
<p>$$ = - {{Gm} \over d}\left\{ {4\sqrt 2 M + (4 + \sqrt 2 )m} \right\}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift | 11,182 |
1l6rh9ui9 | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>An object of mass $$1 \mathrm{~kg}$$ is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be [If, $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ and radius of earth $$=6400 \mathrm{~km}$$ ]</p> | [{"identifier": "A", "content": "48 MJ"}, {"identifier": "B", "content": "24 MJ"}, {"identifier": "C", "content": "36 MJ"}, {"identifier": "D", "content": "12 MJ"}] | ["A"] | null | $\Delta U=U_{f}-U_{i}$
<br/><br/>$
\begin{aligned}
&=-\frac{G M m}{4 R}+\frac{G M m}{R} \\\\
&=\frac{3 G M m}{4 R}=\frac{3}{4} m g R \\\\
&=48 \mathrm{MJ}
\end{aligned}
$ | mcq | jee-main-2022-online-29th-july-evening-shift | 11,183 |
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