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1ldr2al1l | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>If the gravitational field in the space is given as $$\left(-\frac{K}{r^{2}}\right)$$. Taking the reference point to be at $$\mathrm{r}=2 \mathrm{~cm}$$
with gravitational potential $$\mathrm{V}=10 \mathrm{~J} / \mathrm{kg}$$. Find the gravitational potential at $$\mathrm{r}=3 \mathrm{~cm}$$ in SI unit (Given, that... | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "12"}] | ["B"] | null | <p>$$E = - {K \over {{r^2}}}$$</p>
<p>$$\Delta V = - \int\limits_{r = 2\,cm}^{3\,cm} {E.\,dr} $$</p>
<p>$$ = \int\limits_2^3 {{k \over {{r^2}}}dr} $$</p>
<p>$$ = \left[ { - {K \over r}} \right]_2^3 = \left( {{K \over 6}} \right) = {6 \over 6} = 1$$ J/kg</p>
<p>$${V_f} - {V_i} = 1$$</p>
<p>$$ \Rightarrow {V_f} - 10 = ... | mcq | jee-main-2023-online-30th-january-morning-shift | 11,185 |
1ldtzcnjc | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>A body of mass is taken from earth surface to the height h equal to twice the radius of earth (R$$_e$$), the increase in potential energy will be :</p>
<p>(g = acceleration due to gravity on the surface of Earth)</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}mgR_e$$"}, {"identifier": "B", "content": "$$3~mgR_e$$"}, {"identifier": "C", "content": "$$\\frac{1}{3}mgR_e$$"}, {"identifier": "D", "content": "$$\\frac{2}{3}mgR_e$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ledl4rmn/eddbd67a-5fa1-405c-a90c-053d6dba7a7e/79ff16f0-b189-11ed-94f5-239612b3abb5/file-1ledl4rmo.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ledl4rmn/eddbd67a-5fa1-405c-a90c-053d6dba7a7e/79ff16f0-b189-11ed-94f5-239612b3abb5... | mcq | jee-main-2023-online-25th-january-evening-shift | 11,186 |
lgnybh2l | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | A body is released from a height equal to the radius $(\mathrm{R})$ of the earth. The velocity of the body when it strikes the surface of the earth will be
<br/><br/>
(Given $g=$ acceleration due to gravity on the earth.) | [{"identifier": "A", "content": "$\\sqrt{\\frac{g R}{2}}$"}, {"identifier": "B", "content": "$\\sqrt{4 g R}$"}, {"identifier": "C", "content": "$\\sqrt{2 g R}$"}, {"identifier": "D", "content": "$\\sqrt{g R}$"}] | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgrfzyob/3ce53df2-7a14-45dd-8f83-57e6063a144d/ce9901b0-e0c0-11ed-9768-bd0baf44cdee/file-1lgrfzyoc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgrfzyob/3ce53df2-7a14-45dd-8f83-57e6063a144d/ce9901b0-e0c0-11ed-9768-bd0baf44cdee/fi... | mcq | jee-main-2023-online-15th-april-morning-shift | 11,187 |
1lgsxaxnl | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>If $$\mathrm{V}$$ is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:-</p> | [{"identifier": "A", "content": "$$\\frac{3 \\mathrm{~V}}{2}$$"}, {"identifier": "B", "content": "$$\\frac{\\mathrm{V}}{2}$$"}, {"identifier": "C", "content": "$$\\frac{4}{3} \\mathrm{~V}$$"}, {"identifier": "D", "content": "$$\\mathrm{V}$$"}] | ["A"] | null | The gravitational potential (V) due to a sphere of uniform density at a distance r from its center is given by:
<br/><br/>
$$
V(r) = \frac{GM}{2R^3} \left(3R^2 - r^2\right)
$$
<br/><br/>
At the surface of the sphere (r = R), the gravitational potential is:
<br/><br/>
$$
V = \frac{GM}{R}
$$
<br/><br/>
Now, let's find th... | mcq | jee-main-2023-online-11th-april-evening-shift | 11,188 |
1lsgx6elh | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>The gravitational potential at a point above the surface of earth is $$-5.12 \times 10^7 \mathrm{~J} / \mathrm{kg}$$ and the acceleration due to gravity at that point is $$6.4 \mathrm{~m} / \mathrm{s}^2$$. Assume that the mean radius of earth to be $$6400 \mathrm{~km}$$. The height of this point above the earth's su... | [{"identifier": "A", "content": "1600 km"}, {"identifier": "B", "content": "1200 km"}, {"identifier": "C", "content": "540 km"}, {"identifier": "D", "content": "1000 km"}] | ["A"] | null | <p>$$-\frac{G M_E}{R_E+h}=-5.12 \times 10^{-7}$$ .... (i)</p>
<p>$$\frac{G M_E}{\left(R_E+h\right)^2}=6.4$$ ..... (ii)</p>
<p>By (i) and (ii)</p>
<p>$$\Rightarrow h=16 \times 10^5 \mathrm{~m}=1600 \mathrm{~km}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 11,189 |
lv7v4o5n | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>If $$\mathrm{G}$$ be the gravitational constant and $$\mathrm{u}$$ be the energy density then which of the following quantity have the dimensions as that of the $$\sqrt{\mathrm{uG}}$$ :</p> | [{"identifier": "A", "content": "Gravitational potential\n"}, {"identifier": "B", "content": "pressure gradient per unit mass\n"}, {"identifier": "C", "content": "Energy per unit mass\n"}, {"identifier": "D", "content": "Force per unit mass"}] | ["D"] | null | <p>To determine the dimension of the quantity $$\sqrt{uG}$$, we first need to understand the dimensions of both the gravitational constant (G) and the energy density (u).</p>
<p>The gravitational constant $$G$$ has dimensions given by:
<p>$$[G] = M^{-1}L^{3}T^{-2}$$</p>
<p>where $$M$$ stands for mass, $$L$$ for leng... | mcq | jee-main-2024-online-5th-april-morning-shift | 11,190 |
lv7v3v9l | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>Match List I with List II :</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;... | [{"identifier": "A", "content": "(A)-(II), (B)-(I), (C)-(IV), (D)-(III)\n"}, {"identifier": "B", "content": "(A)-(I), (B)-(II), (C)-(III), (D)-(IV)\n"}, {"identifier": "C", "content": "(A)-(III), (B)-(IV), (C)-(I), (D)-(II)\n"}, {"identifier": "D", "content": "(A)-(I), (B)-(IV), (C)-(II), (D)-(III)"}] | ["A"] | null | <p>$$\text { K.E }=\frac{G M m}{2 a} \quad \text{(II)}$$</p>
<p>$$U_G=\frac{-G M m}{a} \quad \text{(I)}$$</p>
<p>$$M . E=\frac{-G M m}{2 a} \quad \text{(IV)}$$</p>
<p>$$\text { and Escape Energy }=\frac{G m}{r} \quad \text{(III)}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift | 11,191 |
lvc57pmu | physics | gravitation | gravitational-potential-and-gravitational-potential-energy | <p>To project a body of mass $$m$$ from earth's surface to infinity, the required kinetic energy is (assume, the radius of earth is $$R_E, g=$$ acceleration due to gravity on the surface of earth):</p> | [{"identifier": "A", "content": "$$1 / 2 m g R_E$$\n"}, {"identifier": "B", "content": "$$4 m g R_E$$\n"}, {"identifier": "C", "content": "$$m g R_E$$\n"}, {"identifier": "D", "content": "$$2 m g R_E$$"}] | ["C"] | null | <p>The kinetic energy required to project a body of mass $m$ from the Earth's surface to infinity, also known as the escape kinetic energy, can be calculated using the concept of gravitational potential energy. The escape velocity $v_e$ is the velocity a body must have to escape the gravitational field of the Earth wit... | mcq | jee-main-2024-online-6th-april-morning-shift | 11,192 |
6qCaJGywpVm4aQ1h | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | If suddenly the gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the satellite will | [{"identifier": "A", "content": "continue to move in its orbit with same velocity"}, {"identifier": "B", "content": "move tangentially to the original orbit with the same velocity"}, {"identifier": "C", "content": "become stationary in its orbit"}, {"identifier": "D", "content": "move towards the earth"}] | ["B"] | null | When gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the centripetal force becomes zero. So the satellite will move tangentially to the original orbit with the same velocity as it has at the instant when gravitational force becomes zero. | mcq | aieee-2002 | 11,193 |
R4pcXzy0527zrAzT | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Two spherical bodies of mass $$M$$ and $$5M$$ & radii $$R$$ & $$2R$$ respectively are released in free space with initial separation between their centers equal to $$12R$$. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is | [{"identifier": "A", "content": "$$2.5$$ $$R$$ "}, {"identifier": "B", "content": "$$4.5$$ $$R$$ "}, {"identifier": "C", "content": "$$7.5$$ $$R$$ "}, {"identifier": "D", "content": "$$1.5$$ $$R$$ "}] | ["C"] | null | <br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265771/exam_images/ur6rvgpqywvuyclxwos4.webp" loading="lazy" alt="AIEEE 2003 Physics - Gravitation Question 182 English Explanation">
<br><br>Let $$t$$ be the time taken for the two masses to collide and $${x_{5M,}}\,{x_M}$$ be ... | mcq | aieee-2003 | 11,195 |
ZcBvISXpJculBaNl | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | The time period of an earth satellite in circular orbit is independent of | [{"identifier": "A", "content": "both the mass and radius of the orbit"}, {"identifier": "B", "content": "radius of its orbit "}, {"identifier": "C", "content": "the mass of the satellite "}, {"identifier": "D", "content": "neither the mass of the satellite nor the radius of its orbit "}] | ["C"] | null | For satellite, gravitational force = centripetal force
<br><br>$$\therefore$$ $${{m{v^2}} \over {R + x}} = {{GmM} \over {{{\left( {R + x} \right)}^2}}}$$
<br><br>$$x=$$ height of satellite from earth surface
<br><br>$$m=$$ mass of satellite
<br><br>$$ \Rightarrow {v^2} = {{GM} \over {\left( {R + x} \right)}}$$ or $$v... | mcq | aieee-2004 | 11,196 |
fKNFrQLaoB5ALOA8 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | A satellite of mass $$m$$ revolves around the earth of radius $$R$$ at a height $$x$$ from its surface. If $$g$$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is | [{"identifier": "A", "content": "$${{g{R^2}} \\over {R + x}}$$ "}, {"identifier": "B", "content": "$${{gR} \\over {R - x}}$$ "}, {"identifier": "C", "content": "$${gx}$$ "}, {"identifier": "D", "content": "$${\\left( {{{g{R^2}} \\over {R + x}}} \\right)^{1/2}}$$ "}] | ["D"] | null | Gravitational force applied on the satellite,
<br><br>= $${{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$$
<br><br>For satellite, gravitational force = centripetal force
<br><br>$$\therefore$$ $${{m{v^2}} \over {\left( {R + x} \right)}} = {{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$$
<br><br>where $$v$$ is the o... | mcq | aieee-2004 | 11,197 |
HX17M0EaT8QqzrLp | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Average density of the earth | [{"identifier": "A", "content": "is a complex function of $$g$$ "}, {"identifier": "B", "content": "does not depend on $$g$$ "}, {"identifier": "C", "content": "is inversely proportional to $$g$$ "}, {"identifier": "D", "content": "is directly proportional to $$g$$ "}] | ["D"] | null | Mass of earth = Volume $$ \times $$ Density of earth($$\rho$$)
<br><br>$$\therefore$$ M = $${{4 \over 3}\pi {R^3}} $$$$ \times $$ $$\rho $$
<br><br>We know, $$g = {{GM} \over {{R^2}}}$$
<br><br>$$ \Rightarrow g = {{G \times \rho \times {4 \over 3}\pi {R^3}} \over {{R^2}}}$$
<br><br>$$ \Rightarrow $$ $$g = {4 \over 3}\... | mcq | aieee-2005 | 11,198 |
4CQFSMKw1f0zwJ0A | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | This question contains Statement - $$1$$ and Statement - $$2$$. of the four choices given after the statements, choose the one that best describes the two statements.
<br/><br/><b>Statement - $$1$$:</b>
<br/><br/>For a mass $$M$$ kept at the center of a cube of side $$'a'$$, the flux of gravitational field passing thro... | [{"identifier": "A", "content": "Statement - $$1$$ is false, Statement - $$2$$ is true"}, {"identifier": "B", "content": "Statement - $$1$$ is true, Statement - $$2$$ is true; Statement - $$2$$ is a correct explanation for Statement - $$1$$"}, {"identifier": "C", "content": "Statement - $$1$$ is true, Statement - $$2$$... | ["B"] | null | Gravitational field $$\overrightarrow g $$ = $$ - {{GM} \over {{r^2}}}$$
<br><br>where, $$M=$$ mass enclosed in the closed surface
<br><br>Gravitational flux through a closed surface is given by
<br><br>$${\left| {\overrightarrow g .d\overrightarrow S } \right|}$$ = $$4\pi {r^2}.{{GM} \over {{r^2}}}$$ = $$4\pi GM$$
<... | mcq | aieee-2008 | 11,199 |
ahlnTnaPo4XNBFY8 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Four particles, each of mass $$M$$ and equidistant from each other, move along a circle of radius $$R$$ under the action of their mutual gravitational attraction. The speed of each particle is : | [{"identifier": "A", "content": "$$\\sqrt {{{GM} \\over R}} $$ "}, {"identifier": "B", "content": "$$\\sqrt {2\\sqrt 2 {{GM} \\over R}} $$ "}, {"identifier": "C", "content": "$$\\sqrt {{{GM} \\over R}\\left( {1 + 2\\sqrt 2 } \\right)} $$ "}, {"identifier": "D", "content": "$${1 \\over 2}\\sqrt {{{GM} \\over R}\\left( {... | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267664/exam_images/irzeijsu6dn5wjuqjdz2.webp" loading="lazy" alt="JEE Main 2014 (Offline) Physics - Gravitation Question 167 English Explanation">
All those particles are moving due to their mutual gravitational attraction.
<br><br>... | mcq | jee-main-2014-offline | 11,200 |
D1B7JPA4oRHdibvJyjLN7 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is $${1 \over 4}$$ the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If t<sub>1 </sub> is the time taken for planet to go over path abc and t<sub>2</sub> for path taken over... | [{"identifier": "A", "content": "t<sub>1</sub> = t<sub>2</sub>"}, {"identifier": "B", "content": "t<sub>1</sub> = 2t<sub>2</sub>"}, {"identifier": "C", "content": "t<sub>1</sub> = 3t<sub>2</sub>"}, {"identifier": "D", "content": "t<sub>1</sub> = 4t<sub>2</sub> "}] | ["C"] | null | Let the area of ellipse = A
<br><br>$$ \therefore $$ Area of abcSa = $${A \over 2} + {A \over 4}$$ = Area of half of the ellipse + Area of the triangle = $${{3A} \over 4}$$
<br><br>Area of adcSa = $${A \over 2} - {A \over 4}$$ = $${A \over 4}$$
<br><br>$$ \therefore $$ $${{{t_1}} \ov... | mcq | jee-main-2016-online-9th-april-morning-slot | 11,201 |
dpTC7VlO8TZX66eCroaTp | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | An astronaut of mass m is working on a satellite orbiting the earth at a distance h from the earth’s surface. The radius of the earth is R, while its mass is M. The gravitational pull F<sub>G</sub> on the astronaut is :
| [{"identifier": "A", "content": "Zero since astronaut feels weightless "}, {"identifier": "B", "content": "0 < F<sub>G</sub> < $${{GMm} \\over {{R^2}}}$$ "}, {"identifier": "C", "content": "$${{GMm} \\over {{{\\left( {R + h} \\right)}^2}}}$$ < F<sub>G</sub> < $${{GMm} \\over {{R^2}}}$$ "}, {"identifier": "D... | null | null | The gravitional pull on the astronaut is,
<br><br>F<sub>G</sub> = $${{GMm} \over {{{\left( {R + h} \right)}^2}}}$$
<br><br>The satellite is moving so fast around the earth that whenever it trying to fall on the earch it is missing the earth and it will keep going arround the earth. On the satelli... | mcqm | jee-main-2016-online-10th-april-morning-slot | 11,202 |
ny8ocbFf8QMDPlOwFsttQ | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | A body of mass m is moving in a circular orbit of radius R about a planet of mass M. At some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius $${R \over 2},$$ and the other mass, in a circular orbit of radius $${3R \over 2}$$. The difference between the final and initial tota... | [{"identifier": "A", "content": "$$ - {{GMm} \\over {2R}}$$ "}, {"identifier": "B", "content": "$$ + {{GMm} \\over {6R}}$$"}, {"identifier": "C", "content": "$${{GMm} \\over {2R}}$$"}, {"identifier": "D", "content": "$$ - {{GMm} \\over {6R}}$$"}] | ["D"] | null | Initially gravitational potenrial energy
<br><br>E<sub>i</sub> = $$-$$ $${{GMm} \over {2R}}$$
<br><br>Final gravitational potential energy
<br><br>E<sub>f</sub> = $$-$$ $${{GM\left( {{m \over 2}} \right)} \over {2\left( {{R \over 2}} \right)}}$$ $$-$$ $${{GM\left( {{m \over 2}} \right)} \over {2\left( {{{3R} \over 2... | mcq | jee-main-2018-online-15th-april-morning-slot | 11,205 |
QIkfT9oveO52Y6u2byKfH | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Take the mean distance of the moon and the sun from the earth to be $$0.4 \times {10^6}$$ km and $$150 \times {10^6}$$ km respectively. Their masses are $$8 \times {10^{22}}$$ kg and $$2 \times {10^{30}}$$ kg respectively. The radius of the earth is $$6400$$ km. Let $$\Delta {F_1}$$ be the difference in the forces exer... | [{"identifier": "A", "content": "$$2$$"}, {"identifier": "B", "content": "$${10^{ - 2}}$$"}, {"identifier": "C", "content": "$$0.6$$"}, {"identifier": "D", "content": "$$6$$"}] | ["A"] | null | As gravitational force of attraction,
<br><br>F = $${{GMm} \over {{R^2}}}$$
<br><br>$$\therefore\,\,\,\,$$ Force of attraction berween earth and moon
<br><br>F<sub>1</sub> = $${{G{M_e}m} \over {r_1^2}}$$
<br><br>Force of attraction between earth and sun,
<br><br>F<sub>2</sub> = $${{GMeMs} \over {r_2^2}}$$
<br><br>$$\... | mcq | jee-main-2018-online-15th-april-morning-slot | 11,206 |
aXBMAIzVxhLTr7gzAc3rsa0w2w9jx7j2kaz | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | The ratio of the weights of a body on the Earth’s surface to that on the surface of a planets is 9 : 4. The mass
of the planet is
$${1 \over 9}$$
th of that of the Earth. If 'R' is the radius of the Earth, what is the radius of the planet ?
(Take the planets to have the same mass density) | [{"identifier": "A", "content": "$${R \\over 9}$$"}, {"identifier": "B", "content": "$${R \\over 2}$$"}, {"identifier": "C", "content": "$${R \\over 3}$$"}, {"identifier": "D", "content": "$${R \\over 4}$$"}] | ["B"] | null | W = mg
<br><br>as m = constant everywhere
<br><br>$$ \therefore $$ W $$ \propto $$ g
<br><br>$${{{g_E}} \over {{g_p}}}$$ = $${9 \over 4}$$
<br><br>We know,
<br><br>$$g = {{GM} \over {{R^2}}}$$
<br><br>$$ \therefore $$ $${{{g_E}} \over {{g_p}}} = {{{M_E}} \over {{M_p}}} \times {{R_p^2} \over {R_E^2}}$$
<br><br>$$ \Right... | mcq | jee-main-2019-online-12th-april-evening-slot | 11,208 |
IjovubYYBDXwSEvtJk3rsa0w2w9jwzj35e6 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational
field of the planet acts on the spaceship, what will be the number of complete revolutions made by the
spaceship in 24 hours around the planet?
<br/><br/>[Given ; Mass of planet = 8 × 10<sup>22</sup> kg, Radius of... | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "11"}] | ["D"] | null | $${{m{V^2}} \over r} = {{GMm} \over {{r^2}}}$$<br><br>
$$V = \sqrt {{{GM} \over r}} $$<br><br>
$$n = {{VT} \over {2\pi r}} = \sqrt {{{GM} \over r}} {T \over {2\pi r}}$$<br><br>
$$ = \left( {\sqrt {{{GM} \over {{r^3}}}} } \right) \times {T \over {2\pi }} = \sqrt {{{6.67 \times {{10}^{ - 11}} \times 8 \times {{10}^{22}}}... | mcq | jee-main-2019-online-10th-april-evening-slot | 11,209 |
BLvNqZnVNi2UZOuJmedUr | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | A solid sphere of mass 'M' and radius 'a' is
surrounded by a uniform concentric spherical
shell of thickness 2a and mass 2M. The
gravitational field at distance '3a' from the
centre will be : | [{"identifier": "A", "content": "$${{GM} \\over {3{a^2}}}$$"}, {"identifier": "B", "content": "$${{2GM} \\over {9{a^2}}}$$"}, {"identifier": "C", "content": "$${{GM} \\over {9{a^2}}}$$"}, {"identifier": "D", "content": "$${{2GM} \\over {3{a^2}}}$$"}] | ["A"] | null | We use Gauss’s Law for gravitation<br><br>
g.4$$\pi $$r<sup>2</sup> = (Mass enclosed) 4$$\pi $$G<br><br>
$$g = {{3M4\pi G} \over {4\pi {{(3a)}^2}}} = {{GM} \over {3{a^2}}}$$
| mcq | jee-main-2019-online-9th-april-morning-slot | 11,210 |
mYzxw4TiukOxKfgLep7k9k2k5i769zr | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | A body A of mass m is moving in a circular orbit
of radius R about a planet. Another body B of
mass
$${m \over 2}$$
collides with A with a velocity which is half $$\left( {{{\overrightarrow v } \over 2}} \right)$$ the instantaneous velocity$${\overrightarrow v }$$
of A.
The collision is completely inelastic. Then, the
... | [{"identifier": "A", "content": "starts moving in an elliptical orbit around\nthe planet."}, {"identifier": "B", "content": "Falls vertically downwards towards the\nplanet"}, {"identifier": "C", "content": "Escapes from the Planet's Gravitational field."}, {"identifier": "D", "content": "continues to move in a circular... | ["A"] | null | Orbital speed for of A is v = $$\sqrt {{{GM} \over R}} $$
<br><br>After collision, let the combined mass moves
with speed v<sub>1</sub>
<br><br>$$ \therefore $$ mv + $${m \over 2}{v \over 2}$$ = $$\left( {{{3m} \over 2}} \right){v_1}$$
<br><br>$$ \Rightarrow $$ v<sub>1</sub> = $${{5v} \over 6}$$
<br><br>Since after col... | mcq | jee-main-2020-online-9th-january-morning-slot | 11,212 |
4I8WG8eRN0d7ejkTUzjgy2xukfrnbi36 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | A satellite is in an elliptical orbit around a planet P. It is observed that the velocity of the satellite
when it is farthest from the planet is 6 times less than that when it is closest to the planet. The
ratio of distances between the satellite and the planet at closest and farthest points is: | [{"identifier": "A", "content": "1 : 2"}, {"identifier": "B", "content": "1 : 3"}, {"identifier": "C", "content": "1 : 6"}, {"identifier": "D", "content": "3 : 4"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266297/exam_images/lhsxksjsnoud9wdnhmx4.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Physics - Gravitation Question 125 English Explanation">
By angular mom... | mcq | jee-main-2020-online-6th-september-morning-slot | 11,213 |
ikQPxx5wpz4uPOxaba1klrholsa | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be : | [{"identifier": "A", "content": "$$\\sqrt {{G \\over 2}(1 + 2\\sqrt 2 )} $$"}, {"identifier": "B", "content": "$$\\sqrt {{G \\over 2}(2\\sqrt 2 - 1)} $$"}, {"identifier": "C", "content": "$$\\sqrt {G(1 + 2\\sqrt 2 )} $$"}, {"identifier": "D", "content": "$${1\\over2}\\sqrt {G(1 + 2\\sqrt 2 )} $$"}] | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxehgyer/c578dd8a-84ff-4cd5-87c2-8f654e73200d/43a2b830-6178-11ec-90ef-97830a7fdb59/file-1kxehgyes.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxehgyer/c578dd8a-84ff-4cd5-87c2-8f654e73200d/43a2b830-6178-11ec-90ef-97830a7fdb59/fi... | mcq | jee-main-2021-online-24th-february-morning-slot | 11,214 |
7JnEoGPaIoBFxShumE1klri98sn | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is : | [{"identifier": "A", "content": "$${1 \\over {2\\pi }}\\sqrt {{{{d^3}} \\over {3Gm}}} $$"}, {"identifier": "B", "content": "$$2\\pi \\sqrt {{{3Gm} \\over {{d^3}}}} $$"}, {"identifier": "C", "content": "$${1 \\over {2\\pi }}\\sqrt {{{3Gm} \\over {{d^3}}}} $$"}, {"identifier": "D", "content": "$$2\\pi \\sqrt {{{{d^3}} \\... | ["D"] | null | The given situation is shown below<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxehbywy/aea2594a-3ff3-49a2-8b56-6e4bb066e705/b8facf10-6177-11ec-97a3-cfb8e5d281d3/file-1kxehbywz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxehbywy/aea2594a-3ff3-49a2-8b56-6e4bb066e7... | mcq | jee-main-2021-online-24th-february-morning-slot | 11,215 |
tRpFJfRFsTl5EOSBbL1klrwwvqy | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F<sub>1</sub>. Now a spherical cavity of radius $$\left( {{R \over 2}} \right)$$ is made in the sphere (as shown in figure) and the force becomes F<sub>2</sub>. The value of F<sub>1</sub> : F<sub>2</sub> is<br/><br/... | [{"identifier": "A", "content": "36 : 25"}, {"identifier": "B", "content": "41 : 50"}, {"identifier": "C", "content": "50 : 41"}, {"identifier": "D", "content": "25 : 36"}] | ["C"] | null | $${g_1} = {{GM} \over {{{(3R)}^2}}} = {{GM} \over {9{R^2}}}$$<br><br>$${g^2} = {{GM} \over {9{R^2}}} - {{G\left( {{M \over 8}} \right)} \over {{{\left( {3R - {R \over 2}} \right)}^2}}}$$<br><br>$$ = {{GM} \over {9{R^2}}} - {{GM} \over {{R^2}50}} = {{41} \over {9 \times 50}}{{GM} \over {{R^2}}}$$<br><br>$${{{g_1}} \over... | mcq | jee-main-2021-online-25th-february-morning-slot | 11,216 |
gsljhrfY7zIYTCsx3S1kltj103i | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | A planet revolving in elliptical orbit has :<br/><br/>A. a constant velocity of revolution.<br/><br/>B. has the least velocity when it is nearest to the sun.<br/><br/>C. its areal velocity is directly proportional to its velocity.<br/><br/>D. areal velocity is inversely proportional to its velocity.<br/><br/>E. to foll... | [{"identifier": "A", "content": "D only"}, {"identifier": "B", "content": "E only"}, {"identifier": "C", "content": "C only"}, {"identifier": "D", "content": "A only"}] | ["B"] | null | <p>According to Kepler’s second law of planetary motion, areal velocity of every planet moving around the sun should remain constant in elliptical orbit.</p> | mcq | jee-main-2021-online-26th-february-morning-slot | 11,217 |
uB8v7xxObniWZRhlP31kmhonmay | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | The maximum and minimum distances of a comet from the Sun are 1.6 $$\times$$ 10<sup>12</sup> m and 8.0 $$\times$$ 10<sup>10</sup> m respectively. If the speed of the comet at the nearest point is 6 $$\times$$ 10<sup>4</sup> ms<sup>$$-$$1</sup>, the speed at the farthest point is : | [{"identifier": "A", "content": "3.0 $$\\times$$ 10<sup>3</sup> m/s"}, {"identifier": "B", "content": "6.0 $$\\times$$ 10<sup>3</sup> m/s"}, {"identifier": "C", "content": "1.5 $$\\times$$ 10<sup>3</sup> m/s"}, {"identifier": "D", "content": "4.5 $$\\times$$ 10<sup>3</sup> m/s"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267811/exam_images/q0j574ymlqyfvbkzluoh.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Morning Shift Physics - Gravitation Question 112 English Explanation">
<br>v<sub>1</sub... | mcq | jee-main-2021-online-16th-march-morning-shift | 11,219 |
doA9Yy306JJCrcWKTC1kmkrmejs | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is : | [{"identifier": "A", "content": "9 T"}, {"identifier": "B", "content": "27 T"}, {"identifier": "C", "content": "12 T"}, {"identifier": "D", "content": "3 T"}] | ["B"] | null | <p>Kepler's Third Law states that the square of the period of a satellite's orbit is proportional to the cube of the semi-major axis of its orbit. This relationship can be written as:</p>
<p>$$T^2 \propto r^3$$</p>
<p>where:</p>
<ul>
<li>$T$ is the orbital period</li>
<li>$r$ is the radius of the circular orbit... | mcq | jee-main-2021-online-18th-march-morning-shift | 11,220 |
7SlPrQN46cTFTLyekk1kmlvpws3 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is $${\overrightarrow L }$$. The magnitude of the areal velocity of the planet is : | [{"identifier": "A", "content": "$${{2L} \\over M}$$"}, {"identifier": "B", "content": "$${{L} \\over 2M}$$"}, {"identifier": "C", "content": "$${{L} \\over M}$$"}, {"identifier": "D", "content": "$${{4L} \\over M}$$"}] | ["B"] | null | <picture><source media="(max-width: 1360px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266005/exam_images/dnysyikteutlzve9auws.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263759/exam_images/hcsdqj7xnq0i6kauinvu.webp"><source media="(max-wi... | mcq | jee-main-2021-online-18th-march-evening-shift | 11,221 |
1krqd0iy8 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Consider a binary star system of star A and star B with masses m<sub>A</sub> and m<sub>B</sub> revolving in a circular orbit of radii r<sub>A</sub> an r<sub>B</sub>, respectively. If T<sub>A</sub> and T<sub>B</sub> are the time period of star A and star B, respectively,<br/><br/>Then : | [{"identifier": "A", "content": "$${{{T_A}} \\over {{T_B}}} = {\\left( {{{{r_A}} \\over {{r_B}}}} \\right)^{{3 \\over 2}}}$$"}, {"identifier": "B", "content": "$${T_A} = {T_B}$$"}, {"identifier": "C", "content": "$${T_A} > {T_B}$$ (if $${m_A} > {m_B}$$)"}, {"identifier": "D", "content": "$${T_A} > {T_B}$$ (if ... | ["B"] | null | <p>In a binary star system, the two stars orbit around a common center of mass. When considering periods of revolution, Kepler's Third Law comes into play. This law states that the square of the period of revolution (T) is proportional to the cube of the semi-major axis (r) of the orbit. It's often written in t... | mcq | jee-main-2021-online-20th-july-evening-shift | 11,222 |
1krumgywu | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | The minimum and maximum distances of a planet revolving around the sun are x<sub>1</sub> and x<sub>2</sub>. If the minimum speed of the planet on its trajectory is v<sub>0</sub> then its maximum speed will be : | [{"identifier": "A", "content": "$${{{v_0}x_1^2} \\over {x_2^2}}$$"}, {"identifier": "B", "content": "$${{{v_0}x_2^2} \\over {x_1^2}}$$"}, {"identifier": "C", "content": "$${{{v_0}x_1^{}} \\over {x_2^{}}}$$"}, {"identifier": "D", "content": "$${{{v_0}x_2^{}} \\over {x_1^{}}}$$"}] | ["D"] | null | Angular momentum conservation equation $${v_0}{x_2} = {v_1}{x_1}$$<br><br>$${v_1} = {{{v_0}{x_2}} \over {{x_1}}}$$ | mcq | jee-main-2021-online-25th-july-morning-shift | 11,223 |
1ks18gqcx | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is : | [{"identifier": "A", "content": "$$\\sqrt {{G \\over {2{R^3}}}} $$"}, {"identifier": "B", "content": "$${1 \\over 2}\\sqrt {{G \\over {{R^3}}}} $$"}, {"identifier": "C", "content": "$${1 \\over {2R}}\\sqrt {{1 \\over G}} $$"}, {"identifier": "D", "content": "$${{2G} \\over {{R^3}}}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264301/exam_images/gq1uq8ad7giyxsvrk6xv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Physics - Gravitation Question 98 English Explanation"><p>The problem desc... | mcq | jee-main-2021-online-27th-july-evening-shift | 11,224 |
1ktmpo7e1 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Four particles each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is :<br/><br/><img src="data:image/png;base64,UklGRi4XAABXRUJQVlA4ICIXAACwVwGdASoAA/sCP4G+12Y2LywnIXApWsAwCWlu/BV4Lq+HZ19fy3/Xeji7bYbYf/nu0Pob7e/27jkYOS... | [{"identifier": "A", "content": "$${1 \\over 2}\\sqrt {{{GM} \\over {R(2\\sqrt 2 + 1)}}} $$"}, {"identifier": "B", "content": "$${1 \\over 2}\\sqrt {{{GM} \\over R}(2\\sqrt 2 + 1)} $$"}, {"identifier": "C", "content": "$${1 \\over 2}\\sqrt {{{GM} \\over R}(2\\sqrt 2 - 1)} $$"}, {"identifier": "D", "content": "$$\\sq... | ["B"] | null | <p>Let us consider the gravitational force acting on each mass M by adjacent particles be F.<br><br>and the gravitational force acting on each mass M diagonally be F<sub>1</sub>.</p>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwoseg2d/8affd279-21c4-435d-ac21-b626c882737f/a1620050-5356-11ec-b443-85... | mcq | jee-main-2021-online-1st-september-evening-shift | 11,225 |
1ktmuaixh | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Two satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hours respectively. The radius of the orbit of nearer satellite is 2 $$\times$$ 10<sup>3</sup> km. The angular speed of the farther satellite as observed from the nearer satellite ... | [] | null | 3 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwosl7k8/b883e894-890a-4a8b-a93e-ee5cd706cd6a/5d7a7790-5357-11ec-b443-85f16d0c41b6/file-1kwosl7ka.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kwosl7k8/b883e894-890a-4a8b-a93e-ee5cd706cd6a/5d7a7790-5357-11ec-b443-85f16d0c41b6/fi... | integer | jee-main-2021-online-1st-september-evening-shift | 11,226 |
1l55jw5xz | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be :</p> | [{"identifier": "A", "content": "$${2 \\over 9}$$ F"}, {"identifier": "B", "content": "$${16 \\over 9}$$ F"}, {"identifier": "C", "content": "$${8 \\over 9}$$ F"}, {"identifier": "D", "content": "F"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5hwg3n0/f13a1ec3-1130-4a3f-980d-b49d7ef40d09/e9c85ac0-01ba-11ed-85a8-43d162d2b7e8/file-1l5hwg3n1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5hwg3n0/f13a1ec3-1130-4a3f-980d-b49d7ef40d09/e9c85ac0-01ba-11ed-85a8-43d162d2b7e8... | mcq | jee-main-2022-online-28th-june-evening-shift | 11,228 |
1l568yvnp | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>Two planets A and B of equal mass are having their period of revolutions T<sub>A</sub> and T<sub>B</sub> such that T<sub>A</sub> = 2T<sub>B</sub>. These planets are revolving in the circular orbits of radii r<sub>A</sub> and r<sub>B</sub> respectively. Which out of the following would be the correct relationship of ... | [{"identifier": "A", "content": "$$2r_A^2 = r_B^3$$"}, {"identifier": "B", "content": "$$r_A^3 = 2r_B^3$$"}, {"identifier": "C", "content": "$$r_A^3 = 4r_B^3$$"}, {"identifier": "D", "content": "$$T_A^2 - T_B^2 = {{{\\pi ^2}} \\over {GM}}\\left( {r_B^3 - 4r_A^3} \\right)$$"}] | ["C"] | null | <p>$${T_A} = 2{T_B}$$</p>
<p>Now $$T_A^2 \propto r_A^3$$</p>
<p>$$ \Rightarrow {\left( {{{{r_A}} \over {{r_B}}}} \right)^3} = {\left( {{{{T_A}} \over {{T_B}}}} \right)^2}$$</p>
<p>$$ \Rightarrow r_A^3 = 4r_B^3$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift | 11,229 |
1l56ued56 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>The distance of the Sun from earth is 1.5 $$\times$$ 10<sup>11</sup> m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be :</p> | [{"identifier": "A", "content": "2.45 $$\\times$$ 10<sup>10</sup> m"}, {"identifier": "B", "content": "1.45 $$\\times$$ 10<sup>10</sup> m"}, {"identifier": "C", "content": "1.45 $$\\times$$ 10<sup>9</sup> m"}, {"identifier": "D", "content": "0.14 $$\\times$$ 10<sup>9</sup> m"}] | ["C"] | null | <p>Diameter = r $$\times$$ $$\delta$$</p>
<p>$$ = 1.5 \times {10^{11}} \times (2000) \times \left( {{1 \over {3600}}} \right) \times \left( {{\pi \over {180}}} \right)$$</p>
<p>$$ = 1.45 \times {10^9}$$ m</p> | mcq | jee-main-2022-online-27th-june-evening-shift | 11,230 |
1l6dy2uti | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>Three identical particles $$\mathrm{A}, \mathrm{B}$$ and $$\mathrm{C}$$ of mass $$100 \mathrm{~kg}$$ each are placed in a straight line with $$\mathrm{AB}=\mathrm{BC}=13 \mathrm{~m}$$. The gravitational force on a fourth particle $$\mathrm{P}$$ of the same mass is $$\mathrm{F}$$, when placed at a distance $$13 \math... | [{"identifier": "A", "content": "21 G"}, {"identifier": "B", "content": "100 G"}, {"identifier": "C", "content": "59 G"}, {"identifier": "D", "content": "42 G"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6tcunfv/d5d44637-c5a5-4a91-b83c-43a82e39ede4/e6feefb0-1bd3-11ed-bd9f-71db3a4811b2/file-1l6tcunfw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6tcunfv/d5d44637-c5a5-4a91-b83c-43a82e39ede4/e6feefb0-1bd3-11ed-bd9f-71db3a4811b2... | mcq | jee-main-2022-online-25th-july-morning-shift | 11,232 |
1ldsamjxf | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.</p> | [{"identifier": "A", "content": "12 hours"}, {"identifier": "B", "content": "3 hours"}, {"identifier": "C", "content": "6 hours"}, {"identifier": "D", "content": "4 hours"}] | ["B"] | null | <p>$$\because {T^2} \propto {R^3}$$</p>
<p>$$\therefore$$ $${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$$</p>
<p>$${{{{24}^2}} \over {T_2^2}} = {{R_1^3} \over {{{\left( {{{{R_1}} \over 4}} \right)}^3}}}$$</p>
<p>$${{{{24}^2}} \over {T_2^2}} = {4^3}$$</p>
<p>$${T_2} = {{24} \over {{2^3}}} = 3$$ hours</p> | mcq | jee-main-2023-online-29th-january-evening-shift | 11,233 |
1ldsnu25e | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>Two particles of equal mass '$$m$$' move in a circle of radius '$$r$$' under the action of their mutual gravitational attraction. The speed of each particle will be :</p> | [{"identifier": "A", "content": "$$\\sqrt{\\frac{G m}{4 r}}$$"}, {"identifier": "B", "content": "$$\\sqrt{\\frac{G m}{2 r}}$$"}, {"identifier": "C", "content": "$$\\sqrt{\\frac{G m}{r}}$$"}, {"identifier": "D", "content": "$$\\sqrt{\\frac{4 G m}{r}}$$"}] | ["A"] | null | $\frac{\mathrm{Gm}^{2}}{4 \mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lei2xrln/9034b0ff-b12e-40cd-b333-548731e8cda9/3e3b95a0-b402-11ed-8ab2-65ca6f80c54f/file-1lei2xrlo.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1... | mcq | jee-main-2023-online-29th-january-morning-shift | 11,234 |
1ldtyi5sp | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>Every planet revolves around the sun in an elliptical orbit :-</p>
<p>A. The force acting on a planet is inversely proportional to square of distance from sun.</p>
<p>B. Force acting on planet is inversely proportional to product of the masses of the planet and the sun.</p>
<p>C. The Centripetal force acting on the ... | [{"identifier": "A", "content": "C and D only"}, {"identifier": "B", "content": "B and C only"}, {"identifier": "C", "content": "A and D only"}, {"identifier": "D", "content": "A and C only"}] | ["C"] | null | <p>A. The force acting on a planet is inversely proportional to the square of the distance from the sun. This is known as the inverse square law and is described by Newton's law of gravitation.</p>
<p>B. Force acting on a planet is inversely proportional to the product of the masses of the planet and the sun. This is ... | mcq | jee-main-2023-online-25th-january-evening-shift | 11,235 |
1ldwrgknp | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>If the distance of the earth from Sun is 1.5 $$\times$$ 10$$^6$$ km. Then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is :</p> | [{"identifier": "A", "content": "$$6\\times10^6$$ km"}, {"identifier": "B", "content": "$$3\\times10^7$$ km"}, {"identifier": "C", "content": "$$6\\times10^7$$ km"}, {"identifier": "D", "content": "$$3\\times10^6$$ km"}] | ["D"] | null | We can use Kepler's third law to solve this problem. Kepler's third law states that the square of the period of revolution of a planet around the Sun is proportional to the cube of its average distance from the Sun. Let $T$ be the period of revolution of the imaginary planet in years, and let $d$ be its average distanc... | mcq | jee-main-2023-online-24th-january-evening-shift | 11,236 |
lgny675b | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | Two identical particles each of mass ' $m$ ' go round a circle of radius $a$ under the action of their mutual gravitational attraction. The angular speed of each particle will be : | [{"identifier": "A", "content": "$\\sqrt{\\frac{G m}{2 a^{3}}}$"}, {"identifier": "B", "content": "$\\sqrt{\\frac{G m}{a^{3}}}$\n"}, {"identifier": "C", "content": "$\\sqrt{\\frac{G m}{8 a^{3}}}$"}, {"identifier": "D", "content": "$\\sqrt{\\frac{G m}{4 a^{3}}}$"}] | ["D"] | null | The gravitational force between two particles of mass $m$ separated by a distance $r$ is given by:
<br/><br/>
$$
F = \frac{Gm^2}{r^2}
$$
<br/><br/>
where $G$ is the gravitational constant. In this problem, the two particles are moving in a circular orbit of radius $a$ under the influence of their mutual gravitational a... | mcq | jee-main-2023-online-15th-april-morning-shift | 11,237 |
1lgyftz9l | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>If the earth suddenly shrinks to $$\frac{1}{64}$$th of its original volume with its mass remaining the same, the period of rotation of earth becomes $$\frac{24}{x}$$h. The value of x is __________.</p> | [] | null | 16 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgzrc0c0/488210fe-6508-4f7a-9a44-b65f6e8b45f8/673423f0-e553-11ed-a972-b5db7753e054/file-1lgzrc0c1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgzrc0c0/488210fe-6508-4f7a-9a44-b65f6e8b45f8/673423f0-e553-11ed-a972-b5db7753e054/fi... | integer | jee-main-2023-online-10th-april-morning-shift | 11,238 |
1lh252xti | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>A planet has double the mass of the earth. Its average density is equal to that of the earth. An object weighing $$\mathrm{W}$$ on earth will weigh on that planet:</p> | [{"identifier": "A", "content": "$$2^{2 / 3} \\mathrm{~W}$$"}, {"identifier": "B", "content": "W"}, {"identifier": "C", "content": "$$2 \\mathrm{~W}$$"}, {"identifier": "D", "content": "$$2^{1 / 3} \\mathrm{~W}$$"}] | ["D"] | null | <p>The weight of an object on a planet is given by the equation $W = mg$, where $m$ is the mass of the object and $g$ is the acceleration due to gravity.</p>
<p>The acceleration due to gravity on a planet is given by the equation $g = \frac{GM}{R^2}$, where $G$ is the gravitational constant, $M$ is the mass of the plan... | mcq | jee-main-2023-online-6th-april-morning-shift | 11,239 |
lsamy2vh | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | A light planet is revolving around a massive star in a circular orbit of radius $\mathrm{R}$ with a period of revolution T. If the force of attraction between planet and star is proportional to $\mathrm{R}^{-3 / 2}$ then choose the correct option : | [{"identifier": "A", "content": "$\\mathrm{T}^2 \\propto \\mathrm{R}^{7 / 2}$"}, {"identifier": "B", "content": "$\\mathrm{T}^2 \\propto \\mathrm{R}^3$"}, {"identifier": "C", "content": "$\\mathrm{T}^2 \\propto \\mathrm{R}^{5 / 2}$"}, {"identifier": "D", "content": "$\\mathrm{T}^2 \\propto \\mathrm{R}^{3 / 2}$"}] | ["C"] | null | <p>To find the correct option for the relationship between the period of revolution T and the radius of the orbit R, we will consider the force of attraction and its proportionality to $\mathrm{R}^{-3 / 2}$.</p>
<p>According to Newton's law of universal gravitation, the force of attraction $F$ between two masses $m... | mcq | jee-main-2024-online-1st-february-evening-shift | 11,240 |
jaoe38c1lse61sz2 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>Four identical particles of mass $$m$$ are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is $$\left(\frac{2 \sqrt{2}+1}{32}\right) \frac{\mathrm{Gm}^2}{L^2}$$, the length of the sides of the square is</p> | [{"identifier": "A", "content": "4L"}, {"identifier": "B", "content": "3L"}, {"identifier": "C", "content": "2L"}, {"identifier": "D", "content": "$$\\frac{L}{2}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lslugnze/9126a5cb-8471-4e9a-876d-d6e7584287ac/438d5ca0-cb3f-11ee-ad47-a16d1086e690/file-6y3zli1lslugnzf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lslugnze/9126a5cb-8471-4e9a-876d-d6e7584287ac/438d5ca0-cb3f-11ee... | mcq | jee-main-2024-online-31st-january-morning-shift | 11,241 |
lv0vxkbr | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>A metal wire of uniform mass density having length $$L$$ and mass $$M$$ is bent to form a semicircular arc and a particle of mass $$\mathrm{m}$$ is placed at the centre of the arc. The gravitational force on the particle by the wire is :</p> | [{"identifier": "A", "content": "$$\\frac{\\mathrm{GmM} \\pi^2}{\\mathrm{~L}^2}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\mathrm{GMm} \\pi}{2 \\mathrm{~L}^2}$$\n"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$\\frac{2 \\mathrm{GmM} \\pi}{\\mathrm{L}^2}$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwkcnf01/62227aa5-9019-4450-8839-4d8058001ff6/afd1b210-199d-11ef-acb3-f132a9e92441/file-1lwkcnf02.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwkcnf01/62227aa5-9019-4450-8839-4d8058001ff6/afd1b210-199d-11ef-acb3-f132a9e92441... | mcq | jee-main-2024-online-4th-april-morning-shift | 11,243 |
lv3vefj4 | physics | gravitation | kepler's-law-and-universal-law-of-gravitation | <p>Two satellite A and B go round a planet in circular orbits having radii 4R and R respectively. If the speed of $$\mathrm{A}$$ is $$3 v$$, the speed of $$\mathrm{B}$$ will be :</p> | [{"identifier": "A", "content": "$$6 v$$\n"}, {"identifier": "B", "content": "$$\\frac{4}{3} v$$\n"}, {"identifier": "C", "content": "$$3 v$$\n"}, {"identifier": "D", "content": "$$12 v$$"}] | ["A"] | null | <p>To solve this question, we will use the fact that for an object in a circular orbit, the centripetal force required to keep the object in orbit is provided by the gravitational force between the object and the planet it is orbiting. This principle gives us the relationship between the speed of the satellite, its orb... | mcq | jee-main-2024-online-8th-april-evening-shift | 11,244 |
itoaTCzLm3bOEG1V | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | One $$kg$$ of a diatomic gas is at a pressure of $$8 \times {10^4}\,N/{m^2}.$$ The density of the gas is $$4kg/{m^3}$$. What is the energy of the gas due to its thermal motion ? | [{"identifier": "A", "content": "$$5 \\times {10^4}\\,J$$ "}, {"identifier": "B", "content": "$$6 \\times {10^4}\\,J$$ "}, {"identifier": "C", "content": "$$7 \\times {10^4}\\,J$$ "}, {"identifier": "D", "content": "$$3 \\times {10^4}\\,J$$ "}] | ["A"] | null | $$Volume\,\, = \,\,{{mass} \over {density}} = {1 \over 4}{m^3}$$
<br>$$K.E = {5 \over 2}PV$$
<br>$$ = {5 \over 2} \times 8 \times {10^4} \times {1 \over 4}$$
<br>$$ = 5 \times {10^4}J$$ | mcq | aieee-2009 | 11,246 |
cvb8OhdJxDNYfqIV | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | Three perfect gases at absolute temperatures $${T_1},\,{T_2}$$ and $${T_3}$$ are mixed. The masses of molecules are $${m_1},{m_2}$$ and $${m_3}$$ and the number of molecules are $${n_1},$$ $${n_2}$$ and $${n_3}$$ respectively. Assuming no loss of energy, the final temperature of the mixture is: | [{"identifier": "A", "content": "$${{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \\over {{n_1} + {n_2} + {n_3}}}$$ "}, {"identifier": "B", "content": "$${{{n_1}T_1^2 + {n_2}T_2^2 + {n_3}T_3^2} \\over {{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}$$ "}, {"identifier": "C", "content": "$${{n_1^2T_1^2 + n_2^2T_2^2 + n_3^2T_3^2} \\over... | ["A"] | null | Number of moles of first gas $$ = {{{n_1}} \over {{N_A}}}$$
<br>Number of moles of second gas $$ = {{{n_2}} \over {{N_A}}}$$
<br>Number of moles of third gas $$ = {{{n_3}} \over {{N_A}}}$$
<br>If there is no loss of energy then
<br>$${P_1}{V_1} + {P_2}{V_2} + {P_3}{V_3} = PV$$
<br>$${{{n_1}} \over {{N_A}}}R{T_1} + {{{n... | mcq | aieee-2011 | 11,247 |
FDAMkN5fSAaDcX32dcjFi | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (C<sub>p</sub> ) and at constant volume (C<sub>v</sub>) is :
| [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "$${7 \\over 2}$$ "}, {"identifier": "C", "content": "$${5 \\over 2}$$"}, {"identifier": "D", "content": "$${7 \\over 5}$$ "}] | ["D"] | null | For ideal gas molecule with 5 degree of freedom,
<br><br>C<sub>v</sub> = $${5 \over 2}$$ R and C<sub>p</sub> = $${7 \over 2}$$ R
<br><br>$$\therefore\,\,\,$$ $${{{C_p}} \over {{C_v}}}$$ = $${{{7 \over 2}R} \over {{5 \over 2}R}}$$ = $${7 \over 5}$$ | mcq | jee-main-2017-online-8th-april-morning-slot | 11,248 |
PGKTmHvWIYIXR6qqsZIsg | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | Two moles of helium are mixed with n moles of hydrogen. If $${{Cp} \over {Cv}} = {3 \over 2}$$ for the mixture, then the value of n is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3 / 2"}] | ["C"] | null | $${{{C_p}} \over {{C_v}}} = {{{f_{mix}} + 2} \over {{f_{mix}}}} = {3 \over 2}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ f<sub>mix</sub> = 4
<br><br>As, f<sub>mix</sub> = $${{{n_1}{f_1} + {n_2}{f_2}} \over {{n_1} + {n_2}}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 4 = $${{2 \times 3 + n \times 5} \over {2 + n}}$$
<br><br>$$ \Ri... | mcq | jee-main-2018-online-16th-april-morning-slot | 11,249 |
0Hs2slbFMnthp8c8bAjFf | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. considering only translational and rotational modes, the total internal energy of the system is : | [{"identifier": "A", "content": "12 RT"}, {"identifier": "B", "content": "20 RT"}, {"identifier": "C", "content": "4 RT"}, {"identifier": "D", "content": "15 RT"}] | ["D"] | null | U $$ = {{{f_1}} \over 2}{n_1}RT + {{{f_2}} \over 2}{n_2}RT$$
<br><br>$$ = {5 \over 2}\left( {3RT} \right) + {3 \over 2} \times 5RT$$
<br><br>U $$ = 15RT$$ | mcq | jee-main-2019-online-11th-january-morning-slot | 11,250 |
OPkaZS430TtlRW6nulQ2f | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | An ideal gas occupies a volume of 2m<sup>3</sup> at a pressure of 3 $$ \times $$ 10<sup>6</sup> Pa. The energy of the gas is : | [{"identifier": "A", "content": "6 $$ \\times $$ 10<sup>4</sup> J"}, {"identifier": "B", "content": "9$$ \\times $$ 10<sup>6</sup> J"}, {"identifier": "C", "content": "3 $$ \\times $$ 10<sup>2</sup> J"}, {"identifier": "D", "content": "10<sup>8</sup> J"}] | ["B"] | null | Energy = $${1 \over 2}$$ nRT = $${f \over 2}$$PV
<br><br>= $${f \over 2}$$ (3 $$ \times $$ 10<sup>6</sup>) (2)
<br><br>= f $$ \times $$ 3 $$ \times $$ 10<sup>6</sup>
<br><br>Considering gas is monoatomic i.e. f = 3
<br><br>E. = 9 $$ \times $$ 10<sup>6</sup> J | mcq | jee-main-2019-online-12th-january-morning-slot | 11,251 |
9nf3gTTSQl39QuM4a0yuj | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | An HCl molecule has rotational, translational
and vibrational motions. If the rms velocity of
HCl molecules in its gaseous phase is $$\overline v $$ , m is
its mass and k<sub>B</sub> is Boltzmann constant, then its
temperature will be : | [{"identifier": "A", "content": "$${{m{{\\overline v }^2}} \\over {5{k_B}}}$$"}, {"identifier": "B", "content": "$${{m{{\\overline v }^2}} \\over {6{k_B}}}$$"}, {"identifier": "C", "content": "$${{m{{\\overline v }^2}} \\over {7{k_B}}}$$"}, {"identifier": "D", "content": "$${{m{{\\overline v }^2}} \\over {3{k_B}}}$$"}] | ["C"] | null | <p>An HCl molecule, being diatomic, has:</p>
<ul>
<li>3 translational degrees of freedom</li>
<li>2 rotational degrees of freedom</li>
<li>2 vibrational degrees of freedom</li>
</ul>
<p>The total number of degrees of freedom is $3 + 2 + 2 = 7$.</p>
<p>According to the equipartition theorem, each degree of freedom contr... | mcq | jee-main-2019-online-9th-april-morning-slot | 11,252 |
RDmQjrHwITVPSW674bjgy2xukf3uj5cb | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | To raise the temperature of a certain mass of gas by 50<sup>o</sup>C at a constant pressure, 160 calories of
heat is required. When the same mass of gas is cooled by 100<sup>o</sup>C at constant volume, 240 calories
of heat is released. How many degrees of freedom does each molecule of this gas have (assume
gas to be i... | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "3"}] | ["A"] | null | $$160 = n{C_p}50$$ ....(i)<br><br>$$240 = n{C_v}100$$ ....(ii)<br><br>Dividing (i) by (ii), we get<br><br>$${{160} \over {240}} = {{{C_p}} \over {{C_v}}} \times {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${{{C_p}} \over {{C_v}}}$$ = $$ {4 \over 3}$$
<br><br>We know,
<br><br>$$ \gamma = {{{C_p}} \over {{C_v}}} = 1 + {2 \... | mcq | jee-main-2020-online-3rd-september-evening-slot | 11,254 |
jafpTQWd84pVQfYxiyjgy2xukfrp43ac | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | Molecules of an ideal gas are known to have three translational degrees of freedom and two
rotational degrees of freedom.The gas is maintained at a temperature of T. The total internal
energy, U of a mole of this gas, and the value of <br/>$$\gamma \left( { = {{{C_p}} \over {{C_v}}}} \right)$$ are given, respectively b... | [{"identifier": "A", "content": "U = $${5 \\over 2}RT$$ and $$\\gamma = {7 \\over 5}$$"}, {"identifier": "B", "content": "U = 5RT and $$\\gamma = {6 \\over 5}$$"}, {"identifier": "C", "content": "U = 5RT and $$\\gamma = {7 \\over 5}$$"}, {"identifier": "D", "content": "U = $${5 \\over 2}RT$$ and $$\\gamma = {6 \\ov... | ["A"] | null | Total degree of freedom (f) = 3 + 2 = 5
<br><br>U = $${{nfRT} \over 2}$$ = $${{5RT} \over 2}$$
<br><br>$$\gamma $$ = $${{{C_p}} \over {{C_v}}}$$ = $$1 + {2 \over f}$$ = $$1 + {2 \over 5}$$ = $${7 \over 5}$$ | mcq | jee-main-2020-online-6th-september-morning-slot | 11,255 |
JPEYJc3DN0sgCEnH4djgy2xukev3hx9u | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | An engine takes in 5 moles of air at 20<sup>o</sup>C and
1 atm, and compresses it adiabatically to
1/10<sup>th</sup> of the original volume. Assuming air to
be a diatomic ideal gas made up of rigid
molecules, the change in its internal energy
during this process comes out to be X kJ. The
value of X to the nearest integ... | [] | null | 46 | For diatomic ideal gas :
<br><br>f = 5
<br><br>$$\gamma $$ = $${7 \over 5}$$
<br><br>T<sub>i</sub>
= T = 273 + 20 = 293 K
<br><br>V<sub>i</sub>
= V
<br><br>V<sub>f</sub> = $${V \over {10}}$$
<br><br>For adiabatic process TV<sup>$$\gamma $$ - 1</sup> = constant
<br><br>$${T_1}V_1^{\gamma - 1} = {T_2}V_2^{\gamma - 1}... | integer | jee-main-2020-online-2nd-september-morning-slot | 11,257 |
5NBpzaifV77DCCQV4pjgy2xukev1nbvn | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | A gas mixture consists of 3 moles of oxygen
and 5 moles of argon at temperature T.
Assuming the gases to be ideal and the oxygen
bond to be rigid, the total internal energy (in
units of RT) of the mixture is : | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "13"}] | ["C"] | null | U $$ = {{{f_1}} \over 2}{n_1}RT + {{{f_2}} \over 2}{n_2}RT$$
<br><br>$$ = {5 \over 2}\left( {3RT} \right) + {3 \over 2} \times 5RT$$
<br><br>$$ \therefore $$ U $$ = 15RT$$ | mcq | jee-main-2020-online-2nd-september-morning-slot | 11,258 |
O373M1zCC6g9sU4ulkjgy2xuketz37vd | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | Match the $${{{C_P}} \over {{C_V}}}$$ ratio for ideal gases with different type of molecules :
<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5p... | [{"identifier": "A", "content": "(A)-(III), (B)-(IV), (C)-(II), (D)-(I)"}, {"identifier": "B", "content": "(A)-(IV), (B)-(II), (C)-(I), (D)-(III)"}, {"identifier": "C", "content": "(A)-(IV), (B)-(I), (C)-(II), (D)-(III)"}, {"identifier": "D", "content": "(A)-(II), (B)-(III), (C)-(I), (D)-(IV)"}] | ["C"] | null | $$\gamma = {C_p}/{C_v}$$<br><br>$${\gamma _A} = 1 + {2 \over 3} = 5/3$$<br><br>$${\gamma _B} = 1 + {2 \over 5} = 7/5$$<br><br>$${\gamma _C} = 1 + {2 \over 7} = 9/7$$<br><br>$${\gamma _D} = 1 + {2 \over 6} = 4/3$$ | mcq | jee-main-2020-online-4th-september-morning-slot | 11,259 |
qDAq65pyHyvDYKnkvR7k9k2k5i891ed | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | Consider two ideal diatomic gases A and B at
some temperature T. Molecules of the gas A are
rigid, and have a mass m. Molecules of the gas
B have an additional vibrational mode, and
have a mass $${m \over 4}$$
. The ratio of the specific heats ($$C_V^A$$ and $$C_V^B$$ ) of gas A and B, respectively is : | [{"identifier": "A", "content": "7 : 9"}, {"identifier": "B", "content": "5 : 7"}, {"identifier": "C", "content": "3 : 5"}, {"identifier": "D", "content": "5 : 9"}] | ["B"] | null | Degree of freedom of a diatomic molecule if
vibration is absent = 5
<br><br>Degree of freedom of a diatomic molecule if
vibration is present = 7
<br><br>$$ \therefore $$ $$C_V^A$$ = $${5 \over 2}R$$
<br><br>and $$C_V^B$$ = $${7 \over 2}R$$
<br><br>$$ \therefore $$ $${{C_V^A} \over {C_V^B}} = {5 \over 7}$$ | mcq | jee-main-2020-online-9th-january-morning-slot | 11,260 |
PesjmeqFYD1rQtwYSH1klt2kqz5 | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | Given below are two statements :<br/><br/>Statement I : In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.<br/><br/>Statement II : In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.<br/><br/>In t... | [{"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."}... | ["C"] | null | The translational kinetic energy & rotational kinetic energy both obey Maxwell's distribution independent of each other.<br><br>T.K.E. of diatomic molecules = $${3 \over 2}kT$$<br><br>R.K.E. of diatomic molecules = $${2 \over 2}kT$$<br><br>So statement I is true but statement II is false. | mcq | jee-main-2021-online-25th-february-evening-slot | 11,262 |
807Tx8PLagOu2pPq951kluljren | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as U $$=$$ 3PV + 4. The gas is : | [{"identifier": "A", "content": "either monoatomic or diatomic."}, {"identifier": "B", "content": "monoatomic only."}, {"identifier": "C", "content": "polyatomic only."}, {"identifier": "D", "content": "diatomic only."}] | ["C"] | null | U = 3PV + 4<br><br>$$ \Rightarrow $$ $${{nf} \over 2}$$RT = 3PV + 4<br><br>$$ \Rightarrow $$ $${{f} \over 2}$$PV = 3PV + 4<br><br>$$ \Rightarrow $$ f = 6 + $${8 \over {PV}}$$<br><br>Since degree of freedom is more than 6 therefore gas is polyatomic. | mcq | jee-main-2021-online-26th-february-evening-slot | 11,263 |
LMJLGCnPVhCdO9GPCd1kmj17vjo | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | A polyatomic ideal gas has 24 vibrational modes. What is the value of $$\gamma$$? | [{"identifier": "A", "content": "1.37"}, {"identifier": "B", "content": "1.30"}, {"identifier": "C", "content": "1.03"}, {"identifier": "D", "content": "10.3"}] | ["C"] | null | f = 3T + 3R + 24V<br><br>= 30<br><br>$$\gamma$$ = 1 + $${2 \over f}$$<br><br>$$\gamma$$ = 1 + $${2 \over 30}$$<br><br>= 1.066<br><br>Nearest Ans. = 1.03 | mcq | jee-main-2021-online-17th-march-morning-shift | 11,264 |
eX8yeE0y1EWPcWf6JJ1kmkahn10 | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | If one mole of the polyatomic gas is having two vibrational modes and $$\beta$$ is the ratio of molar specific heats for polyatomic gas $$\left( {\beta = {{{C_P}} \over {{C_V}}}} \right)$$ then the value of $$\beta$$ is : | [{"identifier": "A", "content": "1.02"}, {"identifier": "B", "content": "1.35"}, {"identifier": "C", "content": "1.2"}, {"identifier": "D", "content": "1.25"}] | ["C"] | null | For polyatomic gas molecule has 3 rotational degrees of freedom,
3 translational degrees of freedom, and 2 vibrational modes.
<br/><br/>So, number of vibrational degrees of freedom = 2 $$ \times $$ 2 = 4
<br/><br/>Degree of freedom of polyatomic gas<br><br>f = T + R + V<br><br>f = 3 + 3 + 4 = 10<br><br>$$\beta = 1 + {... | mcq | jee-main-2021-online-17th-march-evening-shift | 11,266 |
1krqcqw5r | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | The correct relation between the degrees of freedom f and the ratio of specific heat $$\gamma$$ is : | [{"identifier": "A", "content": "$$f = {2 \\over {\\gamma - 1}}$$"}, {"identifier": "B", "content": "$$f = {2 \\over {\\gamma + 1}}$$"}, {"identifier": "C", "content": "$$f = {{\\gamma + 1} \\over 2}$$"}, {"identifier": "D", "content": "$$f = {1 \\over {\\gamma + 1}}$$"}] | ["A"] | null | $$\gamma = 1 + {2 \over f}$$<br><br>$$ \Rightarrow $$ $$f = {2 \over {\gamma - 1}}$$ | mcq | jee-main-2021-online-20th-july-evening-shift | 11,268 |
1krsvkz8e | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | What will be the average value of energy for a monoatomic gas in thermal equilibrium at temperature T? | [{"identifier": "A", "content": "$${3 \\over 2}{k_B}T$$"}, {"identifier": "B", "content": "$${k_B}T$$"}, {"identifier": "C", "content": "$${2 \\over 3}{k_B}T$$"}, {"identifier": "D", "content": "$${1 \\over 2}{k_B}T$$"}] | ["A"] | null | <p>For a monoatomic ideal gas, the average kinetic energy per molecule is determined by the equipartition theorem. This theorem states that the energy is equally distributed among all the available degrees of freedom.</p>
<p>A monoatomic gas has three translational degrees of freedom, corresponding to motion in the x, ... | mcq | jee-main-2021-online-22th-july-evening-shift | 11,269 |
1ktms90wo | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | The temperature of 3.00 mol of an ideal diatomic gas is increased by 40.0$$^\circ$$C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of workdone by the gas is $${x \over {10}}$$. Then the value of x (round... | [] | null | 25 | Given, the number of diatomic moles, n = 3 mol<br/><br/>The increase in temperature of the diatomic mole, <br/><br/>$$\Delta$$T = 40$$^\circ$$C<br/><br/>Now, the degree of freedom<br/><br/>f = linear + rotational + no oscillation<br/><br/>f = 3 + 2 + 0 $$\Rightarrow$$ f = 5<br/><br/>Change in internal energy,<br/><br/>... | integer | jee-main-2021-online-1st-september-evening-shift | 11,270 |
1l56ur4mp | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>According to kinetic theory of gases,</p>
<p>A. The motion of the gas molecules freezes at 0$$^\circ$$C.</p>
<p>B. The mean free path of gas molecules decreases if the density of molecules is increased.</p>
<p>C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.</p>... | [{"identifier": "A", "content": "A and C only"}, {"identifier": "B", "content": "B and C only"}, {"identifier": "C", "content": "A and B only"}, {"identifier": "D", "content": "C and D only"}] | ["B"] | null | <p>According to kinetic theory of gases,</p>
<p>A. The motion of the gas molecules freezes at 0 K.</p>
<p>B. The mean free path decreases on increasing the number density of the molecules as $$\mu = {1 \over {\sqrt 2 \pi n{d^2}}} \Rightarrow \mu \propto {1 \over n}$$.</p>
<p>C. The mean free path increases on increas... | mcq | jee-main-2022-online-27th-june-evening-shift | 11,272 |
1l59m787e | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>The ratio of specific heats $$\left( {{{{C_P}} \over {{C_V}}}} \right)$$ in terms of degree of freedom (f) is given by :</p> | [{"identifier": "A", "content": "$$\\left( {1 + {f \\over 3}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {1 + {2 \\over f}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {1 + {f \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {1 + {1 \\over f}} \\right)$$"}] | ["B"] | null | <p>$${{{C_P}} \over {{C_V}}} = \gamma $$</p>
<p>$${C_V} = \left( {{f \over 2}} \right)R$$ and $${C_P} - {C_V} = R$$</p>
<p>$$ \Rightarrow {{{C_P}} \over C} = {{1 + f/2} \over {f/2}} = 1 + {2 \over f}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift | 11,273 |
1l6i1565s | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>A gas has $$n$$ degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be :</p> | [{"identifier": "A", "content": "$$ \\frac{n}{n+2}$$"}, {"identifier": "B", "content": "$$ \\frac{n+2}{n}$$"}, {"identifier": "C", "content": "$$ \\frac{n}{2n+2}$$"}, {"identifier": "D", "content": "$$ \\frac{n}{n-2}$$"}] | ["A"] | null | <p>$${C_V} = {{nR} \over 2}$$</p>
<p>And $${C_P} = {{nR} \over 2} + R$$</p>
<p>$$ \Rightarrow {{{C_V}} \over {{C_P}}} = {{{{nR} \over 2}} \over {{{nR} \over 2} + R}} = {n \over {n + 2}}$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift | 11,274 |
1l6kn0ylm | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>Which statements are correct about degrees of freedom ?</p>
<p>(A) A molecule with n degrees of freedom has n$$^{2}$$ different ways of storing energy.</p>
<p>(B) Each degree of freedom is associated with $$\frac{1}{2}$$ RT average energy per mole.</p>
<p>(C) A monatomic gas molecule has 1 rotational degree of freed... | [{"identifier": "A", "content": "(B) and (C) only"}, {"identifier": "B", "content": "(B) and (D) only"}, {"identifier": "C", "content": "(A) and (B) only"}, {"identifier": "D", "content": "(C) and (D) only"}] | ["B"] | null | <p>Statement A is incorrect, statement B is correct by equipartition of energy. Statement C is incorrect as monoatomic does not have any rotational degree of freedom and CH<sub>4</sub> is a polyatomic gas so it has 6 degree of freedom. So only B and D are correct.</p> | mcq | jee-main-2022-online-27th-july-evening-shift | 11,275 |
ldo65oa8 | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | Heat energy of $735 \mathrm{~J}$ is given to a diatomic gas allowing the gas to expand at constant pressure.
Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal
energy of the gas will be : | [{"identifier": "A", "content": "$572 \\mathrm{~J}$"}, {"identifier": "B", "content": " $441 \\mathrm{~J}$"}, {"identifier": "C", "content": "$525 \\mathrm{~J}$"}, {"identifier": "D", "content": "$735 \\mathrm{~J}$"}] | ["C"] | null | $\Delta Q=n C_{P} \Delta T= $
<br/><br/>$$n\left( {{f \over 2} + 1} \right)R\Delta T$$
<br/><br/>= $$n\left( {{5 \over 2} + 1} \right)R\Delta T$$
<br/><br/>= $$n\left( {{7 \over 2}} \right)R\Delta T$$
<br/><br/>Given, $\Delta Q$ = 735
<br/><br/>$$ \Rightarrow $$ $$n\left( {{7 \over 2}} \right)R\Delta T$$ = 735
<br/><b... | mcq | jee-main-2023-online-31st-january-evening-shift | 11,277 |
1ldtz1qik | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is :-</p> | [{"identifier": "A", "content": "$$\\frac{9}{2}R$$"}, {"identifier": "B", "content": "$$\\frac{5}{2}R$$"}, {"identifier": "C", "content": "$$\\frac{3}{2}R$$"}, {"identifier": "D", "content": "$$\\frac{7}{2}R$$"}] | ["D"] | null | Diatomic gas molecules have three translational degree of freedom, two rotational degree of freedom \& it is given that it has one vibrational mode so there are two additional degree of freedom corresponding to one vibrational mode, so total degree of freedom $=7$<br/><br/>
$$
\mathrm{C}_{\mathrm{v}}=\frac{\mathrm{fR}}... | mcq | jee-main-2023-online-25th-january-evening-shift | 11,278 |
1ldwr2dow | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>Let $$\gamma_1$$ be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and $$\gamma_2$$ be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio, $$\frac{\gamma_1}{\gamma_2}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{35}{27}$$"}, {"identifier": "B", "content": "$$\\frac{25}{21}$$"}, {"identifier": "C", "content": "$$\\frac{21}{25}$$"}, {"identifier": "D", "content": "$$\\frac{27}{35}$$"}] | ["B"] | null | For monoatomic gas $\gamma_1=\frac{5}{3}$<br/><br/>
For diatomic gas at low temperatures<br/><br/>
$$
\begin{aligned}
& \gamma_2=\frac{7}{5} \\\\
& \therefore \frac{\gamma_1}{\gamma_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21}
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-evening-shift | 11,279 |
1lgp07l1w | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>The mean free path of molecules of a certain gas at STP is $$1500 \mathrm{~d}$$, where $$\mathrm{d}$$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at $$373 \mathrm{~K}$$ is approximately:</p> | [{"identifier": "A", "content": "$$750 \\mathrm{~d}$$"}, {"identifier": "B", "content": "$$1500 \\mathrm{~d}$$"}, {"identifier": "C", "content": "$$\\mathrm{2049~ d}$$"}, {"identifier": "D", "content": "$$1098 \\mathrm{~d}$$"}] | ["C"] | null | The mean free path (λ) of molecules in a gas is given by the formula:
<br/><br/>
$$\lambda = \frac{kT}{\sqrt{2}\pi d^2 P}$$
<br/><br/>
where k is the Boltzmann constant, T is the temperature in Kelvin, d is the diameter of the gas molecules, and P is the pressure.
<br/><br/>
At STP (standard temperature and pressure), ... | mcq | jee-main-2023-online-13th-april-evening-shift | 11,280 |
1lgq223aw | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>The rms speed of oxygen molecule in a vessel at particular temperature is $$\left(1+\frac{5}{x}\right)^{\frac{1}{2}} v$$, where $$v$$ is the average speed of the molecule. The value of $$x$$ will be:</p>
<p>$$\left(\right.$$ Take $$\left.\pi=\frac{22}{7}\right)$$</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "28"}, {"identifier": "D", "content": "27"}] | ["C"] | null | <p>The relationship between the root-mean-square (rms) speed ($$v_{rms}$$) and the average speed ($$v_{avg}$$) of molecules in a gas can be found using the Maxwell-Boltzmann distribution. The rms speed and average speed are related as follows:</p>
$$v_{rms} = \sqrt{\frac{3RT}{M}}$$
$$v_{avg} = \sqrt{\frac{8RT}{\pi M}... | mcq | jee-main-2023-online-13th-april-morning-shift | 11,281 |
1lguxo6ar | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafloride (polyatomic). Arrange these on the basis of their root mean square speed $$\left(v_{\mathrm{rms}}\right)$$ and ... | [{"identifier": "A", "content": "$$\\mathrm{v}_{\\mathrm{rms}}($$ mono $$)=\\mathrm{v}_{\\mathrm{rms}}($$ dia $$)=\\mathrm{v}_{\\mathrm{rms}}($$ poly $$)$$"}, {"identifier": "B", "content": "$$\\mathrm{v}_{\\mathrm{rms}}$$ (mono) $$ > \\mathrm{v}_{\\mathrm{rms}}($$ dia $$) > \\mathrm{v}_{\\mathrm{rms}}$$ (poly)"}, {"id... | ["B"] | null | <p>The root mean square speed ($$v_{rms}$$) of a gas is given by the formula:</p>
<p>$$v_{rms} = \sqrt{\frac{3kT}{m}}$$</p>
<p>where $$k$$ is the Boltzmann constant, $$T$$ is the temperature, and $$m$$ is the molar mass of the gas molecules.</p>
<p>The vessels contain neon (monoatomic), chlorine (diatomic), and uranium... | mcq | jee-main-2023-online-11th-april-morning-shift | 11,282 |
1lgyfa78e | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>Match List I with List II :</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;... | [{"identifier": "A", "content": "(A)-(I), (B)-(III), (C)-(IV), (D)-(II)"}, {"identifier": "B", "content": "(A)-(IV), (B)-(III), (C)-(II), (D)-(I)"}, {"identifier": "C", "content": "(A)-(IV), (B)-(II), (C)-(I), (D)-(III)"}, {"identifier": "D", "content": "(A)-(I), (B)-(IV), (C)-(III), (D)-(II)"}] | ["A"] | null | <p>Monoatomic gases possess only translational degrees of freedom. So, option (I) matches with (A).</p>
<p>Polyatomic gases have translational and rotational degrees of freedom. So, option (II) matches with (D).</p>
<p>Rigid diatomic gases have translational, rotational and vibrational degrees of freedom. So, option (I... | mcq | jee-main-2023-online-10th-april-morning-shift | 11,283 |
jaoe38c1lsd66tqi | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is:</p> | [{"identifier": "A", "content": "29 RT"}, {"identifier": "B", "content": "27 RT"}, {"identifier": "C", "content": "20 RT"}, {"identifier": "D", "content": "21 RT"}] | ["B"] | null | <p>To determine the total internal energy of the gas mixture, we adhere to the equipartition theorem, which dictates that each degree of freedom contributes $$\frac{1}{2} RT$$ to the internal energy per mole, where $R$ is the gas constant and $T$ is the temperature.</p>
<p>An argon atom, being a noble gas, is monoatomi... | mcq | jee-main-2024-online-31st-january-evening-shift | 11,286 |
jaoe38c1lsflyhcu | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>$$N$$ moles of a polyatomic gas $$(f=6)$$ must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of $$N$$ is :</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["C"] | null | <p>$$\mathrm{f}_{\mathrm{eq}}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}$$</p>
<p>For diatomic gas $$\mathrm{f}_{\mathrm{eq}}=5$$</p>
<p>$$\begin{aligned}
& 5=\frac{(\mathrm{N})(6)+(2)(3)}{\mathrm{N}+2} \\
& 5 \mathrm{~N}+10=6 \mathrm{~N}+6 \\
& \mathrm{~N}=4
\end{aligned}$$</... | mcq | jee-main-2024-online-29th-january-evening-shift | 11,287 |
1lsg6302e | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>If three moles of monoatomic gas $$\left(\gamma=\frac{5}{3}\right)$$ is mixed with two moles of a diatomic gas $$\left(\gamma=\frac{7}{5}\right)$$, the value of adiabatic exponent $$\gamma$$ for the mixture is</p> | [{"identifier": "A", "content": "1.35"}, {"identifier": "B", "content": "1.52"}, {"identifier": "C", "content": "1.40"}, {"identifier": "D", "content": "1.75"}] | ["B"] | null | <p>$$\begin{array}{ll}
\mathrm{f}_1=3, & \mathrm{f}_2=5 \\
\mathrm{n}_1=3, & \mathrm{n}_2=2
\end{array}$$</p>
<p>$$\begin{aligned}
& \mathrm{f}_{\text {mixture }}=\frac{\mathrm{n}_1 \mathrm{f}_1+\mathrm{n}_2 \mathrm{f}_2}{\mathrm{n}_1+\mathrm{n}_2}=\frac{9+10}{\mathrm{f}}=\frac{19}{5} \\
& \gamma_{\text {mixture }}=1+\... | mcq | jee-main-2024-online-30th-january-evening-shift | 11,288 |
lv2erjlj | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>The translational degrees of freedom $$\left(f_t\right)$$ and rotational degrees of freedom $$\left(f_r\right)$$ of $$\mathrm{CH}_4$$ molecule are:</p> | [{"identifier": "A", "content": "$$f_t=2$$ and $$f_r=2$$\n"}, {"identifier": "B", "content": "$$f_t=3$$ and $$f_r=3$$\n"}, {"identifier": "C", "content": "$$f_t=3$$ and 4$f_r=2$$\n"}, {"identifier": "D", "content": "$$f_t=2$$ and $$f_r=3$$"}] | ["B"] | null | <p>For non-linear polyatomic molecules, both
translational and rotational degree of freedom have
same value and is equal to 3.</p> | mcq | jee-main-2024-online-4th-april-evening-shift | 11,289 |
lv5gs2yl | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature $$(27^{\circ} \mathrm{C})$$. The ratio of specific heat of gases at constant volume respectively is:</p> | [{"identifier": "A", "content": "$$\\frac{3}{2}$$\n"}, {"identifier": "B", "content": "$$\\frac{3}{5}$$\n"}, {"identifier": "C", "content": "$$\\frac{7}{5}$$\n"}, {"identifier": "D", "content": "$$\\frac{5}{3}$$"}] | ["B"] | null | <p>To find the ratio of specific heats at constant volume ($C_V$) of the gases, we need to understand the degrees of freedom each type of gas molecule has, as this determines their specific heat capacity at constant volume. Degrees of freedom refer to the number of independent ways in which a molecule can store energy.... | mcq | jee-main-2024-online-8th-april-morning-shift | 11,290 |
lvb295f4 | physics | heat-and-thermodynamics | degree-of-freedom-and-law-of-equipartition-of-energy | <p>Energy of 10 non rigid diatomic molecules at temperature $$\mathrm{T}$$ is :</p> | [{"identifier": "A", "content": "35 RT"}, {"identifier": "B", "content": "$$\\frac{7}{2}$$ RT"}, {"identifier": "C", "content": "70 K<sub>B</sub>T"}, {"identifier": "D", "content": "35 K<sub>B</sub>T"}] | ["D"] | null | <p>The energy of a diatomic molecule depends on the degrees of freedom it has. For a non-rigid diatomic molecule, there are more degrees of freedom compared to a rigid diatomic molecule. Specifically, a non-rigid diatomic molecule has translational, rotational, and vibrational degrees of freedom. The translational and ... | mcq | jee-main-2024-online-6th-april-evening-shift | 11,291 |
DaO8lQ0ROBSNJEVc | physics | heat-and-thermodynamics | heat-engine,-second-law-of-thermodynamics-and-carnot-engine | ''Heat cannot by itself flow from a body at lower temperature to a body at higher temperature'' is a statement or consequence of : | [{"identifier": "A", "content": "second law of thermodynamics"}, {"identifier": "B", "content": "conservation of momentum"}, {"identifier": "C", "content": "conservation of mass "}, {"identifier": "D", "content": "first law of thermodynamics "}] | ["A"] | null | The statement "Heat cannot by itself flow from a body at lower temperature to a body at higher temperature" is a statement or consequence of the second law of thermodynamics. This law essentially states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all process... | mcq | aieee-2003 | 11,292 |
BnmZYkFfYDNyoLUBzw1kmlwefq8 | physics | heat-and-thermodynamics | heat-engine,-second-law-of-thermodynamics-and-carnot-engine | An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is S<sub>1</sub> and that of the other part is S<sub>2</sub>. Given that S<sub>1</sub> > S<sub>2</sub>. If the piston is removed then the total entropy of the system will be : | [{"identifier": "A", "content": "S<sub>1</sub> $$-$$ S<sub>2</sub>"}, {"identifier": "B", "content": "$${{{S_1}} \\over {{S_2}}}$$"}, {"identifier": "C", "content": "S<sub>1</sub> $$\\times$$ S<sub>2</sub>"}, {"identifier": "D", "content": "S<sub>1</sub> + S<sub>2</sub>"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267404/exam_images/b0nturloj3mkcjzkj7ez.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Physics - Heat and Thermodynamics Question 195 English Explanation">
<br>... | mcq | jee-main-2021-online-18th-march-evening-shift | 11,295 |
yLUyEKPYz5fH375RlF1krpnjo62 | physics | heat-and-thermodynamics | heat-engine,-second-law-of-thermodynamics-and-carnot-engine | The entropy of any system is given by <br/>$$S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]$$ where $$\alpha$$ and $$\beta$$ are the constants. $$\mu$$, J, k and R are no. of moles, mechanical equivalent of heat, Boltzmann constant and gas constant respectively. <br/>[Take $$S = {{dQ} \ove... | [{"identifier": "A", "content": "$$\\alpha$$ and J have the same dimensions."}, {"identifier": "B", "content": "S and $$\\alpha$$ have different dimensions"}, {"identifier": "C", "content": "S, $$\\beta$$, k and $$\\mu$$R have the same dimensions"}, {"identifier": "D", "content": "$$\\alpha$$ and k have the same dimens... | ["D"] | null | Since, entropy of the system is given by<br/><br/>$$S = {\alpha ^2}\beta \ln \left[ {{{\mu kR} \over {J{\beta ^2}}} + 3} \right]$$ .... (i)<br/><br/>As, $$S = {Q \over {\Delta T}}$$ [given]<br/><br/>$$ \Rightarrow [S] = {{[M{L^2}{T^{ - 2}}]} \over {[K]}}$$ .... (ii)<br/><br/>$$\because$$ Dimensions of Q = [ML<sup>2</su... | mcq | jee-main-2021-online-20th-july-morning-shift | 11,296 |
4nuBvzWOAdZ8p2PV | physics | heat-and-thermodynamics | heat-transfer | If mass-energy equivalence is taken into account, when water is cooled to form ice, the mass of water should | [{"identifier": "A", "content": "increase "}, {"identifier": "B", "content": "remain unchanged "}, {"identifier": "C", "content": "decrease "}, {"identifier": "D", "content": "first increase then decrease "}] | ["C"] | null | When water is cooled to form ice, energy is released from water in the form of heat. As energy is equivalent to mass therefore when water is cooled to ice, its mass decreases. | mcq | aieee-2002 | 11,298 |
lzigye0UC1JjRN6N | physics | heat-and-thermodynamics | heat-transfer | Infrared radiation is detected by | [{"identifier": "A", "content": "spectrometer "}, {"identifier": "B", "content": "pyrometer "}, {"identifier": "C", "content": "nanometer "}, {"identifier": "D", "content": "photometer "}] | ["B"] | null | Pyrometer is used to detect infra-red radiation. | mcq | aieee-2002 | 11,300 |
EZB4fpmf7oBd1OMJ | physics | heat-and-thermodynamics | heat-transfer | If the temperature of the sun were to increase from $$T$$ to $$2T$$ and its radius from $$R$$ to $$2R$$, then the ratio of the radiant energy received on earth to what it was previously will be | [{"identifier": "A", "content": "$$32$$ "}, {"identifier": "B", "content": "$$16$$ "}, {"identifier": "C", "content": "$$4$$ "}, {"identifier": "D", "content": "$$64$$ "}] | ["D"] | null | $$E = \sigma A{T^4};\,\,A \propto {R^2}$$
<br>$$\therefore$$ $$E \propto {R^2}{T^4}$$
<br>$$\therefore$$ $${{{E_2}} \over {{E_1}}} = {{R_2^2T_2^4} \over {R_1^2T_1^4}}$$
<br>$$ \Rightarrow {{{E_2}} \over {{E_1}}}$$
<br>$$ = {{{{\left( {2R} \right)}^2}{{\left( {2T} \right)}^4}} \over {{R^2}{T^4}}}$$
<br>$$ = 64$$ | mcq | aieee-2004 | 11,302 |
E0PASQmkx7bIhneV | physics | heat-and-thermodynamics | heat-transfer | The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $$K$$ and $$2K$$ and thickness $$x$$ and $$4x,$$ respectively, are $${T_2}$$ and $${T_1}\left( {{T_2} > {T_1}} \right).$$ The rate of heat transfer through the slab, in a steady stat... | [{"identifier": "A", "content": "$${2 \\over 3}$$ "}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$1$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$ "}] | ["D"] | null | The thermal resistance
<br/><br>$${x \over {KA}} + {{4x} \over {2KA}} = {{3x} \over {KA}}$$
<br/><br>$$\therefore$$ $${{dQ} \over {dt}} = {{\Delta T} \over {{{3x} \over {KA}}}} = {{\left( {{T_2} - {T_1}} \right)KA} \over {3x}}$$
<br/><br>$$ = {1 \over 3}\left\{ {{{A\left( {{T_2} - {T_1}} \right)K} \over x}} \right\}$$
... | mcq | aieee-2004 | 11,303 |
x0xC6nCqL3gnEicW | physics | heat-and-thermodynamics | heat-transfer | The figure shows a system of two concentric spheres of radii $${r_1}$$ and $${r_2}$$ are kept at temperatures $${T_1}$$ and $${T_2}$$, respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to
<img src="data:image/png;base64,UklGRsQQAABXRUJQVlA4ILgQAADQYwCdASpyA... | [{"identifier": "A", "content": "$$In\\left( {{{{r_2}} \\over {{r_1}}}} \\right)$$ "}, {"identifier": "B", "content": "$${{\\left( {{r_2} - {r_1}} \\right)} \\over {\\left( {{r_1}{r_2}} \\right)}}$$ "}, {"identifier": "C", "content": "$${\\left( {{r_2} - {r_1}} \\right)}$$ "}, {"identifier": "D", "content": "$${{{r_1}{... | ["D"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/5f2Sn8CWo8UlPXQYi/WFW7mXUNheB3yTKB2qIHKOt8mvXz3/FB0HCOZGadEqtdDP0dliM3/image.svg" loading="lazy" alt="AIEEE 2005 Physics - Heat and Thermodynamics Question 355 English Explanation">
<br>Consider a shell of thickness $$(dr)$$ and of radius $$(r)$$ and the... | mcq | aieee-2005 | 11,304 |
2uSaCQcyh0jURp2q | physics | heat-and-thermodynamics | heat-transfer | Assuming the Sun to be a spherical body of radius $$R$$ at a temperature of $$TK$$, evaluate the total radiant powered incident of Earth at a distance $$r$$ from the Sun
<p>Where r<sub>0</sub> is the radius of the Earth and $$\sigma $$ is Stefan's constant.</p> | [{"identifier": "A", "content": "$$4\\pi r_0^2{R^2}\\sigma {{{T^4}} \\over {{r^2}}}$$ "}, {"identifier": "B", "content": "$$\\pi r_0^2{R^2}\\sigma {{{T^4}} \\over {{r^2}}}$$ "}, {"identifier": "C", "content": "$$r_0^2{R^2}\\sigma {{{T^4}} \\over {4\\pi {r^2}}}$$ "}, {"identifier": "D", "content": "$${R^2}\\sigma {{{T^4... | ["B"] | null | Total power radiated by Sun $$ = \sigma {T^4} \times 4\pi {R^2}$$
<br><br>The intensity of power at earth's surface $$ = {{\sigma {T^4} \times 4\pi {R^2}} \over {4\pi {r^2}}}$$
<br><br>Total power received by Earth $$ = {{\sigma {T^4}{R^2}} \over {{r^2}}}\left( {\pi r_0^2} \right)$$ | mcq | aieee-2006 | 11,305 |
qC1cqhv8VOtI3UJZ | physics | heat-and-thermodynamics | heat-transfer | One end of a thermally insulated rod is kept at a temperature $${T_1}$$ and the other at $${T_2}$$. The rod is composed of two sections of length $${L_1}$$ and $${L_2}$$ and thermal conductivities $${K_1}$$ and $${K_2}$$ respectively. The temperature at the interface of the two section is
<img src="data:image/png;base... | [{"identifier": "A", "content": "$${{\\left( {{K_1}{L_1}{T_1} + {K_2}{L_2}{T_2}} \\right)} \\over {\\left( {{K_1}{L_1} + {K_2}{L_2}} \\right)}}$$ "}, {"identifier": "B", "content": "$${{\\left( {{K_2}{L_2}{T_1} + {K_1}{L_1}{T_2}} \\right)} \\over {\\left( {{K_1}{L_1} + {K_2}{L_2}} \\right)}}$$ "}, {"identifier": "C", "... | ["D"] | null | <img class="question-image" src="https://imagex.cdn.examgoal.net/XvX7bIkAwyLVBh5ng/wfroG7CIxzJcXiNzdrBXsySySq4pJ/B2T95LIqr5rzj6WCRHRbmC/image.svg" loading="lazy" alt="AIEEE 2007 Physics - Heat and Thermodynamics Question 351 English Explanation">
<br>$${{{K_1}A\left( {{T_1} - T} \right)} \over {{\ell _1}}} = {{{K_2}A\... | mcq | aieee-2007 | 11,306 |
UMAfd4NQAWsg4LUO | physics | heat-and-thermodynamics | heat-transfer | A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $$\theta $$ along the length $$x$$ of the bar from its hot end is best described by which of the following figures? | [{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264201/exam_images/vtiaornebcqd1niwl7qy.webp\" loading=\"lazy\" alt=\"AIEEE 2009 Physics - Heat and Thermodynamics Question 350 English Option 1\"> "}, {"identifier": "B", "content": "<img class=\... | ["A"] | null | The heat flow rate is given by
<br><br>$${{dQ} \over {dt}} = {{kA\left( {{\theta _1} - \theta } \right)} \over x}$$
<br><br>$$ \Rightarrow {\theta _1} - \theta $$
$$ = {x \over {kA}}{{dQ} \over {dt}} $$
<br><br>$$\Rightarrow \theta $$
$$ = {\theta _1} - {x \over {kA}}{{dQ} \over {dt}}$$
<br><br>where $${\theta _1}$$ is... | mcq | aieee-2009 | 11,307 |
CZBbHg7fWexeDiVe | physics | heat-and-thermodynamics | heat-transfer | Three rods of Copper, Brass and Steel are welded together to form a $$Y$$ shaped structure. Area of cross - section of each rod $$ = 4c{m^2}.$$ End of copper rod is maintained at $${100^ \circ }C$$ where as ends of brass and steel are kept at $${0^ \circ }C$$. Lengths of the copper, brass and steel rods are $$46,$$ $$1... | [{"identifier": "A", "content": "$$1.2$$ $$cal/s$$"}, {"identifier": "B", "content": "$$2.4$$ $$cal/s$$"}, {"identifier": "C", "content": "$$4.8$$ $$cal/s$$"}, {"identifier": "D", "content": "$$6.0$$ $$cal/s$$ "}] | ["C"] | null | Rate of heat flow is given by,
<br><br>$$Q = {{KA\left( {{\theta _1} - {\theta _2}} \right)} \over l}$$
<br><br>Where, $$K=$$ coefficient of thermal conductivity $$l=$$ length of rod and $$A=$$ Area of cross-section of rod
<br><img class="question-image" src="https://imagex.cdn.examgoal.net/HlGlPtIDOXeYoEZVP/OZxophBJu8... | mcq | jee-main-2014-offline | 11,308 |
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