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lv7v4o98 | physics | heat-and-thermodynamics | kinetic-theory-of-gases-and-gas-laws | <p>If the collision frequency of hydrogen molecules in a closed chamber at $$27^{\circ} \mathrm{C}$$ is $$\mathrm{Z}$$, then the collision frequency of the same system at $$127^{\circ} \mathrm{C}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{\\sqrt{3}}{2} \\mathrm{Z}$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{\\sqrt{3}} \\mathrm{Z}$$\n"}, {"identifier": "C", "content": "$$\\frac{3}{4} \\mathrm{Z}$$\n"}, {"identifier": "D", "content": "$$\\frac{4}{3} \\mathrm{Z}$$"}] | ["B"] | null | <p>The collision frequency ($Z$) of gas molecules is proportional to the square root of the absolute temperature ($T$) of the system. Mathematically, it can be represented as $Z \propto \sqrt{T}$. This implies that when the temperature changes, the collision frequency changes as well according to the formula:</p>
<p>$... | mcq | jee-main-2024-online-5th-april-morning-shift | 11,433 |
p9pw0pPuGDcwMcYi | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | Heat given to a body which raises its temperature by $${1^ \circ }C$$ is | [{"identifier": "A", "content": "water equivalent "}, {"identifier": "B", "content": "thermal capacity "}, {"identifier": "C", "content": "specific heat "}, {"identifier": "D", "content": "temperature gradient "}] | ["B"] | null | Heat required for raising the temperature of the whole body by $${1^ \circ }C$$ is called the thermal capacity of the body. | mcq | aieee-2002 | 11,437 |
PdZBrJ3r2oWdvvgs | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | Time taken by a $$836$$ $$W$$ heater to heat one litre of water from $$10{}^ \circ C$$ to $$40{}^ \circ C$$ is | [{"identifier": "A", "content": "$$150$$ $$s$$ "}, {"identifier": "B", "content": "$$100$$ $$s$$ "}, {"identifier": "C", "content": "$$50$$ $$s$$ "}, {"identifier": "D", "content": "$$200$$ $$s$$ "}] | ["A"] | null | $$\Delta Q = mC \times \Delta T$$
<br><br>$$ = 1 \times 4180 \times \left( {40 - 10} \right) = 80 \times 30$$
<br><br>( $$\therefore$$ $$\Delta Q = $$ heat supplied in time $$t$$ for heating $$1L$$ water from $${10^ \circ }C$$ to $${40^ \circ }C$$ )
<br><br>also $$\Delta Q = 836 \times t \Rightarrow t = {{4180 \times 3... | mcq | aieee-2004 | 11,438 |
w4tsdA6ByCH4Xlro | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible ? The surface tension is $$T,$$ density of liquid is $$\rho $$ and $$L$$ is its latent heat of vaporization. | [{"identifier": "A", "content": "$$\\rho L/T$$ "}, {"identifier": "B", "content": "$$\\sqrt {T/\\rho L} $$ "}, {"identifier": "C", "content": "$$T/\\rho L$$ "}, {"identifier": "D", "content": "$$2T/\\rho L$$ "}] | ["D"] | null | When radius is decrease by $$\Delta R,$$
<br>$$4\pi {R^2}\Delta R\rho L = 4\pi T\left[ {{R^2} - {{\left( {R - \Delta R} \right)}^2}} \right]$$
<br>$$ \Rightarrow \rho {R^2}\Delta RL = T\left[ {{R^2} - {R^2} + 2R\Delta R - \Delta {R^2}} \right]$$
<br>$$ \Rightarrow \rho {R^2}\Delta RL = T2R\Delta R\,\,$$ [ $$\Delta R$$ ... | mcq | jee-main-2013-offline | 11,439 |
m8buGfkTMHHXb97h | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm,
filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to
be 75<sup>o</sup>C. T is given by: (Given : room temperature = 30<sup>o</sup>C, specific heat of copper = 0.1... | [{"identifier": "A", "content": "825<sup>o</sup>C "}, {"identifier": "B", "content": "800<sup>o</sup>C"}, {"identifier": "C", "content": "885<sup>o</sup>C"}, {"identifier": "D", "content": "1250<sup>o</sup>C"}] | ["C"] | null | According to principle of calorimetry,
<br><br>Heat lost = Heat gain
<br><br>100 × 0.1( – 75) = 100 × 0.1 × 45 + 170 × 1 × 45
<br><br>10 – 750 = 450 + 7650
<br><br>10 = 1200 + 7650 = 8850
<br><br>T = 885°C | mcq | jee-main-2017-offline | 11,441 |
PBqFAIUbQh46dmcET2nsP | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | In an experiment, a sphere of aluminium of mass 0.20 kg is heated upto 150<sup>o</sup><sup></sup>C. Immediately, it is put into water of volume 150 cc at 27<sup>o</sup>C kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is 40<sup>o</sup>C. The specific heat of aluminium is : (take 4... | [{"identifier": "A", "content": "378 J/kg $$-$$<sup>o</sup>C"}, {"identifier": "B", "content": "315 J/kg $$-$$<sup>o</sup>C"}, {"identifier": "C", "content": "476 J/kg $$-$$<sup>o</sup>C"}, {"identifier": "D", "content": "434 J/kg $$-$$<sup>o</sup>C"}] | ["D"] | null | Let specific heat of aluminium = S,
<br><br>As we know from principle of calorimetry,
<br><br>Q<sub>given</sub> = Q<sub>used</sub>
<br><br>$$\therefore\,\,\,$$ 0.2 $$ \times $$ S $$ \times $$ (150 $$-$$ 40) =<br> 150 $$ \times $$ 1 $$ \times $$ (40 $$-$$ 27) + 25 $$ \times $$ (40$$-$$27)
<br><br>$$ \Rightarrow $$$$\,... | mcq | jee-main-2017-online-8th-april-morning-slot | 11,442 |
DwfLvMCr0LpNvhouT8gC1 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | When 100 g of a liquid A at 100<sup>o</sup>C is added to 50 g of a liquid B at temperature 75<sup>o</sup>C, the temperature of the mixture becomes 90<sup>o</sup>C. The temperature of the mixture, if 100 g of liquid A at 100<sup>o</sup>C is added to 50 g of liquid B at 50<sup>o</sup>C, will be : | [{"identifier": "A", "content": "60<sup>o</sup>C"}, {"identifier": "B", "content": "70<sup>o</sup>C"}, {"identifier": "C", "content": "85<sup>o</sup>C"}, {"identifier": "D", "content": "80<sup>o</sup>C"}] | ["D"] | null | 100 $$ \times $$ S<sub>A</sub> $$ \times $$ [100 $$-$$ 90] = 50 $$ \times $$ S<sub>B</sub> $$ \times $$ (90 $$-$$ 75)
<br><br>2S<sub>A</sub> = 1.5 S<sub>B</sub>
<br><br>S<sub>A</sub> = $${3 \over 4}$$S<sub>B</sub>
<br><br>Now, 100 $$ \times $$ S<sub>A</sub> $$ \times $$ [100 $$-$$ T] = 50 $$ \times $$ S<sub>B</sub> (T ... | mcq | jee-main-2019-online-11th-january-evening-slot | 11,445 |
bfUiMaYwBtFrFmZZ3KdBX | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | A metal ball of mass 0.1 kg is heated upto 500<sup>o</sup>C and dropped into a vessel of heat capacity 800 JK<sup>–1</sup> and containing 0.5 kg water. The initial temperature of water and vessel is 30<sup>o</sup>C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities ... | [{"identifier": "A", "content": "20%"}, {"identifier": "B", "content": "25%"}, {"identifier": "C", "content": "15%"}, {"identifier": "D", "content": "30%"}] | ["A"] | null | 0.1 $$ \times $$ 400 $$ \times $$ (500 $$-$$ T) = 0.5 $$ \times $$ 4200 $$ \times $$ (T $$-$$ 30) + 800 (T $$-$$ 30)
<br><br>$$ \Rightarrow $$ 40(500 $$-$$ T) = (T $$-$$ 30) (2100 + 800)
<br><br>$$ \Rightarrow $$ 20000 $$-$$ 40T = 2900 T $$-$$ 30 $$ \times $$ 2900
<br><br>$$ \Rightarrow $$ &... | mcq | jee-main-2019-online-11th-january-evening-slot | 11,446 |
9scdtfJGNx08W7qOuJB4F | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | A thermally insulated vessel contains 150g of
water at 0°C. Then the air from the vessel is
pumped out adiabatically. A fraction of water
turns into ice and the rest evaporates at 0°C
itself. The mass of evaporated water will be
closest to :
(Latent heat of vaporization of water
= 2.10 × 10<sup>6</sup> J kg<sup>–1</sup... | [{"identifier": "A", "content": "35 g"}, {"identifier": "B", "content": "130 g"}, {"identifier": "C", "content": "20 g"}, {"identifier": "D", "content": "150 g"}] | ["C"] | null | <p>Let x grams of water is evaporated.</p>
<p>According to the principle of calorimetry,</p>
<p>Heat lost by freezing water (that turns into ice) = Heat gained by evaporated water</p>
<p>Given, mass of water = 150 g</p>
<p>$$ \Rightarrow (150 - x) \times {10^{ - 3}} \times 3.36 \times {10^5} = x \times {10^{ - 3}} \tim... | mcq | jee-main-2019-online-8th-april-morning-slot | 11,447 |
pukagGVR02XHr15lnwERI | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | A massless spring (k = 800 N/m), attached with
a mass (500 g) is completely immersed in 1 kg
of water. The spring is stretched by 2 cm and
released so that it starts vibrating. What would
be the order of magnitude of the change in the
temperature of water when the vibrations stop
completely ? (Assume that the water con... | [{"identifier": "A", "content": "10<sup>\u20133</sup> K"}, {"identifier": "B", "content": "10<sup>\u20131</sup> K"}, {"identifier": "C", "content": "10<sup>\u20135</sup>K"}, {"identifier": "D", "content": "10<sup>\u20134 </sup>K"}] | ["C"] | null | By law of conservation of energy<br><br>
$${1 \over 2}k{x^2} = \left( {{m_1}{s_1} + {m_2}{s_2}} \right)\Delta T$$<br><br>
$$\Delta T = {{16 \times {{10}^{ - 2}}} \over {4384}} = 3.65 \times {10^{ - 5}}$$ K | mcq | jee-main-2019-online-9th-april-evening-slot | 11,448 |
EAZwNsvKvmhHFGOaaB3rsa0w2w9jx3pfzz9 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | When M<sub>1</sub> gram of ice at –10<sup>o</sup>C (specific heat = 0.5 cal g<sup>–1</sup>
<sup>o</sup>C<sup>–1</sup>
) is added to M<sub>2</sub> gram of water at 50C, finally
no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g<sup>–1</sup>
is : | [{"identifier": "A", "content": "$${{50{M_2}} \\over {{M_1}}} - 5$$"}, {"identifier": "B", "content": "$${{50{M_2}} \\over {{M_1}}}$$"}, {"identifier": "C", "content": "$${{5{M_2}} \\over {{M_1}}} - 5$$"}, {"identifier": "D", "content": "$${{5{M_1}} \\over {{M_2}}} - 50$$"}] | ["A"] | null | Heat lost = Heat gain<br><br>
$$ \Rightarrow {M_2} \times 1 \times 50 = {M_1} \times 0.5 \times 10 + {M_1}.{L_f}$$<br><br>
$$ \Rightarrow {L_f} = {{50{M_2} - 5{M_1}} \over {{M_1}}}$$<br><br>
$$ = {{50{M_2}} \over {{M_1}}} - 5$$ | mcq | jee-main-2019-online-12th-april-morning-slot | 11,449 |
wwRshZJkaDInCamqZP3rsa0w2w9jx6rc23k | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | One kg of water, at 20<sup>o</sup>C, heated in an electric kettle whose heating element has a mean (temperature
averaged) resistance of 20 $$\Omega $$. The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time
taken for water to evaporate fully, is close to :
[Specific heat of water = 4200 J/(kg <... | [{"identifier": "A", "content": "10 minutes"}, {"identifier": "B", "content": "22 minutes"}, {"identifier": "C", "content": "3 minutes"}, {"identifier": "D", "content": "16 minutes"}] | ["B"] | null | P$$ \times $$t = mS$$\Delta $$t + mL<sub>v</sub>
<br><br>$$ \Rightarrow $$ $${{{V^2}} \over R}t$$ = mS$$\Delta $$t + mL<sub>v</sub>
<br><br>$$ \Rightarrow $$ $${{{{\left( {200} \right)}^2}} \over {20}}t$$ = $$1 \times 4200 \times \left( {100 - 20} \right)$$ + $$1 \times 2260 \times {10^3}$$
<br><br>$$ \Rightarrow $$ 20... | mcq | jee-main-2019-online-12th-april-evening-slot | 11,450 |
JZcpM9fe01VEwHAglwjgy2xukfdquq1c | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | A bullet of mass 5 g, travelling with a speed of
210 m/s, strikes a fixed wooden target. One half
of its kinetic energy is converted into heat in
the bullet while the other half is converted into
heat in the wood. The rise of temperature of
the bullet if the specific heat of its material is
0.030 cal/(g – <sup>o</sup>C... | [{"identifier": "A", "content": "87.5 <sup>o</sup>C"}, {"identifier": "B", "content": "83.3 <sup>o</sup>C"}, {"identifier": "C", "content": "38.4 <sup>o</sup>C"}, {"identifier": "D", "content": "119.2 <sup>o</sup>C"}] | ["A"] | null | $${1 \over 2}m{v^2} \times {1 \over 2} = ms\Delta T$$<br><br>$$ \Rightarrow $$ $$\Delta T = {{{v^2}} \over {4 \times 5}} = {{{{210}^2}} \over {4 \times 30 \times 4.200}}$$<br><br>$$ = 87.5^\circ C$$ | mcq | jee-main-2020-online-5th-september-morning-slot | 11,451 |
JAivoVHixP6yfzWUcGjgy2xuketxsk72 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | The specific heat of water<br/> = 4200 J kg<sup>-1</sup>K<sup>-1</sup> and the latent heat of<br/> ice = 3.4 $$ \times $$ 10<sup>5</sup> J kg<sup>–1</sup>. 100 grams of ice at<br/> 0<sup>o</sup>C is placed in 200 g of water at 25<sup>o</sup>C. The <br/>amount of ice that will melt as the temperature<br/> of water reach... | [{"identifier": "A", "content": "63.8"}, {"identifier": "B", "content": "61.7"}, {"identifier": "C", "content": "69.3"}, {"identifier": "D", "content": "64.6"}] | ["B"] | null | Heat loss by water<br><br>$$Q = {m_w}s\Delta \theta $$<br><br>$$ = \left( {{{200} \over {1000}}} \right).(4200)(25) = 21000\,J$$
<br><br>This heat will absorbed by the ice and let mass $$\Delta $$m<sub>i</sub> got melted.
<br><br>So $$\Delta {m_i}L = 21000$$<br><br>$$\Delta {m_i} = {{21000} \over {3.4 \times {{10}^5}}}... | mcq | jee-main-2020-online-4th-september-morning-slot | 11,453 |
uWx2vVr9cgMamH8hmF7k9k2k5dkln9i | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | Two moles of an ideal gas with $${{{C_P}} \over {{C_V}}} = {5 \over 3}$$
are mixed with 3 moles of another ideal gas
with $${{{C_P}} \over {{C_V}}} = {4 \over 3}$$. The value of $${{{C_P}} \over {{C_V}}}$$ for the
mixture is :
| [{"identifier": "A", "content": "1.50"}, {"identifier": "B", "content": "1.45"}, {"identifier": "C", "content": "1.47"}, {"identifier": "D", "content": "1.42"}] | ["D"] | null | C<sub>p</sub> = $${{{n_1}{C_{{p_1}}} + {n_2}{C_{{p_2}}}} \over {{n_1} + {n_2}}}$$
<br><br>C<sub>v</sub> = $${{{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}} \over {{n_1} + {n_2}}}$$
<br><br>$$\gamma $$<sub>mix</sub> = $${{{C_p}} \over {{C_v}}}$$ = $${{2 \times {5 \over 2}R + 3 \times {8 \over 2}R} \over {2 \times {3 \over 2}R + ... | mcq | jee-main-2020-online-7th-january-morning-slot | 11,455 |
V2KwCTHz3xQrdzKh1Q7k9k2k5hi0ix1 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | Three containers C<sub>1</sub>, C<sub>2</sub> and C<sub>3</sub> have water at
different temperatures. The table below shows
the final temperature T when different amounts
of water (given in litres) are taken from each
containers and mixed (assume no loss of heat
during the process)
<img src="data:image/png;base64,UklGR... | [] | null | 50 | Let containers C<sub>1</sub>, C<sub>2</sub>, C<sub>3</sub> contain water at T<sub>1</sub>, T<sub>2</sub> and T<sub>3</sub> temperatures respectively.
<br><br>Applying law of
calorimetry
<br><br> 1(T<sub>1</sub> – 60) + 2(T<sub>2</sub> – 60) = 0
<br><br>$$ \Rightarrow $$ T<sub>1</sub> + 2T<sub>2</sub> = 180 ....(1)
<br>... | integer | jee-main-2020-online-8th-january-evening-slot | 11,456 |
1krulsom1 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is :<br/><br/><img src="data:image/png;base64,UklGRpYTAABXRUJQVlA4IIoTAAAwdACdASqhAQ... | [{"identifier": "A", "content": "$${8 \\over 3}$$"}, {"identifier": "B", "content": "$${3 \\over 8}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "$${4 \\over 3}$$"}] | ["B"] | null | $${\left( {{{\Delta Q} \over {\Delta t}}} \right)_A} = {\left( {{{\Delta Q} \over {\Delta t}}} \right)_B}$$<br><br>$$m{S_A}{\left( {{{\Delta T} \over {\Delta t}}} \right)_A} = m{S_B}{\left( {{{\Delta T} \over {\Delta t}}} \right)_B}$$<br><br>$$ \Rightarrow $$ $${{{S_A}} \over {{S_B}}} = {{{{\left( {{{\Delta T} \over {\... | mcq | jee-main-2021-online-25th-july-morning-shift | 11,458 |
1ktbmyhe5 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | The temperature of equal masses of three different liquids x, y and z are 10$$^\circ$$C, 20$$^\circ$$C and 30$$^\circ$$C respectively. The temperature of mixture when x is mixed with y is 16$$^\circ$$C and that when y is mixed with z is 26$$^\circ$$C. The temperature of mixture when x and z are mixed will be : | [{"identifier": "A", "content": "28.32$$^\\circ$$C"}, {"identifier": "B", "content": "25.62$$^\\circ$$C"}, {"identifier": "C", "content": "23.84$$^\\circ$$C"}, {"identifier": "D", "content": "20.28$$^\\circ$$C"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267808/exam_images/ejqtv1pmby5kxa3ryzzx.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266484/exam_images/e9kgpr1elviqeyvc987e.webp"><source media="(max-wid... | mcq | jee-main-2021-online-26th-august-evening-shift | 11,459 |
1l56utt40 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | <p>A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is :</p>
<p>(Given : initial temperature of the bullet = 127$$^\circ$$C, Melting point of the bullet = 327$$^\circ$$C, Latent heat of fusion of lead = 2.5 $$\times$$ 10<sup>... | [{"identifier": "A", "content": "125 ms<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "500 ms<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "250 ms<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "600 ms<sup>$$-$$1</sup>"}] | ["B"] | null | <p>$${2 \over 5} \times {1 \over 2}m{v^2} = mL + ms\Delta T$$</p>
<p>$$ \Rightarrow {{{v^2}} \over 5} = 2.5 \times {10^4} + 125 + 200$$</p>
<p>$$ \Rightarrow {{{v^2}} \over 5} = 5 \times {10^4}$$</p>
<p>$$ \Rightarrow v = 500$$ m/s</p> | mcq | jee-main-2022-online-27th-june-evening-shift | 11,461 |
1l58ii9b3 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | <p>A geyser heats water flowing at a rate of 2.0 kg per minute from 30$$^\circ$$C to 70$$^\circ$$C. If geyser operates on a gas burner, the rate of combustion of fuel will be ___________ g min<sup>$$-$$1</sup>.</p>
<p>[Heat of combustion = 8 $$\times$$ 10<sup>3</sup> Jg<sup>$$-$$1</sup>, Specific heat of water = 4.2 Jg... | [] | null | 42 | <p>$$Q = ms\Delta T$$</p>
<p>$${{dQ} \over {dt}} = {\left( {{{dm} \over {dt}}} \right)_{water}}S\Delta T = {\left( {{{dm} \over {dt}}} \right)_{oil}}C$$</p>
<p>$$ \Rightarrow 2 \times 4.2 \times {10^3} \times 40 = {\left( {{{dm} \over {dt}}} \right)_{oil}} \times 8 \times {10^6}$$</p>
<p>$$ \Rightarrow {\left( {{{dm} \... | integer | jee-main-2022-online-26th-june-evening-shift | 11,462 |
1l59m3hnv | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | <p>A copper block of mass 5.0 kg is heated to a temperature of 500$$^\circ$$C and is placed on a large ice block. What is the maximum amount of ice that can melt? [Specific heat of copper : 0.39 J g<sup>$$-$$1</sup> $$^\circ$$C<sup>$$-$$1</sup> and latent heat of fusion of water : 335 J g<sup>$$-$$1</sup>]</p> | [{"identifier": "A", "content": "1.5 kg"}, {"identifier": "B", "content": "5.8 kg"}, {"identifier": "C", "content": "2.9 kg"}, {"identifier": "D", "content": "3.8 kg"}] | ["C"] | null | <p>$$mL = \Delta Q = ms\Delta T$$</p>
<p>$$ \Rightarrow m = {{5 \times 0.39 \times {{10}^3} \times 500} \over {335}}$$</p>
<p>$$ = 2.9$$ kg</p> | mcq | jee-main-2022-online-25th-june-evening-shift | 11,463 |
1l5bc1d28 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | <p>A 100 g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60 ms<sup>$$-$$1</sup>. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail?</p>
<p>[Specific heat capacity of iron = 0.42 Jg<sup>$$-$$1</sup> $$^\circ$$C<sup>$$-$$1</sup>]</p> | [{"identifier": "A", "content": "675$$^\\circ$$C"}, {"identifier": "B", "content": "1600$$^\\circ$$C"}, {"identifier": "C", "content": "16.07$$^\\circ$$C"}, {"identifier": "D", "content": "6.75$$^\\circ$$C"}] | ["C"] | null | <p>$${1 \over 2} \times 1.5 \times {60^2} \times {1 \over 4} = 100 \times 0.42 \times \Delta T$$</p>
<p>$$\Delta T = {{1.5 \times {{60}^2}} \over {8 \times 100 \times 0.42}} = 16.07^\circ C$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift | 11,464 |
1l6f5k912 | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | <p>A block of ice of mass 120 g at temperature 0$$^\circ$$C is put in 300 g of water at 25$$^\circ$$C. The x g of ice melts as the temperature of the water reaches 0$$^\circ$$C. The value of x is _____________.</p>
<p>[Use specific heat capacity of water = 4200 Jkg<sup>$$-$$1</sup>K<sup>$$-$$1</sup>, Latent heat of ice... | [] | null | 90 | <p>Heat lost by water = Heat gained by ice</p>
<p>$$0.3 \times 4200 \times 25 = x \times 3.5 \times {10^5}$$</p>
<p>$$x = {{0.3 \times 4200 \times 25} \over {3.5 \times {{10}^5}}}$$</p>
<p>$$ = 90 \times 100 \times {10^5} \times {10^3}$$ gram = 90 gm</p> | integer | jee-main-2022-online-25th-july-evening-shift | 11,465 |
ldo7fkbb | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | A water heater of power $2000 \mathrm{~W}$ is used to heat water. The specific heat capacity of water is $4200 \mathrm{~J}$ $\mathrm{kg}^{-1} \mathrm{~K}^{-1}$. The efficiency of heater is $70 \%$. Time required to heat $2 \mathrm{~kg}$ of water from $10^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ is _________ s.<br... | [] | null | 300 | <p>The amount of heat energy required to raise the temperature of a substance can be calculated as:</p>
<p>Q = m $$ \times $$ c $$ \times $$ ΔT</p>
<p>where Q is the heat energy required, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.</p>
<p>The time required to... | integer | jee-main-2023-online-31st-january-evening-shift | 11,466 |
1lds9yjwc | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | <p>Heat energy of 184 kJ is given to ice of mass 600 g at $$-12^\circ \mathrm{C}$$. Specific heat of ice is $$\mathrm{2222.3~J~kg^{-1^\circ}~C^{-1}}$$ and latent heat of ice in 336 $$\mathrm{kJ/kg^{-1}}$$</p>
<p>A. Final temperature of system will be 0$$^\circ$$C.</p>
<p>B. Final temperature of the system will be great... | [{"identifier": "A", "content": "A and E only"}, {"identifier": "B", "content": "B and D only"}, {"identifier": "C", "content": "A and C only"}, {"identifier": "D", "content": "A and D only"}] | ["D"] | null | <p>Heat required to raise the temperature of ice to 0$$^\circ$$C is</p>
<p>$$ = {{60} \over {1000}}(2222.3)(12)$$</p>
<p>$$ = 16000.5$$ J</p>
<p>$$ \approx 16$$ kJ</p>
<p>Heat required to melt ice completely</p>
<p>$$ = \left( {{{600} \over {1000}}} \right)(336)$$ kJ</p>
<p>$$ = 201.6$$ kJ</p>
<p>Energy left $$ = (184 ... | mcq | jee-main-2023-online-29th-january-evening-shift | 11,467 |
lvb2957z | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | <p>Given below are two statements:</p>
<p>Statement (I) : Dimensions of specific heat is $$[\mathrm{L}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}]$$.</p>
<p>Statement (II) : Dimensions of gas constant is $$[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1} \mathrm{~K}^{-1}]$$.</p>
<p>In the light of the above statements, choose the mo... | [{"identifier": "A", "content": "Statement (I) is incorrect but statement (II) is correct\n"}, {"identifier": "B", "content": "Both statement (I) and statement (II) are incorrect\n"}, {"identifier": "C", "content": "Both statement (I) and statement (II) are correct\n"}, {"identifier": "D", "content": "Statement (I) is ... | ["D"] | null | <p>To evaluate the veracity of the given statements, we need to understand the physical quantities involved and their dimensional formulas. Specifically, we're looking at the dimensions of specific heat and gas constant.</p>
<p><strong>Specific Heat:</strong></p>
<p>Specific heat (c) is the amount of heat required to... | mcq | jee-main-2024-online-6th-april-evening-shift | 11,469 |
lvc57cty | physics | heat-and-thermodynamics | specific-heat-capacity,-calorimetry-&-change-of-state | <p>The specific heat at constant pressure of a real gas obeying $$P V^2=R T$$ equation is:</p> | [{"identifier": "A", "content": "R"}, {"identifier": "B", "content": "$$C_V+R$$\n"}, {"identifier": "C", "content": "$$C_V+\\frac{R}{2 V}$$\n"}, {"identifier": "D", "content": "$$\\frac{R}{3}+C_V$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \because \quad P V^2=R T \\
& P(2 v d v)+V^2(d P)=R d T
\end{aligned}$$</p>
<p>at $$P=$$ const.</p>
<p>$$P d v=\frac{R d T}{2 V} \quad \text{... (i)}$$</p>
<p>Now, for $$n=1$$</p>
<p>$$\begin{aligned}
& d \theta=d v+d w \\
& C_P d T=C_v d T+P d v \quad \text{... (ii)}
\end{aligned}$$</p>
<p>from ... | mcq | jee-main-2024-online-6th-april-morning-shift | 11,470 |
Bdo5Wo5QGqFeXCm3 | physics | heat-and-thermodynamics | thermodynamics-process | During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $${C_p}/{C_V}$$ for the gas is | [{"identifier": "A", "content": "$${4 \\over 3}$$ "}, {"identifier": "B", "content": "$$2$$ "}, {"identifier": "C", "content": "$${5 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["D"] | null | $$P \propto {T^3} \Rightarrow P{T^{ - 3}} = $$ constant ....$$(i)$$
<br>But for an adiabatic process, the pressure temperature relationship is given by
<br>$${P^{1 - \gamma }}\,\,{T^\gamma } = $$ constant $$ \Rightarrow P{T^{{\gamma \over {1 - \gamma }}}} = $$ constant. ....$$(ii)$$
<br>From $$(i)$$ and $$(ii)$$ $${\g... | mcq | aieee-2003 | 11,471 |
cQ7bgxuWcaSB1sYG | physics | heat-and-thermodynamics | thermodynamics-process | The work of $$146$$ $$kJ$$ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by $${7^ \circ }C.$$ The gas is $$\left( {R = 8.3J\,\,mo{l^{ - 1}}\,{K^{ - 1}}} \right)$$ | [{"identifier": "A", "content": "diatomic "}, {"identifier": "B", "content": "triatomic "}, {"identifier": "C", "content": "a mixture of monoatomic and diatomic "}, {"identifier": "D", "content": "monoatomic"}] | ["A"] | null | $$W = {{nR\Delta T} \over {1 - \gamma }} \Rightarrow - 146000$$
<br>$$ = {{1000 \times 8.3 \times 7} \over {1 - \gamma }}$$
<br> or $$1 - \gamma = - {{58.1} \over {146}} \Rightarrow \gamma $$
<br>$$ = 1 + {{58.1} \over {146}} = 1.4$$
<br>Hence the gas is diatomic. | mcq | aieee-2006 | 11,472 |
Zfn75G8NiR7ukbyD | physics | heat-and-thermodynamics | thermodynamics-process | Two moles of helium gas are taken over the cycle $$ABCD,$$ as shown in the $$P$$-$$T$$ diagram.
<img src="data:image/png;base64,UklGRu4MAABXRUJQVlA4IOIMAADQgACdASplAnwBP4G+2GU2L7knIdHpkyAwCWlu8p8d6a6iN+fn4//1ncFj02iXYzKbQAd2BMsVzDyfe8/cv/dwKRQV+Z7K5F+VyL8rkX5XIvyuRflXAnj9LBdafL5fL5fL5fL5fL5fL5fNb/O3qgFAyTrXgOfmDEGKdZqH... | [{"identifier": "A", "content": "$$300$$ $$R$$ "}, {"identifier": "B", "content": "$$400$$ $$R$$ "}, {"identifier": "C", "content": "$$500$$ $$R$$ "}, {"identifier": "D", "content": "$$200$$ $$R$$ "}] | ["B"] | null | $$A$$ to $$B$$ is an isobaric process. The work done
<br>$$W = nR\left( {{T_2} - {T_1}} \right) = 2R\left( {500 - 300} \right) = 400R$$ | mcq | aieee-2009 | 11,473 |
YWGOIEt58Pw4FQSH | physics | heat-and-thermodynamics | thermodynamics-process | Two moles of helium gas are taken over the cycle $$ABCD,$$ as shown in the $$P$$-$$T$$ diagram.
<img src="data:image/png;base64,UklGRu4MAABXRUJQVlA4IOIMAADQgACdASplAnwBP4G+2GU2L7knIdHpkyAwCWlu8p8d6a6iN+fn4//1ncFj02iXYzKbQAd2BMsVzDyfe8/cv/dwKRQV+Z7K5F+VyL8rkX5XIvyuRflXAnj9LBdafL5fL5fL5fL5fL5fL5fNb/O3qgFAyTrXgOfmDEGKdZqH... | [{"identifier": "A", "content": "$$+414$$ $$R$$ "}, {"identifier": "B", "content": "$$-690$$ $$R$$ "}, {"identifier": "C", "content": "$$+690$$ $$R$$ "}, {"identifier": "D", "content": "$$-414$$ $$R$$ "}] | ["A"] | null | Work done by the system in the isothermal process
<br>$$DA$$ is $$W = 2.303nRT\,{\log _{10}}{{{P_D}} \over {{P_A}}}$$
<br>$$ = 2.303 \times 2R \times 300{\log _{10}}{{1 \times {{10}^5}} \over {2 \times {{10}^5}}} = - 414R.$$
<br>Therefore work done on the gas is $$ + \,414\,R.$$ | mcq | aieee-2009 | 11,474 |
SVw3porCFxNDFXVv | physics | heat-and-thermodynamics | thermodynamics-process | Helium gas goes through a cycle $$ABCD$$ (consisting of two isochoric and isobaric lines) as shown in figure efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas)
<img src="data:image/png;base64,UklGRmwJAABXRUJQVlA4IGAJAACQYgCdASr8AYUBP4HA1ma2MCwnoDVZesAwCWlu4W8DRmNwvj5D/03bj/ncfy49zzfzP+L8MeA... | [{"identifier": "A", "content": "$$15.4\\% $$ "}, {"identifier": "B", "content": "$$9.1\\% $$"}, {"identifier": "C", "content": "$$10.5\\% $$"}, {"identifier": "D", "content": "$$12.5\\% $$"}] | ["A"] | null | Heat given to system $$ = {\left( {n{C_v}\Delta T} \right)_{A \to B}} + {\left( {n{C_p}\Delta T} \right)_{B \to C}}$$
<br>$$ = {\left[ {{3 \over 2}\left( {nR\Delta T} \right)} \right]_{A \to B}} + {\left[ {{5 \over 2}\left( {nR\Delta T} \right)} \right]_{B \to C}}$$
<br>$$ = {\left[ {{3 \over 2} \times {V_0}\Delta P} \... | mcq | aieee-2012 | 11,476 |
qZb3efdEollnE0p1 | physics | heat-and-thermodynamics | thermodynamics-process | One mole of a diatomic ideal gas undergoes a cyclic process $$ABC$$ as shown in figure. The process $$BC$$ is adiabatic. The temperatures at $$A, B$$ and $$C$$ are $$400$$ $$K$$, $$800$$ $$K$$ and $$600$$ $$K$$ respectively. Choose the correct statement :
<img src="data:image/png;base64,UklGRgQLAABXRUJQVlA4IPgKAABwbgC... | [{"identifier": "A", "content": "The change in internal energy in whole cyclic process is $$250$$ $$R.$$ "}, {"identifier": "B", "content": "The change in internal energy in the process $$CA$$ is $$700$$ $$R$$. "}, {"identifier": "C", "content": "The change in internal energy in the process $$AB$$ is - $$350$$ $$R.$$ "... | ["D"] | null | In cyclic process, change in total internal energy is zero.
<br>$$\Delta {U_{cyclic}} = 0$$
<br>$$\Delta {U_{BC}} = n{C_v}\Delta T = 1 \times {{5R} \over 2}\Delta T$$
<p>where, $${C_v} = $$ molar specific heat at constant volume.
<br>For $$BC,$$ $$\Delta T = - 200\,K$$
<br>$$\therefore$$ $$\Delta {U_{BC}} = - 500R$$... | mcq | jee-main-2014-offline | 11,478 |
qDsE9riSZq9oH9xq | physics | heat-and-thermodynamics | thermodynamics-process | Consider a spherical shell of radius $$R$$ at temperature $$T$$. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $$u = {U \over V}\, \propto \,{T^4}$$ and pressure $$p = {1 \over 3}\left( {{U \over V}} \right)$$ . If the shell now undergoes an adiabat... | [{"identifier": "A", "content": "$$T\\, \\propto {1 \\over R}$$ "}, {"identifier": "B", "content": "$$T\\, \\propto {1 \\over {{R^3}}}$$ "}, {"identifier": "C", "content": "$$T\\, \\propto \\,{e^{ - R}}$$ "}, {"identifier": "D", "content": "$$T\\, \\propto \\,{e^{ - 3R}}$$ "}] | ["A"] | null | As, $$P = {1 \over 3}\left( {{U \over V}} \right)$$
<br><br>But $$\,\,\,\,$$ $${U \over V} = KT{}^4$$
<br><br>So, $$\,\,\,\,\,P = {1 \over 3}K{T^4}$$
<br><br>or $$\,\,\,\,{{uRT} \over V} = {1 \over 3}K{T^4}\,\,\,\,$$
<br><br>$$\left[ \, \right.$$ As $$PV = uRT$$ $$\left. \, \right]$$
<br><br>$${4 \over 3}\pi {R^3}{T^3... | mcq | jee-main-2015-offline | 11,479 |
37LVjsorAZZAURzCyuJRO | physics | heat-and-thermodynamics | thermodynamics-process | The ratio of work done by an ideal monoatomic gas to the heat supplied to it
in an isobaric process is : | [{"identifier": "A", "content": "$${3 \\over 5}$$"}, {"identifier": "B", "content": "$${2 \\over 3}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "$${2 \\over 5}$$"}] | ["D"] | null | In an isobaric process,
<br><br>Heat supplied, Q = n C<sub>p</sub> $$\Delta $$ T
<br><br>Work done, w = nR$$\Delta $$T
<br><br>$$ \therefore $$ Ratio = $${w \over Q}$$ = $${{nR\Delta T} \over {n{C_p}\Delta T}}$$
<br><br>= $${R \over {{5 \over 2}R}}$$
<br><br>= $${2 \over... | mcq | jee-main-2016-online-9th-april-morning-slot | 11,481 |
F61d0UOsu7od0SzZ | physics | heat-and-thermodynamics | thermodynamics-process | $$'n'$$ moles of an ideal gas undergoes a process $$A$$ $$ \to $$ $$B$$ as shown in the figure. The maximum temperature of the gas during the process will be : <br/><br/>
<img src="data:image/png;base64,UklGRq4NAABXRUJQVlA4IKINAADw6gCdASoAA58CP4HA3mU2Mb+nIZVJG/AwCWlu4XYBG/Pn9G/7v1ZXE3OEvX3B91eKbhp92wiXon///11Ek0+Vq0vfS... | [{"identifier": "A", "content": "$${{9{P_0}{V_0}} \\over {2nR}}$$ "}, {"identifier": "B", "content": "$${{9{P_0}{V_0}} \\over {nR}}$$"}, {"identifier": "C", "content": "$${{9{P_0}{V_0}} \\over {4nR}}$$"}, {"identifier": "D", "content": "$${{3{P_0}{V_0}} \\over {2nR}}$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7zdgqjq/1421d9c1-c911-42d2-a2c2-dd68b18b6a54/38792770-32ef-11ed-8cf6-c1445513adbd/file-1l7zdgqjr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7zdgqjq/1421d9c1-c911-42d2-a2c2-dd68b18b6a54/38792770-32ef-11ed-8cf6-c1445513adbd/fi... | mcq | jee-main-2016-offline | 11,482 |
BlREiKlyiXtNGn1s2WtAg | physics | heat-and-thermodynamics | thermodynamics-process | For the P-V diagram given for an ideal gas,
<br/><br/><img src="data:image/png;base64,UklGRnIKAABXRUJQVlA4IGYKAABwfgCdASrsAoYBP4G612a2LawnoNCY8sAwCWlu/ELZp5fnZ1+/qh/nu43Hvs7+xWVZDv5Eaq8eX77X8B/7eBaFvbMhFYFr8gKqA/RWBa/ICqgP0VgWvyAqoD9FUmPxPqQ3X+RmgVUB+isC1+QFVAforAtfkBVQHMgFAqxnXMRFDU3SavZkIrAtfkBVQH6KwLX5AVUB+iYD0ZV5mb... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267584/exam_images/oev4wqhpjyjefhr7nrt0.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2017 (Online) 9th April Morning Slot Physics - Heat and Thermodynamics... | ["C"] | null | We know,
<br><br>PV = nRT
<br><br>here R is a constant,
<br><br>assuming n = number of moles does not change, we get,
<br><br>PV = KT, K = constant.
<br><br>Given that, PV = constant,
<br><br>So, T = constant, Hence the process is isothermol.
<br><br>From the graph you can see,
<br><br>Pressure at point 1 is highe... | mcq | jee-main-2017-online-9th-april-morning-slot | 11,483 |
1jQOswd9eND7CNI0xG3rsa0w2w9jwzi4s2x | physics | heat-and-thermodynamics | thermodynamics-process | One mole of ideal gas passes through a process where pressure and volume obey the relation
$$P = {P_0}\left[ {1 - {1 \over 2}{{\left( {{{{V_0}} \over V}} \right)}^2}} \right]$$.
Here P<sub>0</sub> and V<sub>0</sub> are constants. Calculate the change in the temperature of the gas if its
volume changes form V<sub>0</sub... | [{"identifier": "A", "content": "$${3 \\over 4}{{{P_0}{V_0}} \\over R}$$"}, {"identifier": "B", "content": "$${1 \\over 2}{{{P_0}{V_0}} \\over R}$$"}, {"identifier": "C", "content": "$${5 \\over 4}{{{P_0}{V_0}} \\over R}$$"}, {"identifier": "D", "content": "$${1 \\over 4}{{{P_0}{V_0}} \\over R}$$"}] | ["C"] | null | Given $$P = {P_o}\left\{ {1 - {1 \over 2}{{\left( {{{{V_o}} \over V}} \right)}^2}} \right\};$$ ...(i)
<br><br>
As n = 1 mole<br><br>
$$ \therefore $$ PV = nRT = RT
<br><br>$$ \Rightarrow $$ P = $${{RT} \over V}$$ ....(ii)
<br><br>From (i) and (ii), we get
<br><br>$${{RT} \over V} = {P_0}\left[ {1 - {1 \over 2}{{\left( ... | mcq | jee-main-2019-online-10th-april-evening-slot | 11,487 |
db2z1YbN0RodtG5y4i18hoxe66ijvztlnmy | physics | heat-and-thermodynamics | thermodynamics-process | n moles of an ideal gas with constant volume
heat capcity C<sub>V</sub> undergo an isobaric expansion
by certain volume. The ratio of the work done
in the process, to the heat supplied is : | [{"identifier": "A", "content": "$${{nR} \\over {{C_V} - nR}}$$"}, {"identifier": "B", "content": "$${{4nR} \\over {{C_V} - nR}}$$"}, {"identifier": "C", "content": "$${{4nR} \\over {{C_V} + nR}}$$"}, {"identifier": "D", "content": "$${{nR} \\over {{C_V} + nR}}$$"}] | ["D"] | null | w = nR$$\Delta $$T<br>
$$\Delta $$H = (Cv + nR)$$\Delta $$T<br>
$${\omega \over {\Delta H}} = {{nR} \over {{C_v} + nR}}$$ | mcq | jee-main-2019-online-10th-april-morning-slot | 11,488 |
xPbV2zpmA4pNyj10t9uRF | physics | heat-and-thermodynamics | thermodynamics-process | The given diagram shows four processes i.e.,
isochoric, isobaric, isothermal and adiabatic.
The correct assignment of the processes, in the
same order is given by :-
<img src="data:image/png;base64,UklGRqgGAABXRUJQVlA4IJwGAAAQXgCdASrsAlABP4HA3WY2MC4nITD4ssAwCWlu4XYBG/Pz8n98zWh8BbTjsTlTxCWRAqu7j/3D/3b/f90PYACxjDtNU2qnxQ... | [{"identifier": "A", "content": "d a b c"}, {"identifier": "B", "content": "a d b c"}, {"identifier": "C", "content": "d a c b"}, {"identifier": "D", "content": "a d c b"}] | ["A"] | null | Between the isothermal and the adiabatic
processes, P-V graph for adiabatic is steeper<br>
Isochoric $$ \to $$ Process d<br>
Isobaric $$ \to $$ Process a<br>
Adiabatic slope will be more than isothermal so<br>
Isothermal $$ \to $$ Process b<br>
Adiabatic $$ \to $$ Process c<br>
Order $$ \to $$ d a b c | mcq | jee-main-2019-online-8th-april-evening-slot | 11,489 |
xO2KAckBRKc7N6yJvsjgy2xukfl3wcwy | physics | heat-and-thermodynamics | thermodynamics-process | In an adiabatic process, the density of a
diatomic gas becomes 32 times its initial value.
The final pressure of the gas is found to be n
times the initial pressure. The value of n is : | [{"identifier": "A", "content": "128"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "326"}, {"identifier": "D", "content": "$${1 \\over {32}}$$"}] | ["A"] | null | In adiabatic process
<br><br>PV<sup>$$\gamma $$</sup> = constant
<br><br>$$ \Rightarrow $$ $$P{\left( {{m \over \rho }} \right)^\gamma }$$ = constant
<br><br>As mass is constant
<br><br>$$ \therefore $$ P $$ \propto $$ $${{\rho ^\gamma }}$$
<br><br>$$ \Rightarrow $$ $${{{P_f}} \over {{P_i}}} = {\left( {{{{\rho _f}} \ov... | mcq | jee-main-2020-online-5th-september-evening-slot | 11,493 |
23BdQWEDHcd9ZwMZljjgy2xukfg6tbnz | physics | heat-and-thermodynamics | thermodynamics-process | Three different processes that can occur in an
ideal monoatomic gas are shown in the P vs V
diagram. The paths are labelled as A $$ \to $$ B,
A $$ \to $$ C and A $$ \to $$ D. The change in internal
energies during these process are taken as
E<sub>AB</sub>, E<sub>AC</sub> and E<sub>AD</sub> and the work done as W<sub>AB... | [{"identifier": "A", "content": "E<sub>AB</sub> < E<sub>AC</sub> < E<sub>AD</sub>, W<sub>AB</sub> > 0, W<sub>AC</sub> > W<sub>AD</sub>"}, {"identifier": "B", "content": "E<sub>AB</sub> = E<sub>AC</sub> = E<sub>AD</sub>, W<sub>AB</sub> > 0, W<sub>AC</sub> = 0, W<sub>AD</sub> < 0"}, {"identifier": "C", ... | ["B"] | null | For all process,
<br><br>E<sub>AB</sub> = E<sub>AC</sub> = E<sub>AD</sub>
<br><br>W<sub>AB</sub> > 0 as volume increases.
<br><br>W<sub>AC</sub> = 0 as volume constant.
<br><br>W<sub>AD</sub> < 0 as volume decreases. | mcq | jee-main-2020-online-5th-september-morning-slot | 11,494 |
6z2uwDk0IqiA27WtDHjgy2xukf169bn6 | physics | heat-and-thermodynamics | thermodynamics-process | A balloon filled with helium (32<sup>o</sup>C and 1.7 atm.)
bursts. Immediately afterwards the expansion
of helium can be considered as | [{"identifier": "A", "content": "Irreversible adiabatic"}, {"identifier": "B", "content": "Reversible adiabatic"}, {"identifier": "C", "content": "Irreversible isothermal"}, {"identifier": "D", "content": "Reversible isothermal"}] | ["A"] | null | Bursting of helium ballon is irreversible adiabatic because Energy can not be restored. | mcq | jee-main-2020-online-3rd-september-morning-slot | 11,496 |
JFRS6Oi3mT4XfdvGVi7k9k2k5ldlbd5 | physics | heat-and-thermodynamics | thermodynamics-process | Starting at temperature 300 K, one mole of an
<br/>ideal diatomic gas ($$\gamma $$ = 1.4) is first compressed
<br/>adiabatically from volume V<sub>1</sub> to V<sub>2</sub> = $${{{V_1}} \over {16}}$$. It is
<br/>then allowed to expand isobarically to volume
2V<sub>2</sub>. If all the processes are the quasi-static then
... | [] | null | 1818TO1819 | T<sub>1</sub>V<sub>1</sub><sup>$$\gamma $$–1</sup> = T<sub>2</sub>V<sub>2</sub>
<sup>$$\gamma $$–1</sup>
<br><br>$$300 \times {V^{{7 \over 5} - 1}} = {T_2}{\left( {{V \over {16}}} \right)^{{7 \over 5} - 1}}$$
<br><br>$$ \Rightarrow $$ T<sub>2</sub> = 300 × (16)<sup>0.4</sup>
<br><br>Isobaric process
<br><br>V = $${{nRT... | integer | jee-main-2020-online-9th-january-evening-slot | 11,497 |
JKASRVaA0a6bzYXrNb7k9k2k5if0iyr | physics | heat-and-thermodynamics | thermodynamics-process | Which of the following is an equivalent cyclic
process corresponding to the thermodynamic cyclic
given in the figure ? where, 1 $$ \to $$ 2 is adiabatic.
<br/>(Graphs are schematic and are not to scale)
<img src="data:image/png;base64,UklGRlgLAABXRUJQVlA4IEwLAAAQXQCdASqTAVcBPm0wlkkkIqGhIdCJsIANiWlu8T/10fNBdaho/o//Ku0f/... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265551/exam_images/kdbxap74ocvrdyddlu2q.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 9th January Morning Slot Physics - Heat and Thermodynamics... | ["B"] | null | In process 2 to 3 pressure is constant & in
process 3 to 1 volume is constant which is
correct only in option B. | mcq | jee-main-2020-online-9th-january-morning-slot | 11,498 |
vhFFMutJu97cAiYPLA7k9k2k5gw4vvs | physics | heat-and-thermodynamics | thermodynamics-process | A thermodynamic cycle xyzx is shown on a V-T diagram.
<img src="data:image/png;base64,UklGRvwFAABXRUJQVlA4IPAFAABwLgCdASokAc8APm02mEikIyKhIfd5iIANiWlu5TARWcRvzUfCH86/Dbwc/rvcJ+dfyv5R7QB3Rfq/8k/bf5RfpP9E3Sn9f/JHySdZ3/b/Au/kuMw/y/8o5ObuPzp/pz6c39l/DP6B+1/to+a//D/K/6p8i/6lf6P+Mf2L9ivlY/QD3ufsd7NH6wCtbqQW1bY811X/Glci/K5Fmg... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264297/exam_images/xfewcnjd2qgsbpxgbwvh.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 8th January Morning Slot Physics - Heat and Thermodynamics... | ["A"] | null | From the corresponding V-T graph
<br><br>Process x $$ \to $$ y is Isobaric expansion (pressure = constant )
<br><br>Process y $$ \to $$ z is Isochoric (volume = constant)
<br><br>Process z $$ \to $$ x is Isothermal compression (temperature = constant)
<br><br>Therefore, (A) is correct P-V graph. | mcq | jee-main-2020-online-8th-january-morning-slot | 11,499 |
vbAyya031H3KKQdbIm7k9k2k5dtnp2f | physics | heat-and-thermodynamics | thermodynamics-process | A litre of dry air at STP expands adiabatically to a volume of 3 litres. If $$\gamma $$ = 1.40, the work done by air is : (3<sup>1.4</sup> = 4.6555) [Take air to be an ideal gas] | [{"identifier": "A", "content": "60.7 J"}, {"identifier": "B", "content": "100.8 J"}, {"identifier": "C", "content": "90.5 J"}, {"identifier": "D", "content": "48 J"}] | ["C"] | null | $${P_1}V_1^\gamma = {P_2}V_2^\gamma $$
<br><br>$$ \Rightarrow $$ P<sub>2</sub> = P<sub>1</sub>$${\left[ {{{{V_1}} \over {{V_2}}}} \right]^\gamma }$$
<br><br>= 10<sup>5</sup> $$ \times $$ $${\left[ {{1 \over 3}} \right]^{1.4}}$$
<br><br>Work done = $${{{P_1}{V_1} - {P_2}{V_2}} \over {\gamma - 1}}$$
<br><br>= $${{{{10}... | mcq | jee-main-2020-online-7th-january-morning-slot | 11,500 |
Q9ZnZCWSeC2CyGzOuJ1klrh6cbf | physics | heat-and-thermodynamics | thermodynamics-process | Match List I with List II.<br/><br/><table>
<thead>
<tr>
<th></th>
<th>List I</th>
<th></th>
<th>List II</th>
</tr>
</thead>
<tbody>
<tr>
<td>(a)</td>
<td>Isothermal</td>
<td>(i)</td>
<td>Pressure constant</td>
</tr>
<tr>
<td>(b)</td>
<td>Isochoric</td>
<td>(ii)</td>
<td>Temperature constant</td>
</tr>
<tr>
<td>(c)</td... | [{"identifier": "1", "content": "(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)"}, {"identifier": "2", "content": "(a) - (ii), (b) - (iv), (c) - (iii), (d) - (i)"}, {"identifier": "3", "content": "(a) - (iii), (b) - (ii), (c) - (i), (d) - (iv)"}, {"identifier": "4", "content": "(a) - (i), (b) - (iii), (c) - (ii), (d) -... | ["1"] | null | We know that, in isothermal process, $$\Delta$$T = 0<br/><br/>In isochoric process, $$\Delta$$V = 0<br/><br/>In adiabatic process, $$\Delta$$Q = 0<br/><br/>In isobaric process, $$\Delta$$p = 0<br/><br/>So, the correct match is,<br/><br/>A $$\to$$ 2, B $$\to$$ 3, C $$\to$$ 4, D $$\to$$ 1 | mcq | jee-main-2021-online-24th-february-morning-slot | 11,501 |
cnkBd03LdwXQ0FAVCp1klrnprwj | physics | heat-and-thermodynamics | thermodynamics-process | If one mole of an ideal gas at (P<sub>1</sub>, V<sub>1</sub>) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one-half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B $$ \to $$ C)... | [{"identifier": "A", "content": "$$ - {{RT} \\over {2(\\gamma - 1)}}$$"}, {"identifier": "B", "content": "$$RT\\left( {\\ln 2 - {1 \\over {2(\\gamma - 1)}}} \\right)$$"}, {"identifier": "C", "content": "$$RT\\ln 2$$"}, {"identifier": "D", "content": "$$0$$"}] | ["B"] | null | Let p<sub>i</sub>, p<sub>f</sub>, V<sub>i</sub> and V<sub>f</sub> be the initial and final pressure and volume.<br><br>Given, AB is isothermal ($$\Delta$$T = 0),<br><br>BC is isochoric ($$\Delta$$V = 0) and CA is adiabatic ($$\Delta$$Q = 0)<br><br>Since, isothermal work (W<sub>AB</sub>) = $${p_1}{V_1}\ln {{{V_f}} \over... | mcq | jee-main-2021-online-24th-february-evening-slot | 11,503 |
St00Xw55JXUjgYKxKw1klrz73r5 | physics | heat-and-thermodynamics | thermodynamics-process | In a certain thermodynamical process, the pressure of a gas depends on its volume as kV<sup>3</sup>. The work done when the temperature changes from 100$$^\circ$$C to 300$$^\circ$$C will be ___________ nR, where n denotes number of moles of a gas. | [] | null | 50 | $$P = k{v^3}$$<br><br>$$ \Rightarrow $$ $$p{v^{ - 3}} = k$$<br><br>$$ \Rightarrow $$ $$x = - 3$$<br><br>$$w = {{nR({T_1} - {T_2})} \over {x - 1}}$$<br><br>$$ = {{nR(100 - 300)} \over { - 3 - 1}}$$<br><br>$$ = {{nR( - 200)} \over { - 4}}$$<br><br>$$ = 50$$ nR | integer | jee-main-2021-online-25th-february-morning-slot | 11,504 |
pTZsZT5pD2FitHc8y91klt2t99p | physics | heat-and-thermodynamics | thermodynamics-process | Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If V<sub>2</sub> = 2V<sub>1</sub> then the ratio of temperature T<sub>2</sub>/T<sub>1</sub> is :<br/><br/><img src="data:image/png;base64,UklGRswNAABXRUJQVlA4IMANAACwUACdASpMAeUAPm02mEgkIyKhJBRqMIANiWlu6B/I+uVOuqOX6Afxn8WPA3+u/kf1wHhP2s... | [{"identifier": "A", "content": "$$\\sqrt 2 $$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "2"}] | ["A"] | null | From P-V diagram, <br><br>Given PV<sup>1/2</sup> = constant ...... (1)<br><br>We know that<br><br>PV = nRT<br><br>$$P \propto \left( {{T \over V}} \right)$$<br><br>Put in equation (1)<br><br>$$\left( {{T \over V}} \right){(V)^{1/2}}$$ = constant<br><br>$$T \propto {V^{1/2}}$$<br><br>$$ \Rightarrow $$ $${{{T_2}} \over {... | mcq | jee-main-2021-online-25th-february-evening-slot | 11,505 |
6e5DYn4BIsG0hN746b1klunpo3s | physics | heat-and-thermodynamics | thermodynamics-process | The volume V of a given mass of monoatomic gas changes with temperature T according to the relation $$V = K{T^{{2 \over 3}}}$$. The workdone when temperature changes by 90K will be xR. The value of x is _________. [R = universal gas constant] | [] | null | 60 | We know that work done is <br><br>$$W = \int {PdV} $$ .... (1)<br><br>$$ \Rightarrow P = {{nRT} \over V}$$ .... (2)<br><br>$$ \Rightarrow W = \int {{{nRT} \over V}dv} $$ .... (3)<br><br>and given $$V = K{T^{2/3}}$$ .... (4)<br><br>$$ \Rightarrow W = \int {{{nRT} \over {K{T^{2/3}}}}.dv} $$ .... (5)<br><br>$$ \Rightarrow... | integer | jee-main-2021-online-26th-february-evening-slot | 11,506 |
ENeG2yFWa6drdsi2WH1kmlwacmy | physics | heat-and-thermodynamics | thermodynamics-process | For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to (where $$\gamma$$ is the ratio of specific heats) : | [{"identifier": "A", "content": "$$ - {1 \\over \\gamma }{{dV} \\over V}$$"}, {"identifier": "B", "content": "$$ - \\gamma {V \\over {dV}}$$"}, {"identifier": "C", "content": "$$ - \\gamma {{dV} \\over V}$$"}, {"identifier": "D", "content": "$${{dV} \\over V}$$"}] | ["C"] | null | for adiabatic expansion :<br><br>PV<sup>$$\gamma$$</sup> = const.<br><br>$$ \Rightarrow $$ ln P + $$\gamma$$ln v = const.<br><br>$$ \Rightarrow $$ differentiating both sides;<br><br>$${{dp} \over p} + \gamma {{dv} \over v} = 0$$<br><br>$$ \Rightarrow {{dp} \over p} = - \gamma {{dv} \over V}$$ | mcq | jee-main-2021-online-18th-march-evening-shift | 11,509 |
1krpqfdy7 | physics | heat-and-thermodynamics | thermodynamics-process | In the reported figure, heat energy absorbed by a system in going through a cyclic process is ___________ $$\pi$$J.<br/><br/><img src="data:image/png;base64,UklGRugQAABXRUJQVlA4INwQAABwZwCdASqQAUkBPm02l0kkIqKhIRUKIIANiWlu/Gf5jutQzf0m/t/887gv75/XfH/wdekfZ/j/RNfj/2Y/T/2zzs/0vgvwBfyr+c/6HuN+5BAB+Z/1j/k/3TxOtSPw77AH80/on/J... | [] | null | 100 | Consider the given diagram,<br><br> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kypo2x34/fe2299d6-7404-4a25-9b2f-9eea1d921c32/f3b64000-7b6a-11ec-a48e-214c9e0044f4/file-1kypo2x35.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kypo2x34/fe2299d6-7404-4a25-9b2f-9eea1d921c32/f3b... | integer | jee-main-2021-online-20th-july-morning-shift | 11,510 |
1kruk24vq | physics | heat-and-thermodynamics | thermodynamics-process | A monoatomic ideal gas, initially at temperature T<sub>1</sub> is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T<sub>2</sub> by releasing the piston suddenly. If l<sub>1</sub> and l<sub>2</sub> are the lengths of the gas column, before and after t... | [{"identifier": "A", "content": "$${\\left( {{{{l_1}} \\over {{l_2}}}} \\right)^{{2 \\over 3}}}$$"}, {"identifier": "B", "content": "$${\\left( {{{{l_2}} \\over {{l_1}}}} \\right)^{{2 \\over 3}}}$$"}, {"identifier": "C", "content": "$${{{l_2}} \\over {{l_1}}}$$"}, {"identifier": "D", "content": "$${{{l_1}} \\over {{l_2... | ["B"] | null | PV<sup>r</sup> = const.<br><br>TV<sup>r $$-$$ 1</sup> = const.<br><br>$$T{(l)^{{5 \over 3} - 1}}$$ = const.<br><br>$${{{T_1}} \over {{T_2}}} = {\left( {{{{l_2}} \over {{l_1}}}} \right)^{{2 \over 3}}}$$ | mcq | jee-main-2021-online-25th-july-morning-shift | 11,512 |
1kryvqhj8 | physics | heat-and-thermodynamics | thermodynamics-process | In the reported figure, there is a cyclic process ABCDA on a sample of 1 mol of a diatomic gas. The temperature of the gas during the process A $$\to$$ B and C $$\to$$ D are T<sub>1</sub> and T<sub>2</sub> (T<sub>1</sub> > T<sub>2</sub>) respectively.<br/><br/><img src="data:image/png;base64,UklGRoQXAABXRUJQVlA4IHgX... | [{"identifier": "A", "content": "W<sub>AB</sub> = W<sub>DC</sub>"}, {"identifier": "B", "content": "W<sub>AD</sub> = W<sub>BC</sub>"}, {"identifier": "C", "content": "W<sub>BC</sub> + W<sub>DA</sub> > 0"}, {"identifier": "D", "content": "W<sub>AB</sub> < W<sub>CD</sub>"}] | ["B"] | null | Work done in adiabatic process = $${{ - nR} \over {\gamma - 1}}({T_f} - {T_i})$$<br><br>$$\therefore$$ $${W_{AD}} = {{ - nR} \over {\gamma - 1}}({T_2} - {T_1})$$<br><br>and $${W_{BC}} = {{ - nR} \over {\gamma - 1}}({T_2} - {T_1})$$<br><br>$$\therefore$$ $${W_{AD}} = {W_{BC}}$$ | mcq | jee-main-2021-online-27th-july-morning-shift | 11,513 |
1ks0k4p91 | physics | heat-and-thermodynamics | thermodynamics-process | One mole of an ideal gas is taken through an adiabatic process where the temperature rises from 27$$^\circ$$ C to 37$$^\circ$$ C. If the ideal gas is composed of polyatomic molecule that has 4 vibrational modes, which of the following is true? [R = 8.314 J mol<sup>$$-$$1</sup> k<sup>$$-$$1</sup>] | [{"identifier": "A", "content": "work done by the gas is close to 332 J"}, {"identifier": "B", "content": "work done on the gas is close to 582 J"}, {"identifier": "C", "content": "work done by the gas is close to 582 J"}, {"identifier": "D", "content": "work done on the gas is close to 332 J"}] | ["B"] | null | Since, each vibrational mode, corresponds to two degrees of freedom, hence, f = 3 (trans.) + 3 (rot.) + 4 $$ \times $$ 2 (vib.) = 14<br><br> & $$\gamma = 1 + {2 \over f}$$<br><br>$$\gamma = 1 + {2 \over {14}} = {8 \over 7}$$<br><br>$$W = {{nR\Delta T} \over {\gamma - 1}} = - 582$$<br><br>As W < 0. work is do... | mcq | jee-main-2021-online-27th-july-evening-shift | 11,514 |
1l54uy534 | physics | heat-and-thermodynamics | thermodynamics-process | <p>Starting with the same initial conditions, an ideal gas expands from volume V<sub>1</sub> to V<sub>2</sub> in three different ways. The work done by the gas is W<sub>1</sub> if the process is purely isothermal, W<sub>2</sub>, if the process is purely adiabatic and W<sub>3</sub> if the process is purely isobaric. The... | [{"identifier": "A", "content": "W<sub>1</sub> < W<sub>2</sub> < W<sub>3</sub>"}, {"identifier": "B", "content": "W<sub>2</sub> < W<sub>3</sub> < W<sub>1</sub>"}, {"identifier": "C", "content": "W<sub>3</sub> < W<sub>1</sub> < W<sub>2</sub>"}, {"identifier": "D", "content": "W<sub>2</sub> < W<sub>1</sub> < W<sub>3</sub... | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5ghaig9/bdf9a350-82b9-49bf-85d6-7b7d3200a25a/dcc88d90-00f2-11ed-ba34-71a54c393c2b/file-1l5ghaiga.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5ghaig9/bdf9a350-82b9-49bf-85d6-7b7d3200a25a/dcc88d90-00f2-11ed-ba34-71a54c393c2b... | mcq | jee-main-2022-online-29th-june-evening-shift | 11,516 |
1l5693oxw | physics | heat-and-thermodynamics | thermodynamics-process | <p>Given below are two statements :</p>
<p>Statement I : When $$\mu$$ amount of an ideal gas undergoes adiabatic change from state (P<sub>1</sub>, V<sub>1</sub>, T<sub>1</sub>) to state (P<sub>2</sub>, V<sub>2</sub>, T<sub>2</sub>), then work done is $$W = {{\mu R({T_2} - {T_1})} \over {1 - \gamma }}$$, where $$\gamma ... | [{"identifier": "A", "content": "Both Statement I and Statement II are true."}, {"identifier": "B", "content": "Both Statement I and Statement II are false."}, {"identifier": "C", "content": "Statement I is true but Statement II is false."}, {"identifier": "D", "content": "Statement I is false but Statement II is true.... | ["A"] | null | <p>$$W = {{\mu R({T_2} - {T_1})} \over {1 - r}}$$ for a polytropic process for adiabatic process r = $$\gamma$$</p>
<p>$$\Rightarrow$$ Statement I is true.</p>
<p>In an adiabatic process</p>
<p>$$\Delta$$U = $$-$$ $$\Delta$$W</p>
<p>$$\Rightarrow$$ If work is done on the gas</p>
<p>$$\Rightarrow$$ $$\Delta$$W is negati... | mcq | jee-main-2022-online-28th-june-morning-shift | 11,517 |
1l56w4ypt | physics | heat-and-thermodynamics | thermodynamics-process | <p>A diatomic gas ($$\gamma$$ = 1.4) does 400J of work when it is expanded isobarically. The heat given to the gas in the process is __________ J.</p> | [] | null | 1400 | <p>W = nR$$\Delta$$T = 400 J</p>
<p>$$\therefore$$ $$\Delta$$Q = nC<sub>P</sub>$$\Delta$$T</p>
<p>$$ = n \times {7 \over 2}R \times \Delta T = {7 \over 2} \times (400) = 1400$$</p> | integer | jee-main-2022-online-27th-june-evening-shift | 11,518 |
1l5w2lv8a | physics | heat-and-thermodynamics | thermodynamics-process | <p>A sample of monoatomic gas is taken at initial pressure of 75 kPa. The volume of the gas is then compressed from 1200 cm<sup>3</sup> to 150 cm<sup>3</sup> adiabatically. In this process, the value of workdone on the gas will be :</p> | [{"identifier": "A", "content": "79 J"}, {"identifier": "B", "content": "405 J"}, {"identifier": "C", "content": "4050 J"}, {"identifier": "D", "content": "9590 J"}] | ["B"] | null | <p>For monoatomic gas degree of freedom f = 3 and $$\gamma$$ = $${5 \over 3}$$</p>
<p>Here for gas,</p>
<p>Initial pressure (P<sub>1</sub>) = 75 kPa</p>
<p>Initial volume (V<sub>1</sub>) = 1200 cm<sup>3</sup></p>
<p>Final volume (V<sub>2</sub>) = 150 cm<sup>3</sup></p>
<p>Final pressure (P<sub>2</sub>) = ?</p>
<p>For a... | mcq | jee-main-2022-online-30th-june-morning-shift | 11,519 |
1l6gmtw8s | physics | heat-and-thermodynamics | thermodynamics-process | <p>A monoatomic gas at pressure $$\mathrm{P}$$ and volume $$\mathrm{V}$$ is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be :
</p> | [{"identifier": "A", "content": "P"}, {"identifier": "B", "content": "8P"}, {"identifier": "C", "content": "32P"}, {"identifier": "D", "content": "64P"}] | ["C"] | null | <p>$$P{V^\gamma }=$$ constant</p>
<p>$$ \Rightarrow P{V^\gamma } = (P'){\left( {{v \over 8}} \right)^\gamma }$$ where $$\gamma = 5/3$$</p>
<p>$$ \Rightarrow P' = 32P$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift | 11,521 |
1l6p60sh7 | physics | heat-and-thermodynamics | thermodynamics-process | <p>The pressure $$\mathrm{P}_{1}$$ and density $$\mathrm{d}_{1}$$ of diatomic gas $$\left(\gamma=\frac{7}{5}\right)$$ changes suddenly to $$\mathrm{P}_{2}\left(>\mathrm{P}_{1}\right)$$ and $$\mathrm{d}_{2}$$ respectively during an adiabatic process. The temperature of the gas increases and becomes ________ times of ... | [] | null | 4 | <p>$${P_1}V_1^\gamma = {P_2}V_2^2$$</p>
<p>$${{{P_1}} \over {d_1^\gamma }} = {{{P_2}} \over {d_2^\gamma }}$$</p>
<p>$${{{d_1}{T_1}} \over {d_1^\gamma }} = {{{d_2}{T_2}} \over {d_2^\gamma }}$$</p>
<p>$${T_2} = {\left( {{{{d_2}} \over {{d_1}}}} \right)^{\gamma - 1}}{T_1}$$</p>
<p>$$ = {(32)^{{2 \over 5}}}{T_1}$$</p>
<p... | integer | jee-main-2022-online-29th-july-morning-shift | 11,522 |
1l6rhlp1p | physics | heat-and-thermodynamics | thermodynamics-process | <p>A thermodynamic system is taken from an original state D to an intermediate state E by the linear process shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be</p>
<p><img src="data:image/png;base64,UklGRpQPA... | [{"identifier": "A", "content": "$$-$$450 J"}, {"identifier": "B", "content": "450 J"}, {"identifier": "C", "content": "900 J"}, {"identifier": "D", "content": "1350 J"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ltxsegm0/ac01838c-5020-4cfb-bead-5e2b451a5cbe/d3272180-e59c-11ee-af4c-59af6a5599d5/file-6y3zli1ltxsegm1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ltxsegm0/ac01838c-5020-4cfb-bead-5e2b451a5cbe/d3272180-e59c-11ee-af... | mcq | jee-main-2022-online-29th-july-evening-shift | 11,523 |
1ldr1avjg | physics | heat-and-thermodynamics | thermodynamics-process | <p>Heat is given to an ideal gas in an isothermal process.</p>
<p>A. Internal energy of the gas will decrease.</p>
<p>B. Internal energy of the gas will increase.</p>
<p>C. Internal energy of the gas will not change.</p>
<p>D. The gas will do positive work.</p>
<p>E. The gas will do negative work.</p>
<p>Choose the cor... | [{"identifier": "A", "content": "B and D only"}, {"identifier": "B", "content": "C and E only"}, {"identifier": "C", "content": "A and E only"}, {"identifier": "D", "content": "C and D only"}] | ["D"] | null | <p>Isothermal process $$\Delta T=0$$</p>
<p>$$\Delta U=\frac{f}{2}nR\Delta T$$</p>
<p>$$\Delta U=0$$</p>
<p>No change in internal energy</p>
<p>$$\Delta Q=\Delta W$$ (1$$^{st}$$ law)</p>
<p>$$\Delta Q=+\mathrm{ve}$$</p>
<p>$$\Delta W=+\mathrm{ve}$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift | 11,526 |
1ldwr4z7g | physics | heat-and-thermodynamics | thermodynamics-process | <p>In an Isothermal change, the change in pressure and volume of a gas can be represented for three different temperature; $$\mathrm{T_3 > T_2 > T_1}$$ as :</p> | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldxibc4n/a162e46b-cd7c-443c-87c6-e65693b13bd5/eae8b370-a8b1-11ed-a3d7-5b81580a6b38/file-1ldxibc4o.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1ldxibc4n/a162e46b-cd7c-443c-87c6-e65693b13bd5/eae... | ["B"] | null | <p>For isothermal process P-V graph is rectangular
hyperbola</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le4a6508/fc7ff9da-93d6-4b9e-a21b-08960bfeac2c/8ddfb580-ac6b-11ed-8b11-cb59760cfd30/file-1le4a6509.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le4a6508/fc7ff9da... | mcq | jee-main-2023-online-24th-january-evening-shift | 11,528 |
1ldydw41j | physics | heat-and-thermodynamics | thermodynamics-process | <p>1 g of a liquid is converted to vapour at 3 $$\times$$ 10$$^5$$ Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm$$^3$$ during this phase change, then the increase in internal energy in the process will be :</p> | [{"identifier": "A", "content": "4800 J"}, {"identifier": "B", "content": "4320 J"}, {"identifier": "C", "content": "432000 J"}, {"identifier": "D", "content": "4.32 $$\\times$$ 10$$^8$$ J"}] | ["B"] | null | Work done = P$\Delta$V<br/><br/>
= 3 × 10<sup>5</sup> × 1600 × 10<sup>–6</sup><br/><br/>
= 480 J<br/><br/>
Only 10% of heat is used in work done.<br/><br/>
Hence $\Delta$Q = 4800 J
The rest goes in internal energy, which is 90% of
heat.<br/><br/>
Change in internal energy = 0.9 × 4800 = 4320 J | mcq | jee-main-2023-online-24th-january-morning-shift | 11,529 |
lgnyjy7p | physics | heat-and-thermodynamics | thermodynamics-process | A thermodynamic system is taken through cyclic process. The total work done in the process is :<br/><br/>
<img src="data:image/png;base64,UklGRn4OAABXRUJQVlA4IHIOAAAQ7ACdASoAA58CP4G+2GY2LyynINBZWsAwCWlu4XEU6mNwvj6M9IJfZ0P9G/mrieIP/iqqY///WGSfrIohxZ3yil7izvlFL3FnfKKXuLO+SIj4JZ+aMm5VmBBFaTJcUaJUfBLPzRk3MGS4n4RIQTEfBLPzRh... | [{"identifier": "A", "content": "$100 \\mathrm{~J}$"}, {"identifier": "B", "content": "Zero"}, {"identifier": "C", "content": "$300 \\mathrm{~J}$"}, {"identifier": "D", "content": "$200 \\mathrm{~J}$"}] | ["C"] | null | On $\mathrm{P}-\mathrm{V}$ scale area of loop $=$ work done<br/><br/>
$$
\begin{aligned}
& \Rightarrow \mathrm{W}=+\frac{1}{2}(2) \times 300 \\\\
& \mathrm{~W}=300 \mathrm{~J}
\end{aligned}
$$ | mcq | jee-main-2023-online-15th-april-morning-shift | 11,530 |
1lgswlsgn | physics | heat-and-thermodynamics | thermodynamics-process | <p>The Thermodynamic process, in which internal energy of the system remains constant is</p> | [{"identifier": "A", "content": "Isobaric"}, {"identifier": "B", "content": "Isochoric"}, {"identifier": "C", "content": "Adiabatic"}, {"identifier": "D", "content": "Isothermal"}] | ["D"] | null | If the temperature (T) remains constant, the internal energy (U) also remains constant, since the internal energy of an ideal gas depends only on its temperature.
<br/><br/>
In this case, the thermodynamic process in which the internal energy of the system remains constant is an isothermal process. Isothermal processes... | mcq | jee-main-2023-online-11th-april-evening-shift | 11,531 |
lsamui7s | physics | heat-and-thermodynamics | thermodynamics-process | A diatomic gas $(\gamma=1.4)$ does $200 \mathrm{~J}$ of work when it is expanded isobarically. The heat given to the gas in the process is : | [{"identifier": "A", "content": "$800 \\mathrm{~J}$"}, {"identifier": "B", "content": "$600 \\mathrm{~J}$"}, {"identifier": "C", "content": "$700 \\mathrm{~J}$"}, {"identifier": "D", "content": "$850 \\mathrm{~J}$"}] | ["C"] | null | $\begin{aligned} & \gamma=1+\frac{2}{\mathrm{f}}=1.4 \Rightarrow \frac{2}{\mathrm{f}}=0.4 \\\\ & \Rightarrow \mathrm{f}=5 \\\\ & \mathrm{~W}=\mathrm{nR} \Delta \mathrm{T}=200 \mathrm{~J} \\\\ & \mathrm{Q}=\left(\frac{\mathrm{f}+2}{2}\right) \mathrm{nR} \Delta \mathrm{T} \\\\ & =\frac{7}{2} \times 200=700 \mathrm{~J}\en... | mcq | jee-main-2024-online-1st-february-evening-shift | 11,534 |
jaoe38c1lscpchzv | physics | heat-and-thermodynamics | thermodynamics-process | <p>During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio of $$\frac{\mathrm{Cp}}{\mathrm{Cv}}$$ for the gas is :</p> | [{"identifier": "A", "content": "$$\\frac{7}{5}$$\n"}, {"identifier": "B", "content": "$$\\frac{3}{2}$$\n"}, {"identifier": "C", "content": "$$\\frac{9}{7}$$\n"}, {"identifier": "D", "content": "$$\\frac{5}{3}$$"}] | ["B"] | null | <p>For an adiabatic process, the following relation holds:</p>
<p>$$P V^{\gamma} = \text{constant}$$</p>
<p>where P is the pressure, V is the volume, and $$\gamma = \frac{C_p}{C_v}$$.</p>
<p>We are given that the pressure is proportional to the cube of the absolute temperature: </p>
<p>$$P \propto T^3$$.</p>
<p>Us... | mcq | jee-main-2024-online-27th-january-evening-shift | 11,535 |
jaoe38c1lse64544 | physics | heat-and-thermodynamics | thermodynamics-process | <p>The given figure represents two isobaric processes for the same mass of an ideal gas, then</p>
<p><img src="data:image/png;base64,UklGRigNAABXRUJQVlA4IBwNAABw4wCdASoAA64CP4HA3GW2Ma2poTcI0sAwCWlu/DqYhWVHZ1+/sf2hf7zy/gKegztdxX0EH2IZg0Wv0H7/7Qfv8KUeszEq+/6zMSr7/rMxKvv+szEq+/6bQTqSEMmsQGhwcBpPIex0XthY5Og/ygLYWOToP8oC2Fj... | [{"identifier": "A", "content": "$$P_2>P_1$$\n"}, {"identifier": "B", "content": "$$P_1>P_2$$\n"}, {"identifier": "C", "content": "$$P_1=P_2$$\n"}, {"identifier": "D", "content": "$$P_2 \\geq P_1$$"}] | ["B"] | null | <p>The two isobaric processes depicted in the figure show changes in the volume of an ideal gas with temperature under constant pressure conditions. <br/><br/>Isobaric processes follow the equation $$PV = nRT$$, where P is pressure, V is volume, n is the number of moles of the gas, R is the ideal gas constant, and T is... | mcq | jee-main-2024-online-31st-january-morning-shift | 11,536 |
1lsg5luua | physics | heat-and-thermodynamics | thermodynamics-process | <p>Choose the correct statement for processes A & B shown in figure.</p>
<p><img src="data:image/png;base64,UklGRtYLAABXRUJQVlA4IMoLAACQ7gCdASoAA/0CP4HA22W2MK2nITH5AsAwCWlu4XaVJmNwvHRge1gBOl+txxs7FP//+xofc2Tjam1NqbU2ptTam1NqbU2ptTam00JP7iWEtG7BZ3ITvCL3SdCo4dONqbU2ptTam1NgSPcSvHfOkfkCoPbRpHLJFU8HK9OHTjam1NqbU2ptTam1... | [{"identifier": "A", "content": "$$P V=k$$ for process $$B$$ and $$A$$.\n"}, {"identifier": "B", "content": "$$\\frac{P^{\\gamma-1}}{T^\\gamma}=k$$ for process $$B$$ and $$T=k$$ for process $$A$$.\n"}, {"identifier": "C", "content": "$$\\frac{T^\\gamma}{P^{\\gamma-1}}=k$$ for process $$A$$ and $$P V=k$$ for process $$B... | null | null | <p>Steeper curve (B) is adiabatic</p>
<p>Adiabatic $$\Rightarrow \mathrm{PV}^v=$$ const.</p>
<p>Or $$\mathrm{P}\left(\frac{\mathrm{T}}{\mathrm{P}}\right)^v=$$ const.</p>
<p>$$\frac{\mathrm{T}^v}{\mathrm{P}^{v-1}}=\text { const. }$$</p>
<p>Curve (A) is isothermal</p>
<p>$$\mathrm{T}=$$ const.</p>
<p>$$\mathrm{PV}=$$ con... | mcqm | jee-main-2024-online-30th-january-evening-shift | 11,538 |
1lsgd7jbd | physics | heat-and-thermodynamics | thermodynamics-process | <p>Two thermodynamical processes are shown in the figure. The molar heat capacity for process A and B are $$\mathrm{C}_{\mathrm{A}}$$ and $$\mathrm{C}_{\mathrm{B}}$$. The molar heat capacity at constant pressure and constant volume are represented by $$\mathrm{C_P}$$ and $$\mathrm{C_V}$$, respectively. Choose the corre... | [{"identifier": "A", "content": "$$\\mathrm{C_P>C_B>C_A>C_V}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{C}_{\\mathrm{P}}>\\mathrm{C}_{\\mathrm{V}}>\\mathrm{C}_{\\mathrm{A}}=\\mathrm{C}_{\\mathrm{B}}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{C}_{\\mathrm{A}}=0$$ and $$\\mathrm{C}_{\\mathrm{B}}=\\infty$$\n... | ["A"] | null | <p>For process $$\mathrm{A}$$</p>
<p>$$\begin{aligned}
& \log P=\gamma \log \mathrm{V} \Rightarrow \mathrm{P}=\mathrm{V}^\gamma,(\gamma>1) \\
& P V^{-\gamma}=\text { Constant }
\end{aligned}$$</p>
<p>$$C_A=C_V+\frac{R}{1+\gamma}$$ ..... (i)</p>
<p>Likewise for process $$\mathrm{B} \rightarrow P V^{-1}=$$ Constant</p>
<... | mcq | jee-main-2024-online-30th-january-morning-shift | 11,539 |
luxweree | physics | heat-and-thermodynamics | thermodynamics-process | <p>A real gas within a closed chamber at $$27^{\circ} \mathrm{C}$$ undergoes the cyclic process as shown in figure. The gas obeys $$P V^3=R T$$ equation for the path $$A$$ to $$B$$. The net work done in the complete cycle is (assuming $$R=8 \mathrm{~J} / \mathrm{mol} \mathrm{K}$$):</p>
<p><img src="data:image/png;base6... | [{"identifier": "A", "content": "$$-20$$J"}, {"identifier": "B", "content": "205J"}, {"identifier": "C", "content": "225J"}, {"identifier": "D", "content": "20J"}] | ["B"] | null | <p>For $A$ to $B$ :</p>
<p>Given $\mathrm{PV}^3=\mathrm{RT} \Rightarrow \mathrm{P}=\frac{\mathrm{RT}}{\mathbf{V}^3}$</p>
<p>Work done $\mathrm{W}_{\mathrm{AB}}=\int \mathrm{PdV}$</p>
<p>$\begin{aligned} & =\int \frac{R T}{V^3} d V=R T\left[\frac{V^{-2}}{-2}\right]_{V_1=2}^{V_B=4} \\\\ & =-\frac{R T}{2}\left[\frac{1}... | mcq | jee-main-2024-online-9th-april-evening-shift | 11,540 |
luy9cm2x | physics | heat-and-thermodynamics | thermodynamics-process | <p>The volume of an ideal gas $$(\gamma=1.5)$$ is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is :</p> | [{"identifier": "A", "content": "$$\\frac{4}{5}$$\n"}, {"identifier": "B", "content": "$$\\frac{8}{5 \\sqrt{5}}$$\n"}, {"identifier": "C", "content": "$$\\frac{2}{\\sqrt{5}}$$\n"}, {"identifier": "D", "content": "$$\\frac{16}{25}$$"}] | ["B"] | null | <p>To find the ratio of the initial pressure to the final pressure of an ideal gas undergoing an adiabatic process, we can use the adiabatic equation for an ideal gas, which relates pressure (P) and volume (V) as follows:</p>
<p>$P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}$</p>
<p>Here, $P_{1}$ and $V_{1}$ are the init... | mcq | jee-main-2024-online-9th-april-morning-shift | 11,541 |
luyit8ee | physics | heat-and-thermodynamics | thermodynamics-process | <p>A sample of 1 mole gas at temperature $$T$$ is adiabatically expanded to double its volume. If adiab constant for the gas is $$\gamma=\frac{3}{2}$$, then the work done by the gas in the process is :</p> | [{"identifier": "A", "content": "$$\\mathrm{R} \\mathrm{T}[2+\\sqrt{2}]$$\n"}, {"identifier": "B", "content": "$$\\mathrm{RT}[2-\\sqrt{2}]$$\n"}, {"identifier": "C", "content": "$$\\frac{\\mathrm{R}}{\\mathrm{T}}[2-\\sqrt{2}]$$\n"}, {"identifier": "D", "content": "$$\\frac{T}{R}[2+\\sqrt{2}]$$"}] | ["B"] | null | <p>For an adiabatic process, the work done by the gas can be found using the formula:</p>
<p>$$ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} $$</p>
<p>Given that the volume is doubled ($$V_2 = 2V_1$$) and the adiabatic constant $$\gamma = \frac{3}{2}$$, we can manipulate the ideal gas law and the adiabatic process relati... | mcq | jee-main-2024-online-9th-april-morning-shift | 11,542 |
lv2ersc0 | physics | heat-and-thermodynamics | thermodynamics-process | <p>A sample of gas at temperature $$T$$ is adiabatically expanded to double its volume. Adiabatic constant for the gas is $$\gamma=3 / 2$$. The work done by the gas in the process is:</p>
<p>$$(\mu=1 \text { mole })$$</p> | [{"identifier": "A", "content": "$$R T[2 \\sqrt{2}-1]$$\n"}, {"identifier": "B", "content": "$$R T[2-\\sqrt{2}]$$\n"}, {"identifier": "C", "content": "$$R T[1-2 \\sqrt{2}]$$\n"}, {"identifier": "D", "content": "$$R T[\\sqrt{2}-2]$$"}] | ["B"] | null | <p>$$\begin{aligned}
& w=\frac{-n R}{\gamma-1}(\Delta T) \\
&=\frac{-R}{1 / 2}\left(\frac{T}{\sqrt{2}}-T\right) \\
&=2 R\left(\frac{\sqrt{2} T-T}{\sqrt{2}}\right) \\
&=R T(2-\sqrt{2}) \\
& \therefore \quad T V_\gamma^{-1}=\text { cons. } \\
& T V_\gamma^{-1}=T_f(2 V)^{\gamma-1} \\
& T_f=\frac{T}{\sqrt{2}}
\end{aligned}... | mcq | jee-main-2024-online-4th-april-evening-shift | 11,543 |
lv3vec1d | physics | heat-and-thermodynamics | thermodynamics-process | <p>A diatomic gas $$(\gamma=1.4)$$ does $$100 \mathrm{~J}$$ of work in an isobaric expansion. The heat given to the gas is :</p> | [{"identifier": "A", "content": "150 J"}, {"identifier": "B", "content": "490 J"}, {"identifier": "C", "content": "350 J"}, {"identifier": "D", "content": "250 J"}] | ["C"] | null | <p>To find the heat given to a diatomic gas during an isobaric (constant pressure) expansion, we can use the formula that relates the work done by the gas, the heat added to the system, and the change in the internal energy of the system. The first law of thermodynamics states that:</p>
<p>$$\Delta Q = \Delta U + W$$<... | mcq | jee-main-2024-online-8th-april-evening-shift | 11,544 |
lv5gt50w | physics | heat-and-thermodynamics | thermodynamics-process | <p>Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The relation between the ratio $$\frac{V_a}{V_d}$$ and the ratio $$\frac{V_b}{V_c}$$ is:</p>
<p><img src="data:image/png;base64,UklGRuwQAABXRUJQVlA4IOAQAADw9gCdASoAA2sCP4G61mW2Lb+nITD54/AwCWlu+F+7fi1fL+LMkCY3C9Xzr... | [{"identifier": "A", "content": "$$\\frac{V_a}{V_d} \\neq \\frac{V_b}{V_c}$$\n"}, {"identifier": "B", "content": "$$\\frac{V_a}{V_d}=\\left(\\frac{V_b}{V_c}\\right)^{-1}$$\n"}, {"identifier": "C", "content": "$$\\frac{V_a}{V_d}=\\frac{V_b}{V_c}$$\n"}, {"identifier": "D", "content": "$$\\frac{V_a}{V_d}=\\left(\\frac{V_b... | ["C"] | null | <p>$$\begin{aligned}
\text { (1) } P_a V_a & =P_b V_b \quad \text{.... (i)}\\
P_c V_c & =P_d V_d \quad \text{.... (ii)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
\text { (2) } P_a V_a^{\gamma-1} & =P_d V_d^{\gamma-1} \quad \text{.... (iii)}\\
P_b V_b^{\gamma-1} & =P_c V_c^{\gamma-1} \quad \text{.... (iv)}
\end{aligned}$... | mcq | jee-main-2024-online-8th-april-morning-shift | 11,545 |
lv7v4rfw | physics | heat-and-thermodynamics | thermodynamics-process | <p>The heat absorbed by a system in going through the given cyclic process is :</p>
<p><img src="data:image/png;base64,UklGRmAOAABXRUJQVlA4IFQOAABQ0gCdASoAAyECP4HA22U2MK2nIbN5IsAwCWlu/E84+ajHZ1+frt/urXmzx7qZhIN9//cZH/+9QhL76x9aJT31j60SnvrH1olPfWPrRKe+k9JWQ8VbsASWpDoBdMI4JYdWzrGqIOg4G+Lk3pvym1b4pSFvTCEAulSNaVH1pUfWlR9aV... | [{"identifier": "A", "content": "61.6 J"}, {"identifier": "B", "content": "431.2 J"}, {"identifier": "C", "content": "19.6 J"}, {"identifier": "D", "content": "616 J"}] | ["A"] | null | <p>$$\begin{aligned}
\Delta Q & =\Delta U+w \\
& =\pi(140)^2 \times 10^3 \times 10^{-6} \\
& =61.6 \mathrm{~J}
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift | 11,546 |
lv9s20vb | physics | heat-and-thermodynamics | thermodynamics-process | <p>During an adiabatic process, if the pressure of a gas is found to be proportional to the cube of its absolute temperature, then the ratio of $$\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}$$ for the gas is :</p> | [{"identifier": "A", "content": "$$\\frac{5}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{3}{2}$$\n"}, {"identifier": "C", "content": "$$\\frac{7}{5}$$\n"}, {"identifier": "D", "content": "$$\\frac{9}{7}$$"}] | ["B"] | null | <p>To begin with, we're told that during an adiabatic process, the pressure $P$ of a gas is directly proportional to the cube of its absolute temperature $T$, that is, $P \propto T^3$. From this, we can express the relation as $P = kT^3$, where $k$ is a constant.</p>
<p>In an adiabatic process, $PV^\gamma = \text{cons... | mcq | jee-main-2024-online-5th-april-evening-shift | 11,547 |
msBvERiBooMmhO9m | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | <b>Statement - 1:</b> The temperature dependence of resistance is usually given as $$R = {R_0}\left( {1 + \alpha \,\Delta t} \right).$$ The resistance of wire changes from $$100\Omega $$ to $$150\Omega $$ when its temperature is increased from $${27^ \circ }C$$ to $${227^ \circ }C$$. This implies that $$\alpha = 2.5 ... | [{"identifier": "A", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is the correct explanation of Statement - 1"}, {"identifier": "B", "content": "Statement - 1 is true, Statement - 2 is true; Statement - 2 is not the correct explanation of Statement - 1"}, {"identifier": "C", "content": "State... | ["C"] | null | The relation $$R = {R_0}\left( {1 + \alpha \,\Delta t} \right)$$ is valid for small values of $$\Delta t$$ and $${R_0}$$ is resistance at $${0^ \circ }C$$ and also $$\left( {R - {R_0}} \right)$$ should be much smaller than $${R_0}.$$ So, statement $$(1)$$ is wrong but statement $$(2)$$ is correct. | mcq | aieee-2009 | 11,548 |
8JDX8fI8MpvmmvRm | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | A wooden wheel of radius $$R$$ is made of two semicircular part (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area $$S$$ and length $$L.$$ $$L$$ is slightly less than $$2\pi R.$$ To fit the ring on the wheel, it is heated so that its temperature rises by $$\Delta T$$ a... | [{"identifier": "A", "content": "$$2\\pi SY\\alpha \\Delta T$$ "}, {"identifier": "B", "content": "$$SY\\alpha \\Delta T$$ "}, {"identifier": "C", "content": "$$\\pi SY\\alpha \\Delta T$$ "}, {"identifier": "D", "content": "$$2SY\\alpha \\Delta T$$ "}] | ["D"] | null | $$\gamma = {{F/S} \over {\Delta L/L}} \Rightarrow \Delta L = {{FL} \over {SY}}$$
<br>$$\therefore$$ $$L\alpha \Delta T = {{FL} \over {SY}}$$
<br>$$\left[ \, \right.$$ as $${\Delta L = L\alpha \Delta T}$$ $$\left. \, \right]$$
<br>$$\therefore$$ $$F = SY\alpha \Delta T$$
<br>$$\therefore$$ The ring is pressing the whee... | mcq | aieee-2012 | 11,549 |
6YFVwjLK0Brr3s1U | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | A pendulum clock loses $$12$$ $$s$$ a day if the temperature is $${40^ \circ }C$$ and gains $$4$$ $$s$$ a day if the temperature is $${20^ \circ }C.$$ The temperature at which the clock will show correct time, and the co-efficient of linear expansion $$\left( \alpha \right)$$ of the metal of the pendulum shaft are re... | [{"identifier": "A", "content": "$${30^ \\circ }C;\\,\\,\\alpha = 1.85 \\times {10^{ - 3}}/{}^ \\circ C$$ "}, {"identifier": "B", "content": "$${55^ \\circ }C;\\,\\,\\alpha = 1.85 \\times {10^{ - 2}}/{}^ \\circ C$$"}, {"identifier": "C", "content": "$${25^ \\circ }C;\\,\\,\\alpha = 1.85 \\times {10^{ - 5}}/{}^ \\cir... | ["C"] | null | Time lost/gained per day $$ = {1 \over 2} \propto \Delta \theta \times 86400$$ second
<br><br/>$$12 = {1 \over 2}\alpha \left( {40 - \theta } \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br/>$$4 = {1 \over 2}\alpha \left( {\theta - 20} \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left... | mcq | jee-main-2016-offline | 11,550 |
OQw0DGU5ITOsOm5I | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | An external pressure P is applied on a cube at 0<sup>o</sup>C so that it is equally compressed from all sides. K is the
bulk modulus of the material of the cube and $$\alpha$$ is its coefficient of linear expansion. Suppose we want to
bring the cube to its original size by heating. The temperature should be raised by: | [{"identifier": "A", "content": "$${P \\over {3\\alpha K}}$$ "}, {"identifier": "B", "content": "$${P \\over {\\alpha K}}$$"}, {"identifier": "C", "content": "$${3 \\alpha \\over {P K}}$$"}, {"identifier": "D", "content": "3PK$$\\alpha$$"}] | ["A"] | null | As we know, Bulk modulus
<br><br>K = $${{\Delta P} \over {\left( {{{ - \Delta V} \over V}} \right)}}$$
<br><br>$$ \Rightarrow $$ $${{{\Delta V} \over V} = {P \over K}}$$
<br><br>V = V<sub>0</sub>(1 + $$\gamma $$$$\Delta $$t)
<br><br>$${{{\Delta V} \over {{V_0}}} = \gamma \Delta t}$$
<br><br>$$ \therefore $$ $${{P \over... | mcq | jee-main-2017-offline | 11,552 |
jCMGKOq1PZGleFvceyXci | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | A compressive force, F is applied at the two ends of a long thin steel rod. It is heated, simultaneously, such that its temperature increases by $$\Delta $$T. The net change in its length is zero. Let $$\ell $$ be the length of the rod, A its area of cross-section,Y its Young’s modulus, and $$\alpha $$ its coefficient ... | [{"identifier": "A", "content": "$$\\ell $$<sup>2</sup> Y$$\\alpha $$ $$\\Delta $$T"}, {"identifier": "B", "content": "$$\\ell $$A Y$$\\alpha $$ $$\\Delta $$T"}, {"identifier": "C", "content": "A Y$$\\alpha $$ $$\\Delta $$T"}, {"identifier": "D", "content": "$${{AY} \\over {\\alpha \\,\\Delta T}}$$ "}] | ["C"] | null | Because of thermal expansion, change in length
<br><br> ($$\Delta $$$$\ell $$) = $$\ell $$ $$\alpha $$ $$\Delta $$T . . . . .(1)
<br><br>Because of compressive force, the compansion is $$\Delta $$$$\ell $$ ' ,
<br><br>$$\therefore\,\,\,$$ Young's Modulus (y) = $${{{F \... | mcq | jee-main-2017-online-8th-april-morning-slot | 11,553 |
PkfQ4yGV8oKsws68z31oe | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | A steel rail of length 5 m and area of cross section 40cm<sup>2</sup>
is prevented from expanding along its length while the temperature rises
by 10<sup>o</sup>C. If coefficient of linear expansion and Young’s modulus of steel are 1.2×10<sup>−5</sup> K<sup>−1</sup> and 2×10<sup>11</sup> Nm<sup>−2</sup> respectively,... | [{"identifier": "A", "content": "2 $$ \\times $$ 10<sup>7</sup> N"}, {"identifier": "B", "content": "1 $$ \\times $$ 10<sup>5</sup> N"}, {"identifier": "C", "content": "2 $$ \\times $$ 10<sup>9</sup> N"}, {"identifier": "D", "content": "3 $$ \\times $$ 10<sup>$$-$$5</sup> N"}] | ["B"] | null | Young's modulus (Y) = $${{{F \over A}} \over {{{\Delta L} \over L}}}$$
<br><br>as $${{{\Delta L} \over L}}$$ = $$\alpha $$ $$\Delta $$$$\theta $$
<br><br>$$\therefore\,\,\,$$ Y = $${{F \over {A\alpha \Delta \theta }}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ F = YA$$\alpha $$$$\Delta $$$$\theta $$
<br><br>= 2 $$ \ti... | mcq | jee-main-2017-online-9th-april-morning-slot | 11,554 |
F0SevcTd4K0G7oXKPhPGI | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | Temperature difference of 120<sup>o</sup>C is maintained between ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB and length $${{3L} \over 2},$$ is connected across AB (see figure). In steady state, temperature difference between P and Q will be close to :
<br/><br/><img src="d... | [{"identifier": "A", "content": "45<sup>o</sup>C"}, {"identifier": "B", "content": "75<sup>o</sup>C"}, {"identifier": "C", "content": "60<sup>o</sup>C"}, {"identifier": "D", "content": "35<sup>o</sup>C"}] | ["A"] | null | We know,
<br><br>Resistance, R = $${{\rho L} \over A}$$
<br><br>$$ \therefore $$ R $$ \propto $$ L
<br><br>So, Resistance is directly proportional to lengt5h of the rod.
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266569/exam_images/tgrwnrbirrroynqm8lkk.webp" style="max-wid... | mcq | jee-main-2019-online-9th-january-morning-slot | 11,555 |
D3NcFbtZorNToyDMnYLtC | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | A rod, of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion $$\alpha $$/<sup>o</sup>C. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when its temperature rises b... | [{"identifier": "A", "content": "$${F \\over {A\\alpha \\Delta T}}$$"}, {"identifier": "B", "content": "$${F \\over {A\\alpha (\\Delta T - 273)}}$$"}, {"identifier": "C", "content": "$${F \\over {2A\\alpha \\Delta T}}$$"}, {"identifier": "D", "content": "$${{2F} \\over {A\\alpha \\Delta T}}$$"}] | ["A"] | null | We know,
<br><br>Young's Modulus, Y = $${{Stress} \over {Strain}}$$
<br><br>Stress = $${F \over A}$$
<br><br>Strain = $${{\Delta l} \over l}$$
<br><br>$$ \therefore $$ Y = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$
<br><br>We also know,
<br><br>$$l$$<sub>f</sub> = $$l$$<sub>i</sub> (1 + $$\alpha $$... | mcq | jee-main-2019-online-9th-january-morning-slot | 11,556 |
xixUBcsElOF06DLf7WOnG | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | Two rods A and B of identical dimensions are at temperature 30<sup>°</sup>C. If A is heated upto 180<sup>o</sup>C and B upto T<sup>o</sup>C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4 : 3, then the value of T is | [{"identifier": "A", "content": "200<sup>o</sup>C"}, {"identifier": "B", "content": "270<sup>o</sup>C"}, {"identifier": "C", "content": "230<sup>o</sup>C"}, {"identifier": "D", "content": "250<sup>o</sup>C"}] | ["C"] | null | $$\Delta {\ell _1} = \Delta {\ell _2}$$
<br><br>$$\ell {\alpha _1}\Delta {T_1} = \ell {\alpha _2}\Delta {T_2}$$
<br><br>$${{{\alpha _1}} \over {{\alpha _2}}} = {{\Delta {T_1}} \over {\Delta {T_2}}}$$
<br><br>$${4 \over 3} = {{T - 30} \over {180 - 30}}$$
<br><br>$$T = {230^o}C$$ | mcq | jee-main-2019-online-11th-january-evening-slot | 11,557 |
3WjYxzZgrMlBhhRb9OnF4 | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | A thermometer graduated according to a linear scale reads a value x<sub>0</sub> when in contact with boiling water, and x<sub>0</sub>/3 when in contact with ice. What is the temperature of an object in <sup>o</sup>C, if this thermometer in the contact with the object reads x<sub>0</sub>/2 ? | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "35"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "40"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266271/exam_images/jdbhvhkyml7kjln1m8dq.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 11th January Evening Slot Physics - Heat and Thermodynamics Question 301 English Explanation">
<... | mcq | jee-main-2019-online-11th-january-evening-slot | 11,558 |
ZFeFZkrk542zaNIPSejgy2xukf250xxc | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | A bakelite beaker has volume capacity of 500 cc at 30<sup>o</sup>C. When it is partially filled with Vm
volume
(at 30<sup>o</sup>C) of mercury, it is found that the unfilled volume of the beaker remains constant as
temperature is varied. If $$\gamma $$<sub>(beaker)</sub> = 6 × 10<sup>–6</sup> <sup>o</sup>C<sup>–1</sup... | [] | null | 20 | $$\Delta $$V = V$$\gamma $$$$\Delta $$T
<br><br>$$ \therefore $$ V<sub>1</sub>$$\gamma $$<sub>1</sub> = V<sub>2</sub>$$\gamma $$<sub>2</sub>
<br><br>$$ \Rightarrow $$ 500 $$ \times $$ 6 $$ \times $$ 10<sup>-6</sup> = V<sub>m</sub>
$$ \times $$ 1.5 $$ \times $$ 10<sup>-4</sup>
<br><br>$$ \Rightarrow $$ V<sub>m</sub> = ... | integer | jee-main-2020-online-3rd-september-morning-slot | 11,560 |
QZOG01f63XnOZPoM2vjgy2xukfl319dk | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | Two different wires having lengths L<sub>1</sub> and L<sub>2</sub>,
and respective temperature coefficient of linear
expansion $$\alpha $$<sub>1</sub> and $$\alpha $$<sub>2</sub>, are joined end-to-end.
Then the effective temperature coefficient of
linear expansion is : | [{"identifier": "A", "content": "$$2\\sqrt {{\\alpha _1}{\\alpha _2}} $$"}, {"identifier": "B", "content": "$$4{{{\\alpha _1}{\\alpha _2}} \\over {{\\alpha _1} + {\\alpha _2}}}{{{L_2}{L_1}} \\over {{{\\left( {{L_2} + {L_1}} \\right)}^2}}}$$"}, {"identifier": "C", "content": "$${{{\\alpha _1} + {\\alpha _2}} \\over 2}$$... | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264856/exam_images/ufmjvq9rrjlit0rmao2y.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Physics - Heat and Thermodynamics Question 236 English Explanation">
<b... | mcq | jee-main-2020-online-5th-september-evening-slot | 11,561 |
yyC60k4DnxkwfO9w1xjgy2xukexrvjsd | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | When the temperature of a metal wire is
increased from 0<sup>o</sup>C to 10<sup>o</sup>C, its length
increases by 0.02%. The percentage change in
its mass density will be closest to : | [{"identifier": "A", "content": "0.008"}, {"identifier": "B", "content": "0.06"}, {"identifier": "C", "content": "0.8"}, {"identifier": "D", "content": "2.3"}] | ["B"] | null | Given, $${{\Delta L} \over L}$$ = 0.02%
<br><br>We know, $$\Delta $$L = L$$\alpha $$$$\Delta $$T
<br><br>$$ \Rightarrow $$ $${{\Delta L} \over L} = \alpha \Delta T$$ = 0.02
<br><br>Also, $$\beta $$ = 2$$\alpha $$
<br><br>$$ \Rightarrow $$ $$\beta \Delta T = 2\alpha \Delta T$$ = 0.04
<br><br>Density($$\rho $$) = $${M \o... | mcq | jee-main-2020-online-2nd-september-evening-slot | 11,562 |
uj2YyPhGGVg1OVm5Yi7k9k2k5duluhl | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | A non-isotropic solid metal cube has coefficients of linear expansion as :<br/> 5 $$ \times $$ 10<sup>-5</sup>/<sup>o</sup>C along the x-axis and 5 $$ \times $$ 10<sup>-6</sup>/<sup>o</sup>C along the y and the z-axis. If the coefficient of volume expansion of the solid is C $$ \times $$ 10<sup>-6</sup>/<sup>o</sup>C t... | [] | null | 60 | $$\gamma $$ = $$\alpha $$<sub>x</sub> + $$\alpha $$<sub>y</sub> + $$\alpha $$<sub>z</sub>
<br><br>$$ \Rightarrow $$ C $$ \times $$ 10<sup>–6</sup> = 5 × 10<sup>–5</sup> + 5 × 10<sup>–6</sup> + 5 × 10<sup>–6</sup>
<br><br>$$ \Rightarrow $$ C $$ \times $$ 10<sup>–6</sup> = 50 × 10<sup>–6</sup> + 10 × 10<sup>–6</sup>
<br>... | integer | jee-main-2020-online-7th-january-morning-slot | 11,563 |
nOINpzjUPZIIlg9XAV1klrhg8al | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is '$$\alpha$$'. The metal sheet is heated uniformly, by a small temperature $$\Delta$$T, so that its new temperature is T + $$\Delta$$T. Calculate the increase in the volume of t... | [{"identifier": "A", "content": "3a<sup>3</sup>$$\\alpha$$$$\\Delta$$T"}, {"identifier": "B", "content": "4$$\\pi$$a<sup>3</sup>$$\\alpha$$$$\\Delta$$T"}, {"identifier": "C", "content": "$${{4 \\over 3}}$$$$\\pi$$a<sup>3</sup>$$\\alpha$$$$\\Delta$$T"}, {"identifier": "D", "content": "4a<sup>3</sup>$$\\alpha$$$$\\Delta$... | ["A"] | null | We know that, $$\gamma = 3\alpha $$ .... (i)<br/><br/>where, $$\alpha$$ is the coefficient of linear expansion and $$\gamma$$ is the coefficient of volume expansion. <br/><br/>We know that,<br/><br/>$${{\Delta V} \over V} = \gamma \Delta T$$<br/><br/>$$ \Rightarrow {{\Delta V} \over V} = 3\alpha \Delta T$$ [from Eq. (... | mcq | jee-main-2021-online-24th-february-morning-slot | 11,564 |
JisOssqBPNfh4cC2bO1klrxlnfl | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.<br/><br/>Assertion A : When a rod lying freely is heated, no thermal stress is developed in it.<br/><br/>Reason R : On heating, the length of the rod increases.<br/><br/>In the light of the above statements, choose th... | [{"identifier": "A", "content": "A is true but R is false"}, {"identifier": "B", "content": "A is false but R is true"}, {"identifier": "C", "content": "Both A and B are true but R is NOT the correct explanation of A"}, {"identifier": "D", "content": "Both A and R are true and R is the correct explanation of A"}] | ["C"] | null | When a rod is free and it is heated then there is no thermal stress produced in it.
<br><br>The rod will expand due to increase in temperature.
<br><br>So both A & R are true. | mcq | jee-main-2021-online-25th-february-morning-slot | 11,565 |
6fCurmHX8XxbNl9yVO1kmipketj | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | A bimetallic strip consists of metals A and B. It is mounted rigidly as shown. The metal A has higher coefficient of expansion compared to that of metal B. When the bimetallic strip is placed in a cold bath, it will :<br/><br/><img src="data:image/png;base64,UklGRuIBAABXRUJQVlA4INYBAABQEQCdASqAAGQAPm0ylEakIyIhMJEoAIANi... | [{"identifier": "A", "content": "Neither bend nor shrink"}, {"identifier": "B", "content": "Bend towards the left"}, {"identifier": "C", "content": "Not bend but shrink"}, {"identifier": "D", "content": "Bend towards the right"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266224/exam_images/osrl6nlj9kkuepupag9d.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Physics - Heat and Thermodynamics Question 206 English Explanation">
<br>... | mcq | jee-main-2021-online-16th-march-evening-shift | 11,566 |
1krsxywxj | physics | heat-and-thermodynamics | thermometry-and-thermal-expansion | The area of cross-section of a railway track is 0.01 m<sup>2</sup>. The temperature variation is 10$$^\circ$$C. Coefficient of liner expansion of material of track is 10<sup>$$-$$5</sup>/$$^\circ$$C. The energy stored per meter in the track is ____________ J/m.<br/><br/>(Young's modulus of material of track is 10<sup>1... | [] | null | 05 | As the tracks won't be allowed to expand linearly, the rise in temperature would lead to developing thermal stress in track.<br><br>$${{(Stress)} \over y} = \alpha \Delta T$$ or $$\sigma = Y\alpha \Delta T$$<br><br>Energy stored per unit volume = $${1 \over 2}{\sigma \over Y}$$<br><br>$$\Rightarrow$$ Energy stored pe... | integer | jee-main-2021-online-22th-july-evening-shift | 11,567 |
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