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NEET Practice Questions (MCQs) with Answers & Solutions

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If the door of a refrigerator is kept open, then which of the following is true ?

Room is cooled

Room is heated

Room is either cooled or heated

Room is neither cooled nor heated

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Explanation

In a refrigerator, the heat dissipated in the atmosphere is more than that taken from the cooling chamber, therefore the room is heated if the door of a refrigerator is kept open.

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In a cyclic process, the internal energy of the gas -

Increases

Decreases

Remains constant

Becomes zero

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Explanation

Internal energy is a state function.

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For a reversible process, necessary condition is -

In the whole cycle of the system, the loss of any type of heat energy should be zero

That the process should be too fast

That the process should be slow so that the working substance should remain in thermal and mechanical equilibrium with the surroundings

The loss of energy should be zero and it should be quasistatic

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Explanation

For a reversible process $\int \frac{d Q}{T} = 0$

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A Carnot's engine used first an ideal monoatomic gas then an ideal diatomic gas. If the source and sink temperature are 411°C and 69°C respectively and the engine extracts 1000 J of heat in each cycle, then area enclosed by the PV diagram is -

100 J

300 J

500 J

700 J

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{(273 + 69)}{(273 + 411)} = 0 .5$

⇒ Work done $= η \times Q = 0 .5 \times 1000 = 500 J$

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The temperature of sink of Carnot engine is 27°C. Efficiency of engine is 25%. Then temperature of source is -

227°C

327°C

127°C

27°C

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}}$ ⇒ $\frac{25}{100} = 1 - \frac{300}{T_{1}}$ ⇒ $\frac{1}{4} = 1 - \frac{300}{T_{1}}$

$T_{1} = 400 K = 127 ° C$

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The temperature of reservoir of Carnot's engine operating with an efficiency of 70% is 1000K. The temperature of its sink is -

300 K

400 K

500 K

700 K

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}}$ ⇒ $\frac{70}{100} = 1 - \frac{T_{2}}{1000}$

⇒ T₂ = 300 K

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In a Carnot engine, when T₂ = 0°C and T₁ = 200°C, its efficiency is η₁ and when T₁ = 0°C and T₂ = –200°C, its efficiency is $η_{2}$ , then what is η₁ /η₂ ?

0.577

0.733

0.638

Can not be calculated

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}} = \frac{T_{1} - T_{2}}{T_{1}}$

$\Rightarrow η_{1} = \frac{(473 - 273)}{473} = \frac{200}{473}$ and $η_{2} = \frac{273 - 73}{273} = \frac{200}{273}$

So required ratio $\frac{η_{1}}{η_{2}} = \frac{273}{473} = 0 .577$

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The efficiency of Carnot's engine operating between reservoirs, maintained at temperatures 27°C and –123°C, is

50%

24%

0.75%

0.4%

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{(273 - 123)}{(273 + 27)} = 1 - \frac{150}{300} = \frac{1}{2} = 50 %$

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A Carnot engine operates between 227°C and 27°C . Efficiency of the engine will be -

$\frac{1}{3}$

$\frac{2}{5}$

$\frac{3}{4}$

$\frac{3}{5}$

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{300}{500} = \frac{2}{5}$

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Efficiency of a Carnot engine is 50% when temperature of outlet is 500 K. In order to increase efficiency up to 60% keeping temperature of intake the same what is temperature of outlet ?

200 K

400 K

600 K

800 K

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Explanation

$η = 1 - \frac{T_{2}}{T_{1}}$ ⇒ $\frac{1}{2} = 1 - \frac{500}{T_{1}}$ ⇒ $\frac{500}{T_{1}} = \frac{1}{2}$ …..(i)

$\frac{60}{100} = 1 - \frac{T_{2}'}{T_{1}}$ ⇒ $\frac{T_{2}'}{T_{1}} = \frac{2}{5}$ …..(ii)

Dividing equation (i) by (ii), $\frac{500}{T_{2}'} = \frac{5}{4}$ ⇒ $T_{2} = 400 K$

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