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JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

Option 2 : 99°C, 37°C

**Concept:**

Carnot engine evaluates that maximum possible efficiency in heat engine during the conversion process of heat and work, it can work between two reservoirs, can possess.

The efficiency of the Carnot engine is given by the formula:

\({\rm{\eta }} = \left( {1 - \frac{{{{\rm{T}}_{\rm{C}}}}}{{{{\rm{T}}_{\rm{H}}}}}} \right)\)

Where,

T_{C} = Cold temperature (temperature of sink)

T_{H} = Hot temperature (temperature of source)

**Calculation:**

The efficiency before reducing temperature is given as:

\( \Rightarrow \frac{1}{6} = \left( {1 - \frac{{{{\rm{T}}_{\rm{C}}}}}{{{{\rm{T}}_{\rm{H}}}}}} \right)\)

\( \Rightarrow \frac{{{{\rm{T}}_{\rm{C}}}}}{{{{\rm{T}}_{\rm{H}}}}} = 1 - \frac{1}{6}\)

\( \Rightarrow \frac{{{{\rm{T}}_{\rm{C}}}}}{{{{\rm{T}}_{\rm{H}}}}} = \frac{{6 - 1}}{6}\)

\(\therefore \frac{{{{\rm{T}}_{\rm{C}}}}}{{{{\rm{T}}_{\rm{H}}}}} = \frac{5}{6}\)

The efficiency after reducing temperature is given as:

\( \Rightarrow 2\left( {\frac{1}{6}} \right) = \left( {1 - \frac{{{{\rm{T}}_{\rm{C}}} - 62}}{{{{\rm{T}}_{\rm{H}}}}}} \right)\)

\( \Rightarrow \frac{1}{3} = \left( {1 - \frac{{{{\rm{T}}_{\rm{C}}} - 62}}{{{{\rm{T}}_{\rm{H}}}}}} \right)\)

\( \Rightarrow \frac{1}{3} = \frac{{{{\rm{T}}_{\rm{H}}} - {{\rm{T}}_{\rm{C}}} + 62}}{{{{\rm{T}}_{\rm{H}}}}}\)

\( \Rightarrow \frac{1}{3} = 1 - \frac{{{{\rm{T}}_{\rm{C}}}}}{{{{\rm{T}}_{\rm{H}}}}} + \frac{{62}}{{{{\rm{T}}_{\rm{H}}}}}\)

\( \Rightarrow \frac{1}{3} = 1 - \frac{5}{6} + \frac{{62}}{{{{\rm{T}}_{\rm{H}}}}}\)

\( \Rightarrow \frac{{62}}{{{{\rm{T}}_{\rm{H}}}}} = \frac{1}{3} - 1 + \frac{5}{6}\)

\( \Rightarrow \frac{{62}}{{{{\rm{T}}_{\rm{H}}}}} = \frac{{2 - 6 + 5}}{6}\)

\( \Rightarrow \frac{{62}}{{{{\rm{T}}_{\rm{H}}}}} = \frac{{7 - 6}}{6}\)

\( \Rightarrow \frac{{62}}{{{{\rm{T}}_{\rm{H}}}}} = \frac{1}{6}\)

∴ T_{H }= 372 K

From options, the temperature should be in Kelvin.

⇒ T_{H }= 372 K - 273

∴ T_{H }= 99°C

Thus, we have obtained the temperature of the sink.

Now, temperature of the source is:

\( \Rightarrow \frac{{{{\rm{T}}_{\rm{C}}}}}{{372}} = \frac{5}{6}\)

\( \Rightarrow {{\rm{T}}_{\rm{C}}} = \frac{{5 \times 372}}{6}\)

⇒ T_{C }= 310 K

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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