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NEET PHYSICSThermodynamicsMedium

Question

A Carnot engine whose sink is at 300 K300 \text{ K} has an efficiency of 40%40\%. By how much should the temperature of the source be increased to increase its efficiency by 50%50\% of its original efficiency?

A

275 K275 \text{ K}

B

325 K325 \text{ K}

C

250 K250 \text{ K}

D

380 K380 \text{ K}

Step-by-Step Solution

The efficiency (η\eta) of a Carnot engine is given by the formula: η=1T2T1\eta = 1 - \frac{T_2}{T_1}, where T1T_1 is the source temperature and T2T_2 is the sink temperature.

Step 1: Calculate the initial source temperature (T1T_1). Given: η1=40%=0.4\eta_1 = 40\% = 0.4, T2=300 KT_2 = 300 \text{ K}. 0.4=1300T10.4 = 1 - \frac{300}{T_1} 300T1=10.4=0.6\frac{300}{T_1} = 1 - 0.4 = 0.6 T1=3000.6=500 KT_1 = \frac{300}{0.6} = 500 \text{ K}

Step 2: Calculate the new efficiency (η2\eta_2). The efficiency is increased by 50%50\% of its original value. Increase =50% of 0.4=0.5×0.4=0.2= 50\% \text{ of } 0.4 = 0.5 \times 0.4 = 0.2. New Efficiency η2=0.4+0.2=0.6\eta_2 = 0.4 + 0.2 = 0.6 (or 60%60\%).

Step 3: Calculate the new source temperature (T1T_1'). Keeping the sink temperature (T2T_2) constant at 300 K300 \text{ K}: 0.6=1300T10.6 = 1 - \frac{300}{T_1'} 300T1=10.6=0.4\frac{300}{T_1'} = 1 - 0.6 = 0.4 T1=3000.4=750 KT_1' = \frac{300}{0.4} = 750 \text{ K}

Step 4: Calculate the increase in temperature. ΔT=T1T1=750 K500 K=250 K\Delta T = T_1' - T_1 = 750 \text{ K} - 500 \text{ K} = 250 \text{ K}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSThermodynamicscarnotengineefficiencyshouldtemperature

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