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NEET PHYSICSThermal Properties of MatterMedium

Question

A body cools from a temperature 3T3T to 2T2T in 10 minutes10\text{ minutes}. The room temperature is TT. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes10\text{ minutes} will be:

A

74T\frac{7}{4}T

B

32T\frac{3}{2}T

C

43T\frac{4}{3}T

D

TT

Step-by-Step Solution

According to the average form of Newton's law of cooling: T1T2t=K(T1+T22Ts)\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right) where TsT_s is the surrounding (room) temperature.

For the first 10 minutes10\text{ minutes}: 3T2T10=K(3T+2T2T)\frac{3T - 2T}{10} = K \left( \frac{3T + 2T}{2} - T \right) T10=K(5T2T)\frac{T}{10} = K \left( \frac{5T}{2} - T \right) T10=K(3T2)\frac{T}{10} = K \left( \frac{3T}{2} \right) ... (i)

For the next 10 minutes10\text{ minutes}, let the final temperature be TT': 2TT10=K(2T+T2T)\frac{2T - T'}{10} = K \left( \frac{2T + T'}{2} - T \right) 2TT10=K(2T+T2T2)\frac{2T - T'}{10} = K \left( \frac{2T + T' - 2T}{2} \right) 2TT10=K(T2)\frac{2T - T'}{10} = K \left( \frac{T'}{2} \right) ... (ii)

Dividing equation (i) by (ii), we get: T102TT10=K(3T2)K(T2)\frac{\frac{T}{10}}{\frac{2T - T'}{10}} = \frac{K \left( \frac{3T}{2} \right)}{K \left( \frac{T'}{2} \right)} T2TT=3TT\frac{T}{2T - T'} = \frac{3T}{T'} T=3(2TT)T' = 3(2T - T') T=6T3TT' = 6T - 3T' 4T=6T4T' = 6T T=6T4=32TT' = \frac{6T}{4} = \frac{3}{2}T

Therefore, the temperature of the body at the end of next 10 minutes10\text{ minutes} will be 32T\frac{3}{2}T.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Thermal Properties of Matter. 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.

PHYSICSThermal Properties of Mattertemperatureminutestemperatureassumenewtons

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