Q. A body takes 10 min to cool from 60$℃$ to 50$℃$. If the temperature of surroundings is 25$℃$ and 527$℃$ respectively. The ratio of energy radiated by $P\ and\ Q\ is$

A) 48$℃$
B) 46$℃$
C) 49$℃$
D) 42.85$℃$
Explanation

(d) According to Newton’s law
$\frac{\theta_{1} - \theta_{2}}{t} = K\left\lbrack \frac{\theta_{1} + \theta_{2}}{2} - \theta_{0} \right\rbrack$
$\therefore\ \frac{60 - 50}{10} = K\left\lbrack \frac{60 + 50}{2} - 25 \right\rbrack$ ….(i)
Let $\theta$ be the temperature after another 10 min
$\therefore\ \frac{50 - \theta}{10} = K\left\lbrack \frac{\theta + 50}{2} - 25 \right\rbrack$ ….(ii)
Dividing Eq.(i) by Eq. (ii), we get
$\frac{10}{50 - \theta}\ = \ \frac{30 \times 2}{\theta}\ \therefore\ \theta = 42.85℃$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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