Question

A solid sphere of radius $4\text{cm}$ and a hollow sphere of same material and having outer radius $4\text{cm}$ and inner radius $2\text{cm}$ are heated to the same temperature and allowed to cool in the same room. The ratio of the rate of fall of temperatures is

• A. $7:8$
• B. $8:5$
• C. $2:1$
• D. $4:1$

Answer 1

The correct option is A $7:8$

According to the Newton's law of cooling,
$m{c}_p\frac{dT}{dt}=-hA(T-T_0)$
where all the terms have standard meanings.
Since the material of both the spheres is same, all the conditions and above mentioned terms will be same except $m$
and $\frac{dT}{dt}\propto \frac{1}{m}$
Hence
$\frac{\frac{d{T}_1}{dt}}{\frac{d{T}_2}{dt}}=\frac{m_2}{m_1}=\frac{\rho \times \frac{4}{3}\pi (4^3-2^3)}{\rho \times \frac{4}{3}\pi \times 4^3}=\frac{56}{64}=\frac{7}{8}$