Question

The radius and one-fourth wall thickness of the other are filled with ice. If the time taken for complete melting on ice in the larger sphere is $25$ and for smaller one is $16$ minute, the ratio to thermal conductivities of the materials of larger spheres to that of smaller sphere is

• A. $4:5$
• B. $5:4$
• C. $25:8$
• D. $8:25$

Answer 1

Answer: (D)

$8:25$

Radius of small sphere $=r$
Thickness of small sphere $=t$
Radius of bigger sphere $=2r$
Thickness of bigger spere $=t/4$
Mass of ice melted $=$ (volume of sphere) $\times$ (density of ice)
Let $K_1$ and $K_2$ be the thermal conductivities of larger and smaller sphere.
For bigger sphere,
$\frac{K_{1}4\pi (2r{)}^2\times 100}{t/4}=\frac{\frac{4}{3}\pi (2r{)}^3\rho L}{25\times 60}$
For smaller space,
$\frac{K_2\times 4\pi r^2\times 100}{t}=\frac{\frac{4}{3}\pi r^3\rho L}{16\times 60}$
$\therefore \frac{K_1}{K_2}=\frac{8}{25}$.