Question

There are two hollow spheres made of different materials, one with double the radius and one fourth wall thickness of the other are filled with ice. If the time taken for melting the ice completely in larger sphere is $25$ minutes and that for smaller sphere is $16$ minutes, the ratio of the thermal conductivity of the larger sphere to the smaller sphere is :

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

Answer 1

The correct option is D $8:25$

Smaller sphere

Radius $\to \frac{R}{2}$. wall thickness $\to 4d$

Larger sphere

Radius $\to R$, wall thickness $\to d$

Vol of ice in smaller sphere $=\frac{4}{3}\pi {(\frac{R}{2}-4d)}^3$

$m_S=$ Mass of ice in smaller sphere $={\rho }_{ice}\times \frac{4}{3}\pi {(\frac{R}{2}-4d)}^3$

$m_L=$ Mass of ice in larger sphere $={\rho }_{ice}\times \frac{4}{3}\pi (r-d{)}^3$

Heat required to melt $m_S=m_{s}L=\frac{4}{3}\pi {\left(\frac{R}{2}4d\right)}^3\times {\rho }_{ice}\times L=Q_s$ ...(i)

Heat required to melt $m_L=m_{l}L=\frac{4}{3}\pi (R-d{)}^3\times {\rho }_{ice}\times L=Q_L$ ...(ii)

$\frac{Q}{t}=\frac{KA}{x}\left(\Delta T\right)$

$\frac{Q_s}{16}=\frac{K_{s}4\pi {\left(\frac{R}{2}\right)}^2}{4d}\Delta T$ ...(for small sphere) ...(1)

$\frac{A_L}{25}=\frac{K_{2}4\pi R^2}{d}\Delta T$ ...(for large sphere) ....(2)

Dividing eq(1) by eq(2)

$\therefore \frac{Q_L}{25}\times \frac{16}{Q_s}=\left(\frac{K_L}{K_S}\right)\frac{4\pi R^2\Delta T\times 4d}{4\pi {\left(\frac{R}{2}\right)}^2\Delta T\times d}$

$\therefore \frac{Q_L}{Q_S}\times \frac{16}{25}=\left(\frac{K_L}{K_S}\right)\times 16$

but $\frac{Q_L}{Q_S}=\frac{(R-d{)}^3}{{(\frac{R}{2}-4d)}^3}$

here $R>>d$,

$\therefore \frac{Q_L}{Q_S}=\frac{R^3}{{\left(\frac{R}{2}\right)}^3}=8$

$\therefore \frac{8}{25}=\frac{K_L}{K_S}$