Two uniform solid spheres made of copper have radii 15cm and 20cm...

For two bodies made up of same material, rate of loss of heat will depend only on their surface area for same temperature change.

$∴ \dfrac{{\left(\dfrac{dQ}{dt}\right)}_{1}}{{\left(\dfrac{dQ}{dt}\right)}_{2}} = \dfrac{{A}_{1}}{{A}_{2}} = {r}_{1}^{2}{r}_{2}^{2} \ldots \ldots \left(i\right)$

Ratio of initial rates of cooling$= \dfrac{{\left(\dfrac{d\theta}{dt}\right)}_{1}}{{\left(\dfrac{d\theta}{dt}\right)}_{2}}$

As, $\dfrac{dQ}{dt} = mc\dfrac{d\theta}{dt}$

$∴ \dfrac{{m}_{1}{c}_{1} ( d\theta / dt{)}_{1}}{{m}_{2}{c}_{2} ( d\theta / dt{)}_{2}} = {r}_{1}^{2}{r}_{2}^{2}$ ...[from (i)]

since, both the spheres are made up of copper,

${c}_{1} = {c}_{2} = c$.

$∴ \dfrac{{\left(\dfrac{d\theta}{dt}\right)}_{1}}{{\left(\dfrac{d\theta}{dt}\right)}_{2}} = \dfrac{{r}_{1}^{2} \times {m}_{2}}{{r}_{2}^{2} \times {m}_{1}}$

As, spheres have same densities,

$\dfrac{{m}_{1}}{{m}_{2}} = \dfrac{{V}_{1}}{{V}_{2}} = {r}_{1}^{3}{r}_{2}^{3}$

$∴ \dfrac{( d\theta / dt{)}_{1}}{( d\theta / dt{)}_{2}} = {r}_{1}^{2}{r}_{2}^{2} \times {r}_{2}^{3}{r}_{1}^{3} = \dfrac{{r}_{2}}{{r}_{1}} = \dfrac{20}{15} = \dfrac{4}{3} = 1 . 33$