You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stefan-Boltzman's Law

Two solid sph...

Question

Two solid spheres, of radii $R_{1}$ and $R_{2}$ are made of the same material and have similar surface. The spheres are raised to the same temperature and then allowed to cool under identical conditions.Assuming Spheres to be perfect conductors of heat, their initial rates of loss of heat are

R_{1}^{2} / R_{2}^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

R_{1} / R_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{2} / R_{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R_{2}^{2} / R_{1}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $R_{1}^{2} / R_{2}^{2}$
From stefan's law, we know that:
$Q = σ A T^{4} \times e$
where, area of sphere $A = 4 π R^{2}$
Here, temperature, thermal conductivity and emissivity of material are same.
So, we get the relation:
Initial rate of cooling is given as: $\frac{Q_{1}}{Q_{2}} = \frac{R_{1}^{2}}{R_{2}^{2}}$

Suggest Corrections

Similar questions

r_{1}

r_{2}

(r_{1} + r_{2})

2 \sqrt{r_{1}^{2} + r_{2}^{2} - r_{1} r_{2}}

R_{1}

R_{2}

ρ_{1}

ρ_{2}

C_{1}

C_{2}

r_{1}

r_{2} (r_{1} = 1.5 r_{2})

1 K

Two conducting spheres of radii $r_{1}$ and r₂ are at the same potential. Their charges are in the ratio of

R_{1}

R_{2}

\frac{n}{1}

Join BYJU'S Learning Program