The temperature of a spherical black body is inversely proportional to its radius. If its radius is doubled,then the power radiating from it will be

  • A
    doubled
  • B
    $\frac{1}{4}$ times of initial value
  • C
    halved
  • D
    four times of initial value

Explore More

Similar Questions

Two spheres of the same material have radii $1 \; m$ and $4 \; m$ and temperatures $4000 \; K$ and $2000 \; K$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

$A$ black body is at a temperature of $500 \; K$. It emits energy at a rate which is proportional to:

$A$ black body at a high temperature $T \ K$ radiates energy at the rate $E \ W/m^2$. When the temperature falls to $(T/2) \ K$,the radiated energy will be:

$A$ black sphere has radius $R$ whose rate of radiation is $E$ at temperature $T$. If the radius is made half and the temperature is made $4T$,the rate of radiation will be: (in $E$)

The temperature of a body is increased from $-73^{\circ}C$ to $327^{\circ}C$. The ratio of energy emitted per second is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo