If the temperature of a hot body is increased by `50%`, then the increase in the quantity of emitted heat radiation will be

Options

A. `125%`
B. `200%`
C. `300%`
D. `400%`

Answer

D

Explanation

We know that
`Q prop T^(4)`
Hence, `(Q_(1))/(Q_(2))=((T_(1))/(T_(2)))^(4)`
`implies" "(Q_(1))/(Q_(2))=((T)/(T+T//2))^(4)`
`implies" "(Q_(1))/(Q_(2))=(16)/(81)`
`implies" "Q_(2)=(81)/(16)Q_(1)`
% increase in energy
`=(Q_(2)-Q_(1))/(Q_(1))xx100=400%`