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JEE Advanced 2019
Paper-1 2019
Question
A current carrying wire heats a metal rod. The wire provides a constant power (P) to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature (T) in the metal rod changes with time (t) as :
$ \mathrm{T}(\mathrm{t})=\mathrm{T}_0\left(1+\beta \mathrm{t}^{1 / 4}\right) $
where $\beta$ is a constant with appropriate dimension while $\mathrm{T}_0$ is a constant with dimension of temperature. The heat capacity of the metal is :
Select the correct option:
A
$\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_0\right)^3}{\beta^4 \mathrm{~T}_0^4}$
B
$\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_0\right)}{\beta^4 \mathrm{~T}_0^2}$
C
$\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_0\right)^4}{\beta^4 \mathrm{~T}_0^5}$
D
$\frac{4 \mathrm{P}\left(\mathrm{T}(\mathrm{t})-\mathrm{T}_0\right)^2}{\beta^4 \mathrm{~T}_0^3}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
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Question Tags
JEE Advance
Physics
Easy
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