JEE Advanced 2026 · Paper 2 · Physics Q2

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A nuclear reactor starts producing a radioactive nuclide $X$ from $t=0$ , at a constant rate of $\alpha$ per second. Each decay of $X$ produces energy $E_0$ , which is utilized to heat a liquid of mass $m$ and specific heat $s$ . Assuming no heat loss from the liquid and taking $\lambda$ as the decay constant of $X$ , the rate of increase in the temperature of the liquid is:

\dfrac{\alpha E_0}{m\,s}\left(1-e^{-\lambda t}\right)

\dfrac{\alpha E_0}{m\,s}\left(e^{\lambda t}-1\right)

\dfrac{\lambda E_0}{m\,s}\left(1-e^{-\lambda t}\right)

\dfrac{E_0}{m\,s}\left(\alpha-\lambda e^{-\lambda t}\right)

Answer and worked solution

Official answer: A

Number of nuclei. $\dfrac{dN}{dt}=\alpha-\lambda N$ with $N(0)=0$ gives
$N=\frac\alpha\lambda\left(1-e^{-\lambda t}\right).$

Heating. Power delivered $=\lambda N E_0=\alpha E_0\left(1-e^{-\lambda t}\right)=ms\,\dfrac{dT}{dt}$ , so
$\frac{dT}{dt}=\frac{\alpha E_0}{ms}\left(1-e^{-\lambda t}\right).$

Topics:Radioactive decay Calorimetry and specific heat

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JEE Advanced 2026 · Paper 2, Physics question 2