JEE Advanced 2026 — Paper 2 — PHY-I solution

JEE Advanced2026Paper 2PHY-I

A nuclear reactor starts producing a radioactive nuclide $X$ from $t = 0$ , at a constant rate of $α$ 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 $λ$ as the decay constant of $X$ , the rate of increase in the temperature of the liquid is:

A
$\frac{α E _{0}}{m s} (1 - e^{- λ t})$
B
$\frac{α E _{0}}{m s} (e^{λ t} - 1)$
C
$\frac{λ E _{0}}{m s} (1 - e^{- λ t})$
D
$\frac{E _{0}}{m s} (α - λ e^{- λ t})$

Solution

Step 1 — Equation for the number of $X$ nuclei. Production at rate $α$ and decay at rate $λ N$ give

$\frac{d N}{d t} = α - λ N$ .

Step 2 — Solve with $N (0) = 0$ .

$N (t) = \frac{α}{λ} (1 - e^{- λ t})$ .

Step 3 — Rate of decays. The number of decays per unit time equals $λ N$ , so the energy deposited per unit time is

$P = E_{0} λ N = E_{0} α (1 - e^{- λ t})$ .

Step 4 — Temperature rate. With $P = m s \frac{d T}{d t}$ ,

$\frac{d T}{d t} = \frac{α E _{0}}{m s} (1 - e^{- λ t})$ , which is option (A).

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