JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Modern Physics

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:

Answer: (A)

Solution

The rate equation for the nuclide is

\frac{d N}{d t} = α - λ N

. Solving this differential equation gives

N = \frac{α}{λ} (1 - e^{- λ t})

.

The number of decayed nuclei is

N_{y} = α t - \frac{α}{λ} (1 - e^{- λ t})

.

The energy released heats the liquid:

E_{0} N_{y} = m s Δ T

.

Differentiating:

E_{0} \frac{d N _{y}}{d t} = m s \frac{d T}{d t}

.

Since

\frac{d N _{y}}{d t} = α - α e^{- λ t}

, we get

\frac{d T}{d t} = \frac{E _{0} α}{m s} (1 - e^{- λ t})

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Single correct option