JEE Physics 2026 Question with Solution

Given: Time $t = 60 \, \text{s}$, battery emf $V = 9 \, \text{V}$, and internal resistance $r = 1 \, \Omega$.

Find: The heat generated between A and B in $1 \, \text{minute}$.

The extracted solution works backward from the answer key because the circuit data and the stated internal resistance lead to an inconsistency if interpreted directly.

From the answer key,

$$P = \frac{H}{t} = \frac{1620}{60} = 27 \, \text{W}$$

If this heat is generated in the external circuit between A and B, then

$$I = \frac{V}{R_{ext} + r} = \frac{9}{R_{ext} + 1}$$

and

$$P = I^2 R_{ext} = \left(\frac{9}{R_{ext}+1}\right)^2 R_{ext}$$

So,

$$27 = \frac{81R_{ext}}{(R_{ext}+1)^2}$$

which gives

$$(R_{ext}+1)^2 = 3R_{ext}$$ $$R_{ext}^2 - R_{ext} + 1 = 0$$

This quadratic has no real root, so the solution notes that the data appears inconsistent.

It then uses the intended total resistance as

$$R_{total} = 3 \, \Omega$$

Hence,

$$P = \frac{V^2}{R} = \frac{9^2}{3} = 27 \, \text{W}$$

Therefore,

$$H = Pt = 27 \times 60 = 1620 \, \text{J}$$

So the heat generated is $1620 \, \text{J}$ and the correct option is B.

Note: The solution itself indicates a discrepancy in the circuit interpretation and effectively backs out the intended equivalent resistance from the answer key.

Given: The provided solution concludes that the intended equivalent resistance of the complete circuit is $3 \, \Omega$.

Find: Heat generated in $60 \, \text{s}$.

Use electrical power:

$$P = \frac{V^2}{R}$$

Substitute the given battery voltage:

$$P = \frac{9^2}{3} = \frac{81}{3} = 27 \, \text{W}$$

Now heat produced in time $t$ is

$$H = Pt$$

Therefore,

$$H = 27 \times 60 = 1620 \, \text{J}$$

Thus, the required heat generated is $1620 \, \text{J}$.