JEE Physics 2025 Question with Solution

Given:

• Length $L = 0.1 \, \text{m}$
• Diameter $d = 0.0005 \, \text{m}$, so radius $r = 0.00025 \, \text{m}$
• Temperature $T = 1727^\circ \text{C} = 2000 \, \text{K}$
• Power radiated $P = 94.2 \, \text{W}$

Find: $x$ if emissivity $\varepsilon = \frac{x}{8}$

Using Stefan–Boltzmann law,

$$P = \varepsilon \sigma A T^4$$

So,

$$\varepsilon = \frac{P}{\sigma A T^4}$$

For the cylindrical wire, the lateral surface area is

$$A = 2\pi rL = 2 \times 3.14 \times 0.00025 \times 0.1 = 1.57 \times 10^{-4} \, \text{m}^2$$

Also,

$$T^4 = (2000)^4 = 16 \times 10^{12}$$

Now,

$$\varepsilon = \frac{94.2}{6.0 \times 10^{-8} \times 1.57 \times 10^{-4} \times 16 \times 10^{12}} = \frac{94.2}{150.72} \approx 0.625$$

Thus,

$$\varepsilon = 0.625 = \frac{5}{8}$$

Since $\varepsilon = \frac{x}{8}$,

$$x = 8 \times 0.625 = 5$$

Therefore, the value of $x$ is $5$.

Given: $L = 10 \, \text{cm}$, $d = 0.5 \, \text{mm}$, $T = 2000 \, \text{K}$, $P = 94.2 \, \text{W}$

Find: $x$ in $\varepsilon = \frac{x}{8}$

Use

$$P = \epsilon \sigma A T^4$$

Substituting directly as shown,

$$94.2 = \epsilon \times (6 \times 10^{-8}) \times (3.14) \times (0.5) \times (10^{-3}) \times (10 \times 10^{-2}) \times (2000)^4$$

Solving gives

$$\epsilon = \frac{5}{8}$$

Hence,

$$\frac{x}{8} = \frac{5}{8}$$

So,

$$x = 5$$

Therefore, the value of $x$ is $5$.