JEE Physics 2025 Question with Solution

Given: Length of the sheet $L = 5 \, \text{mm}$, breadth of the sheet $B = 2.5 \, \text{mm}$, pitch of screw gauge $= 0.75 \, \text{mm}$, and number of circular scale divisions $= 15$.

Find: The value of $x$ if the maximum fractional error in area is $\frac{x}{100}$.

First, calculate the least count of the screw gauge:

$$\text{Least count} = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.75}{15} = 0.05 \, \text{mm}$$

The absolute error in each linear measurement is therefore $0.05 \, \text{mm}$.

Fractional error in length:

$$\frac{\Delta L}{L} = \frac{0.05}{5} = 0.01$$

Fractional error in breadth:

$$\frac{\Delta B}{B} = \frac{0.05}{2.5} = 0.02$$

Since area $A = L \times B$, the maximum fractional error in area is the sum of the fractional errors:

$$\left(\frac{\Delta A}{A}\right)_{\text{max}} = \frac{\Delta L}{L} + \frac{\Delta B}{B} = 0.01 + 0.02 = 0.03$$

Now compare with the given form:

$$0.03 = \frac{x}{100}$$

So,

$$x = 3$$

Therefore, the value of $x$ is $3$.

Given: $L = 5 \, \text{mm}$, $B = 2.5 \, \text{mm}$, screw gauge pitch $= 0.75 \, \text{mm}$, and circular scale divisions $= 15$.

Find: Maximum fractional error in area written as $\frac{x}{100}$.

For a screw gauge, the least count is:

$$\text{L.C.} = \frac{0.75}{15} = 0.05 \, \text{mm}$$

Thus,

$$\Delta L = 0.05 \, \text{mm}, \qquad \Delta B = 0.05 \, \text{mm}$$

Now compute the fractional errors separately:

$$\frac{\Delta L}{L} = \frac{0.05}{5} = \frac{1}{100}$$ $$\frac{\Delta B}{B} = \frac{0.05}{2.5} = \frac{2}{100}$$

For the product of two measured quantities,

$$A = L \times B \quad \Rightarrow \quad \frac{\Delta A}{A} = \frac{\Delta L}{L} + \frac{\Delta B}{B}$$

Substituting values:

$$\frac{\Delta A}{A} = \frac{1}{100} + \frac{2}{100} = \frac{3}{100}$$

Hence, in the form $\frac{x}{100}$, we get $x = 3$.

Therefore, the final answer is $3$.