QJEEM24Phfb87: JEE Main 2024 Physics - Class 11 JEEM24Phfb87 Solution: While measuring diameter of wire using screw gauge the…

To determine the diameter of the wire using a screw gauge, we employ the formula:

$\text{Total reading} = \text{MSR} + (\text{CSR} \times \text{LC})$

where:

• MSR (Main Scale Reading) is the value read directly from the main scale in mm.


• CSR (Circular Scale Reading) is the number of divisions observed on the circular scale.


• LC (Least Count) is the value of one division on the circular scale, calculated as $\frac{\text{Pitch}}{\text{Number of divisions on the circular scale}}$ .

Given:

• Main Scale Reading (MSR) = $1 \mathrm{~mm}$


• Circular Scale Reading (CSR) = 42 divisions


• Pitch of screw gauge = $1 \mathrm{~mm}$


• Number of divisions on circular scale = 100

First, we find the Least Count (LC):

$LC = \frac{\text{Pitch}}{\text{Number of divisions on the circular scale}} = \frac{1 \mathrm{~mm}}{100} = 0.01 \mathrm{~mm}$

Then we calculate the total measurement of the diameter of the wire:

$\text{Total reading} = \text{MSR} + (\text{CSR} \times \text{LC}) = 1 \mathrm{~mm} + (42 \times 0.01 \mathrm{~mm}) = 1 \mathrm{~mm} + 0.42 \mathrm{~mm} = 1.42 \mathrm{~mm}$

Given that the diameter of the wire is also represented as $\frac{x}{50} \mathrm{~mm}$ , we can equate this to our found total reading:

$1.42 \mathrm{~mm} = \frac{x}{50} \mathrm{~mm}$

Solving for $x$ :

$1.42 = \frac{x}{50}$

$x = 1.42 \times 50$

$x = 71$

Therefore, the value of $x$ is 71, which corresponds to Option B: 71.