JEE Main 2026 — Physics Mechanics

Least count of the screw gauge is given by:

$$\text{LC} = \dfrac{\text{Pitch}}{\text{Number of divisions on circular scale}}$$LC=Number of divisions on circular scalePitch​

$$\text{LC} = \dfrac{0.1}{100} = 0.001 \text{ mm}$$LC=1000.1​=0.001 mm

When the two studs are in contact, the $5^{\text{th}}$5th division of the circular scale coincides with the reference line. This indicates a positive zero error.

$$\text{Zero error} = +5 \times \text{LC} = +5 \times 0.001 = +0.005 \text{ mm}$$Zero error=+5×LC=+5×0.001=+0.005 mm

For the measurement of the sphere's diameter:

$$\text{Main scale reading (MSR)} = 5 \text{ mm}$$Main scale reading (MSR)=5 mm

$$\text{Circular scale reading (CSR)} = 50$$Circular scale reading (CSR)=50

$$\text{Measured diameter} = \text{MSR} + \text{CSR} \times \text{LC}$$Measured diameter=MSR+CSR×LC

$$\text{Measured diameter} = 5 + 50 \times 0.001 = 5.050 \text{ mm}$$Measured diameter=5+50×0.001=5.050 mm

The true diameter is obtained by subtracting the zero error from the measured value:

$$\text{True diameter} = \text{Measured diameter} - \text{Zero error}$$True diameter=Measured diameter−Zero error

$$\text{True diameter} = 5.050 - 0.005 = 5.045 \text{ mm}$$True diameter=5.050−0.005=5.045 mm

Answer: $5.045$5.045