JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 11)
Explanation
Determine one main scale division (msd):
$ 300 \, \text{msd} = 15 \, \text{cm} $
Therefore,
$ 1 \, \text{msd} = \frac{15}{300} \, \text{cm} = 0.05 \, \text{cm} $
Determine one vernier scale division (vsd):
$ 25 \, \text{vsd} = 24 \, \text{msd} $
Therefore,
$ 1 \, \text{vsd} = \frac{24}{25} \, \text{msd} $
Calculate the least count (LC):
The least count is given by the difference between one main scale division and one vernier scale division:
$ \text{LC} = 1 \, \text{msd} - 1 \, \text{vsd} $
Substitute the expression for 1 vsd:
$ \text{LC} = 1 \, \text{msd} - \frac{24}{25} \times 1 \, \text{msd} $
This simplifies to:
$ \text{LC} = \frac{1}{25} \times 1 \, \text{msd} $
Calculate the LC in cm:
Substituting the value of 1 msd:
$ \text{LC} = \frac{1}{25} \times 0.05 \, \text{cm} = 0.002 \, \text{cm} $
Hence, the least count of the traveling microscope is 0.002 cm.
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