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Physics - NLM Question with Solution | TestHub

PhysicsNLMMiscellaneousMedium2 minQB

PhysicsMediumnumerical

Passage / Comprehension

I am a vernier caliper. I have a task to measure the length of a pencil as shown in the figure. The vernier constant of my vernier scale is $0.05$ times the value of one division of the main scale.

My vernier scale has $n$ divisions and coincides with $(n - 4)$ divisions of the main scale. While measuring the length of the pencil, the $\frac{n}{5}^{t h}$ division of the Vernier scale exactly coincides with a division of the main scale. Neglect any zero error.

The length of the pencil is $____$ cm (In nearest integers)

Answer:

8.00

Solution:

L.C $= \frac{1}{20} \times 0.1 = 0.005 cm$ Let 1 V.S.D $= x \Rightarrow \frac{1}{20} = 1 - x \Rightarrow x = 0.95 mm$

$(n - 4) \times 1 = n \times 0.95 \Rightarrow n = 80 Length of pencil = 8.4 + 0.005 \times (\frac{80}{5}) = 8.48 c m$

Stream:JEESubject:PhysicsTopic:NLMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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