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Experimental Physics

JEE Main level, single correct, medium

27 of 70

Question 27 of 70

In an experiment to find out the diameter of wire using screw gauge, the following observation were noted:

(a) Screw moves $0.5mm$ on main scale in one complete rotation
(b) Total divisions on circular scale $=50$
(c) Main scale reading is $2.5mm$
(d) ${45}^{th}$ ${}^{}$ division of circular scale is in the pitch line
(e) Instrument has $0.03mm$ negative error Then the diameter of wire is :

Show answer and solution

Answer: (C) $2.98mm$

From the data given in the question,
Pitch of the screw gauge

=0.5mm

.
Total divisions on the circular scale

n=50

.
Therefore, the least count of the screw gauge is

LC=\dfrac{\text{pitch}}{n}=\dfrac{0.5}{50}=0.01mm

Now,

MSR=2.5mm

CSR=45

Diameter reading

=

MSR+LC\times CSR-

zero error

\Rightarrow d=2.5+0.45-\left(-0.03\right)

\Rightarrow d=2.98mm

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