Question

Two full turns of the circular scale of a screw gauge cover a distance of $1mm$ on its main scale. The total number of divisions of $1mm$on its main scale. The total number of divisions on the circular scale is $50$. Further, it is found that the screw guage has a zero error of $-0.03mm$. While measuring the diameter of a thin wire, a student notes the main scale reading of $3mm$and the number of circular scale divisions in line with the main scale as $35$. The diameter of the wire is

• A. $3.32mm$
• B. $3.73mm$
• C. $3.67mm$
• D. $3.88mm$

Answer 1

The correct option is D

$3.88mm$

Step 1: Given;

Main scale reading = $3mm$

circular scale reading= $35$

Total number of divisions on the circular scale= $50$

Zero error= $-0.03mm$ (negative sign means that the zero of the vernier scale lies before the zero of the main scale.)

Step 2: Formula used:

$L.C.=\frac{pitch}{totalnumberofdivisionsonthecircularscale}$

$Reading=[Mainscalereading+circularscalereading\times L.C.]-(zeroerror)$

Step 3: Find Pitch;

$2$ rounds = $1mm$

$1$ round = $\frac{1}{2}=0.5mm$

$\Rightarrow$Pitch= $0.5mm$

Step 4: Find the Least count (L.C.);

$L.C.=\frac{0.5}{50}mm=0.01mm$

Step 5: Find the reading of the diameter of the wire;

$Reading=[3+35\times 0.01]-(-0.03)=3.88mm$

Hence, The diameter of the wire is $3.88mm$.

Thus, option D is correct.