JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 27)
In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by $$0.5 \mathrm{~mm}$$ on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while $$46^{\text {th }}$$ division the circular scale coincide with the reference line. The diameter of the wire is ______________ $$\times 10^{-2} \mathrm{~mm}$$.
Answer
220
Explanation
Least count of screw gauge $$ = {{0.5} \over {100}}$$ mm $$ = {{1} \over {200}}$$ mm
Zero error of screw gauge $$ = +{{6} \over {200}}$$ mm $$ = +{{3} \over {100}}=0.03$$ mm
Reading of screw gauge $$ = 4\times0.5+{{46} \over {200}}$$ mm
$$ = 2+{{23} \over {100}}$$ mm $$=2.23$$ mm
So diameter of wire $$=2.23$$ mm $$-~0.03$$ mm
$$=2.20$$ mm
$$=220\times10^{-2}$$ mm
Comments (0)
