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A steel wire of diameter $0.5 mm$ and Young's modulus $2 \times 10^{11} N m ^{-2}$ carries a load of mass $M$. The length of the wire with the load is $1.0 m$. A vernier scale with $10$ divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count $1.0 mm$, is attached. The $10$ divisions of the vernier scale correspond to $9$ divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $1.2 kg$, the vernier scale division which coincides with a main scale division is. . . . Take $g =10 m s ^{-2}$ and $\pi=3.2$.
$3$
$5$
$8$
$9$
Solution
Diameter $ d =0.5 mm$
$Y =2 \times 10^{11} N / m ^2$
Strain $=\frac{\text { Stress }}{ Y }$
$\frac{\Delta L }{ L }=\frac{ F }{ AY }$
$\Delta L =\frac{ FL }{ AY }=\frac{1.2 \times 10 \times 1}{3.2 \times \frac{\left(0.5 \times 10^{-3}\right)^2}{4} \times 2 \times 10^{11}}$
$\Delta L =\frac{12}{1.6 \times 0.25 \times 10^5}=30 \times 10^{-5}=0.3 mm$
$L.C.$ of Vernier scale $=0.1 mm$ Number of division which coincide with main scale $=3$.
