A steel wire of diameter 0.5 text{ mm} and Yo | vedclass

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(A) The change in length $\Delta L$ is given by the formula $\Delta L = \frac{FL}{AY}$.Given: $F = mg = 1.2 	imes 10 = 12 	ext{ N}$,$L = 1.0 	ext{

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$3$

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(A) The change in length $\Delta L$ is given by the formula $\Delta L = \frac{FL}{AY}$.Given: $F = mg = 1.2 	imes 10 = 12 	ext{ N}$,$L = 1.0 	ext{ m}$,$d = 0.5 	imes 10^{-3} 	ext{ m}$,$Y = 2 	imes 10^{11} 	ext{ N m}^{-2}$,and $\pi = 3.2$.Area $A = \pi r^2 = \pi \left(\frac{d}{2}ight)^2 = 3.2 	imes \frac{(0.5 	imes 10^{-3})^2}{4} = 0.8 	imes 0.25 	imes 10^{-6} = 0.2 	imes 10^{-6} 	ext{ m}^2$.Calculating $\Delta L$: $\Delta L = \frac{12 	imes 1.0}{0.2 	imes 10^{-6} 	imes 2 	imes 10^{11}} = \frac{12}{0.4 	imes 10^5} = 30 	imes 10^{-5} 	ext{ m} = 0.3 	ext{ mm}$.The Least Count $(LC)$ of the vernier scale is $LC = 	ext{Main Scale Division} - 	ext{Vernier Scale Division} = 1.0 	ext{ mm} - \frac{9}{10} 	imes 1.0 	ext{ mm} = 0.1 	ext{ mm}$.The reading on the vernier scale is given by $\Delta L = n 	imes LC$,where $n$ is the vernier division coinciding with the main scale.$0.3 	ext{ mm} = n 	imes 0.1 	ext{ mm} \implies n = 3$.

Similar Questions

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