Overall changes in volume and radii of a unif | vedclass

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(D) The volume of a cylinder is given by $V = \pi r^2 L$.
Taking the logarithmic derivative,we get $\frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac

Vedclass · Accepted Answer

$4.08 	imes 10^8 \ N/m^2$

Vedclass · Answer

(D) The volume of a cylinder is given by $V = \pi r^2 L$.
Taking the logarithmic derivative,we get $\frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac{\Delta L}{L}$.
Given: $\frac{\Delta V}{V} = 0.2 \% = 0.002$ and $\frac{\Delta r}{r} = -0.002 \% = -0.00002$ (since radius decreases when stretched).
Substituting these values: $0.002 = 2(-0.00002) + \frac{\Delta L}{L}$.
$0.002 = -0.00004 + \frac{\Delta L}{L} \implies \frac{\Delta L}{L} = 0.00204$.
Stress $\sigma = Y 	imes 	ext{strain} = Y 	imes \frac{\Delta L}{L}$.
$\sigma = (2.0 	imes 10^{11}) 	imes (0.00204) = 4.08 	imes 10^8 \ N/m^2$.
Note: The provided options seem to contain a typo in the exponent or magnitude. Based on the calculation,the result is $4.08 	imes 10^8 \ N/m^2$.

Overall changes in volume and radii of a uniform cylindrical steel wire are $0.2 \%$ and $0.002 \%$ respectively when subjected to some suitable force. Calculate the longitudinal tensile stress acting on the wire. (Given: Young's modulus $Y = 2.0 \times 10^{11} \ N/m^2$)

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