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(B) The vertical deflection $x$ of a cantilever rod due to a load $F = mg$ at its end is related to the shear modulus $\eta$ by the formula:$\eta = \f

Vedclass · Accepted Answer

$0.02$

Vedclass · Answer

(B) The vertical deflection $x$ of a cantilever rod due to a load $F = mg$ at its end is related to the shear modulus $\eta$ by the formula:$\eta = \frac{F L}{A x} \Rightarrow x = \frac{mg L}{A \eta}$Where $L = 6 	ext{ cm} = 6 	imes 10^{-2} 	ext{ m}$, radius $r = 2 	ext{ cm} = 2 	imes 10^{-2} 	ext{ m}$, $m = 400 \pi 	ext{ kg}$, $g = 10 	ext{ m/s}^2$, and $\eta = 3 	imes 10^{10} 	ext{ N/m}^2$.The cross-sectional area $A = \pi r^2 = \pi (2 	imes 10^{-2})^2 = 4 \pi 	imes 10^{-4} 	ext{ m}^2$.Substituting these values into the formula:$x = \frac{(400 \pi 	imes 10) 	imes (6 	imes 10^{-2})}{(4 \pi 	imes 10^{-4}) 	imes (3 	imes 10^{10})}$$x = \frac{4000 \pi 	imes 6 	imes 10^{-2}}{12 \pi 	imes 10^6}$$x = \frac{24000 \pi 	imes 10^{-2}}{12 \pi 	imes 10^6} = \frac{240 \pi}{12 \pi 	imes 10^6} = 20 	imes 10^{-6} 	ext{ m}$$x = 20 	imes 10^{-3} 	ext{ mm} = 0.02 	ext{ mm}$.Thus, the correct option is $B$.

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