A metal wire of length 0.5; m and cross-secti | vedclass

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(D) The tension $T$ in the wire provides the necessary centripetal force for the circular motion: $T = \frac{mv^{2}}{\ell}$.Given $m = 10\; kg$ and $\

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$50$

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(D) The tension $T$ in the wire provides the necessary centripetal force for the circular motion: $T = \frac{mv^{2}}{\ell}$.Given $m = 10\; kg$ and $\ell = 0.5\; m$,we have $T = \frac{10 	imes v^{2}}{0.5} = 20v^{2}$.The maximum tension the wire can withstand is determined by the breaking stress: $T_{\max} = 	ext{Breaking Stress} 	imes 	ext{Area}$.$T_{\max} = (5 	imes 10^{8}\; N/m^{2}) 	imes (10^{-4}\; m^{2}) = 5 	imes 10^{4}\; N$.Equating the tension to the maximum breaking force: $20v^{2} = 5 	imes 10^{4}$.$v^{2} = \frac{5 	imes 10^{4}}{20} = 0.25 	imes 10^{4} = 2500$.$v = \sqrt{2500} = 50\; m/s$.

List-$I$	List-$II$
$A$. Young's Modulus	$I$. $\frac{Ad}{\Delta L}$
$B$. Compressibility	$II$. $\frac{FL}{A\Delta L}$
$C$. Bulk Modulus	$III$. $-\frac{1}{\Delta P}(\frac{\Delta V}{V})$
$D$. Poisson's Ratio	$IV$. $-\frac{\Delta D/D}{\Delta L/L}$

Similar Questions

Match List-$I$ with List-$II$:

List-$I$ List-$II$

$A$. Young's Modulus $I$. $\frac{Ad}{\Delta L}$

$B$. Compressibility $II$. $\frac{FL}{A\Delta L}$

$C$. Bulk Modulus $III$. $-\frac{1}{\Delta P}(\frac{\Delta V}{V})$

$D$. Poisson's Ratio $IV$. $-\frac{\Delta D/D}{\Delta L/L}$

Mark the wrong statement.

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