A string of length 0.314 text{ m} and Young's | vedclass

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(B) The total extension $\Delta L$ is the sum of extensions of both strings: $\Delta L = \Delta L_A + \Delta L_B = \frac{F L_A}{Y_A A} + \frac{F L_B}{

Vedclass · Accepted Answer

$2$

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(B) The total extension $\Delta L$ is the sum of extensions of both strings: $\Delta L = \Delta L_A + \Delta L_B = \frac{F L_A}{Y_A A} + \frac{F L_B}{Y_B A}$.Given $F = mg = 0.8 	imes 10 = 8 	ext{ N}$.For string $A$: $L_A = 0.314 	ext{ m}$, $Y_A = 2 	imes 10^{10} 	ext{ N/m}^2$.For string $B$: $L_B = 2 L_A = 0.628 	ext{ m}$, $Y_B = 2 Y_A = 4 	imes 10^{10} 	ext{ N/m}^2$.The cross-sectional area $A = \pi r^2 = 3.14 	imes (0.2 	imes 10^{-3} 	ext{ m})^2 = 3.14 	imes 4 	imes 10^{-8} = 1.256 	imes 10^{-7} 	ext{ m}^2$.Calculating $\Delta L_A$: $\Delta L_A = \frac{8 	imes 0.314}{2 	imes 10^{10} 	imes 1.256 	imes 10^{-7}} = \frac{2.512}{2512} = 0.001 	ext{ m} = 1 	ext{ mm}$.Calculating $\Delta L_B$: $\Delta L_B = \frac{8 	imes 0.628}{4 	imes 10^{10} 	imes 1.256 	imes 10^{-7}} = \frac{5.024}{5024} = 0.001 	ext{ m} = 1 	ext{ mm}$.Total extension $\Delta L_{total} = \Delta L_A + \Delta L_B = 1 	ext{ mm} + 1 	ext{ mm} = 2 	ext{ mm}$.

Column $I$	Column $II$
$A$. Hooke's law	$1$. Tangential strain
$B$. Shearing strain	$2$. Temporary loss of elastic property
$C$. Bulk strain	$3$. Elastic limit
$D$. Elastic fatigue	$4$. $3$ times the linear strain

Similar Questions

Which of the following statements is incorrect?

$A$ metal wire is clamped between two vertical walls. At $20 ^o C$,the unstrained length of the wire is exactly equal to the separation between the walls. If the temperature of the wire is decreased,the graph between elastic energy density $(u)$ and temperature $(T)$ of the wire is:

Match the following:
Column $I$ Column $II$
$A$. Hooke's law $1$. Tangential strain
$B$. Shearing strain $2$. Temporary loss of elastic property
$C$. Bulk strain $3$. Elastic limit
$D$. Elastic fatigue $4$. $3$ times the linear strain

$A$ wire can be broken by applying a load of $200\, N$. The force required to break another wire of the same length and same material,but double in diameter,is .......... $N$.

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