A wire of cross-sectional area A,modulus of e | vedclass

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(C) For the equilibrium of the mass in the vertical direction,the vertical components of the tension $T$ in the wire must balance the weight of the ma

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$1$

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(C) For the equilibrium of the mass in the vertical direction,the vertical components of the tension $T$ in the wire must balance the weight of the mass:$2T \sin 	heta = mg$Given $m = 2 	ext{ kg}$,$g = 10 	ext{ m/s}^2$,and $	heta = \frac{1}{100} 	ext{ rad}$. Using the small angle approximation $\sin 	heta \approx 	heta$:$2T 	heta = 20$$T = \frac{10}{	heta} = \frac{10}{1/100} = 1000 	ext{ N}$The original length of the wire is $2L = 2 	ext{ m}$,so $L = 1 	ext{ m}$.The new length of the wire is $2 \sqrt{L^2 + x^2}$,where $x = L 	an 	heta \approx L 	heta$.The change in length $\Delta L$ is:$\Delta L = 2 \sqrt{L^2 + x^2} - 2L = 2L \left( \sqrt{1 + 	an^2 	heta} - 1 ight) \approx 2L \left( 1 + \frac{	an^2 	heta}{2} - 1 ight) = L 	an^2 	heta \approx L 	heta^2$$\Delta L = 1 	imes (1/100)^2 = 10^{-4} 	ext{ m}$Using Young's modulus $Y = \frac{T/A}{\Delta L / (2L)}$:$2 	imes 10^{11} = \frac{1000 / A}{10^{-4} / 2}$$2 	imes 10^{11} = \frac{2000}{A 	imes 10^{-4}}$$A = \frac{2000}{2 	imes 10^{11} 	imes 10^{-4}} = \frac{1000}{10^7} = 10^{-4} 	ext{ m}^2$Thus,the value of $A$ is $1 	imes 10^{-4} 	ext{ m}^2$.

Column $-I$	Column $-II$
$(a)$ When temperature is raised,Young's modulus of a body	$(i)$ Zero
$(b)$ Young's modulus for air	$(ii)$ Infinite
	$(iii)$ Decreases
	$(iv)$ Increases

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Column $-II$ is related to Column $-I$. Join them appropriately:

Column $-I$ Column $-II$

$(a)$ When temperature is raised,Young's modulus of a body $(i)$ Zero

$(b)$ Young's modulus for air $(ii)$ Infinite

$(iii)$ Decreases

$(iv)$ Increases

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