A steel rod has a radius of 50 mm and a leng | vedclass

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(C) The formula for Young's modulus $Y$ is given by $Y = \frac{FL}{A \Delta L}$.Given values: Force $F = 400 \ kN = 400 	imes 10^3 \ N$, Length $L =

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$2 	imes 10^{11} \ N/m^2$

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(C) The formula for Young's modulus $Y$ is given by $Y = \frac{FL}{A \Delta L}$.Given values: Force $F = 400 \ kN = 400 	imes 10^3 \ N$, Length $L = 2 \ m$, Radius $r = 50 \ mm = 50 	imes 10^{-3} \ m$, Elongation $\Delta L = 0.5 \ mm = 0.5 	imes 10^{-3} \ m$.The area of cross-section $A = \pi r^2 = \pi 	imes (50 	imes 10^{-3})^2 = \pi 	imes 2500 	imes 10^{-6} = 2.5 \pi 	imes 10^{-3} \ m^2 \approx 7.85 	imes 10^{-3} \ m^2$.Substituting these values into the formula:$Y = \frac{400 	imes 10^3 	imes 2}{(2.5 \pi 	imes 10^{-3}) 	imes (0.5 	imes 10^{-3})}$$Y = \frac{800 	imes 10^3}{1.25 \pi 	imes 10^{-6}} = \frac{800}{1.25 \pi} 	imes 10^9 \approx \frac{640}{3.14} 	imes 10^9 \approx 203.8 	imes 10^9 \approx 2 	imes 10^{11} \ N/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

$A$ steel rod has a radius of $50 \ mm$ and a length of $2 \ m$. It is stretched along its length with a force of $400 \ kN$. This causes an elongation of $0.5 \ mm$. Find the (approximate) Young's modulus of steel from this information.

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Young's moduli of two wires $A$ and $B$ are in the ratio $7 : 4$. Wire $A$ is $2\, m$ long and has radius $R$. Wire $B$ is $1.5\, m$ long and has radius $2\, mm$. If the two wires stretch by the same length for a given load,then the value of $R$ is close to ......... $mm$.

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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