(A) The stress at a distance $x$ from the free end is $\sigma = \frac{(\rho A x) g}{A} = \rho g x$.The elongation $d\ell$ in a small element $dx$ is g

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$Y = \frac{\rho g L^2}{2 \ell}$

(A) The stress at a distance $x$ from the free end is $\sigma = \frac{(\rho A x) g}{A} = \rho g x$.The elongation $d\ell$ in a small element $dx$ is given by $d\ell = \frac{\sigma dx}{Y} = \frac{\rho g x dx}{Y}$.Integrating from $x = 0$ to $x = L$,the total elongation $\ell$ is:$\ell = \int_0^L \frac{\rho g x}{Y} dx = \frac{\rho g}{Y} \left[ \frac{x^2}{2} \right]_0^L = \frac{\rho g L^2}{2Y}$.Rearranging for Young's modulus $Y$,we get:$Y = \frac{\rho g L^2}{2 \ell}$.

Class 11 Physics Mechanical Properties of Solids Young’s Modulus

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$A$ thick brass wire of length $L$ and density $\rho$ is suspended from a rigid support. Due to its own weight,$\ell$ is the increase in length. The Young's modulus $Y$ of the brass wire in terms of density is $(g = \text{acceleration due to gravity})$

MHT CET 2020 All MHT CET

A
$Y = \frac{\rho g L^2}{2 \ell}$
B
$Y = \frac{\rho g L^2}{4 \ell}$
C
$Y = \frac{\rho g L}{\ell}$
D
$Y = \frac{\rho g L^2}{\ell}$

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$A$ thick brass wire of length $L$ and density $\rho$ is suspended from a rigid support. Due to its own weight,$\ell$ is the increase in length. The Young's modulus $Y$ of the brass wire in terms of density is $(g = \text{acceleration due to gravity})$

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$5 \,m$ long aluminium wire $(Y = 7 \times 10^{10} \,N/m^2)$ of diameter $3 \,mm$ supports a $40 \,kg$ mass. In order to have the same elongation in a copper wire $(Y = 12 \times 10^{10} \,N/m^2)$ of the same length under the same weight, the diameter should be (in $mm$):

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