JEE MAIN - Physics (2024 - 6th April Evening Shift - No. 30)
A wire of cross sectional area A, modulus of elasticity $$2 \times 10^{11} \mathrm{~Nm}^{-2}$$ and length $$2 \mathrm{~m}$$ is stretched between two vertical rigid supports. When a mass of $$2 \mathrm{~kg}$$ is suspended at the middle it sags lower from its original position making angle $$\theta=\frac{1}{100}$$ radian on the points of support. The value of A is ________ $$\times 10^{-4} \mathrm{~m}^2$$ (consider $$x<<\mathrm{L}$$ ).
(given : $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$$)
_6th_April_Evening_Shift_en_30_1.png)
Explanation
$$\begin{aligned} \Delta I_{\text {spring }} & =\left(\sqrt{L^2+x^2}\right)-L \\ & =\frac{x^2}{2 L} \\ T=k \Delta I & =\frac{k n^2}{2 L} \end{aligned}$$
_6th_April_Evening_Shift_en_30_2.png)
$$\begin{array}{rl} &\Rightarrow 2 T \theta=m g \\ & 2\left(\frac{Y A}{l}\right) \frac{x^2}{2 l} \theta=m g \\ & A =\frac{m g}{\theta^3 Y} \\ & =\frac{20}{10^{-6} \times 2 \times 10^{11}} \\ & =10 \times 10^{-5} \\ & =1 \times 10^{-4} \end{array}$$
Comments (0)
