JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 20)
A string of area of cross-section $$4 \mathrm{~mm}^{2}$$ and length $$0.5 \mathrm{~m}$$ is connected with a rigid body of mass $$2 \mathrm{~kg}$$. The body is rotated in a vertical circular path of radius $$0.5 \mathrm{~m}$$. The body acquires a speed of $$5 \mathrm{~m} / \mathrm{s}$$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is _________ $$ \times 10^{-5}$$.
(use Young's modulus $$10^{11} \mathrm{~N} / \mathrm{m}^{2}$$ and $$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$$)
Explanation
$$A = 4 \times {10^{ - 6}}$$ m2
$$l = 0.5$$ m
$$m = 2$$ kg
$${v_b} = 5$$ m/s
$${T_b} = mg + m\left( {{{V_b^2} \over l}} \right)$$
$$ = 20 + 2 \times {{25} \over {{1 \over 2}}} = 120$$ N
$${{\Delta l} \over l} = {{{T_b}} \over A} \times {1 \over Y} = {{120} \over {4 \times {{10}^{ - 6}}}} \times {10^{ - 11}} = 30 \times {10^{ - 5}}$$
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