JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 21)
A metal wire of length $$0.5 \mathrm{~m}$$ and cross-sectional area $$10^{-4} \mathrm{~m}^{2}$$ has breaking stress $$5 \times 10^{8} \,\mathrm{Nm}^{-2}$$. A block of $$10 \mathrm{~kg}$$ is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be _________ $$\mathrm{ms}^{-1}$$.
Answer
50
Explanation
$\mathrm{T}=\frac{\mathrm{mv}^2}{\ell}=\frac{10 \times \mathrm{v}^2}{0.5}=20 \mathrm{v}^2$
$\mathrm{~T}_{\max }=$ Breaking stress $\times$ Area
$=5 \times 10^8 \times 10^{-4}=5 \times 10^4$
$20 \mathrm{~V}^2=5 \times 10^4$
$\mathrm{~V}=\sqrt{\frac{1}{4} \times 10^4}=50 \mathrm{~m} / \mathrm{s}$
$\mathrm{~T}_{\max }=$ Breaking stress $\times$ Area
$=5 \times 10^8 \times 10^{-4}=5 \times 10^4$
$20 \mathrm{~V}^2=5 \times 10^4$
$\mathrm{~V}=\sqrt{\frac{1}{4} \times 10^4}=50 \mathrm{~m} / \mathrm{s}$
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