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QJEEM22Ph1c30: JEE Main 2022 Physics - Class 11 JEEM22Ph1c30 Solution: A metal wire of length and cross-sectional area has breaking…

A metal wire of length 0.5 m0.5 \mathrm{~m} and cross-sectional area 10−4 m210^{-4} \mathrm{~m}^{2} has breaking stress 5×108 Nm−25 \times 10^{8} \,\mathrm{Nm}^{-2} . A block of 10 kg10 \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 _________ ms−1\mathrm{ms}^{-1} .

Step-by-Step Verified Solution

T=mv2ℓ=10×v20.5=20v2\mathrm{T}=\frac{\mathrm{mv}^2}{\ell}=\frac{10 \times \mathrm{v}^2}{0.5}=20 \mathrm{v}^2

 Tmax⁡=\mathrm{~T}_{\max }= Breaking stress ×\times Area

=5×108×10−4=5×104=5 \times 10^8 \times 10^{-4}=5 \times 10^4

20 V2=5×10420 \mathrm{~V}^2=5 \times 10^4

 V=14×104=50 m/s\mathrm{~V}=\sqrt{\frac{1}{4} \times 10^4}=50 \mathrm{~m} / \mathrm{s}

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