Mechanical Properties of Solids

JEE Main level, numerical answer, medium

43 of 118

Question 43 of 118

A metal wire of length 0.5m0.5m and cross-sectional area 10−4m2{10}^{-4}{m}^{2} has breaking stress 5×108Nm−25\times {10}^{8}N{m}^{-2}. A block of 10kg10kg is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be _____ ms−1m{s}^{-1}.

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Answer: 50

Here in horizontal circular motion, the only centripetal force is tension. So,T=mω2r=mv2l=10×v20.5=20v2T=m{\omega }^{2}r=\dfrac{m{v}^{2}}{l}=\dfrac{10\times {v}^{2}}{0.5}=20{v}^{2}
Now, Tmax⁡={T}_{\max }= Breaking stress ×\times Area
⇒20v2=5×108×10−4\Rightarrow 20{v}^{2}=5\times {10}^{8}\times {10}^{-4}
⇒20v2=5×104\Rightarrow 20{v}^{2}=5\times {10}^{4}
Thus, the maximum linear velocity of block is
v=14104=50ms−1v=\sqrt{\dfrac{1}{4}{10}^{4}}=50m{s}^{-1}

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