Mechanical Properties of Solids

JEE Main level, numerical answer, medium

36 of 118

Question 36 of 118

A metal block of mass mm is suspended from a rigid support through a metal wire of diameter 14mm14mm. The tensile stress developed in the wire under equilibrium state is 7×105Nm–27\times {10}^{5}N{m}^{–2}. The value of mass mm is ______ kgkg.
(Take g=9.8ms−2g=9.8m{s}^{-2} and π=227\pi =\dfrac{22}{7})

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Answer: 11

The data given is
T=7×105Nm−2r=7×10−3mT=7\times {10}^{5}N{m}^{-2} r=7\times {10}^{-3}m
Using the equation
mgA=T⇒mg=TA⇒mg=Tπr2\dfrac{mg}{A}=T \Rightarrow mg=TA \Rightarrow mg=T\pi {r}^{2}
⇒mg=7×105×227×72×10−6\Rightarrow mg=7\times {10}^{5}\times \dfrac{22}{7}\times {7}^{2}\times {10}^{-6}
⇒mg=49×2210\Rightarrow mg=\dfrac{49\times 22}{10}
⇒m=49×2298=11kg\Rightarrow m=\dfrac{49\times 22}{98}=11kg

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