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JEE Main 20236 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2023 · 6 April · Evening Shift · Physics · Mechanical Properties of Solids

A metal block of mass m is suspended from a rigid support through a...

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})

Correct answer

0

Step-by-step solution

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 ⇒ mg = TA ⇒ mg = T\pi{r}^{2}

⇒mg=7×105×227×72×10−6⇒ mg = 7 \times {10}^{5} \times \dfrac{22}{7} \times {7}^{2} \times {10}^{- 6}

⇒mg=49×2210⇒ mg = \dfrac{49 \times 22}{10}

⇒m=49×2298=11kg⇒ m = \dfrac{49 \times 22}{98} = 11kg

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