JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Elasticity & SHM

A metal block of mass

m

is suspended from a rigid support through a metal wire of diameter 14 mm. The tensile stress developed in the wire under equilibrium state is

7 \times 1 0^{5} Nm^{- 2}

. The value of mass

m

is ______ kg.
(Take,

g = 9.8 ms^{- 2}

and

π = \frac{22}{7}

)

Answer: 11

Solution

Tensile stress is

σ = \frac{F}{A} = \frac{m g}{π D ^{2} /4} = \frac{4 m g}{π D ^{2}}

, so

m = \frac{π D ^{2} σ}{4 g}

. With

D = 14 \times 1 0^{- 3}

σ = 7 \times 1 0^{5}

N m

^{- 2}

π = \frac{22}{7}

and

g = 9.8

m s

^{- 2}

m = \frac{\frac{22}{7} \times ( 14 \times 1 0 ^{- 3} ) ^{2} \times 7 \times 1 0 ^{5}}{4 \times 9.8}

. The mass is

11

kg.

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer