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A steel wire of diameter 0.8 mm and leng...

A steel wire of diameter 0.8 mm and length 1m is clamped firmly at two points P and Q which are one metre apart and in the same horizontal plane. A body is hung from the middle point of the wire, such that the middle point sags one cm lower from the original position as shown in figure. Calculate the mass of the body. `Y = 2xx 10^(11) N//m^(2)`.

Text Solution

Verified by Experts

Let T be the tension in each segment of the wire and `/_ ORQ = /_ ORP = theta`
`2T cos theta = mg `
Stress `= ( T)/( A)`, T = Stress `xxA`
`T = Y xx` Strain `xx `A
`A = pi d^(2) //4`
From the triangle POR, `PR = sqrt( PO^(2) + OR^(2)) = sqrt(2500 +1) = 50.01 `cm
Final length of the wire ` = 50.01 + 50.01 = 100.02cm`
Original length `= 100cm`
Change in length = 0.02cm
Strain `= ( 0.02)/( 100) = 2 xx 10^(-4)`
`T = Y xx ` Strain `xx A`
`= 2xx 10^(11) xx 2 xx 10^(-4) xx 3.14 xx (( 0.8xx 10^(-3))^(2))/( 4)`
= 20.096N
`m = ( 2T cos theta )/( g) =( 2 xx 20.096 xx 1)/(9.8 xx 50.01 ) = 0.082 kg`
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