Steel wire of length 'L' at C is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from C to C to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is C, Young's modulu

Question

Steel wire of length 'L' at $40 °$ C is suspended from the ceiling and then a mass 'm' is hung from its free end. The wire is cooled down from $40 °$ C to $30 °$ C to regain its original length 'L'. The coefficient of linear thermal expansion of the steel is $10^{- 5} / °$ C, Young's modulus of steel is $10^{11} {N/m}^{2}$ and radius of the wire is 1 mm. Assume that $L ≫$ diameter of the wire. Then the value of $' m'$ in kg is nearly [2011]

Smart Achievers · Accepted Answer

(3)

$We know that, Y = \frac{F / A}{Δ l / l}$

$∴$ $Y = \frac{m g / A}{Δ l / l} = \frac{m g l}{A Δ l} \dots (i)$

Also, $Δ l = l α Δ T \dots (i i)$

From (i) and (ii)

$Y = \frac{m g l}{A l α Δ T} = \frac{m g}{A α Δ T}$

$∴$ $m = \frac{Y A α Δ T}{g} = \frac{10^{11} \times π {(10^{- 3})}^{2} \times 10^{- 5} \times 10}{10} = π \approx 3$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (3) We know that, Y=F/AΔl/l ∴ Y=mg/AΔl/l=mglAΔl ⋯(i) Also, Δl=lαΔT ⋯(ii) From (i) and (ii) Y=mglAlαΔT=mgAαΔT ∴ m=YAαΔTg=1011×π(10-3)2×10-5×1010=π≈3

Want Us to Email you About Special Offers & Updates?