A steel wire of length 3 . 2 m Y S = 2 . 0 × 10 11 N m - 2 and a copper wire of length 4 . 4 m Y C = 1 . 1 × 10 11 N m - 2 , both of radius 1 . 4 mm are connected end to end. When stretched by a load,

Question

A steel wire of length 3 . 2 m Y S = 2 . 0 × 10 11 N m - 2 and a copper wire of length 4 . 4 m Y C = 1 . 1 × 10 11 N m - 2 , both of radius 1 . 4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1 . 4 mm . The load applied, in Newton, will be: (Given π = 22 7 )

TestHub · Accepted Answer

When two wires are connected end to end, the tension in them will be same.
Change in length using Hooke's law can be found as,
$Y = \frac{F l}{A ∆ l} \Rightarrow Δ l = \frac{F l}{A Y}$
Total change in length can be written as,
$Δ l = Δ l_{1} + Δ l_{2}$
$\Rightarrow Δ l = \frac{F l_{1}}{A_{1} Y_{1}} + \frac{F l_{2}}{A_{2} Y_{2}}$
$\Rightarrow F = \frac{Δ l}{\frac{l_{1}}{A_{1} Y_{1}} + \frac{l_{2}}{A_{2} Y_{2}}} = \frac{1.4 \times 10^{- 3}}{\frac{3.2}{π {(1.4 \times 10^{- 3})}^{2} \times 2.0 \times 10^{11}} + \frac{4.4}{π {(1.4 \times 10^{- 3})}^{2} \times 1.1 \times 10^{11}}} \Rightarrow F = 1.54 \times 10^{2} = 154 N$

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