A 1 m long steel wire of cross-sectional area 1 mm2 is extended by 1 mm- If Y - 2 - 1011 Nm-2- then the work done is

The correct option is B 0.2 J L = 1 m A = 1 mm^2 = 1 × 10^ 6 m^2 Δ L = 1 mm = (1 × 10^ 3)m Y = 2 × 10^11 N/m^2 Y = F × LA ×Δ L ...

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Standard XII

Physics

COE in SHM

A 1 m long ...

Question

A $1 m$ long steel wire of cross-sectional area $1 m m^{2}$ is extended by $1 m m$ . If $Y = 2 \times 10^{11} N m^{- 2}$ , then the work done is

0.1 J

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0.2 J

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0.3 J

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0.4 J

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Solution

The correct option is B $0.2 J$
$L = 1 m$
$A = 1 m m^{2} = 1 \times 10^{- 6} m^{2}$
$Δ L = 1 m m = (1 \times 10^{- 3}) m$
$Y = 2 \times 10^{11} N / m^{2}$
$Y = \frac{F \times L}{A \times Δ L}$
$F = \frac{Y A Δ L}{L}$
$F = \frac{2 \times 10^{11} \times 1 \times 10^{- 6} \times 1 \times 10^{- 3}}{1}$
$F = 200 N$
Here, work done $= F \times S$ $S = Δ L$ (change in length)
$= 200 \times 1 \times 10^{- 3}$
$= 0.2 J$

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A 1m long steel wire of cross-sectional area 1mm2 is extended by 1mm. If Y=2×1011Nm−2, then the work done is

The correct option is B 0.2JL=1mA=1mm2=1×10−6m2ΔL=1mm=(1×10−3)mY=2×1011N/m2Y=F×LA×ΔLF=YAΔLLF=2×1011×1×10−6×1×10−31F=200NHere, work done =F×S S=ΔL(change in length)=200×1×10−3=0.2J

A $1 m$ long steel wire of cross-sectional area $1 m m^{2}$ is extended by $1 m m$ . If $Y = 2 \times 10^{11} N m^{- 2}$ , then the work done is