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Standard XII

Physics

V-t to a-t Graph

A 10 mm diame...

Question

A 10 mm diameter mild steel bar of 1.5 m is stressed by a weight of 120 N dropping freely through a distance of 20 mm before commencing to strech the bar. The maximum instantaneous elongation in bar $E = 2 \times 10^{5} M P a$ .

1.95 mm

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0.688 mm

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2.037 mm

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3.03 mm

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Solution

The correct option is B 0.688 mm
$d = 10 m m; l = 1500 m m; W = 120 N; h = 20 m m$

$E = 2 \times 10^{5} M P a$ $A = \frac{π}{4} d^{2} = 78.54 m m^{2}$

$(σ_{i n s t})_{m a x} = \frac{W}{A} [1 + \sqrt{1 + \frac{2 A E h}{W L}}]$

$= \frac{120}{78.54} ⎡ ⎣ 1 + \sqrt{1 + \frac{78.54 \times 2 \times 10^{5} \times 20}{120 \times 1500}} ⎤ ⎦$

$= 91.86 M P a$

$∴ δ L_{i n s t} = \frac{σ_{i n s t} L}{E} = \frac{91.86 \times 1500}{2 \times 10^{5}} = 0.688 m m$

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A 10 mm diameter mild steel bar of 1.5 m is stressed by a weight of 120 N dropping freely through a distance of 20 mm before commencing to strech the bar. The maximum instantaneous elongation in bar E=2×105 MPa.

The correct option is B 0.688 mmd=10 mm;l=1500 mm;W=120 N;h=20 mm E=2×105 MPa A=π4d2=78.54 mm2 (σinst)max=WA[1+√1+2AEhWL] =12078.54⎡⎣1+√1+78.54×2×105×20120×1500⎤⎦ =91.86 MPa ∴δLinst=σinstLE=91.86×15002×105=0.688 mm

A 10 mm diameter mild steel bar of 1.5 m is stressed by a weight of 120 N dropping freely through a distance of 20 mm before commencing to strech the bar. The maximum instantaneous elongation in bar $E = 2 \times 10^{5} M P a$ .