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Q.

What mass must be suspended from the free end of a tungsten wire of length 50cm and diameter 0.25mm to stretch it by 1mm (Y of tungsten =3.448×1011Nm−2 )

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a

3.38335kg

b

4.445kg

c

1.445kg

d

2.445kg

answer is A.

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Detailed Solution

y=mgℓAe⇒m=y(πr2)egℓm=3.448×1011×π(0.125)2×10-3×10-610×50×10-2=3.38335 kg

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What mass must be suspended from the free end of a tungsten wire of length 50cm and diameter 0.25mm to stretch it by 1mm (Y of tungsten =3.448×1011Nm−2 )