A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10-2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Asked by Abhisek | 1 year ago |  272

##### Solution :-

Given,

Water pressure at the bottom, p = 1000 atm

= 1000 x 1.013 x 105 Pa

p = 1.01 x 108 Pa

Initial volume of the steel ball, V = 0.30 m3

We know, bulk modulus of steel, B = 1.6 × 1011 Nm–2

Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.

Bulk modulus, B

$$\dfrac{p}{(∆V/V)}$$

∆V  =  $$\dfrac{pV}{B}$$

$$\dfrac{1.01 × 10^8 × 0.30}{(1.6 × 10^{11 }) }$$

=  1.89 × 10-4 m3

Hence, volume of the ball changes by 1.89 × 10-4 m3 on reaching the bottom of the trench.

Answered by Abhisek | 1 year ago

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