Given,

Pressure acting on the glass plate, p = 10 atm

= 10 × 1.013 × 10^{5} Pa

We know,

Bulk modulus of glass, B = 37 × 10^{9} Nm^{–2}

= Bulk modulus, B = p / (∆V/V)

Where,

\( \dfrac{∆V}{V}\) = Fractional change in volume

\( \dfrac{∆V}{V}\) = \( \dfrac{p}{B}\)

= \( \dfrac{10 × 1.013 × 10^5}{(37 × 10^9)}\)

= 2.73 × 10 ^{-4}

Therefore, the fractional change in the volume of the glass plate is 2.73 × 10^{–4}.

A mild steel wire of cross-sectional area 0.60 x 10 ^{-2} cm^{2} and length 2 m is stretched ( not beyond its elastic limit ) horizontally between two columns. If a 100g mass is hung at the midpoint of the wire, find the depression at the midpoint.

The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of the water. The water pressure at the bottom of the trench is about 1.1 × 10^{8} Pa. A steel ball of initial volume 0.32 m^{3 }is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches the bottom?

Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 10^{7 }Pa? Assume that each rivet is to carry one-quarter of the load.

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.

A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in the figure. The cross-sectional areas of wires A and B are 1.0 mm^{2 }and 2.0 mm^{2}, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.