Explanation

(d) Let L be the length of each side of cube. Initial volume of cube $= L^{3}.$ When each side of cube decreases by 2%, the new length
$L^{'} = L - \frac{2L}{100} = \frac{98L}{100}$
New volume$\ = {L^{'}}^{3} = (98L/100)^{3}$
$\therefore Change\ in\ volume$
$\mathrm{\Delta}V = L^{3} - (98L/100)^{3}$
$= L^{3}\left\lceil 1 - \left( 1 - \frac{2}{100} \right)^{3} \right\rceil$
$= L^{3}\left\lceil 1 - \left( 1 - \frac{6}{100} + \ \ldots \right) \right\rceil$
(from binomial expension)
$= L^{3}\left\lbrack \frac{6}{100} \right\rbrack = \frac{{6L}^{3}}{100}$
$\therefore\ \ Bulk\ strain = \frac{\mathrm{\Delta}V}{V} = \frac{{6L}^{3}/100}{L^{3}} = 0.06$