Explanation

(b) $V.T.$ graph is a straight line passing through origin.
Hence, $V \propto T$ or $P =$ constant
$\therefore\ \mathrm{\Delta}Q = nC_{P}\mathrm{\Delta}T$ and $\mathrm{\Delta}U = nC_{V}\mathrm{\Delta}T$
Also $\mathrm{\Delta}W = \mathrm{\Delta}Q - \mathrm{\Delta}U = \mu(C_{P} - C_{V})\mathrm{\Delta}T$
$\therefore\frac{\mathrm{\Delta}Q}{\mathrm{\Delta}W} = \frac{nC_{P}\mathrm{\Delta}T}{n(C_{P} - C_{V})\mathrm{\Delta}T} = \frac{C_{P}}{C_{P} - C_{V}} = \frac{1}{1 - \frac{C_{V}}{C_{P}}}$
$\frac{C_{V}}{C_{P}} = \frac{3}{5}$ for helium gas. Hence $\frac{\mathrm{\Delta}Q}{\mathrm{\Delta}W} = \frac{1}{1 - 3/5} = \frac{5}{2}$