JEE Main 2024 — 9 April, Shift 1 — Question 79

Heat of neutralization of strong acid–strong base is constant per mole of $\mathrm{H^+}$ neutralised.

$\mathrm{HCl}$ provides 1 mole $\mathrm{H^+}$ per mole.

$\mathrm{H_2SO_4}$ provides 2 moles $\mathrm{H^+}$ per mole.

Equal volume and same molarity ⇒ equal moles of acid.

Thus heat released by $\mathrm{H_2SO_4}$ is double, hence

$$\mathrm{y = 2x}$$ $$\mathrm{\frac{y}{x} = 2}$$