JEE Main 2015 · 10 Apr · Shift 1, Physics question 11

Each molecule delivers momentum $2mv_x$ per collision, at a rate $\frac{v_x}{2L}$, so the average force is $\dfrac{mv_x^2}L\propto T$. That gives $q=1$.

The key gives (3), $q=\frac12$, which is what you get by taking the force per impact, $\propto v\propto\sqrt T$. The solution follows the key and records the other reading.