AIIMS 2015 Physics State Government Schemes MCQ Question

Step 1: Find the height of the nozzle exitThe nozzle has a length $L$ and is inclined at an angle $\alpha = 30^\circ$ above the reference line $XX'$. The vertical height ($y_0$) of the nozzle's exit mouth above the line $XX'$ is:$y_0 = L \sin\alpha = 2 \times \sin(30^\circ) = 2 \times 0.5 = 1\text{ m}$Step 2: Find the velocity of efflux ($v$)The depth of the exit nozzle below the free surface of the water is $h - y_0$. Applying Torricelli's Law, the velocity of the water leaving the nozzle is:v = \sqrt{2g(h - L \sin\alpha)}$$$$v^2 = 2g(10 - 1) = 2g(9) = 18gStep 3: Determine the maximum height reached by the water projectileOnce the water leaves the nozzle, it acts as a projectile launched at an angle $\alpha = 30^\circ$. The maximum vertical height ($H_{\text{proj}}$) it can reach above its launch point is given by:$H_{\text{proj}} = \frac{v^2 \sin^2\alpha}{2g}$Substitute $v^2 = 18g$ and $\sin(30^\circ) = 0.5$ into the equation:$H_{\text{proj}} = \frac{18g \times (0.5)^2}{2g} = \frac{18 \times 0.25}{2} = 2.25\text{ m}$Step 4: Calculate total height above the reference line $XX'$The total height attained above the line $XX'$ is the initial exit height plus the maximum height of the projectile path:$H_{\text{total}} = y_0 + H_{\text{proj}} = 1\text{ m} + 2.25\text{ m} = 3.25\text{ m}$Rounding to one decimal place gives $3.2\text{ m}$, which matches option (d).