Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide

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Question

Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. A hole is made at position x = a (0 < a < L) and height y = h (measured from the bottom). Water streams out and hits the ground at position x = b. The maximum horizontal range (b − a) of the water jet is achieved when (H is the total height of water in the tank):

Accepted Answer

Correct Answer: A — h = H/2 and b − a = HStep-by-step Solution:Speed of efflux from a hole at height h (Torricelli's theorem): v = √(2g(H − h))The water jet is a projectile launched horizontally from height h above the ground.Time to fall to ground: h = (1/2)gt² → t = √(2h/g)Horizontal range: b − a = v × t = √(2g(H−h)) × √(2h/g) = √[4h(H−h)] = 2√[h(H−h)]To maximize (b−a), maximize h(H−h):d/dh [h(H−h)] = H − 2h = 0 → h = H/2Maximum range: b − a = 2√[(H/2)(H/2)] = 2√(H²/4) = 2 × H/2 = HTherefore: h = H/2, b − a = H ✓Why Option B is wrong: h = H/4 is not the optimal height for maximum range.Why Option C is wrong: h = H/3 is also sub-optimal.Why Option D is wrong: h = H/2 is correctly identified, but b − a = H/2 is wrong (maximum range is H, not H/2).Key Concepts from Mechanical Properties of Fluids (NCERT Class 11, Chapter 10):Torricelli's theorem: Speed of efflux through a hole at depth D below the surface = √(2gD). Here D = H − h.This is derived from Bernoulli's equation: P + (1/2)ρv² + ρgh = constant.The horizontal range problem combines Torricelli's theorem with projectile motion.The range is maximized at h = H/2 — by symmetry, a hole at h and one at H−h give the same range. Only h = H/2 is its own symmetric point.

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