A particle initially at rest at point A(0, 0) slides down from a frictionless wire in a vertical plane to another point \(B(y_{0}, z_{0})\) under the influence of gravity. Then according to principle of variation, the minimum time \((\tau)\) taken by particle can be evaluated using the expression :
(where, \(z' = \frac{dz}{dy}\))
MH SET - 2025, SET Exam - 2025,
Answer: A
The time is \(\frac{1}{\sqrt{2g}}\int \frac{\sqrt{1+z'^{2}}}{\sqrt{z}}\,dy\).
This is the brachistochrone set-up, with \(z\) measured downward from the starting point. Starting from rest, energy conservation gives the speed after falling a height \(z\), and the arc length element is written in terms of \(z' = dz/dy\):
\[ \begin{aligned} v &= \sqrt{2gz} \\ ds &= \sqrt{1 + z'^{2}}\,dy \\ \tau &= \int\frac{ds}{v} \\ &= \frac{1}{\sqrt{2g}}\int\frac{\sqrt{1+z'^{2}}}{\sqrt{z}}\,dy \end{aligned} \]
The calculus of variations then minimises this functional. The arc length always has \(1 + z'^{2}\) (Pythagoras), so the options with \(1 - z'^{2}\) are wrong, and the speed depends on the height fallen \(z\), not on the slope \(z'\), which rules out \(\sqrt{z'}\) and \(\sqrt{zz'}\) in the denominator.
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