NEET 2025 — Physics solution

NEET2025Physics

A bob of heavy mass $m$ is suspended by a light string of length $ℓ$ . The bob is given a horizontal velocity $v_{0}$ as shown in figure. If the string gets slack at some point P making an angle $θ$ from the horizontal, the ratio of the speed $v$ of the bob at point P to its initial speed $v_{0}$ is:

A
$(sin θ)^{1/2}$
B
$(\frac{1}{2 + 3 sin θ})^{1/2}$
C
$(\frac{cos θ}{2 + 3 sin θ})^{1/2}$
D
$(\frac{sin θ}{2 + 3 sin θ})^{1/2}$

Solution

At the point where the string goes slack, tension $T_{P} = 0$ . The component of gravity along the string supplies the centripetal force: $m g sin θ = \frac{m v _{P}^{2}}{ℓ}$ , giving $v_{P}^{2} = g ℓ sin θ$ . ...(i)

Conservation of energy from the lowest point to P (height risen $= ℓ + ℓ sin θ$ ):

$\frac{1}{2} m v_{0}^{2} = m g ℓ (1 + sin θ) + \frac{1}{2} m v_{P}^{2}$ .

$v_{0}^{2} = 2 g ℓ (1 + sin θ) + g ℓ sin θ = g ℓ (2 + 3 sin θ)$ . ...(ii)

Dividing (i) by (ii): $\frac{v _{P}^{2}}{v _{0}^{2}} = \frac{sin θ}{2 + 3 sin θ}$ , hence $\frac{v _{P}}{v _{0}} = (\frac{sin θ}{2 + 3 sin θ})^{1/2}$ .

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →