JEE Main 2025 Jan 22 Shift 2, Physics Q20: Work Energy Theorem And Its Applications

JEE Main2025Jan 22, Shift 2Physics

A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point A is 10 m/s. The ratio of its kinetic energies at point B and C is :

(Take acceleration due to gravity as 10 m/s $^{2}$ )

A
$\frac{2 + 3}{3}$
B
$\frac{2 + 2}{3}$
C
$\frac{3 + 3}{2}$
D
$\frac{3 - 2}{2}$

Solution

Take point A at the bottom of the circle as the reference for height. From the figure, B is located such that its height above A is $h_{B} = R - R cos 3 0^{\circ} = R (1 - \frac{3}{2})$ , and C is located above the centre with $h_{C} = R + R sin 3 0^{\circ} = \frac{3 R}{2}$ (since the radius OC tilts 30° above the horizontal in the configuration shown).

Using energy conservation between A and B:

$\frac{1}{2} m v_{A}^{2} = \frac{1}{2} m v_{B}^{2} + m g h_{B}$

$100 = v_{B}^{2} + 2 g R (1 - \frac{3}{2}) = v_{B}^{2} + 40 (1 - \frac{3}{2})$

$v_{B}^{2} = 100 - 40 + 203 = 60 + 203$ .

Energy conservation between A and C:

$100 = v_{C}^{2} + 2 g \cdot \frac{3 R}{2} = v_{C}^{2} + 60 \Rightarrow v_{C}^{2} = 40$ .

Ratio of kinetic energies:

$\frac{K _{B}}{K _{C}} = \frac{v _{B}^{2}}{v _{C}^{2}} = \frac{60 + 20 3}{40} = \frac{3 + 3}{2}$ .

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JEE Main 2025 Jan 22 Shift 2, Physics Q20: Work Energy Theorem And Its Applications

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