A body of m kg slides from rest along the curve of vertical circle from point $A$ to $B$ in friction less path. The velocity of the body at B is : (given, $\mathrm{R}=14 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ and $\sqrt{2}=1.4$ )

(B) $21.9 \mathrm{~m} / \mathrm{s}$

(A) $19.8 \mathrm{~m} / \mathrm{s}$

(C) $16.7 \mathrm{~m} / \mathrm{s}$

(D) $10.6 \mathrm{~m} / \mathrm{s}$

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Work, Power & Energy

A body of m kg slides from rest along the curve of vertical circle from point

A

B

in friction less path. The velocity of the body at B is :

(given,

R = 14 m, g = 10 m / s^{2}

and

2 = 1.4

)

Answer: (B)

Point

A

is on the circular track at

4 5^{\circ}

from the topmost point (measured from the centre), and

B

is the lowest point of the circle. The body starts from rest at

A

. Height dropped from

A

B

h = R + R cos 4 5^{\circ} = R (1 + cos 4 5^{\circ})

By the work–energy theorem (frictionless path),

m g h = \frac{1}{2} m v_{B}^{2}

v_{B}^{2} = 2 g R (1 + cos 4 5^{\circ})

Using

R = 14 m

g = 10 m/s^{2}

and

cos 4 5^{\circ} = 1/ 2 = 1/1.4 \approx 0.714

v_{B}^{2} = 2 (10) (14) (1.714) = 480 ⟹ v_{B} \approx 21.9 m/s

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option