JEE Main 2022 · 27 Jul · Shift 2 · Physics Q59

← JEE Main 2022 · 27 Jul · Shift 2

PhysicsInteger+4 / -1Question 59 of 90

A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should by unbinding the strings to achieve a speed of 4 ms $^{-1}$ , is ______ cm. (take g = 10 ms $^{-2}$ )

A horizontal solid cylinder hanging from a ceiling by two vertical strings (tension T each) wound around it.

Your answer

Answer and worked solution

Official answer: 120

Unwinding cylinder. $mg-2T=ma$ and $2TR=\frac12mR^2\alpha$ give $a=\frac{2g}3$ .

Distance $s=\frac{v^2}{2a}=\frac{16}{40/3}=1.2$ m.

Answer: 120

Topics:Torque and rotational dynamics Rolling motion

← Previous question Next question →

JEE Main 2022 · 27 Jul · Shift 2, Physics question 59