JEE Main 2021 (Online) 1st September Evening Shift
Numerical
A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in vertical plane. The upper end is pushed so that the rod falls under gravity, ignoring the friction due to clamping at its lower end, the speed of the free end of rod when it passes through its lowest position is ____________ ms$$-$$1. (Take g = 10 ms$$-$$2)
Your Input ________
Answer
Correct Answer is 6
Explanation
by energy conservation $$mgl = {1 \over 2}I{\omega ^2} = {1 \over 2}{{m{l^2}{\omega ^2}} \over 3}$$
$$ \Rightarrow \omega = \sqrt {{{6g} \over l}} $$
As we know the relation between the linear speed and angular speed,
$$v = \omega r = \omega l = \sqrt {6gl} $$
$$v = \sqrt {6 \times 10 \times .6} $$ = 6 m/s
Hence, the speed of the free end of the rod when it passes through its lowest position is 6 m/s.
2
JEE Main 2021 (Online) 27th July Evening Shift
Numerical
In the given figure, two wheels P and Q are connected by a belt B. The radius of P is three times as that of Q. In case of same rotational kinetic energy, the ratio of rotational inertias $$\left( {{{{I_1}} \over {{I_2}}}} \right)$$ will be x : 1. The value of x will be _____________.
A solid disc of radius 20 cm and mass 10 kg is rotating with an angular velocity of 600 rpm, about an axis normal to its circular plane and passing through its centre of mass. The retarding torque required to bring the disc at rest in 10 s is ____________ $$\pi$$ $$\times$$ 10$$-$$1 Nm.