JEE MAIN - Physics (2022 - 28th June Morning Shift - No. 18)
A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60$$^\circ$$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ____________ ms$$-$$1. (if g = 10 m/s2)
_28th_June_Morning_Shift_en_18_1.png)
Answer
5
Explanation
$${1 \over 2}m{v^2} = mgl(1 - \cos \theta )$$
$$ \Rightarrow v = \sqrt {2gl(1 - \cos \theta )} $$
$$ = \sqrt {2 \times 10 \times 2.5 \times {1 \over 2}} $$
$$ = 5$$ m/s
Comments (0)
