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QJEEM22Ph7170: JEE Main 2022 Physics - Class 11 JEEM22Ph7170 Solution: A pendulum is suspended by a string of length 250 cm. The mass…

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)

JEE Main 2022 (Online) 28th June Morning Shift Physics - Simple Harmonic Motion Question 59 English

Step-by-Step Verified Solution

12mv2=mgl(1−cos⁡θ){1 \over 2}m{v^2} = mgl(1 - \cos \theta )

⇒v=2gl(1−cos⁡θ)\Rightarrow v = \sqrt {2gl(1 - \cos \theta )}

=2×10×2.5×12= \sqrt {2 \times 10 \times 2.5 \times {1 \over 2}}

=5= 5 m/s

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