Kinematics Question 550
Question: A pendulum bob on a 2 m string is d it’splaced 60o from the vertical and then released.What is the speed of the bob as it passes through the lowest point in it’s path [JIPMER 1999]
Options:
A) $ \sqrt{2}m/s $
B) $ \sqrt{9.8}m/s $
C) 4.43 m/s
D)$ 1/\sqrt{2}m/s $
Show Answer
Answer:
Correct Answer: C
Solution:
$ v=\sqrt{2gl(1-\cos \theta )}=\sqrt{2\times 9.8\times 2(1-\cos 60{}^\circ )} $
$ =4.43m/s $
BETA
