MCQ
A pendulum bob on a $2\, m$ string is displaced $60^{°}$ from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path .......... $m/s$
- A$\sqrt 2$
- B$\sqrt {9.8}$
- ✓$4.43$
- D$1/\sqrt 2 $
✓
Answer
Correct option: C.
$4.43$
c
(c)$v = \sqrt {2gl(1 - \cos \theta )} = \sqrt {2 \times 9.8 \times 2(1 - \cos 60^\circ )} $$ = 4.43\,m/s$
(c)$v = \sqrt {2gl(1 - \cos \theta )} = \sqrt {2 \times 9.8 \times 2(1 - \cos 60^\circ )} $$ = 4.43\,m/s$
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