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13.Oscillations
medium
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 $g=10\,m / s ^{2}$ )
A
$5$
B
$1$
C
$2$
D
$7$
(JEE MAIN-2022)
Solution
$V _{\max }=\sqrt{2 gh }$
The speed will be highest at the lowest position.
$h =\left(\ell-\ell \cos 60^{\circ}\right)=\frac{\ell}{2}$
$V _{\max }=\sqrt{2 \times g \times \frac{\ell}{2}}=\sqrt{10 \times 2.5}=5\,m / s$
Standard 11
Physics
