Question

# A body crosses the topmost point of a vertical circle with critical speed. What will be its centripetal acceleration when the string is horizontal.

A

g

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B

2g

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C

3g

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D

6g

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Solution

## The correct option is C 3gStep1: Given dataThe object at the top of the vertical circle has a speed equal to critical speed.Step2: Formula usedThe critical speed at the top of a vertical circle, ${v}_{c}=\sqrt{gl}\left[whereg=accelerationduetogravity,l=radiusoftheverticalcircle\right]$At a point in a horizontal circle, when the string becomes horizontal, the speed of the object, ${v}^{2}={u}^{2}+2gh\left[wherev=finalspeed,u=initialspeed,g=acceleartionduetogravity,h=changeinheight\right]$Step3: Calculating the final speed.According to the given data, the speed of the object at the topmost point $=\sqrt{gl}$This will act as the initial speed for when the string becomes horizontal i.e. $u=\sqrt{gl}$Therefore the speed of the object at this point will be ${v}^{2}={\left(\sqrt{gl}\right)}^{2}+2gh$The change in height of the object from the initial point is equal to, ${v}^{2}=gl+2gl=3gl$ Step4: Calculating centripetal acceleration${a}_{c}=\frac{{v}^{2}}{l}\phantom{\rule{0ex}{0ex}}{a}_{c}=\frac{3gl}{l}\phantom{\rule{0ex}{0ex}}{a}_{c}=3g$The centripetal acceleration when the string becomes horizontal is 3g.Hence, option C is the correct answer.

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