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Asked by jhajuhi19 | 07 Sep, 2021, 06:59: PM
When the bob is at C , we have T - mg = m ( v2 / ) ,  Hence  T  = m g + m ( v2 / ) ..........................(1)

where T is tension force in the string , m is mass of bob, g is acceleration due to gravity .

The term  m ( v2 / ) is centripetal force in radial direction , v is linear speed of bob and  is the length of string

When the bob is at A , we have T   = m ( v2 / )  ................................(2)

When the bob is at B , we have T = 0 (given in this case )  and  mg = m ( v2 / )  .............................(3)

Hence Tension is maximum when the bob is at C , reduced to zero in this case at B .
Tension force will be in between at A , i.e., greater than zero but less than the value at C .

If the bob is revolving about the point O in uniform angular speed , then its radial acceleration is same .

Since at B , we get from eqn.(3) , radial acceleration  ( v2 / ) = g  , then radial acceleration will be same at all point

Hence radial acceleration at A , ( v2 / ) = g  = 10 m/s2

Answered by Thiyagarajan K | 07 Sep, 2021, 08:22: PM
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