Contact

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

9321924448 / 9987178554

Mon to Sat - 10 AM to 7 PM

Asked by Umesh 1st December 2017, 9:43 AM
Let the Tension in the two strings be T1 and T2. Resolved vertical component balances the weight of the ball.
Resolved horizontal component provides centripetal force.

Hence for vertical direction T1 cos30 + T2 cos45 = mg

(√3/2) T1 + (1/√2) T2 = 5×9.8 = 49

multiply the above eqn. by 2√2, we will get
√6T1 + 2 T2 = 49 .................................................(1)

Horizontal direction:- T1 sin30 + T2 sin45 = mv2/r = 5v2/1.6= 3.125v2

(1/2)T1 + (1/√2) T2 = 3.125v2
multiply by 2√2, we will get

√2T1 + 2T2 = 3.125v2  .........................................(2)

(1) - (2) gives (√6-√2) T1 = 49 - 3.125v2 ......................(3)
To make LHS of (3) positive 3.125v2 < 49.  Hence v < 3.96

Answered by Expert 1st December 2017, 4:08 PM
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10

You have rated this answer 3/10