The acceleration of block A is a upwards and the acceleration of block C is f downwards. Find the acceleration of block B. Assume that the string is inextensible and all the pulleys are smooth and light.
A
12(f−a), Upwards
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B
12(a+f), Downwards
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C
12(a+f), Upwards
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D
12(f−a), Downwards
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Solution
The correct option is A12(f−a), Upwards
lA, lB and lC are the intercepts of the same string.
Since the string is inextensible, lA+2lB+lC = constant
Double differentiating above equation w.r.t time t, we get equation in terms of acceleration, −aA−2aB+aC=0
(Assuming increase in length as positive)
Since aA=a,aC=f ⇒aB=f−a2 (Upwards)