In the figure shown below, the block on the frictionless surface is having acceleration a towards left and the other block is having acceleration a′. Find the relation between the accelerations of the two blocks.
A
a=a′
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B
a=2a′
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C
a=a′2
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D
a=32a′
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Solution
The correct option is Aa=a′
Let l1,l2&l3 be the intercepts of the same string. ∴l1+l2+l3=constant --- (i)
Also l2 is fixed as the two pulleys 1 & 2 are fixed ∴l2=constant --- (ii)
Differentiating (i) w.r.t. time v1+v2−v3=0 --- (iii) (∵l3 is decreasing)
Now differentiating (ii) w.r.t. time v2=0 --- (iv)
Putting (iv) in (iii) v1−v3=0
Now differentiating again w.r.t. time ⇒a1−a3=0 ⇒a1=a3 ⇒a=a′