Q.

A mass of M kg  is suspended by a weight less string the horizontal force that requires to displace it until the string making an angle 45° with the initial vertical direction is

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a

Mg(2−1)

b

Mg(2+1)

c

Mg2

d

Mg2

answer is A.

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Detailed Solution

work done by horizontal force is W= F×ℓ sin45°            Work done by horizontal force =Fℓ2---------(1)          gain in PE of pendulum = Mgh PE=Mg(ℓ−ℓ cos45°) PE=Mgℓ(1-12)             PE =Mgℓ  (2−12)-----------(2)        substitute equation (1) in (2) ∴   Fℓ2  =   Mgℓ(2−12)        ⇒     F    =    Mg (2−1)
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