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Two-point masses of 0.3kg and 0.7kg are fixed at the ends of a rod which is of length 1.4m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum is located at a distance of

(a) 0.42 m from the mass of 0.3kg

(b) 0.70 m from the mass of 0.7kg

(c) 0.98m from the mass of 0.3kg

(d) 0.98m from the mass of 0.7kg

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Work-energy theorem,

     Wallforces​=\(1\over2\)Iw2      

Given:  w  is constant

So required work done to be minimum implies that I must be minimum.

Let the rotational axis passes through O.

I=0.3(x2)+0.7(1.4−x)2  

For I to be minimum,        \(dI\over dx\)​=0

⟹0.3×2(x)−0.7×2(1.4−x)=0

⟹x=0.98m

hence, correct answer is 0.98 from the mass of 0.3 kg

Rotational motion

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