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Two discs of same mass and same thickness have densities as 7gm/cm cube and 51 gm/cm cube.The ratio of thier moment of inertia about their central axes is what?????
Moment of inertia of a circular disc about an axis passing through its centre of gravity and perpendicular to its plane (central axis) is given by I = ½ mr2 .We know that masses are same for both the discs. So we have to find the radii of the two discs. Let m be the mass and b be the thickness of both the discs.Volume of disc 1 = mass/density = m/d1 = Area x thickness = πr12 x bVolume of disc 2 = mass/density = m/d2 = Area x thickness = πr22 x br12 = m/(πbd1 )r22 = m/(πbd2 )Ratio of the M.I.s of the two discs = I1/I2 = ½ mr12/(½ mr22 ). = r12 / r22 = [m/(πbd1 )] / [m/(πbd2 )] = d2/d1I1 : I2 = d2 : d1 Hence I1 : I2 = 51 :3
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