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Let M be the mass and L be the length of a thin uniform rod. In first case, axis of rotation is passing through centre and perpendicular to the length of the rod. In second case, axis of rotation is passing through one end and perpendicular to the length of the rod. The ratio of radius of gyration in first case to second case isA. 1B. `1/2`C. `1/4`D. `1/8` |
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Answer» Correct Answer - B According to question, Moment of inertia of rod whose axis is passing through centre and perpendicular to the rod is given by `l=(ML^2)/12` …(i) in terms of radius of gyration `l=MK^2` …(ii) Comparing Eqs. (i) and (ii) , we get `MK_1^2=(ML^2)/12` Rightarrow `K_1=L/(2sqrt3` In second case, moment of inertia when axis is passing through one of the end is given by `l=(ML^2)/3` Similarly in terms of radius of gyration `l= M_2^2` From Eqs. (iv) and (v), we get `(ML^2)/3=MK_2^2` `K_2=L/sqrt3` Again taking the ratio of K, and K, from Eqs.(iv) and (vi), We have, `K_1/K_2=(Lxxsqrt3)/(2sqrt3xxL)=1/2` |
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