Let ' $\mathrm{M}$ ' be the mass and ' $\mathrm{L}$ ' 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) $\frac{1}{2}$C) $\frac{1}{4}$D) $\frac{1}{8}$

Home » Let ‘ ‘ be the mass and ‘ ‘ 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) C) D)

Let ‘ $\mathrm{M}$ ‘ be the mass and ‘ $\mathrm{L}$ ‘ 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 is
A) 1
B) $\frac{1}{2}$
C) $\frac{1}{4}$
D) $\frac{1}{8}$

Physics

Let ‘ $\mathrm{M}$ ‘ be the mass and ‘ $\mathrm{L}$ ‘ 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 is
A) 1
B) $\frac{1}{2}$
C) $\frac{1}{4}$
D) $\frac{1}{8}$

Answer is (B)
M.I of rod whose axis of rotation is passing through center and perpendicular to the plane of rod is $\mathrm{I}=\frac{\mathrm{ML}^{2}}{12} \quad$ and $\quad \mathrm{I}=\mathrm{MK}_{1}^{2}$ (where K.J radius of gravitation)
$\therefore \quad M K_{1}^{2} \Rightarrow \quad K_{1}=\frac{L}{2 \sqrt{3}}$
When axis of rotation of rod is passing through one and of rod, then
$I=M K_{2}^{2}=\frac{M L^{2}}{3} \Rightarrow K_{2}=\frac{L}{\sqrt{3}}$
Taking ratios of $(1)$ and $(2)$ we get
$\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{\mathrm{L}}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\mathrm{~L}}=\frac{1}{2} \quad \Rightarrow \frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{1}{2}$

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