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(A) $metre$ scale suspended from one end acts as a physical pendulum.The time period $T$ of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{m

Vedclass · Accepted Answer

$1.64$

Vedclass · Answer

(A) $metre$ scale suspended from one end acts as a physical pendulum.The time period $T$ of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$.For a uniform rod of length $L$ and mass $m$ suspended from one end,the moment of inertia $I$ about the axis of rotation is $I = \frac{mL^2}{3}$.The distance $d$ from the center of mass to the pivot point is $d = \frac{L}{2}$.Substituting these values into the formula:$T = 2\pi \sqrt{\frac{mL^2/3}{mg(L/2)}} = 2\pi \sqrt{\frac{L^2/3}{gL/2}} = 2\pi \sqrt{\frac{2L}{3g}}$.Given $L = 1 \ m$ and taking $g = 9.8 \ m/s^2$:$T = 2\pi \sqrt{\frac{2 	imes 1}{3 	imes 9.8}} = 2\pi \sqrt{\frac{2}{29.4}} = 2\pi \sqrt{0.068} \approx 2 	imes 3.14159 	imes 0.2608 \approx 1.64 \ \sec$.

$A$ $metre$ scale is suspended vertically from a horizontal axis passing through one end of it. Its time period would be ....... $\sec$

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