A uniform rod of length 1.8 m suspended by an end is made to undergo small oscillations. Find the length of the simple pendulum having the mass and time period equal to that of the rod.

Options

A. 3.6 m
B. 1.2 m
C. 2.4 m
D. 4.2 m

Answer

B

Explanation

Given length of the rod, l=1.8m
Time period of the oscillation of uniform rod with length l pivoted at one end is given by
`T=2pisqrt((I)/(mgR))" "......(i)`
where, R is the distance of the point of plvot to the centre of the gravity and I is moment of intertia of the rod.
`:.R=(1)/(2)andI=(ml^(2))/(3)`
`:.` From Eq. (i),
`T=2pisqrt((ml^(2))/((3)/(mg""(l)/(2))))=2pisqrt((2)/(3).(l)/(g))=2pisqrt((2)/(3)xx(1.8)/(g))`
`:.T=2pisqrt((1.2)/(g))`
Time period of the simple pendulum of length l' is given by
`T=2pisqrt((l')/(g))` (Given, T'=T)
`2pisqrt((l)/(g))=2pisqrt((1.2)/(g))`
`impliesl'=1.2m`
Hence, the length of the simple pendulum is 1.2 m,