A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is `(g = 10 m//s^(2))`

Options

A. `60^@`
B. `30^@`
C. `45^@`
D. zero

Answer

C

Explanation

Given radius , r=10m
speed , v =`10ms^(-1)`
length ,l =1.0m
Centrifugal force on the rod ,F =`(mv^(2))/(r)` along BF,
let `theta`be the angle ,which the rod makes with the vertical , force acting on the rod , are shown in figure .
Resolving mg and F into two rectangular componnents , We have forces parallel to the rod
`mgcostheta+(mv^(2))/(r)sintheta=T`
Force perpendicular to the rod
`mgsintheta+(mv^(2))/(r)costheta`
The rod would be balanced if `mgsintheta+(mv^(2))/(r)costheta`
`rArr " "mgsintheta=(mv^(2))/(r)costheta`
This gives `tantheta=(v^(2))/(rg)=(10^(2))/(10xx10)=1 =tan45^@`
Here,` " "theta=45^@`