A heavy particle hanging from a fixed point by a light inextensible string of length `l` is projected horizonally with speed `sqrt(gl)` . Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle.

Answer

C

Explanation

Let `T=mg` at `theta` as slow in figure.
`h=l(1-costheta)` ...(i)
Applying conservation of mechanical energy beweet points `A` and `B` , we get
`(1)/(2)m(u^(2)-v^(2))=mgh`
Here, `u^(2)=gl` ...(ii)
and `v=speed` of particle in position `B`
:. `v^(2)=u^(2)-2gh`
Further, `T-mgcostheta=(mv^(2))/(l)` or `mg-mgcostheta=(mv^(2))/(l)` `(T=mg)`
or `v^(2)=gl(1-costheta)`
Substitution value of `v^(2),u^(2)` and `h` from Eqs. `(iv),(ii)` and `(i)` on Eq. `(iii)` , we get
`gl(1-costheta)=gl-2gl(1-costheta)` or `costheta=(2)/(3)` or `theta=cos^(-1)((2)/(3))`
Substituting `costheta=(2)/(3)` in Eq. `(iv)` , we get
`v=sqrt((gl)/(3))`