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.

Options

A. `sqrtgl`
B. `sqrt2gl`
C. `sqrt3gl`
D. `sqrt gl/3`

Answer

`(4)`

Explanation

Let T = mgat an angle `theta `as shown in the figure and `h = l(1 -cos theta) a (i) `
Applying law of conservation of mechanical energy between the points A and B,
`we get 1/2 m(u^(2) - v^(2)) = mgh`
`Here u^(2) = g l a (ii) `
`and v^(2) = u^(2) - 2gh a (iii)`
Further `T - mg cos theta = (mv)^(2)/l (T = mg)`
`v^(2) = g l(1 - cos theta ) a(iv)`
From equations (i), (ii) , (iii) and (iv)
we have `gl(1-cos theta) = g l - 2gl(1- cos theta)`
`cos theta = 2/3, theta = cos^(-1) (2/3)`
From equation (iv) `v = sqrt((gl)/(3))`