A particle of mass `m` is suspended by a string of length `l` from a fixed rigid support. A sufficient horizontal velocity `=sqrt(3gl)` is imparted to it suddenly. Calculate the angle made by the string with the vertical when the accekleration of the particle is inclined to the string by `45^(@)` .

A particle of mass `m` is suspended by a string of length `l` from a f

1.	A particle of mass `m` is suspended by a string of length `l` from a fixed rigid support. A sufficient horizontal velocity `=sqrt(3gl)` is imparted to it suddenly. Calculate the angle made by the string with the vertical when the accekleration of the particle is inclined to the string by `45^(@)` .
Answer» Correct Answer - A::B `h=l(l-costheta)` `v_^(2)v_(0)^(2)-2gh=3gl-2gl(l-costheta)=gl(1+2costheta)` At `45^(@)` means radial and tangential components of acceleration are equal. :. `(v^(2))/(l)=gsintheta` or `1+2costheta=sintheta` Solving the equation we get, `theta=90^(@)or(pi)/(2)`

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A particle of mass `m` is suspended by a string of length `l` from a fixed rigid support. A sufficient horizontal velocity `=sqrt(3gl)` is imparted to it suddenly. Calculate the angle made by the string with the vertical when the accekleration of the particle is inclined to the string by `45^(@)` .

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