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One end of a straight uniform 1m long ba...

One end of a straight uniform 1m long bars pivoted on horizontal table. It is released from rest when it makes an angle `30^(@)` from the horizontal (see figure ) . Velocity of end point of the rod when it hits the table is given as `sqrt( n ) ms^(-1)`,where n is an integer.The value of n is `"_______________"`

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Loss of gravitation potential energy = Gain in kinetic energy
`:. Mg""(l)/(2) sin 30^(@) = ( I)/(2) Iomega^(2) ` ` :' mg ""(l)/( 2) ((1)/(2))= ( 1)/( 2)((ml^(2))/( 3)) omega^(2)` `rArr ( mgl)/( 4) =( ml^(2))/( 6) omega^(2) rarr = omega = sqrt(( 6g)/( 4l))`
Velocity of end point `=omega l = sqrt(( 6gl )/(4)) = sqrt((60)/( 4)) = sqrt(15 ) m//s ` `:. n = 15 `
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