1nm=___Gm

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The correct Answer is:
B
Let x be the displacement of the plank towards left. Now the centre of gravity is also displaced through x.
`R_1+R_2=mg`
Taking moment about g we get
`R_1(l/2-x)=R_2(1/2-x)`
`=(mg-R_1)(l/2+x)…………1 `ltbr.gt `S, R_1(l/2-x)=(mg-R_1)(l/2+x)`
`rarr R_1 1/2-R_1x=`
`mgl/2-R_1x+mgx-R_1l/2`
L `rarr R-1l/2+R_1l/2=mg(x+1/2)`
`rarr R_1(l/2+l/2)=mg((2x+1)/2)`
`rarr R_1l=mg(2x+1)
R_1=mg((1+2x)/(2l)` .........i
Now, `F_1=muR_1=mumg((1+2x))/(2l)`
similarly `F_2=muR=mumg((1+2x))/(2l)`
similarlty `F_2=muR_2=muMg((1+2x))/(2l)`
sin `F_1gtF_2`
`rarrF_1-F_2=ma-(2mu(mg)/l)x`
`a/x=2mu=(mg)/l=omega^2`
`rarr =omegamusqrt(g/l)`
`: Time period `=2pisqrt(l/(2mug))`
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