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A simple pendulum with solid metal bob has a time period T . What will be the period of the same pendulum if it is made to oscillate in a non-viscous liquid of density one-tenth of the metal of the bob?
T = 2pi(L/g)1/2 ...........1
finally weight of body is reduced in water = (dbVbg - dLVbg)
db,dL are dencity is bob & liquid , Vb = volume of bob
mgeff = Vbg(db-dL)
dL = db/10 (given)
so , mgeff = Vbdbg(9/10)
geff = 9g/10
time period = T1 = 2pi(L/geff)1/2 ................2
divide 2 by 1
T1/T = (10/9)1/2
T1 = (10/9)1/2T
approve my ans if u like it
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