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 = (d_bV_bg - d_LV_bg)

d_b,d_L are dencity is bob & liquid , V_b = volume of bob

mg_eff = V_bg(d_b-d_L)

d_L = d_b/10          (given)

so , mg_eff  = V_bd_bg(9/10)

       g_eff = 9g/10  

time period = T¹ = 2pi(L/g_eff)^1/2           ................2

divide 2 by 1

T¹/T = (10/9)^1/2

T¹ = (10/9)^1/2T

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