A brass sphere is hung from one end of a massless and inextensible thread. When the sphere is set into oscillation, it oscillates with a time period T. If now the sphere is dipped completely into a non-viscous liquid, then what will be the time period of its oscillation? (density of the liquid is 1/10th of that of brass)

A brass sphere is hung from one end of a massless and inextensible thr

1.	A brass sphere is hung from one end of a massless and inextensible thread. When the sphere is set into oscillation, it oscillates with a time period T. If now the sphere is dipped completely into a non-viscous liquid, then what will be the time period of its oscillation? (density of the liquid is 1/10th of that of brass)
Answer» Solution :LET VOLUME of the sphere be V, density of brass be `RHO`, density of the LIQUID be `rho.`. `therefore` Apparent weight of the sphere when immersed in the liquid = real weight - weight of displaced liquid = `Vrhog-Vrho.g=VG(rho-rho.)` `therefore` Apparent acceleration due to gravity of the sphere immersed in the liquid, `g.=("apparent weight")/("mass")=(Vg(rho-rho.))/(Vrho)=g(1-(rho.)/rho)` According to the problem, `(rho.)/rho=1/10`, hence `g.=g(1-1/10)=9/10g`. In the case of a simple pendulum, `Tprop1/sqrtg`, so, if the time period of oscillation of the sphere, when immersed in the liquid, is T., then `(T.)/T=sqrt(g/(g.))=sqrt(10/9)or,T.=sqrt10/3T`.