Answer:2.69 cm/s
Explanation:
We know that drag force
Fd=6 πμ r v
μ=Dynamic viscosity
r=radius
v=terminal velocity
Bouncy force
Fb= ρ V g
[tex]rho_w=density\ of\ water[/tex]
V=volume
At the condition of terminal velocity
when going down ( aluminum)
[tex]Fb+ Fd= m g[/tex]
[tex]Fd_1 = m g - Fb_1[/tex]---------1
when going up ( air)
[tex]Fb_2= mg +Fd[/tex]
[tex]Fd_2= Fb_2 - mg[/tex]-----------2
m =mass of object
[tex]m= density\times volume[/tex]
[tex]\rho_a=density\ of\ Aluminium[/tex]
[tex]\rho _b=density\ of\ bubble[/tex]
Divide 1 and 2
[tex]\dfrac{v_1}{v_2}=\dfrac{mg-F_b_1}{F_b_2-mg}[/tex]
[tex]\dfrac{4.58}{v_2}=\dfrac{\rho _a-\rho _w}{\rho _w-\rho _b}[/tex]
divide by [tex]\rho _w[/tex] in R.H.S
[tex]\dfrac{4.58}{v_2}=\dfrac{2.7-1}{1.-0.0012}[/tex]
[tex]v_2=\dfrac{4.58}{1.7}=2.69\ cm/s[/tex]