A uniform solid sphere of relative density 5 is released in water filled in a long vertical tube. Its terminal velocity acheived is V. If another uniform solid sphere of same material but double the radius is released in the same water then the terminal velocity achieved will be.

force (F) = 6piRkV                 (V is terminal velocity)

F_i/F_f = (R_i/R_f).(V_i/V_f)   ..........1              (i , f denotes initial & final)

F = vg(p - p_w)             (v is volume)

F₁/F₂ = v₁/v₂                        (v = 4/3 piR³ )

F₁/F₂ = (R_i/R_f)²       ...............2

equating 1 & 2

(R_i/R_f)² = (V_i/V_f)

final radius is double so

 V_i/V_f = 1/4

 V_f = 4V

approve if u like my ans