MCQ
A ball rises to surface at a constant velocity in a liquid whose density is $4$ times greater than that of the material of the ball. The ratio of the force of friction acting on the rising ball and its weight is
  • $3 : 1$
  • B
    $4 : 1$
  • C
    $1 : 3$
  • D
    $1 : 4$

Answer

Correct option: A.
$3 : 1$
a
$\mathrm{F}_{\mathrm{B}}=\mathrm{F}_{\mathrm{v}}+\mathrm{F}_{\mathrm{G}}$

$\mathrm{F}_{\mathrm{v}}=3 \rho \mathrm{vg}$

$\frac{\mathrm{F}_{\mathrm{v}}}{\mathrm{F}_{\mathrm{G}}}=\frac{3 \mathrm{pvg}}{\rho \mathrm{vg}}$

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