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Step 1: Identify Forces Acting on the Ball
When the ball is dropped into glycerin, three main forces act on it:
The weight of the ball $Mg$ (acting downward)
The buoyant force $F_{B}$ due to glycerin (acting upward)
The viscous force $F_{v}$ (also acting upward, opposing the motion)
Step 2: Apply the Condition of Constant (Terminal) Velocity
As the ball reaches terminal velocity, its acceleration becomes zero. Hence, the net force on the ball must be zero. Mathematically,
$Mg = F_{B} + F_{v}$
Step 3: Express the Buoyant Force
According to the principle of buoyancy (Archimedes’ principle),
$F_{B} = \rho_{\text{fluid}} \times V \times g$
Here, $\rho_{\text{fluid}}$ is the density of glycerin, which is given as $\frac{d}{2}$.
The volume of the ball is $V$, and $g$ is the acceleration due to gravity. Thus,
$F_{B} = \frac{d}{2} \cdot V \cdot g$
Step 4: Relate Mass of the Ball to Its Density
The mass of the ball is given by $M$. The ball’s density is $d$, so
$M = d \cdot V \implies d \cdot V = M$
Hence, $dVg = Mg$.
Step 5: Calculate the Viscous Force
From the condition $Mg = F_{B} + F_{v}$, we get
$F_{v} = Mg - F_{B}$
Substituting $F_{B} = \frac{d}{2} V g$ and using $dVg = Mg$, we have
$F_{v} = dVg - \frac{d}{2}Vg = \frac{d}{2}Vg = \frac{Mg}{2}$
Step 6: State the Final Result
Therefore, the viscous force acting on the ball at terminal velocity is
$F_{v} = \frac{Mg}{2}$
