Solution
Let the density of the fluid be $\rho$. The density of the piston material is $\eta \rho = 3\rho$.
Narrow Arm: Piston thickness $t$, Area $a$.
Wider Arm: Piston thickness $T$, Area $A$.
Let the reference level be the bottom of the piston in the narrow arm.
Pressure at the reference level (from narrow arm): $$ P_{ref} = P_{atm} + \text{Pressure due to piston weight} = P_{atm} + \frac{(a \cdot t \cdot 3\rho)g}{a} = P_{atm} + 3t\rho g $$
Case 1: Minimum Height ($h_{min}$)
The wider piston is at its lowest possible equilibrium point. It has a tendency to sink further, so static friction $f$ acts upwards.
Pressure at the bottom of the wide piston (height $h_{min}$ above ref) is $P_{ref} – \rho g h_{min}$.
Force balance on wide piston:
$$ (P_{ref} – \rho g h_{min})A + f = Mg + P_{atm}A $$
Substituting $M = A \cdot T \cdot 3\rho$ and dividing by $A$:
$$ P_{atm} + 3t\rho g – \rho g h_{min} + \frac{f}{A} = 3T\rho g + P_{atm} $$
$$ 3t\rho g – \rho g h_{min} + \frac{f}{A} = 3T\rho g \quad \dots (1) $$
Case 2: Maximum Height ($h_{max}$)
The piston is at its highest point; it has a tendency to move up, so friction $f$ acts downwards.
$$ (P_{ref} – \rho g h_{max})A = Mg + P_{atm}A + f $$
$$ 3t\rho g – \rho g h_{max} = 3T\rho g + \frac{f}{A} \quad \dots (2) $$
Solving:
Add equations (1) and (2) to eliminate friction ($f/A$):
$$ (3t\rho g – \rho g h_{min}) + (3t\rho g – \rho g h_{max}) = 6T\rho g $$
$$ 6t – (h_{min} + h_{max}) = 6T $$
$$ T = t – \frac{h_{min} + h_{max}}{6} $$
Calculation:
Given $t = 5.0 \text{ cm}$, $h_{min} = 1.0 \text{ cm}$, $h_{max} = 2.0 \text{ cm}$, $\eta = 3$ (so $2\eta = 6$).
$$ T = 5.0 – \frac{1.0 + 2.0}{6} = 5.0 – 0.5 = 4.5 \text{ cm} $$
Answer: The thickness of the piston in the wider arm is 4.5 cm.