Question

A uniform wooden rod (specific gravity = 0.6, diameter = 4 cm, length = 8 m) is hinged at A at waterline. A lead sphere (specific gravity = 11.4) is attached at the free end to keep the rod submerged horizontally. Find the radius of the lead ball (in cm, round to 2 decimals).

Hint:

Always equate buoyancy–weight moment of the rod with the net downward moment of the ball when the rod is hinged at one end.

Answer 1

Correct Answer: 3.48

Step 1: Compute weights and buoyant forces.
Rod density: \[ \rho_w = 1000,\quad \rho_{rod} = 0.6\rho_w = 600\ \text{kg/m}^3 \] Rod volume: \[ V_{rod} = \pi(0.02)^2(8) = 0.01005\ \text{m}^3 \] Rod weight: \[ W_{rod} = \rho_{rod} g V_{rod} = 600g(0.01005) \] Rod buoyancy: \[ B_{rod} = \rho_w g V_{rod} = 1000g(0.01005) \] Net upward moment acting at midpoint (4 m from hinge): \[ M_{net} = (B_{rod}-W_{rod})(4) = (400g)(0.01005)(4) \] Step 2: Moment balance with lead ball.
Let radius of ball = \(R\) m. Ball volume: \[ V_b = \frac{4}{3}\pi R^3 \] Ball density: \[ \rho_b = 11.4\times 1000 = 11400\ \text{kg/m}^3 \] Ball weight: \[ W_b = \rho_b g V_b \] Ball buoyancy: \[ B_b = 1000g V_b \] Net downward force of ball: \[ F_b = W_b - B_b = (11400 - 1000)g V_b = 10400 g V_b \] Moment balance about hinge (rod length = 8 m): \[ M_{net} = F_b (8) \] Substituting: \[ 400g(0.01005)(4) = 10400g \left(\frac{4}{3}\pi R^3\right)(8) \] Cancel \(g\), simplify: \[ 16.08 = 332800\pi R^3 \] \[ R^3 = \frac{16.08}{332800\pi} \approx 2.39\times10^{-5} \] \[ R = (2.39\times10^{-5})^{1/3} \approx 0.0364\ \text{m} \] Convert to cm: \[ R = 3.64\ \text{cm} \] Rounded to 2 decimals: \[ \boxed{3.64\ \text{cm}} \]