JAMB - Physics (2003 - No. 2)

A test tube of radius 1.0cm is loaded to 8.8g. If is placed upright in water, find the depth to which it would sink
2.8cm
5.2cm
25.5cm
28.0cm

Explanation

To find the depth to which a test tube sinks when placed upright in water, we apply the principle of buoyancy.

Given:
- Radius of the test tube, \( r = 1.0 \, \text{cm} = 0.01 \, \text{m} \)
- Mass of the test tube, \( m = 8.8 \, \text{g} = 0.0088 \, \text{kg} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

 Calculate the weight of the test tube
The weight \( W \) of the test tube is given by:
\(W = mg = 0.0088 \, \text{kg} \times 10 \, \text{m/s}^2 = 0.088 \, \text{N}\)

 Calculate the buoyant force
The buoyant force \( F_b \) is given by:
\(F_b = \rho g V_{\text{displaced}} = \rho g \pi r^2 h\)

 But, the buoyant force is equal to the weight:
\(\rho g \pi r^2 h = W\)
Substituting the known values:
\(1000 \times 10 \times \pi \times (0.01)^2 \times h = 0.088\)

 Solve for \( h \)
Rearranging gives:
\(h = \frac{0.088}{1000 \times 10 \times \pi \times (0.01)^2}\)

Calculating:
\(h = \frac{0.088}{1000 \times 10 \times \pi \times 0.0001} = \frac{0.088}{3.142} \approx 0.028 \, \text{m}\) ≈ 2.8cm

 

 

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