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A test tube of radius 1.0cm is loaded to 8.8g. If is placed upright in...

Physics
JAMB 2003

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

  • A. 2.8cm
  • B. 5.2cm
  • C. 25.5cm
  • D. 28.0cm
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Correct Answer: Option A
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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