Question

A glass tube of circular cross-section is closed at one end. This end is weighted and the tube floats vertically in water heavy end down. How far below the water surface is the end of the tube? Given Outer radius of the tube 0.14 cm, mass of weighted tube 0.2 gm, surface tension of water 73 dyne/cm and g = 980 cm/$se{c}^2$

• A. 4.31 cm
• B. 2.31 cm
  
• C. 8.62 cm
  
• D. 2.15 cm
  

Answer 1

The correct option is A 4.31 cm

Let $l$ be the length of the ube inside water The forces acting on the tube are
(i) Upthrust of water acting upwards $=\pi r^{2}l\times 1\times 980$ $=\frac{22}{7}\times (0.14{)}^{2}l\times 980=60.368l$ dyne
(ii) Weight of the system acting downward = mg = 0.2 x 980 = 196 dyne
(iii) Force of surface tension acting downwards $=2\pi rT=2\times \frac{22}{7}\times 0.14\times 73=64.24$ dyne
Since the tube is in equilibrium the upward force is balanced by the downward forces That is
60.36$l$=196 + 64.24 = 260.24 $\therefore$ $l=\frac{260.24}{60.368}=4.31cm$