Two narrow bores of diameter 5.0 mm and 8.0 mm are joined together to form a U - shaped tube open at the both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube?
[ Take surface tension of water $T = 7.3 \times 10^{- 2} {Nm}^{- 1}$ , angle of contact =0, g = 10ms^-2 and density of water $= 1.0 \times 10^{3} {kgm}^{- 3}$ ]

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2.19 mm

3.62 mm

4.97 mm

5.34 mm

answer is A.

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Detailed Solution

Given,

$d_{1} = 5 mm, d_{2} = 8 mm$

The difference in capillary rise between the two arms of the U-tube will match the level difference between them.

The capillary rise h = $\frac{2 Tcosθ}{rρg}$

$h_{1} = \frac{2 \times 7.3 \times 10^{- 2}}{2.5 \times 10^{- 3} \times 10^{3} \times 10} = 5.84 mm$

$h_{2} = \frac{2 \times 7.3 \times 10^{- 2}}{4 \times 10^{- 3} \times 10^{3} \times 10} = 3.65 mm$

The level difference is

$Δh = h_{1} - h_{2} = 5.84 - 3.65 = 2.19 mm$

Hence the correct answer is 2.19mm.

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