One end of a uniform glass capillary tube of radius r = 0.025cm is immersed in a water to a depth h =1cm. The excess pressure (in N /m2) required to blow an air bubble out of the tube is ST of water =7×10−2Nm−1, water density =103kgm−3,g=10ms−2

∆P=hρg+2Tr⇒∆P=10−2×103×10+2×7×10−225×10−3×10−2⇒∆P=0.001×105+0.0056×105⇒∆P=0.0066×105 Nm−2

One end of a uniform glass capillary tube of radius r = 0.025cm is immersed in a water to a depth h =1cm. The excess pressure (in N /m2) required to blow an air bubble out of the tube is ST of water =7×10−2Nm−1, water density =103kgm−3,g=10ms−2

∆P=hρg+2Tr⇒∆P=10−2×103×10+2×7×10−225×10−3×10−2⇒∆P=0.001×105+0.0056×105⇒∆P=0.0066×105 Nm−2

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One end of a uniform glass capillary tube of radius r = 0.025cm is immersed in a water to a depth h =1cm. The excess pressure (in N /m²) required to blow an air bubble out of the tube is $(ST of water = 7 \times 10^{- 2} {Nm}^{- 1}, water density = 10^{3} {kgm}^{- 3}, g = 10 {ms}^{- 2})$

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$1.0018 \times 10^{5}$

$0.0066 \times 10^{5}$

$0.0048 \times 10^{5}$

$1.0033 \times 10^{5}$

answer is B.

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Detailed Solution

$\begin{array}{l} ∆ P = hρg + \frac{2 T}{r} \\ \Rightarrow ∆ P = 10^{- 2} \times 10^{3} \times 10 + \frac{2 \times 7 \times 10^{- 2}}{25 \times 10^{- 3} \times 10^{- 2}} \\ \Rightarrow ∆ P = 0.001 \times 10^{5} + 0.0056 \times 10^{5} \\ \Rightarrow ∆ P = 0 .0066 \times 10^{5} {Nm}^{- 2} \end{array}$