The following observations were taken for determining surface tension of water by capillary tube method - Diameter- D - 1-25 - 10-2m- Rise of water in capillary- h - 1-45 - 10-2m- Taking g - 9-80 m-s2 and using the relation T - rhg-2 - 103 N-m- What is the possible error in surface tension- T -

The correct option is B 1.5% From the question, T = rhg2× 10^3 Δ TT = #160;Δ rr + Δ hh #160; Δ TT = 0.011.25 + 0.011.45 Δ TT = 0.0 ...

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Standard VII

Physics

Inter-Conversion of Temperature in Different Scales

The following...

Question

The following observations were taken for determining surface tension of water by capillary tube method : Diameter, $D = 1.25 \times 10^{- 2} m$ , Rise of water in capillary, $h = 1.45 \times 10^{- 2} m$ . Taking $g = 9.80 m / s^{2}$ and using the relation $T = (r h g / 2) \times 10^{3} N / m$ . What is the possible error in surface tension, $T$ ?

0.15 %

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1.5 %

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15 %

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2.4 %

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Solution

The correct option is B $1.5 %$
From the question,
$T = \frac{r h g}{2} \times 10^{3}$
$\frac{Δ T}{T} = \frac{Δ r}{r} + \frac{Δ h}{h}$
$\frac{Δ T}{T} = \frac{0.01}{1.25} + \frac{0.01}{1.45}$
$\frac{Δ T}{T} = 0.01489$
In percentage, $\frac{Δ T}{T} = 1.5 %$

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Q. The following observations were taken for determining surface tension

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diameter of capillary,

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rise of water,

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and the simplified relation.

T = \frac{r h g}{2} \times 10^{3} N / m

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and the simplified relation

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The following observations were taken for determining surface tension of water by capillary tube method : Diameter, D=1.25×10−2m, Rise of water in capillary, h=1.45×10−2m. Taking g=9.80m/s2 and using the relation T=(rhg/2)×103N/m. What is the possible error in surface tension, T ?

The correct option is B 1.5%From the question,T=rhg2×103ΔTT=Δrr+ΔhhΔTT=0.011.25+0.011.45ΔTT=0.01489In percentage, ΔTT=1.5%

The correct option is B $1.5 %$
From the question,
$T = \frac{r h g}{2} \times 10^{3}$
$\frac{Δ T}{T} = \frac{Δ r}{r} + \frac{Δ h}{h}$
$\frac{Δ T}{T} = \frac{0.01}{1.25} + \frac{0.01}{1.45}$
$\frac{Δ T}{T} = 0.01489$
In percentage, $\frac{Δ T}{T} = 1.5 %$