The following observations were taken for determining surface tension T of water by capillary method: diameter of capillary, D = 1.25 × 10-2 m rise of water, h = 1.45 × 10-2 m. Using g = 9.80 m/s2 and the simplified relation T = (rhg/2) × 103 N/m , the possible error in surface tension is closest to :

Question

The following observations were taken for determining surface tension T of water by capillary method: diameter of capillary, D = 1.25 × 10-2 m rise of water, h = 1.45 × 10-2 m. Using g = 9.80 m/s2 and the simplified relation T = (rhg/2) × 103 N/m , the possible error in surface tension is closest to : (A) 0.15 % (B) 1.5 % (C) 2.4 % (D) 10 %

Tardigrade Admin · Accepted Answer

(Δ T/T) ×100 = (Δ D/D) ×100 + (Δ h/h) ×100 = (0.01/1.25) ×100 + (0.01/1.45) ×100 = (100/125) + (100/145) = 0.8 + 0.689 = 1.489 ≃ 1.5 %

$0.15%$

$1.5%$

$2.4%$

$10%$

$\frac{Δ T}{T} \times 100 = \frac{Δ D}{D} \times 100 + \frac{Δ h}{h} \times 100$
$= \frac{0.01}{1.25} \times 100 + \frac{0.01}{1.45} \times 100$
$= \frac{100}{125} + \frac{100}{145}$
$= 0.8 + 0.689$
$= 1.489$
$≃ 1.5%$

Q. The following observations were taken for determining surface tension T of water by capillary method : diameter of capillary, D=1.25×10−2m rise of water, h=1.45×10−2m. Using g=9.80m/s2 and the simplified relation T=2rhg​×103N/m , the possible error in surface tension is closest to :

0.15%

1.5%

2.4%

10%

TΔT​×100=DΔD​×100+hΔh​×100 =1.250.01​×100+1.450.01​×100 =125100​+145100​ =0.8+0.689 =1.489 ≃1.5%

$0.15%$

$1.5%$

$2.4%$

$10%$

$\frac{Δ T}{T} \times 100 = \frac{Δ D}{D} \times 100 + \frac{Δ h}{h} \times 100$
$= \frac{0.01}{1.25} \times 100 + \frac{0.01}{1.45} \times 100$
$= \frac{100}{125} + \frac{100}{145}$
$= 0.8 + 0.689$
$= 1.489$
$≃ 1.5%$