The following observations were taken for determining surface tension T of water by capillary method : diameter of capillary, $D=1.25 \times 10^{-2} \mathrm{~m}$ rise of water, $\mathrm{h}=1.45 \times 10^{-2} \mathrm{~m}$. Using $\mathrm{g}=9.80 \mathrm{~m} / \mathrm{s}^{2}$ and the simplified relation $\mathrm{T}=\frac{\mathrm{rhg}}{2} \times 10^{3} \mathrm{~N} / \mathrm{m}$, the possible error in surface tension is closest to :

JEE Main · 2017 · 02 Apr 2017 · Offline · Units & Measurement

The following observations were taken for determining surface tension T of water by capillary method :
diameter of capillary,

D = 1.25 \times 1 0^{- 2} m

rise of water,

h = 1.45 \times 1 0^{- 2} m

.

Using

g = 9.80 m / s^{2}

and the simplified relation

T = \frac{rhg}{2} \times 1 0^{3} N / m

, the possible error in surface tension is closest to :

Answer: (D)

Surface tension here is a product,

T = \frac{r h g}{2}

, so the fractional errors simply add:

\frac{Δ T}{T} = \frac{Δ r}{r} + \frac{Δ h}{h}

, with

g

taken as exact. Because

r = \frac{D}{2}

, the first term is just

\frac{Δ D}{D}

. Using a least count of

1 0^{- 4}

\frac{Δ D}{D} = \frac{1 0 ^{- 4}}{1.25 \times 1 0 ^{- 2}} \approx 0.8%

and

\frac{Δ h}{h} = \frac{1 0 ^{- 4}}{1.45 \times 1 0 ^{- 2}} \approx 0.7%

. Adding, the error is close to

1.5%

JEE Main · 2017 · 02 Apr 2017 · Offline · Single correct option