Question

A capillary tube of uniform bore is dipped vertically in water, which rises by 7 cm in the tube find the radius of capillary if surface tension rises of water is 70 dyne/cm​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• Rise in tube, h = 7 cm

= 7 × $\sf 10^{-2}\;m$

• Surface Tension, T = 70 dyne/cm

= 70 × $\sf 10^{-3}\;N/m$

$\;\;\underline{\textbf{\textsf{ To Find :-}}}$⠀⠀⠀⠀⠀⠀⠀

• Radius of capillary tube, r

⠀⠀⠀⠀⠀⠀⠀

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

⠀⠀$\underline{\:\textsf{ We know that :}}$⠀⠀⠀⠀⠀

$\;{\boxed{\sf{\purple{h = \dfrac{2T\;cos\;\theta}{rpg}}}}}$

⠀⠀⠀⠀⠀⠀⠀

$\sf Where,$

• Density of water, p = 1 × $\sf 10^{34}$ g/m³

• g = 9.8 m/s²

• Angle of constant, $\sf \theta$ = 0⁰

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\underline{\:\textsf{ Now, put the given values in the formula :}}$⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀

$\;{\boxed{\sf{\purple{h = \dfrac{2T\;cos\;\theta}{rpg}}}}}$

⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow \sf 7 \times 10^{-2} = \dfrac{2 \times 70 \times 10^{-3} \times cos\;0^0}{r \times 1 \times 10^{34} \times 9.8}$

⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow \sf r = \dfrac{2 \times 70 \times 10^{-3} \times cos\;0^0}{7 \times 10^{-2} \times 1 \times 10^3 \times 9.8}$

⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow \sf r = 2 \times 10^{-4}\;m$

⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow \sf r = 0.2 \times 10^{-3}$

⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow {\underline{\boxed{\sf{\purple{0.2\;m}}}}}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{ Radius of capillary tube is </p><p>\textbf{ 0.2m}}}.$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

⠀⠀⠀⠀⠀⠀⠀

Answer 2

Explanation:

0.2 m is the answer........