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

GivEn:

• Rise in tube, h = 7 cm = 7 × $\sf 10^{-2}\;m$
• Surface Tension, T = 70 dyne/cm = 70 × $\sf 10^{-3}\;N/m$

⠀⠀⠀⠀⠀⠀⠀

To find:

• Radius of capillary tube?

⠀⠀⠀⠀⠀⠀⠀

SoluTion:

☯ Let radius of capillary tube be r.

⠀⠀⠀⠀⠀⠀⠀

Here,

• Density of water, p = 1 × $\sf 10^{34}$ g/m³
• Acceleration due to gravity, g = 9.8 m/s²
• Angle of constant, $\theta$ = 0⁰

⠀⠀⠀⠀⠀⠀⠀

Therefore,

The rise in tube is given by,

⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

$\;\;\;\;\;\;\;\small\sf \underline{Putting\;values\;:}$

⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

$:\implies{\underline{\boxed{\sf{\pink{0.2\;m}}}}}\;\bigstar$

⠀⠀⠀⠀⠀⠀⠀

$\therefore$ Hence, Radius of capillary tube is 0.2 m.

Answer 2

Explanation:

hope it will help you ✌️❤️❤️