Explanation
By using Bernoulli’s principle
The velocity of the liquid coming out is given as,
$$v = \sqrt {2gh} $$
The mass of the liquid will be
$$m = \rho AV$$
The change in momentum will be
$$P = mv$$
The force will be proportional of $${v^2}$$ which in turn is proportional to $$h$$
The difference in pressure is given as,
$$\Delta P = \rho gh$$
$$0.1 \times {10^5} = \rho gh$$
The velocity of water flow is given as,
$$v = \sqrt {2g \times \dfrac{{0.1 \times {{10}^5}}}{{{{10}^3} \times g}}} $$
$$v = \sqrt {20} \;{\rm{m/s}}$$
Given , angle of contact $$(θ) = 140°$$
Radius of tube $$(r) = 1 mm = 10⁻³ m$$
Surface tension $$(S) = 0.465 N/m$$
Density of mercury$$ (ρ) = 13.6 × 10³ kg/m³$$
Height of liquid rise or fall due to surface tension $$(h)$$
$$ = 2S cos θ/rρg = 2 × 0.465 × cos 140°/1 × 10⁻³ × 13.6 × 10³ × 9.8$$
$$= 2 × 0.465 × (- 0.7660) / 10⁻³ × 13.6 × 10³ × 9.8 $$
$$= - 5.34 mm $$
Hence, the mercury level will depressed by $$5.34 mm.$$
A glass capillary is put in a trough containing one of these for liquids it is observed that the meniscus is convex. the liquid in the trough is Ethyl alcohol.
$$v_{1}=$$ velocity of surface of liquid
$$v_{2}=$$ velocity of liquid far om orifice
Acc. to continuity theorem.
$$A v_{1}=a v_{2} \Rightarrow \dfrac{A v_{1}}{a}=v_{2}$$
$$\Rightarrow \dfrac{a v_{2}}{A}=v_{1}$$
Acc. to Bernoulli's Theorem
$$P_{0}+\rho g h+\dfrac{1}{2} \rho v_{1}^{2}=P_{0}+\rho g(0)+\dfrac{1}{2} \rho v_{L}^{2}$$
$$\rho g h+\dfrac{1}{2} \rho\left[\dfrac{a^{2} v_{1}^{2}}{A^{2}}-v_{1}^{2}\right]=0$$
$$\dfrac{v_{1}^{2}}{2}\left[\dfrac{A^{2}-a^{2}}{A^{2}}\right]=g h$$
$$v_{1}=\sqrt{2 g h} \sqrt{\dfrac{A^{2}}{A^{2}-a^{2}}}$$
Answer: $$ \sqrt{2 g h} \sqrt{ \dfrac{A^{2}}{A^{2}-a^{2}}}$$
Please disable the adBlock and continue. Thank you.