In a horizontal flow the pressure of water is reduced when it comes to narrow pipe from wide while flowing. This is explained using :

the principle of continuity (Av = const.)
the Bernoulli theorem (p + ¹⁄₂ρ v²= const.)
the principle of continuity and the Bernoulli’s theorem together
none of the above

Solution

The mechanical energy of an ideal fluid remain constant.

Thus p + ¹⁄₂ρ v²= const.,

Also av = const.

The wetability of a surface by a liquid depends primarily on

surface tension
density
angle of contact between the surface and the liquid
viscosity

Solution

Wetability of a surface by a liquid primarily depends on angle of contact between the surface and liquid.If angle of contact is acute liquids wet the solid and vice-versa.

The terminal velocity v_r of a small steel ball of radius r falling under gravity through a column of a viscous liquid of coefficient of viscosity η depends on mass of the ball m,acceleration due to gravity g, coefficient of viscosity η and radius r. Which of the following relations is dimensionally correct ?

v_r α \(\frac{mgr}{\eta }\)
v_r α mgηr
v_r α \(\frac{mg}{r\eta }\)
v_r α \(\frac{\eta mg}{r}\)

Solution

Note that according to Stoke’s law

6πηrv_r= mg

Hence, the valid relation is v_r α mg/rη

A U-Tube contains some liquid moves horizontally with acceleration a,if separation between two limbs is x then difference between their liquid level y is

^ax⁄_2g
^ax⁄_g
^2ax⁄_g
^a⁄_xg

In a surface tension experiment with a capillary tube, water rises upto 0.1 m. If the same experiment is repeated in an artificial satellite revolving around the earth, then the water will rise in the capillary upto a height of

0.1 m
9.8 m
0.98 m
full length of capillary tube

A tank has a small hole at its botom of area of cross-sectiona. Liquid is being poured in the tank at the rate Vm³/s,the maximum level of liquid in the container will be(Area of tank A)

\(\frac{V}{gaA}\)
\(\frac{V^{2}}{2gAa}\)
\(\frac{V^{2}}{gAa}\)
\(\frac{V}{2gaA}\)

Two liquids of densities d₁ and d₂ are flowing in identical capillary tubes uder the same pressure difference. If t₁ and t₂ are time taken for the flow of equal quantities (mass) of liquids, then the ratio of coefficient of viscosity of liquids must be

\(\frac{d_{1}t_{1}}{d_{2}t_{2}}\)
\(\frac{t_{1}}{t_{2}}\)
\(\frac{d_{1}}{d_{2}}\frac{t_{2}}{t_{1}}\)
\(\sqrt{\frac{d_{1}t_{1}}{d_{2}t_{2}}}\)

Two soap bubbles(surface tension T) coalesce to form a big bubble under isothermal condition. If in this process the change in volume be V and the surface are abe S, then the correct relation is (P is atmospheric pressure)

PV +TS = 0
3PV + 4TS = 0
3PV + TS = 0
4PV + 3TS = 0

Figure shows a capillary rise H. If the air is blown through the horizontal tube in the direction as shown then rise in capillary tube will be

= H
> H
< H
zero

Solution

Due to increase in velocity,pressure will be low above the surface of water.

A sphere of brass released in a long liquid column attains a terminial speed v₀. If the terminal speed attained by sphere of marble of the same radius and released in the same liquid is nv₀, then the value of n will be (Given : The specific gravities of brass,marbles and the liquid are 8.5, 2.5 and 0.8respectively.)

⁵⁄₁₇
¹⁷⁄₇₇
¹¹⁄₃₁
¹⁷⁄₅

Solution

For the same radius, terminal velocity is directly proportional to density difference.

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