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Language English हिंदी

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Water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2P, the rate of flow becomes

$4 Q$

$Q$

$\frac{Q}{4}$

$\frac{Q}{8}$

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Explanation

(d)

$V = \frac{π (∆ P) r^{4}}{8 η l} ∴ V \propto (∆ P) r^{4} (η and l are constants) ∴ \frac{V_{2}}{V_{1}} = (\frac{∆ P_{2}}{∆ P_{1}}) {(\frac{r_{2}}{r_{1}})}^{4} = 2 \times {(\frac{1}{2})}^{4} = \frac{1}{8} ∴ V_{2} = \frac{Q}{8}$

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Water is flowing in a pipe of diameter 4 cm with a velocity 3 m/s. The water then enters into a tube of diameter 2 cm. The velocity of water in the other pipe is

3 m/s

6 m/s

12 m/s

8 m/s

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Explanation

(c)

$a_{1} v_{1} = a_{2} v_{2} \Rightarrow \frac{v_{2}}{v_{1}} = \frac{a_{1}}{a_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} \Rightarrow v_{2} = 3 \times (2)^{2} = 12 m / s$

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What is the velocity v of a metallic ball of radius r falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body ? (The densities of metal and of liquid are $ρ$ and $σ$ respectively, and the viscosity of the liquid is $η$ ).

$\frac{r^{2} g}{9 η} (ρ - 2 σ)$

$\frac{r^{2} g}{9 η} (2 ρ - σ)$

$\frac{r^{2} g}{9 η} (ρ - σ)$

$\frac{2 r^{2} g}{9 η} (ρ - σ)$

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Explanation

(c)

$W h e n t h e a c c e l e r a t i o n o f t h e b a l l i s \frac{g}{2}; F o r c e e q u a t i o n : M g - B - F = M a V ρ g - V σ g - 6 π η r v = V ρ \frac{g}{2} \Rightarrow 6 π η r v = \frac{V ρ g}{2} - V σ g \Rightarrow 6 π η r v = \frac{V g}{2} (ρ - 2 σ) = \frac{4}{3} {πr}^{3} \frac{g}{2} (ρ - 2 σ) \Rightarrow v = \frac{r^{2} g}{9 η} (ρ - 2 σ)$

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Consider the following equation of Bernouilli’s theorem.

$P + \frac{1}{2} {ρV}^{2} + ρgh = K (constant)$

The dimensions of K/P are the same as that of which of the following

Thrust

Pressure

Angle

Viscosity

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Explanation

$P + \frac{1}{2} {ρV}^{2} + ρgh = K (constant)$

According to principle of homogenecity dimension of K=dimension of $ρ$

Therefore, $\frac{K}{P}$ is dimensionless

out of given options only angle is dimensionless.

So, the dimension of $\frac{K}{P}$ are the same as that of angle.

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An incompressible fluid flows steadily through a cylindrical pipe which has radius 2r at point A and radius r at B further along the flow direction. If the velocity at point A is v, its velocity at point B is

v/2

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Explanation

$A r e a \times v e l o c i t y = co n s \tan t {πr}^{2} \times v = constant i . e ., {r_{1}}^{2} v_{1} = {r_{2}}^{2} v_{2} given, r_{1} = 2 r_{1} r_{2} = r and v_{1} = v Hence, {(2 r)}^{2} \times v = r^{2} . v_{2} \Rightarrow v_{2} = 4 v$

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A block of ice floats on a liquid of density 1.2 $g / c m^{3}$ in a beaker. Then level of liquid when ice completely melts-

Remains same

Rises

Lowers

(a), (b) or (c)

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Explanation

(b) The volume of liquid displaced by floating ice $V_{D} = \frac{M}{σ_{L}}$

Volume of water formed by melting ice, $V_{F} = \frac{M}{σ_{W}}$

If $σ_{L} > σ_{W}$ , then, $\frac{M}{σ_{L}} < \frac{M}{σ_{W}} i . e . V_{D} < V_{F}$

i.e. volume of liquid displaced by floating ice will be lesser than water formed and so the level if liquid will rise.

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A vessel of area of cross-section A has liquid to a height H. There is a hole at the bottom of vessel having area of cross-section a. The time taken to decrease the level from $H_{1}$ to $H_{2}$ will be

$\frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H_{1}} - \sqrt{H_{2}}]$

$\sqrt{2 gh}$

$\sqrt{2 gh (H_{1} - H_{2})}$

$\frac{A}{a} \sqrt{\frac{g}{2}} [\sqrt{H_{1}} - \sqrt{H_{2}}]$

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Explanation

At height h the rate of discharge of liquid through hole of bottom = $\sqrt{2 g h} \times a$

Let in time dt level of water in tank decreases by dh.

Equating volume

$A d h = d t \sqrt{2 g h} \times a \Rightarrow d t = \frac{A}{a \sqrt{2 g}} \frac{d h}{\sqrt{h}}$

Now on inegrating

$\int d t = \frac{A}{a \sqrt{2 g}} \int_{H_{2}}^{H_{1}} \frac{d h}{h} \Rightarrow t = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H_{1}} - \sqrt{H_{2}}]$

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A liquid does not wet the sides of a solid if the angle of contact is

Zero

Obtuse (More than 90°)

Acute (Less than 90°)

90°

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Explanation

(b)

We know that the liquids make the solid substances wet when they get into physical contact with each other. This depends on the contact angle between the liquid and solid surfaces. When the liquid does not wet the sides of a solid object, it means that the angle of contact is more than $90 °$ but less than $180 °.$

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The meniscus of mercury in the capillary tube is

Convex

Concave

Plane

Uncertain

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Explanation

Convex meniscus is formed because of cohesion and adhesion.

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At which of the following temperatures, the value of surface tension of water is minimum ?

$4 ° C$

$25 ° C$

$50 ° C$

$75 ° C$

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Explanation

(d) Surface tension decreases with increase in temperature.

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Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.