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The Working of an atomizer depends upon

Bernoulli's theorem

Boyle's law

Archimedes principle

Newton's law of motion

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Explanation

(a)

Atomizer works on the principle of liquid flow, It follows bernaulis principle because here, horizontal air passing causes the vertices liquid to come out.

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The pans of a physical balance are in equilibrium. Air is blown under the right hand pan; then the right hand pan will

Move up

Move down

Move erratically

Remain at the same level

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Explanation

(b) According to Bernoulli's theorem.

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According to Bernoulli's equation

$\frac{P}{ρg} + h + \frac{1}{2} \frac{v^{2}}{g} = constant$

The terms A, B and C are generally called respectively:

Gravitational head, pressure head and velocity head

Gravity, gravitational head and velocity head

Pressure head, gravitational head and velocity head

Gravity, pressure and velocity head

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Explanation

(c)

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A sniper fires a rifle bullet into a gasoline tank making a hole 53.0 m below the surface of gasoline. The tank was sealed at 3.10 atm. The stored gasoline has a density of 660 ${kgm}^{- 3}$ . The velocity with which gasoline begins to shoot out of the hole is

27.8 ${ms}^{- 1}$

41.0 ${ms}^{- 1}$

9.6 ${ms}^{- 1}$

19.7 ${ms}^{- 1}$

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Explanation

The velocity of the gasoline shooting out of the hole can be calculated using Bernoulli's equation. The pressure difference between the inside of the tank and the atmospheric pressure provides the driving force for the flow, and the velocity depends on the height of the gasoline column.

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A rectangular vessel when full of water takes 10 minutes to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water

9 minute

7 minute

5 minute

3 minute

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Explanation

(b) Time taken to be emptied for h height, $t = \sqrt{\frac{2 h}{g}}$

and for $\frac{h}{2}$ height , $t = \sqrt{\frac{2 h / 2}{g}} = \sqrt{\frac{h}{g}} ∴ \frac{t'}{t} = \frac{1}{\sqrt{2}} \Rightarrow t' = \frac{t}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 7 minute$

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A streamlined body falls through air from a height h on the surface of a liquid. If d and D(D > d) represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is

$\sqrt{\frac{2 h}{g}}$

$\sqrt{\frac{2 h}{g} \cdot \frac{D}{d}}$

$\sqrt{\frac{2 h}{g} \cdot \frac{d}{D}}$

$\sqrt{\frac{2 h}{g}} (\frac{d}{D - d})$

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Explanation

(d) Upthrust – weight of body = apparent weight

VDg-Vdg = Vda,

Where a = retardation of body $∴ a = (\frac{D - d}{d}) g$

The velocity gained after fall from h height in air,

$v = \sqrt{2 gh}$

Hence, time to come in rest,

$t = \frac{v}{a} = \frac{\sqrt{2 gh} \times d}{(D - d) g} = \sqrt{\frac{2 h}{g}} \times \frac{d}{(D - d)}$

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A large tank of cross-section area A is filled with water to a height H. A small hole of area 'a' is made at the base of the tank. It takes time $T_{1}$ to decrease the height of water to $\frac{H}{η} (η > 1)$ ; and it takes $T_{2}$ time to take out the rest of water. If $T_{1} = T_{2}$ , then the value of $η$ is

$2 \sqrt{2}$

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Explanation

(c) $t = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H_{1}} - \sqrt{H_{2}}] Now, T_{1} = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{H} - \sqrt{\frac{H}{η}}] and T_{2} = \frac{A}{a} \sqrt{\frac{2}{g}} [\sqrt{\frac{H}{η}} - 0]$

According to problem

$T_{1} = T_{2} ∴ \sqrt{H} - \sqrt{\frac{H}{η}} = \sqrt{\frac{H}{η}} - 0 \Rightarrow \sqrt{H} = 2 \sqrt{\frac{H}{η}} \Rightarrow η = 4$

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As the temperature of water increases, its viscosity

Remains unchanged

Decreases

Increases

Increases or decreases depending on the external pressure

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Explanation

(b)

As the temperature of a liquid increases, the energy of its molecules increases which increases the movement of molecules.so the liquid becomes more fluid. thus viscosity of liquid decreases.

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A small drop of water falls from rest through a large height h in air; the final velocity is

$\propto \sqrt{h}$

$\propto h$

$\propto (1 / h)$

Almost independent of h

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Explanation

(d)

$T e r m i n a l v e l o c i t y v = \frac{2 r^{2} g (σ - ρ)}{9 η}$

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The rate of flow of liquid in a tube of radius r, length l, whose ends are maintained at a pressure difference P is $V = \frac{{πQPr}^{4}}{ηl}$ where $η$ is coefficient of the viscosity and Q is

$\frac{1}{8}$

$\frac{1}{16}$

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Explanation

(b) In this formula, Q=1/8.

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