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Physics MCQs for NEET — Practice Questions with Answers

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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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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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