Physics NCERT Exemplar Solutions Class 12th Chapter Ten Class 12th Physics Mechanical Properties of Fluids

A cube of metal is subjected to a hydrostatic pressure 4GPa. The percentage change in the length of the side of the cube is close to: (Given bulk modulus of metal, $B = 8 * 10^{10}$ $P a)$

Option 1 - <p> 0.6</p>

Option 2 - <p>5</p>

Option 3 - <p>1.67</p>

Option 4 - <p>20(Fluids)</p>

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Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 2

Detailed Solution:

Sol.

$\begin{array}{r} \end{array}$ $\begin{array}{r} ↑_{1}^{?} (1) ε = 3 \\ (2) ε - I r = 2.5 V \\ \Rightarrow I r = 0.5 \\ Now, IR = 2.5 \\ \Rightarrow \frac{R}{r} = 5 . \\ \Rightarrow \frac{P_{R}}{R_{r}} = \frac{I^{2} R}{I^{2} r} = \frac{R}{r} = 5 \\ \Rightarrow P_{r} = \frac{0.5}{5} = 0.1 \end{array}$

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A water drop of radius $1 μ m$ falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 × 10^-⁵ Nsm^-² and its density is negligible as compared to that of water 10⁶ gm^-³. Terminal velocity of the water drop is:

(Take acceleration due to gravity = 10 ms^-²)

Alok Kumar Singh

NLM 1

$ρ_{ω} \frac{4}{3} π r^{3} g = 6 π η r V_{T}$

$\frac{ρ_{ω} 4}{3} π r^{2} g =$ $6 π η v_{T}$

$V_{T} = \frac{4}{3} \frac{ρ_{ω} π r^{2} g}{6 π η}$

$= \frac{2}{9} \times 1 0^{3} \times \frac{{(1 0^{- 6})}^{2} \times 10}{1.8 \times 1 0^{- 5}}$

$= 0.1234 \times 1 0^{- 3}$

vT = 123.4 × 10^-6m/sec

If p is the density and $η$ is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number Re is:

Kindly consider the following answer

$R e = \frac{ρ v d}{η}$

Two capacitors having C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is not connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium, is

After switch ‘S’ is closed

$Q_{1} + Q_{2} = C_{1} V$ - (1)

Using KVL

$\frac{Q_{1}}{C_{1}} - \frac{Q_{2}}{C_{2}} = 0$

$Q_{1} = Q_{2} \frac{C_{1}}{C_{2}}$ - (2)

from (1) & (2)

$\Rightarrow Q_{2} [\frac{C_{1} + C_{2}}{C_{2}}] = C_{1} V \Rightarrow Q_{2} = (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) V$

A water drop of radius 1cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm, then the gain in surface energy upto first decimal place will be:

(Given π = 3.14)

$\frac{4}{3} π R^{3} = 729 \frac{4}{3} π r^{3}$

R³ = 729r³

$R = (729) \frac{1}{3} (r) (\frac{1}{3})$

R = 9r

$Δ u = T (4 π r^{2}) \times 729 - T \times 4 π R^{2}$

$= T \times 4 π \times 8 R^{2}$

$\begin{array}{l} = 75.39 \times 1 0^{- 5} J \\ Δ u = 7.5 \times 1 0^{- 4} J \end{array}$

Consider a cylindrical tank of radius 1m is filled with water. The top surface of water is at 15m front the bottom of the cylinder. There is a hole on the wall of a cylinder at a height of 5m from the bottom. A force of 5 × 10⁵N is applied an the top surface of water using a piston. The speed of ifflux from the hole will be:

(given atmospheric pressure P_A = 1.01 × 10⁵Pa, density of water $ρ_{w} = 1000 k g / m^{3}$  and gravitational acceleration g = 10m/s²)

$ρ_{ω} = 1000 k g / m^{3}$

G = 10 m/s²

Using bernaulli’s eqn

$\frac{5 \times 1 0^{5}}{π} + ρ g \times 10 = ρ_{0} + \frac{1}{2} ρ V_{e}^{2}$

$= \sqrt[]{318} ? 17.8 m / s$

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Physics NCERT Exemplar Solutions Class 12th Chapter Ten 2025

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