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Mechanical Properties of Fluids

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5 months ago

Mechanical Properties of Fluids Class 11th

A spherical soap bubble of radius 3cm is formed inside another spherical soap bubble of radius 6cm. If the internal pressure of the smaller bubble of radius 3cm in the above system is equal to the internal pressure of the another single soap bubble of radius r cm.

The value of r is___________.

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Vishal Baghel

Contributor-Level 10

$P_{2} - P_{06} = \frac{4 T}{6} & P_{1} - P_{2} = \frac{4 T}{3}$

$P_{1} - P_{0} = \frac{4 T}{2} = \frac{4 T}{r}$

r = 2cm

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5 months ago

Mechanical Properties of Fluids Class 11th

A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5cm. Another wire of the same material of length 4L and radius 4r pulled by a force 4F under same conditions. The increases in length of this wire is _________cm.

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Vishal Baghel

Contributor-Level 10

$Δ L_{1} = \frac{F L}{A Y} = \frac{F L}{π r^{2} Y} = 5 c m$

$Δ L_{2} = \frac{4 F 4 L}{π 16 r^{2} y}$ = $\frac{F L}{π r^{2} Y} = 5 c m$

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5 months ago

Mechanical Properties of Fluids Class 11th

The diameter of an air bubble which was initially 2 mm, rises steadily through a solution of density 1750 kg m^-³ at the rate of 0.35 cms^-¹. The coefficient of viscosity of the solution is_________ poise (in nearest integer). (the density of air is negligible).

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Vishal Baghel

Contributor-Level 10

d₁ = 2mm = 2r

$ρ = 1750 k g / m^{3}$ coeff of viscosity H

$\frac{d}{d t} (2 r) = 0.35 c m / s$

here, $(\frac{4}{3} π r^{3}) ρ g = 6 π n r v$

$\Rightarrow η = \frac{4 r^{2} ρ g}{3 * 6 v} = \frac{2 r^{2} ρ g}{9 v}$

$\frac{350}{9 * 35} = \frac{10}{9} ? 1$

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5 months ago

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

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)

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alok kumar singh

Contributor-Level 10

NLM 1

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

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

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

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

$= 0.1234 * 1 0^{- 3}$

vT = 123.4 * 10^-6m/sec

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New answer posted

5 months ago

Mechanical Properties of Fluids physics ncert exemplar solutions class 12th chapter two

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:

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Vishal Baghel

Contributor-Level 10

Kindly consider the following answer

$R e = \frac{ρ v d}{η}$

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5 months ago

Mechanical Properties of Fluids physics ncert exemplar solutions class 12th chapter two

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 R_e is:

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New answer posted

5 months ago

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

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

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Payal Gupta

Contributor-Level 10

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)

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

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5 months ago

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

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)

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Vishal Baghel

Contributor-Level 10

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

R³ = 729r³

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

R = 9r

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

$= T * 4 π * 8 R^{2}$

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

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5 months ago

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

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)

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Vishal Baghel

Contributor-Level 10

$ρ_{ω} = 1000 k g / m^{3}$

G = 10 m/s²

Using bernaulli's eqn

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

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

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5 months ago

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

A pressure-pump has a horizontal tube of cross sectional area 10cm2 for the outflow of water at a speed of 20 m/s. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is:

[given : density of water = 1000 kg/m³]

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Vishal Baghel

Contributor-Level 10

A = 10 cm², V = 20 m/s

$F = \frac{d p}{d t} = ρ A V^{2} = 1 0^{3} * 10 * 1 0^{- 4} * 400$

= 400 N.

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