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Physics NCERT Exemplar Solutions Class 12th Chapter Ten

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3 months ago

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

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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3 months ago

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

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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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Ten Physics Wave Optics

Using Young's double slit experiment, a monochromatic light of wavelength 5000 $\overset{o}{A}$ produces fringes of fringe width 0.5mm. If another monochromatic light of wavelength 6000 $\overset{o}{A}$ is used and the separation between the slits is doubled, then the new fringe width will be:

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alok kumar singh

Contributor-Level 10

$β_{1} = \frac{λ_{1} D_{1}}{d_{1}}$

$β_{2} = \frac{λ_{2} D_{2}}{d_{2}}$

$\frac{β_{1}}{β_{2}} = \frac{λ_{1}}{d_{1}} * \frac{d_{2}}{λ_{2}}$

$\frac{0.5}{β_{2}} = \frac{5000 * d_{1}}{d_{1} * 600}$

$β_{2} = \frac{0.5 * 6}{10}$

2 = 0.3mm

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3 months ago

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

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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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Ten Wave Optics

In Young's double slit experiment, the fringe width is 12mm. If the entire arrangement is placed in water refractive index $\frac{4}{3}$ , then the fringe width becomes (in mm):

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Vishal Baghel

Contributor-Level 10

$β = 12 m m = 12 * 1 0^{- 3} m$

$\begin{array}{l} β' = \frac{β}{μ} = \frac{12 * 1 0^{- 3}}{4 / 3} = 9 * 1 0^{- 3} m \\ β' = 9 m m \end{array}$

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3 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Ten Moving Charges and Magnetism

An infinitely long hollow conducting cylinder with radius R carries a uniform current along its surface. Choose the correct representation of magnetic field (B) as a function of radial distance (r) from the axis of cylinder.

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Payal Gupta

Contributor-Level 10

Using Ampere's law for long hollow cylinder carrying current on its surface

B_in = 0

$B_{o u t} = \frac{μ_{0} i}{2 π r}$

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Ten Wave Optics

A beam of plane polarized light of large cross-sectional area and uniform intensity of $3.3 {W m}^{- 2}$ falls normally on a polarizer (cross sectional area $3 * 10^{- 4} m^{2}$ ) which rotates about its axis with an angular speed of $31.4 r a d / s$ . The energy of light passing through the polariser per revolution, is close to:

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alok kumar singh

Contributor-Level 10

$E = (I) (t) (A) ?{c o s}^{2} ? θ?$
$\Rightarrow (3.3) [\frac{2 π}{31.4}] [3 * 10^{- 4}] * \frac{1}{2}$

$\Rightarrow 0.99 * 10^{- 4}$

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Ten Wave Optics

Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum:

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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4 months ago

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

One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture of constant volume is $\frac{α^{2}}{4} R J / m o l K;$ then the value of will be________.

(Assume that the given diatomic gas has no vibrational mode.)

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Vishal Baghel

Contributor-Level 10

$C_{v} = \frac{α^{2} R}{4} J / m o l^{- k}$

As, Cv (mix) = $\frac{1 * \frac{3}{2} R + 3 * \frac{5}{2} R}{4} = \frac{9 R}{4} = \frac{α^{2} R}{4}$

$∴ α = 3$

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4 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Ten Wave Optics

Orange light of wavelength $6000 * 10^{- 10} m$ illuminates a single slit of width $0.6 * 10^{- 4} m$ . the maximum possible number of diffraction minima produced on both sides of the central maximum is

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alok kumar singh

Contributor-Level 10

Sol. Angular momentum conservation:

$\begin{array}{r} \Rightarrow I_{1} ω_{1} + I_{2} ω_{2} = (I_{1} + I_{2}) ω_{f} \\ \Rightarrow \frac{{M R}^{2}}{2} ω_{o} = (\frac{{M R}^{2}}{2} + \frac{{M R}^{2}}{8}) ω_{f} \\ \Rightarrow ω_{f} = \frac{4}{5} ω_{o} \\ \Rightarrow {K E}_{final} = \frac{1}{2} (I_{1} + I_{2}) ω_{f}^{2} = \frac{{M R}^{2} ω^{2}}{5} \\ \Rightarrow K E_{initial} = \frac{1}{2} I, ω_{0}^{2} = \frac{{M R}^{2} ω^{2}}{4} \\ \Rightarrow % loss \Rightarrow 20 % . \end{array}$

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