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Physics NCERT Exemplar Solutions Class 12th Chapter Six

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Electromagnetic Waves

An expression for a dimensionless quantity P is given by $P = \frac{α}{β} l o g_{e} (\frac{k t}{β t});$ where α and β are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of will be

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Payal Gupta

Contributor-Level 10

$P = \frac{α}{β} l o g_{e} (\frac{k t}{β x})$

$\frac{k t}{β x}$ = Dimensionless

$β = \frac{k t}{x} = \frac{[M L^{2} T^{- 2} k^{- 1}] [k]}{[L]}$

$\frac{α}{β} =$ dimensionless

= dimensionless of

= MLT²

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Physics NCERT Exemplar Solutions Class 12th Chapter Six The p-Block Elements

Production of iron in blast furnace follows the following equation

$F e_{3} O_{4} (s) + 4 C O (g) \to 3 F e (I) + 4 C O_{2} (g)$

When 4.640 kg of Fe₃O₄ and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is:

[Given : Molar Atomic mass (g mo^-1); Fe = 56

Molar Atomic mass (g mol^-1); O = 16

Molar Atomic mass (g mol^-1); C = 12]

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alok kumar singh

Contributor-Level 10

Moles of Fe₃O₄ = $\frac{4.64 * 1 0^{3}}{232} = 20$

Moles of CO = $\frac{2.52 * 1 0^{3}}{28} = 90$

So limiting reagent = Fe₃O₄

So moles of Fe formed = 60

Weight of Fe = 60 * 56 = 3360

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Work, Energy and Power

A particle of mass 500 gm is moving in a straight line with velocity = The work done by the net force during its displacement from x = 0 to x = 4m is:

(Take b = 0.25m^-3/2s^-1).

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alok kumar singh

Contributor-Level 10

$ω = Δ k . E = \frac{1}{2} m [v_{t}^{2} - v_{i}^{2}]$

$= \frac{1}{2} m {[b * 4^{5 / 2}]}^{2} - \frac{1}{2} m {[0]}^{2}$

$= \frac{1}{2} * 0.5 [0.2 5^{2} * 4^{5}] = 2^{4} = 16 J$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Electromagnetic Induction

In a coil of resistance $8 Ω,$ the magnetic flux due to an external magnetic field varies with time as $ϕ = \frac{2}{3} (9 - t^{2}) .$ The value of total heat produced in the coil, till the flux becomes zero, will be_______J.

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alok kumar singh

Contributor-Level 10

$? = \frac{2}{3} (9 - t^{2}) = 0$

t = 3 sec

$E m f = \frac{- d ?}{d t} = \frac{4 t}{3}$

$l = \frac{E m f}{R} = \frac{\frac{4}{3} t}{8} = \frac{t}{6}$

Heat = $\int l^{2} R d t = \int_{0}^{3} \frac{t^{2}}{36} * 8 d t = 2 J$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Class 12th

A velocity selector consists of electric field $\vec{E} = E \hat{k}$ and magnetic field $\vec{B} = B \hat{j}$ with B = 12 mT. The value of E required for and electron of energy 728 eV moving along the positive x-axis to pass undeflected is:

(Given. Mass of electron = 9.1 * 1031kg)

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alok kumar singh

Contributor-Level 10

$K E = 728 * 1.6 * 1 0^{- 19} = \frac{1}{2} * 9.1 * 1 0^{- 31} * V^{2}$

V = 16 * 106 m/s

For Lorentz force to be zero

eE = eVB

$\Rightarrow E = v B = 16 * 1 0^{6} * 12 * 1 0^{- 3}$

= 192 * 103 V/m

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Class 12th

Two concentric circular loops of radii r₁ = 30cm and r₂ = 50cm are placed in X – Y plane as shown in the figure. A current I = 7A is flowing through them in the direction as shown in figure. The net magnetic momentum of this system of two circular loops is approximately:

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alok kumar singh

Contributor-Level 10

$\vec{M} = I A_{1} - I A_{2}$

$\Rightarrow M = - 7 * \frac{22}{7} * [\frac{25}{100} - \frac{9}{100}] \hat{k}$

$\Rightarrow \vec{M} = \frac{- 7}{2} \hat{k} A m^{2}$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Magnetic Effect of Electric Current

B_X and B_Y are the magnetic fields at the centre of two coils X and Y respectively each carrying equal current. If coil X has 200 turns and 20 cm radius and coil Y has 400 turns and 20 cm radius, the ratio of B_X and B_Y is:

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Vishal Baghel

Contributor-Level 10

$B = N \frac{μ_{0} l}{2 μ}$

$B α \frac{N}{μ}$

$\frac{β_{x}}{β_{y}} = \frac{N_{x}}{r_{x}} * \frac{r_{y}}{N_{y}}$

$\frac{B_{x}}{B_{y}} = \frac{200}{20} * \frac{20}{400} = \frac{1}{2}$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Magnetic Effect of Electric Current

A charge particle is moving in uniform magnetic field $(2 \hat{i} + 3 \hat{j})$ T. If it has an acceleration of $(α \hat{i} - 4 \hat{j}) m / s^{2}$ , then the value of will be:

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Work, Energy and Power

A body mass 8kg and another of mass 2kg are moving with equal kinetic energy. Thee ratio of their respective momenta will be:

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alok kumar singh

Contributor-Level 10

$K E = \frac{ρ^{2}}{2 m}$

$\Rightarrow ρ^{2} α m$

$\Rightarrow {(\frac{P_{1}}{P_{2}})}^{2} = \frac{8}{2} = \frac{4}{1}$

$\Rightarrow \frac{P_{1}}{P_{2}} = \frac{2}{1}$

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2 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Electromagnetic Induction

An aeroplane, with its wings spread 10m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 * 10^-4Wb/m² and the angle of dip is 60°. The emf induced between the tips of the plane wings will be

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Payal Gupta

Contributor-Level 10

B_v = B sin 60°

$B_{v} = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2}$

$E m f = B_{v} * v * l = 2.5 * 1 0^{- 4} * \frac{\sqrt[]{3}}{2} * 180 * \frac{5}{18} * 1 = 108.25 * 1 0^{- 3} v o l t s$

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