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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A sinusoidal wave y(t) = 40 sin (10 * 10⁶pt) is amplitude modulated by another sinusoidal wave x(t) = 20sin(1000πt). The amplitude of minimum frequency component of modulated signal is:

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Vishal Baghel

Contributor-Level 10

$C_{m} (t) = A_{C} s i n ω_{C} t + \frac{μ A_{C}}{2} [c o s (ω_{C} - ω_{m}) t - c o s (ω_{C} + ω_{m}) t]$

$μ = \frac{A m}{A_{C}}$

$A_{C} = 40$

$μ = \frac{1}{2}$

So amplitude of minimum frequency = $\frac{μ A_{C}}{2}$

= 10

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11 months ago

Physics Moving Charges and Magnetism

The I-V characteristics of p-n junction diode in forward bias is shown in the figure. The ration of dynamic resistance, corresponding to forward bias voltage of 2 V and 4 V respectively, is

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Payal Gupta

Contributor-Level 10

Dynamic Resistance at 2v

= $\frac{Δ v}{Δ I} = \frac{2.1 - 2}{(10 - 5) * 1 0^{- 3}} = \frac{0.1}{5} * 1 0^{3}$

Dynamic Resistance at 4 v

= $\frac{Δ v}{Δ I} = \frac{4.2 - 4}{(250 - 200)} = \frac{0.2}{50 * 1 0^{- 3}} = \frac{0.2}{50} * 1 0^{3}$

$\frac{R_{2 v}}{R_{4 v}} = \frac{0.1 * 1 0^{3} * 50}{5 * 0.2 * 1 0^{3}} = 5 : 1$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

The positive feedback is required by an amplifier to act an oscillator. The feedback here means:

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Vishal Baghel

Contributor-Level 10

A portion of the output power is returned back to the input.

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A radioactive nucleus can decay by two different processes. Half-life for the first process is 3.0 hours while it is 4.5 hours for the second process. The effective half-life of the nucleus will be:

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Vishal Baghel

Contributor-Level 10

$\frac{1}{t} = \frac{1}{t_{1}} + \frac{1}{t_{2}}$

$\frac{1}{t} = \frac{1}{3} = \frac{1}{4.5}$

$= \frac{4.5 + 3}{3 * 4.5}$

t = $\frac{3 * 4.5}{7.5} = \frac{9}{5}$

t = 1.8 hrs

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A metal surface is illuminated by a radiation of wavelength 4500 . The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90° with the magnetic field. If it starts revolving in a circular path of radius 2nm, the work function of the metal is approximately:

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Vishal Baghel

Contributor-Level 10

$k = \frac{h c}{λ} - ?$

As radius, r = $\frac{m v}{e B} = \frac{\sqrt[]{2 k m}}{e B}$

$\begin{array}{l} \Rightarrow {(r e B)}^{2} = 2 k m \\ \Rightarrow ? = \frac{h c}{λ} - \frac{{(r e B)}^{2}}{2 m} \end{array}$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

For a specific wavelength 670nm of light coming from a galaxy moving with velocity v, the observed wavelength is 670.7nm.

The value of v is:

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Vishal Baghel

Contributor-Level 10

$\frac{Δ λ}{λ} = \frac{v}{c}$

$v = \frac{Δ λ}{λ} C$

= $\frac{0.7 * 1 0^{- 9}}{670 * 1 0^{- 9}} * 3 * 1 0^{8}$

= $3.13 * 1 0^{5}$

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11 months ago

Physics Dual Nature of Radiation and Matter

An electron with speed $υ$ and a photon with speed c have the same de-Broglie wavelength. If the kinetic energy and momentum of electron are E_e and p_e and that of photon are E_ph and p_ph respectively. Which of the following is correct?

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Payal Gupta

Contributor-Level 10

$λ_{e} = \frac{h}{m v} = \frac{h}{p_{e}}$ - (1)

$λ_{P h} = \frac{h}{P_{P h}}$ - (2)

A/C to question

$λ_{e} = λ_{P h}$

$P_{e} = P_{P h} \Rightarrow \frac{P_{P}}{P_{P h}} = 1$ - (3)

$k . E_{e} = E_{e} = \frac{1}{2} m v^{2}$

$= \frac{1}{2} P e v$ - (4)

$k . E_{P h} = E_{P h} = m c^{2}$

= mc c

q = P_Ph c- (5)

$(4) \div (5)$

$\frac{E_{e}}{E_{P h}} = \frac{v}{2 c}$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

Which is the correct ascending order of wavelengths?

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Vishal Baghel

Contributor-Level 10

If frequency is night wavelength is low.

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

A metallic conductor of length 1m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5rad s^-1. If the horizontal component of earth's magnetic field is 0.2 * 10^-4T, then emf induced between the two ends of the conductor is:

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Vishal Baghel

Contributor-Level 10

emf induced = $B_{H} \frac{ω l^{2}}{2}$

$\begin{array}{l} = 0.2 * 1 0^{- 4} * 5 * \frac{1^{2}}{2} \\ = 50 μ v \end{array}$

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11 months ago

Physics physics ncert exemplar solutions class 12th chapter two

Two coils of self inductance L₁ and L₂ are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be:

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Vishal Baghel

Contributor-Level 10

$L_{e q} = L_{1} + L_{2} - 2 M$

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