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Physics NCERT Exemplar Solutions Class 12th Chapter Six

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5 months ago

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

A fighter jet is flying horizontally at a certain with a speed of 200ms^-¹. When it passes directly overhead ad anti-aircraft gun, a bullet is fired from the gun, at an angle q with the horizontal, to hit the jet. If the bullet speed is 400m/s, the value of q will be _____________.

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Payal Gupta

Contributor-Level 10

B fighter jet

A anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos = 200

$\begin{array}{l} c o s θ = \frac{1}{2} \\ θ = 60 ° \end{array}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Six Communication Systems

Choose the correct statement for amplitude modulation

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Payal Gupta

Contributor-Level 10

In amplitude modulation, the amplitude of the career signal is varied in according with the modulating signal.

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Physics NCERT Exemplar Solutions Class 12th Chapter Six 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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5 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six 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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5 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Electric Charges and Fields

In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size $\frac{λ}{100},$ where $λ$ is the wavelength of the wave in free space. The phenomenon, which happens there will be

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Payal Gupta

Contributor-Level 10

If size of object is very small as compare to wave length of EM wave in free space then, scattering will happen.

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5 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Electrostatic Potential and Capacitance

If Electric field intensity of a uniform plane electro magnetic wave is given as $E = - 301.6 s i n (k z - ω t) {\hat{a}}_{x} + 452.4 s i n (k z - ω t) \hat{a} t \frac{V}{m} .$

Then, magnetic intensity 'H' of this wave in Am^-¹ will be:

[Given : Speed of light in vacuum c = 3 * 10⁸ ms^-¹, Permeability of vacuum $μ_{0} = 4 π * 1 0^{- 7} N A^{- 2}$ ]

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Payal Gupta

Contributor-Level 10

Since

$\vec{E} . \vec{B} = 0 (\vec{E} ⊥ \vec{B})$

$& \frac{E_{0}}{B_{0}} = C = 3 * 1 0^{8} m / s e c$

for option D

$\frac{E_{0}}{B_{0}} = \frac{\sqrt[]{301 . 6^{2} + 452 . 4^{2}}}{1 0^{- 6} \sqrt[]{1 . 5^{2} + 1^{2}}} \begin{matrix} = 301.6 * 1 0^{6} & = 3.01 * 1 0^{8} m / s e c \end{matrix}$

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Physics NCERT Exemplar Solutions Class 12th Chapter Six Thermodynamics

A proton and an alpha particle of the same velocity enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the radii of the circular paths described by the alpha particle and proton is

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Payal Gupta

Contributor-Level 10

$r = \frac{m v}{q B}$

$r_{α} = \frac{m_{α} v}{q_{α} B}$

$r_{P} = \frac{m_{P} v}{q_{P} B}$

$\frac{r_{α}}{r_{P}} = \frac{m_{α} q_{P}}{m_{P} q_{α}} = \frac{m_{α}}{m_{P}} (\frac{q_{P}}{q_{α}})$

$= \frac{4 m}{m} (\frac{q}{2 q}) = 2 : 1$

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

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is

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Payal Gupta

Contributor-Level 10

Volume = constant

$A l = C$

$A d l + l d A = 0$

$\frac{d l}{l} + \frac{d A}{A} = 0$

$\frac{d R}{R} * 100 = (\frac{d l}{l} - \frac{d A}{A}) * 100$

= 0.4 (0.4)

= 0.8 %

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5 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Nuclei

The magnetic flux through a coil perpendicular to its plane is varying according to the relation $ϕ = (5 t^{3} + 4 t^{2} + 2 t - 5)$ Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2s will be

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Payal Gupta

Contributor-Level 10

$V = \frac{- d ?}{d t} = - (15 t^{2} + 8 t + 2)$

at t = 2 sec, V = (15 * 4 + 8 * 2 + 2)

= 78 V

$I = \frac{V}{R} = \frac{78}{5} = 15.6 A$

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5 months ago

Physics NCERT Exemplar Solutions Class 12th Chapter Six Electromagnetic Induction

Given below two statements : One is labeled as Assertion (A) and other is labeled as Reason (R).

Assertion (A) : Non-polar materials do not have any permanent dipole moment.

Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negativne charge distribution .

In the light of above statements, choose the most appropriate answer from the options given below

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Payal Gupta

Contributor-Level 10

In non-polar molecules, centre of +ve charge coincides with centre of –ve charge, Hence, net dipole moment becomes zero.

When non-polar material is placed in external field, centre of charge does not coincide, hence give non-zero moment.

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