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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

The velocity of sound in a gas, in which two wavelengths 4.08m and 4.16m produce 40 beats in 12s, will be

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Payal Gupta

Contributor-Level 10

$| f_{1} - f_{2} | = \frac{40}{12} = \frac{10}{3}$

$\Rightarrow v (\frac{1}{λ_{1}} - \frac{1}{λ_{2}}) = \frac{10}{3}$

$\Rightarrow$ $V$ $=$ $\frac{10}{3}$ $*$ $\frac{λ_{1} λ_{2}}{λ_{2} - λ_{1}}$ $=$ $\frac{10}{3}$ $*$ $\frac{4.08 * 4.16}{. 08}$ $=$ $\frac{10}{3}$ $*$ $\frac{408 * 4.16}{8}$ $=$ $7$ $0$ $7$ $.$ $2$ $m$ $/$

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6 months ago

Physics Communication Systems

Match List – I with List – II

	List – I		List – II
A.	Television signal	I.	03 KHz
B.	Radio signal	II.	20 KHz
C.	High Quality Music	III.	02 MHz
D.	Human speech	IV.	06 MHz

Choose the correct answer from the options given below:

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Payal Gupta

Contributor-Level 10

Using factual data.

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

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

Assertion A : n-p-n transistor permits more current that a p-n-p transistor.

Reason R : Electrons have greater mobility as a change carrier.

Choose the correct answer form the options given below

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Payal Gupta

Contributor-Level 10

Using factual data.

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

For using a multimeter to identify diode from electrical compounds, choose the correct statement out of the following about the diode

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Payal Gupta

Contributor-Level 10

Diode conducts current only in one direction.

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Fourteen

In the following circuit, the correct relation between output (Y) and inputs A and B will be

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Payal Gupta

Contributor-Level 10

From Truth table $Y = \bar{A B}$ .

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Thirteen

The Q-value of a nuclear reaction and kinetic energy of the projectile particle, K_p are related as

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Payal Gupta

Contributor-Level 10

X + p → Y + b

$Q = K_{Y} + K_{b} - K_{P}$

$Q + K_{P} = K_{Y} + K_{b}$

$\Rightarrow Q + K_{P} > 0$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

THE de Broglie wavelength for an electron and a photon are $λ_{e} a n d λ_{p}$ respectively. For the same kinetic energy of electron and photon, which of the following presents the correct relation between the de Broglie wavelengths of two

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Payal Gupta

Contributor-Level 10

$λ_{e} = \frac{h}{P_{e}} = \frac{h}{\sqrt[]{2 m_{e} K_{e}}} \Rightarrow λ_{e}^{2} = \frac{h^{2}}{2 m_{e} k_{e}} \Rightarrow K_{e} = \frac{h^{2}}{2 m_{e} λ_{e}^{2}}$ - (i)

$λ_{P} = \frac{h}{P_{P}} = \frac{h c}{K_{P}} \Rightarrow K_{P} = \frac{h c}{λ_{P}}$ - (ii)

K_e = K_P

$\Rightarrow λ_{P} α λ_{e}^{2}$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a alight of wavelength 2440 $\overset{o}{A}$ is used to see the object will be

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Payal Gupta

Contributor-Level 10

Resolving power, = $\frac{d}{1.22 λ} = \frac{24.4 * 1 0^{- 2}}{1.22 * 2440 * 1 0^{- 10}}$

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6 months ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The refracting angle of a prims is A and refractive index of the material of the prism is cot (A/2). Then the angle of minimum deviation will be

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Payal Gupta

Contributor-Level 10

At minimum deviation –

$δ = 2 i - A \Rightarrow i = \frac{δ + A}{2}, r = \frac{A}{2}$

Using snell's low : - $s i n i = μ = s i n r$

$\Rightarrow \frac{δ + A}{2} = \frac{π - A}{2} \Rightarrow δ = π - 2 A$

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6 months ago

Physics Electromagnetic Waves

A radar sends an electromagnetic single of electric (E₀) = 2.25 V/m and magnetic field (B₀) = 1.5 * 10^-⁸ T which strikes a target on line of sight at a distance of 3km in a medium. After that, a part of signal (echo) reflects back towards the radar with same velocity and by same path. If the signal was transmitted at time t = 0 from radar, then after how much time echo will reach to the radar?

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Payal Gupta

Contributor-Level 10

$E_{0} = 2.25 \frac{v}{m}$ Speed of declromagnetic signal, $v = \frac{E_{0}}{B_{0}} = \frac{2.25 v / m}{1.5 * 1 0^{- 8} T}$

$B_{0} = 1.5 * 1 0^{- 8} T \Rightarrow v = \frac{225}{150} * 1 0^{8} = 1.5 * 1 0^{8} m / s$

Time to reach each to the same

radar, $t = \frac{2 * 3 * 1 0^{3} m}{1.5 * 1 0^{8} m / s} = 4 * 1 0^{- 5} S$

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