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Class 12th Dual Nature of Radiation and Matter

An a particle and a proton are accelerated form rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be:

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alok kumar singh

Contributor-Level 10

$k_{α} = 2 e V \Rightarrow P_{α} = \sqrt[]{2 K_{α} m_{α}}$

k_P = eV -> P_P = $\sqrt[]{2 K_{P} m_{P}}$

$∴ \frac{P_{α}}{P_{P}} = \sqrt[]{\frac{K_{α} m_{α}}{K_{P} m_{P}}} = \sqrt[]{\frac{2 * (4 m_{P})}{m_{P}}} \begin{matrix} \sqrt[]{8} : 1 & 2 \sqrt[]{2} : 1 \end{matrix}$

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5 months ago

Class 12th Communication Systems

In the case of amplitude modulation to avoid distortion the modulation index $(μ)$ should be:

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Vishal Baghel

Contributor-Level 10

Based on Theoretical data (modulation index, $μ \leq 1)$

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Class 12th Dual Nature of Radiation and Matter

An a particle and a proton are accelerated form rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be:

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New answer posted

5 months ago

Class 12th Physics

Identify the solar cell characteristics from the following options:

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Vishal Baghel

Contributor-Level 10

V-l characteristics graph is plotted in 4th quadrant for solar cell.

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5 months ago

Class 12th Physics

The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8} t h$ of its original mass to disintegrate will be:

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Vishal Baghel

Contributor-Level 10

$t_{\frac{1}{2}} = 60 d a y s$

at t = 0 $\frac{m}{m_{0}} \to$ 0

$a t t \to \frac{m}{m_{0}} (1 - \frac{7}{8})$

= $\frac{m}{8 m_{0}} \to \frac{7}{8} (\frac{m}{m_{0}})$

$\Rightarrow \frac{1}{2^{3}} = \frac{1}{2^{(\frac{t}{60})}} \Rightarrow \frac{t}{60} = 3$

t = 180 days

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Class 12th Dual Nature of Radiation and Matter

The equation $λ = \frac{1.227}{x} n m$ can be used to find the de-Brogli wavelength of an electron. In this equation x stands for:

Where m = mass of electron

P = momentum of electron

K = kinetic energy of electron

V = Accelerating potential in volts for electron

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Vishal Baghel

Contributor-Level 10

Kindly consider the following answer

x = $\sqrt[]{V}$

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5 months ago

Class 12th Physics

In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:

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Vishal Baghel

Contributor-Level 10

Angular magnification, $θ = | \frac{f_{0}}{f_{e}} |$ $\frac{_{}}{_{}}$

Length of tuber L = $| f_{0} | + | f_{e} |$ $_{}_{}$

$| f_{0} | + | f_{e} | = 30 c m$

$A s, | \frac{f_{0}}{f_{e}} | = 2$

$| f_{0} | = 2 | f_{e} |$

$\Rightarrow | f_{0} | + \frac{| f_{0} |}{2} = 30 \Rightarrow | f_{0} | = \frac{2}{3} * 30 = 20 c m$

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Class 12th Physics Wave Optics

An unpolarised light beam of intensity 2l₀ is passed through a polaroid P and then through another polaroid Q which is oriented is such a way that its passing axis makes an angle of 30° relative to that of P. The intensity of the emergent light is

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alok kumar singh

Contributor-Level 10

$I = I_{0} c o s^{2} 30 °$

$= I_{0} {(\frac{\sqrt[]{3}}{2})}^{2} = \frac{3}{4} I_{0}$

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Class 12th Physics

As shown in the figure, after passing through the medium 1. The speed of light $υ_{2}$ in medium 2 will be:

(Given c = 3 * 10⁸ ms^-¹)

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Vishal Baghel

Contributor-Level 10

In a medium, speed of light, $V = \frac{C}{\sqrt[]{μ_{r} ε_{r}}} = \frac{3 * 1 0^{8} m / s}{\sqrt[]{1 * 9}} = 1 * 1 0^{8} m / s$

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Class 12th Physics

A circuit element X when connected to an a.c. supply of peak voltage 100V given a peak current of 5A which is in phase with the voltage. A second element Y when connected to the same a.c. supply also gives the same value of peak current which lags behind the voltage by If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

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alok kumar singh

Contributor-Level 10

v = 100 sin $ω t$

$i_{0} = \frac{100}{R_{x}} = 5 \Rightarrow R_{x} = 20 Ω$

i = $5 s i n (ω t - \frac{π}{2})$

i = 100 5sin $ω t$ &nb

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v = 100 sin $ω t$

$i_{0} = \frac{100}{R_{x}} = 5 \Rightarrow R_{x} = 20 Ω$

i = $5 s i n (ω t - \frac{π}{2})$

i = 100 5sin $ω t$

$5 = \frac{100}{R_{y}} \Rightarrow R_{y} = \frac{100}{5} = 20 Ω$

When x and y both are connected in series :-

v = 100sin $ω t$

tanθ = 1 = q = 45°

R₀ = $\sqrt[]{R_{x}^{2} + R_{y}^{2}} = 20 \sqrt[]{2} Ω$

$∴ \overset{o}{l_{0}} = \frac{v_{0}}{R_{0}} = \frac{100}{20 \sqrt[]{2}} = \frac{5}{\sqrt[]{2}} A .$

$∴ l_{m s} = \frac{l_{0}}{\sqrt[]{2}}$

$= \frac{5}{\sqrt[]{2} \cdot \sqrt[]{2}}$

$= \frac{5}{2} A = \frac{5}{2} A$

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