Class 12th

An object of mass 1 kg is taken to a height form the surface of earth which is equal to three times that radius of earth. The gain in potential energy of the object will be

[If, g = 10ms^-2 and radius of earth = 6400km]

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Contributor-Level 10

m = 1kg ; $Δ U = \frac{- G m_{e} m}{(R + h)} - (\frac{G m_{e} m}{R})$

$Δ U = \frac{G M m}{R} - \frac{G M m}{R + h}$

$= m g_{0} R - m g_{0} \frac{R}{(1 + \frac{h}{R})} = m g_{0} {1 - \frac{1}{(1 + \frac{3 R}{R})}}$

$= m g_{0} R (\frac{3}{4})$

$= \frac{3}{4} * 1 * 10 * 6400 k m$

$= 48000 * 1 0^{3} J .$

$= 48 M J .$

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11 months ago

Class 12th Physics Nuclei

Read the following statements:

(A) Volume of the nucleus is directly proportional to the mass number.

(B) Volume of the nucleus is independent of mass number.

(C) Density of the nucleus is directly proportional to the mass number.

(D) Density of nucleus is directly proportional to the cube root of the mass number.

(E) Density of the nucleus is independent of the mass number.

Choose the correct option from the following options

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Class 12th Dual Nature of Radiation and Matter

Contributor-Level 10

Radius, R = R₀A^1/3

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11 months ago

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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Class 12th Communication Systems

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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11 months ago

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

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Class 12th Dual Nature of Radiation and Matter

Contributor-Level 10

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

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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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11 months ago

Identify the solar cell characteristics from the following options:

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Contributor-Level 10

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

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11 months ago

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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Class 12th Dual Nature of Radiation and Matter

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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11 months ago

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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Contributor-Level 10

Kindly consider the following answer

x = $\sqrt[]{V}$

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11 months ago

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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Class 12th Physics Wave Optics

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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11 months ago

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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