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Class 12th Physics

The vertical component of the earth's magnetic field is 6 * 10^-5 T at any place where the angle of dip is 37°. The earth's resultant magnetic field at that place will be (Given tan37° = $\frac{3}{4}$ )

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alok kumar singh

Contributor-Level 10

B_v = 6 * 10^-5 T

B_v = B_N sin37°

$∴ B_{N} = \frac{B_{v}}{s i n 37 °} = \frac{6 * 1 0^{- 5}}{3 / 5} = 1 0^{- 4} T$

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Class 12th Physics

A thermodynamic system is taken from an original state D to an intermediate sate E by the linear pressure shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be

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alok kumar singh

Contributor-Level 10

W_total= W_D_®E + W_E->_F = Area of triangle DEF

= $\frac{1}{2} * 3 * 300 = 450 J .$

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Class 12th Physics

A ball is released from a height h. If t₁ and t₂ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t₁ and t₂.

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alok kumar singh

Contributor-Level 10

$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

h = $\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

from (i) and (ii) :-

$\frac{1}{2} g t_{1}^{2} = \frac{g}{4} {(t_{1} + t_{2})}^{2}$

$2 t_{1}^{2} = {(t_{1} + t_{2})}^{2}$

$\sqrt[]{2} t_{1} = t_{1} + t_{2}$

$t_{1} = \frac{t_{2}}{\sqrt[]{2} - 1} = (\sqrt[]{2} + 1) t_{2}$

$t_{2} = (\sqrt[]{2} - 1) t_{1}$

...more

$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

h = $\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

from (i) and (ii) :-

$\frac{1}{2} g t_{1}^{2} = \frac{g}{4} {(t_{1} + t_{2})}^{2}$

$2 t_{1}^{2} = {(t_{1} + t_{2})}^{2}$

$\sqrt[]{2} t_{1} = t_{1} + t_{2}$

$t_{1} = \frac{t_{2}}{\sqrt[]{2} - 1} = (\sqrt[]{2} + 1) t_{2}$

$t_{2} = (\sqrt[]{2} - 1) t_{1}$

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Class 12th Physics

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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alok kumar singh

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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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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alok kumar singh

Contributor-Level 10

Radius, R = R₀A^1/3

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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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8 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

8 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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8 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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