Physics Ncert Solutions Class 12th

Contributor-Level 10

$d = \sqrt[]{2 R h}$

$d = \sqrt[]{2 * 6400 * h * 1 0^{- 3}} (h i n m)$

Area = $π d^{2}$

$= (π * 2 * 6400 * h * 1 0^{- 3})$ km²

h = 150m

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The energy band gap of semiconducting material to produce violet (wavelength = 4000 ) LED is ________ eV. (Round off to the nearest integer).

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Contributor-Level 10

$E_{9} = \frac{h c}{λ} = \frac{1242}{λ (n m)} = \frac{1242}{400} = 3.105$

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The volume charge density of a sphere of radius 6m is 2 $μ C c m^{- 3} .$ The number of lines of force per unit surface area coming out from the surface of the sphere is _________ * 10¹⁰ NC^-1.

[Given : Permittivity of vacuum $\in_{0} = 8.85 * 1 0^{- 12} C^{2} N^{- 1} - m^{- 2}]$

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Contributor-Level 10

No. of electric field line per unit area = electric field

$E = \frac{ρ r}{3 \in_{0}} f o r r = R$

$E = 45 * 1 0^{10} N / C$

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A capacitor of capacitance 500 $μ F$ is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. Then maximum current in LC circuit will be_______ A.

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Contributor-Level 10

C = 500 $μ F,$ V = 100 v, L = 50 mH

In this LC – oscillation

q = q₀ cos $ω t$

$i = \frac{- d q}{d t} = q_{0} ω s i n ω t ω = \frac{1}{\sqrt[]{2 c}} = \frac{1}{\sqrt[]{50 * 1 0^{- 3} * 5 * 1 0^{- 4}}}$

= $\frac{1000}{5} = 200$

So, i_max = $q_{0} ω = 500 * 1 0^{6} * 100 * 200$

= 10A

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A 8 V Zener diode along with a series resistance R is connected across a 20V supply (as shown in the figure). If the maximum Zener current is 25 mA, then the minimum value of R will be_______ $Ω$ .

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Contributor-Level 10

$i_{(z e n e r - m a x)} = 25 m A$

20 – i_max R – 8 = 0

i_max R = 12

At minimum zener current $(μ A) : -$

$20 - i_{m i n} R - i_{m i n} R_{L} = 0$

$\frac{R}{R_{L}} = \frac{12}{8} = \frac{3}{2}$

$l_{m i n} R = 12$

$i_{m i n} R_{L} = 8$

At maxm zener current –

$20 - i_{m a x} R - 8 = 0$

$i_{L} = O {a s i_{z} m a x^{m} = 25 m A}$

i_maxR = 12v

25 * 103 R = 12

$R = \frac{12 * 1 0^{3}}{25} = 12 * 40 = 480 Ω$

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The metallic bob of simple pendulum has the relative density. 5 The time period of this pendulum is 10 s. If the metallic bob is immersed in water, then the new time period becomes $5 \sqrt[]{x} s .$ The value of x will be_______.

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Contributor-Level 10

g_eff = g – $(\frac{ρ_{w}}{ρ_{b}}) g$

$T = 2 π \sqrt[]{\frac{l}{g}}$

$T' = 2 π \sqrt[]{\frac{l}{g_{e f f}}}$

$\Rightarrow T' = \sqrt[]{\frac{5}{4}} T$

= $\sqrt[]{\frac{5}{4}} * 10$

= $5 \sqrt[]{5} s e c$

So, x = 5

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The X-Y plane be taken as the boundary between two transparent media M₁ and M₂. M₁ in $z \geq 0$ has a refractive index of $\sqrt[]{2}$ and M₂ with Z < 0 has refractive index of $\sqrt[]{3} .$ A ray of light travelling in M₁ along the direction given by the vector $\vec{P} = 4 \sqrt[]{3} \hat{i} - 3 \sqrt[]{3} \hat{j} - 5 \hat{k},$ is incident on the plane of separation. The value of difference between the angle of incident in M₁ and the angle of refraction in M₂ will be_______ degree.

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Contributor-Level 10

For angle of incidence 'i' :-

cos I = $\frac{5}{\sqrt[]{48 + 27 + 25}} = \frac{5}{\sqrt[]{100}}$

=> I = 60°

Using snell's law :-

$μ_{1} s i n i = μ_{2} s i n r$

sin r = $\frac{\sqrt[]{2}}{\sqrt[]{3}} s i n 60 ° = \sqrt[]{\frac{2}{3}} * \frac{\sqrt[]{3}}{2} = \frac{1}{\sqrt[]{2}}$

$∴ r = 45 °$

So, difference, I – r = 60° - 45² = 15°

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A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be __________hours.

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Contributor-Level 10

t_1/2= 2hr 30 min

Radiation intensity a disintegrations / sec (Activity)

As, $A = \frac{A_{0}}{{2^{t}}^{\frac{t}{2}}}$

For safe working, $A = \frac{A_{0}}{64} \Rightarrow \frac{A_{0}}{64} = \frac{A_{0}}{2^{\frac{t}{t_{\frac{1}{2}}}}}$

$\Rightarrow t = 6 t_{\frac{1}{2}} = 6 * 2.5 h r$

= 15

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The current in a coil of self inductance 2.0 H is increasing according to I = 2sin(t²)A. The amount of energy spent during the period when current changes from 0 to 2 A is…………J.

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Contributor-Level 10

Given, L = 2H

l = 2 sin (t²)

$u = \int_{0}^{2} L i d i = \frac{L}{2} {[i^{2}]}_{0}^{2} = \frac{L}{2} [4 - 0]$

$u = \frac{2}{2} * 4 = 4 J$

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A circular coil of 1000 turns each with area 1m² is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07 T. The maximum voltage generation will be ________V.

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