Physics Ncert Solutions Class 12th

A ray of light is incident at an angle of incidence $60 °$ on the glass slab of refractive index $\sqrt[]{3} .$ After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is $4 \sqrt[]{3}$ cm. The thickness of the glass slab is _______cm.

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$d = 4 \sqrt[]{3}$ cm (Lateral shift)

By Snell's law

$μ_{a i r} s i n 60 ° = μ_{g} s i n θ$

$\Rightarrow 1 * \frac{\sqrt[]{3}}{2} = \sqrt[]{3} s i n θ$

$s i n θ = \frac{1}{2}$

= 30°

$t a n 30 ° = \frac{d}{t} = \frac{4 \sqrt[]{3}}{t}$

$\Rightarrow \frac{1}{\sqrt[]{3}} = \frac{4 \sqrt[]{3}}{t}$

t = 12 cm

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In the given circuit, the value of current I_Lwill be _________mA. (When R_L = 1 k $Ω$ )

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$I_{L} = \frac{V}{R_{L}} = \frac{5}{1 k Ω}$

$l_{L} = 5 * 1 0^{- 3} A = 5 m A$

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An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is 5.0 mm, it can radiate a signal of minimum frequency of ________ GHz.

(Given µ_r = 1 for dielectric medium)

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$v = \frac{C}{\sqrt[]{μ_{r} ε_{r}}} = \frac{3 * 1 0^{8}}{\sqrt[]{6.25 * 1}} = \frac{3 * 1 0^{8}}{2.5} = 1.25 * 1 0^{8} m / s e c$

$A s f λ = f (5 * 1 0^{- 3} * 4) = 1.25 * 1 0^{8} \Rightarrow f = 6.25 G H z$

So, $f \approx 6 .$

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An electric bulb is rated as 200W. What will be the peak magnetic field at 4m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

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I_R = I_B (due to magnetic field)

(due to Radiation)

$\frac{P}{A} * E f f i c i e n c y = \frac{B_{0}^{2}}{2 μ_{0}} c$

$\Rightarrow \frac{200}{4 π * 4^{2}} * \frac{3.5}{100} = \frac{B_{0}^{2}}{2 * 4 π * 1 0^{- 7}} * 3 * 1 0^{8}$

$\Rightarrow B_{0} = 1.71 * 1 0^{- 8} T$

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The soft-iron is a suitable material for making an electromagnet. This is because soft iron has

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High permeability and low retentivity

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In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emef's of two cells respectively is 3:2, the difference in the balancing length of the potentiometer wire in above two cases will be __________ cm.

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$E α l$

$\Rightarrow \frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}}$

$\Rightarrow \frac{3}{2} = \frac{75}{l_{2}}$

$l_{2} = \frac{75 * 2}{3}$

$l_{2} = 50 c m$

$& l_{1} = 75 c m$

$Δ l = l_{1} - l_{2} = 25 c m$

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When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $υ_{1} .$ When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $υ_{2} .$ If $υ_{2} = x υ_{1},$ the value of x will be __________.

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hu = hu₀ + K.E

Cases u = 2u₀

h2u₀ = hu₀ + K.E₁

K.E₁ = hu₀

$\frac{1}{2} m v_{1}^{2} = h υ_{0}$

$v_{1}^{2} = \frac{2 h υ_{0}}{m}$

$v_{1} = \sqrt[]{\frac{2 h υ_{0}}{m}}$ - (1)

Now, cases 2

h 5u₀ = hu₀ + k.E₂

k.E₂ = 4hu₀

$\frac{1}{2} m v_{2}^{2} = 4 h υ_{0}$

v₂ = $\sqrt[]{\frac{8 h υ_{0}}{m}}$ - (2)

$\frac{v_{2}}{v_{1}} = \sqrt[]{\frac{8}{2} = 2}$

v₂ = 2v₁

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A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10µA and the collector current changes by 1.5 mA. The load resistance is 5 kΩ. The voltage gain of the transistor will be ________.

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I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_{B} = \frac{V_{B}}{I_{B}} = \frac{10 * 1 0^{- 3}}{10 * 1 0^{- 6}} = 1 0^{3} Ω$

$A_{v} = (\frac{Δ l_{C}}{Δ l_{B}}) * (\frac{R_{C}}{R_{B}})$

= $\frac{1.5 * 1 0^{- 3}}{10 * 1 0^{- 6}} * \frac{5 * 1 0^{3}}{1 0^{3}}$

A_v = 750

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Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is ________ cm.

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Physics Ncert Solutions Class 12th Electrostatics

Contributor-Level 10

f₁ = 15 cm

$P_{1} = \frac{1}{15} * 100 = \frac{100}{15} D$

$P_{3} = \frac{100}{15} D$

$\frac{1}{f^{2}} = (μ_{2} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$= (1.25 - 1) * \frac{- 2}{R}$

$\frac{1}{f_{2}} = \frac{0.25 * - 2}{15}$

$\frac{1}{f_{2}} = \frac{- 1}{30} c m^{- 1}$

$\begin{array}{l} p_{2} = \frac{1}{f_{2}} = - \frac{100}{30} \\ P_{R} = P_{1} + P_{2} + P_{3} \\ P_{R} = \frac{100}{15} - \frac{100}{30} + \frac{100}{15} \\ P_{R} = 10 D \end{array}$

$P_{R} = \frac{1}{f_{R}}$

$f_{R} = \frac{1}{10} * 100 c m$

$f_{R} = 10 c m$

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A parallel plate capacitor is formed by two plates each of area 30πcm² separated by 1 mm. A material of dielectric strength 3.6 * 10⁷ Vm^-¹ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 * 10^-⁶ C, the value of dielectric constant of the material is [Use ] $\frac{1}{4 π ε_{0}} = 9 * 1 0^{9} N m^{2} C^{- 2}$

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