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Class 12th Physics Wave Optics

In young's double slit experiment performed using a monochromatic light of wavelength $λ,$ when a glass plate $(μ = 1.5)$ of thickness $x λ$ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be:

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alok kumar singh

Contributor-Level 10

Þ AC = d + t $(μ - 1) \to$ path length

BC = d

path different = AC – BC = $n λ$

$t (μ - 1) = μ λ$

for contal maxima

$x λ (1.5 - 1) = n λ$

0.5x = x

n = 0, 1,2- -

n = 0 Not possible

Now, n = 1

0.5x = 1

x = 2

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11 months ago

Class 12th Physics

An EM wave propagating in x-direction has wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm^-1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum:

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alok kumar singh

Contributor-Level 10

Given, $λ = 8 m m, v = 8 m m, v = 3 * 1 0^{8}$

$\vec{E} = ε_{0} s i n [k x - ω t] \hat{j}$

$k = \frac{2 π}{λ} = \frac{2 π}{8 * 1 0^{- 3}} = \frac{π}{4} * 1 0^{3}$

$\vec{E} = 60 s i n [k (x - \frac{ω}{k} t)] \hat{j}$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - v t]] \hat{j} [? \frac{ω}{k} = v]$

$\vec{E} = 60 s i n [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{j} v / m$

Now $\frac{E_{0}}{B_{0}} = C$

B0 = $\frac{60}{3 * 1 0^{8}} = \frac{20}{1 0^{8}} = 2 * 1 0^{- 7}$ $\vec{B} = 2 * 1 0^{- 7} [\frac{π}{4} * 1 0^{3} [x - 3 * 1 0^{8} t]] \hat{k} T$

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11 months ago

Class 12th Physics Moving Charges and Magnetism

Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2 * 10^-6N, then the value of x is approximately:

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alok kumar singh

Contributor-Level 10

$\frac{F}{l} = \frac{μ_{0}}{2 π} \frac{i_{1} i_{2}}{d}$

$2 * 1 0^{- 6} = \frac{4 π * 1 0^{- 7}}{2 π} * \frac{x * x}{d}$

x² = 2x = $\sqrt[]{2}$ = 1.4x = 1.4

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11 months ago

Class 12th Physics

A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be:

(Assume the coil to be short circuited.)

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alok kumar singh

Contributor-Level 10

$P = \frac{E^{2}}{R} = \frac{{(N A \frac{d B}{d t})}^{2} * 4 A C}{ρ l}$

$p' = {(\frac{N A}{2} \frac{d B}{d t})}^{2} * 4 A C$

$p' = 2 P$

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11 months ago

Class 12th Physics Moving Charges and Magnetism

The space inside a straight current carrying solenoid is filled with a magnetic material having magnetic susceptibility equal to 1.2 * 10^-5. What is fractional increase in the magnetic field inside solenoid with respect to air medium inside the solenoid?

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alok kumar singh

Contributor-Level 10

$χ (S u s c e p t i b i l i t y) = 1.2 * 1 0^{- 5}$

fractional change = $= \frac{Δ B}{B} = \frac{B i n s i d e - B o u t}{B o u t}$

$= \frac{μ_{m} H - μ_{0} H}{μ_{0} H} = \frac{(μ_{0} μ_{r} - μ_{0}) H}{μ_{0} H}$ $[? μ_{m} = b_{0} μ_{0}]$

$= μ_{r} - 1$

= $= (1 + χ) - 1 = χ = 1.2 * 1 0^{- 5}$

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11 months ago

Class 12th Physics

Resistance of the wire is measured as at 10°C and 30°C respectively. Temperature co-efficient of resistance of the material of the wire is:

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alok kumar singh

Contributor-Level 10

R_T = R₀ $(1 + α Δ T)$

at t = 10°C, R_T = 2 $Ω$

2 = R₀ (1 + [10 – T0])- (1)

at T = 30°C, R_T= 3 $Ω$

3 = $R_{0} [1 + α (30 - T_{0})]-$ (2)

from (1)

$2 = R_{0} (1 + α 10)-$ (3)

from (2) 3 = R₀ (1 + 30) - (4)

$(4) \div (3)$

$\frac{3}{2} = \frac{1 + 30 α}{1 + 10 α}$

3 + 30 = 2T 60α = $\frac{1}{30} = 0.033$

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11 months ago

Class 12th Physics

Two point charges A and B of magnitude + 8 * 10^-6 and -8 * 10^-6C respectively are placed at a distance d apart. The electric field at the middle point O between the charges is 6.4 * 10⁴ NC^-1. The distance 'd' between the point charges A and B is:

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alok kumar singh

Contributor-Level 10

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{- 6}]}{d^{2}} * 4$

E₂ (done to q₂) = $\frac{k q_{2}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{- 6}]}{d^{2}} * 4$

Enet = E₁ + E₂

$= 2 [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{- 6}}{d^{2}}]$

Given Enet = 6.4 * 104

$\Rightarrow 2 * [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{- 6}}{d^{2}}] = 6.4 * 1 0^{4}$

d² = 9 * 10^-6 + 6

d= 3m

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11 months ago

Class 12th Chemistry

The number of oxygen's present in a nucleotide formed from a base, that is present only in RNA is_______________.

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Payal Gupta

Contributor-Level 10

Consider the following image

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11 months ago

Class 12th Chemistry

Compound 'P' on nitration with dil. HNO₃ yields two isomers (A) and (B). These isomers can be separated by steam distillation. Isomers (A) and (B) show the intramolecular and intermolecular hydrogen bonding respectively. Compound (P) on reaction with conc. HNO₃ yields a yellow compound 'C'___________

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Payal Gupta

Contributor-Level 10

Intermolecular H- bonding and intra-molecular H- bonding producing compound may be the phenol derivatives.

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11 months ago

Class 12th Physics

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alok kumar singh

Contributor-Level 10

E₁ (done to q₁) = $\frac{k, q_{1}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

E₂ (done to q₂) = $\frac{k ? q_{2}}{{(d / 2)}^{2}} = \frac{[9 * 1 0^{9} * 8 * 1 0^{? 6}]}{d^{2}} * 4$

Enet = E₁+ E₂

$= 2 [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}]$

Given Enet = 6.4 * 104

$? 2 * [\frac{4 * 9 * 1 0^{9} * 8 * 1 0^{? 6}}{d^{2}}] = 6.4 * 1 0^{4}$

d² = 9 * 10^-6 + 6

d= 3m

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