Class 12th

Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T? is connected to T?. As the current in R = 6Ω attains a maximum value of steady state level, T? is disconnected from T? and immediately connected to T?. Potential drop across r = 3Ω resistor immediately after T? is connected to T? is _______ V. (Round off to the Nearest Integer)

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Class 12th Physics Ray Optics and Optical Instruments

Contributor-Level 10

When T? & T? are connected, at steady state current, I becomes
I = 6V/6Ω = 1A
Now when T? & T? are connected, the current through the inductor, just after connecting, remains same. So
V? Ω = 1A * 3Ω = 3volt.

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4 months ago

A prism of refractive index n? and another prism of refractive index n? are stuck together (as shown in the figure). n? and n? depend on λ, the wavelength of light, according to the relation:
n? = 1.2 + (10.8 * 10?¹?)/λ² and n? = 1.45 + (1.8 * 10?¹?)/λ²
The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be _______ nm.

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Class 12th Electromagnetic Induction

Contributor-Level 10

Light rays travel undeviated if the refractive index of the medium does not change for any value of angle of incidence.
i.e. n? = n?
1.2 + 10.8*10? ¹? /λ² = 1.45 + 1.8*10? ¹? /λ²
9.0*10? ¹? /λ² = 0.25
λ² = 36*10? ¹?
λ = 6*10? m = 600 nm

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4 months ago

A circular conducting coil of radius 1 m being heated by the change of magnetic field B passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2μΩ. The magnetic field is slowly switched off such that its magnitude changes in time as B = (4/π) * 10?³ T [1 - ( t/100 )]. The energy dissipated by the coil before the magnetic field is switched off completely is E = _______ mJ.

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Contributor-Level 9

ε = |dΦ/dt| = A|dB/dt|
Given B (t), dB/dt = (4/π) * 10? ³ * (-1/100)
ε = (π * 1²) * | (4/π) * 10? ³ * (-1/100)|
= 4 * 10? V
To find when B=0:
B = 0 ⇒ 1 - t/100 = 0
⇒ t = 100 second
Energy Dissipated, E = P * t = (ε²/R) * t
E = (4*10? )² / (2*10? ) * 100 = 80mJ

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4 months ago

A transistor is connected in common emitter circuit configuration, then collector supply voltage is 10 V and the voltage drop across a resistor of 1000Ω in the collector is 0.6 V. If the current gain factor (β) is 24, then the base current is _______ µA. (Round off to the Nearest Integer)

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Class 12th Physics Magnetism and Matter

Contributor-Level 10

As we know, β = Ic/Ib = 24 (Given)
R = 1000Ω, ΔV = 0.6V, Ic = ΔV/R = 0.6/1000 = 6*10? A
So, Ib = Ic/β = (6*10? )/24 = 25*10? A
or Ib = 25µA

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4 months ago

The value of aluminium susceptibility is 2.2 * 10?. The percentage increase in the magnetic field if space within a current carrying toroid is filled with Aluminium is x/10?. Then the value of x is _______.

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Contributor-Level 9

We know, μ? = 1 + χ?
B? ∝ μ?
B ∝ μ
(ΔB/B? ) = (B-B? )/B? = (μ-μ? )/μ? = μ/μ? - 1 = μ? - 1 = χ?
% (ΔB/B) = χ? * 100 = 2.2 * 10? * 100 = 22/10?
(Note: The value for χ? in the source image appears to have a typo as 2.2 * 10? , it has been corrected to 2.2 * 10? to match the final answer.)

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4 months ago

A 0.07 H inductor and a 12Ω resistor are connected in series to a 220 V, 50 Hz ac source. The approximate current in the circuit and the phase angle between current and source voltage are respectively. [Take π as 22/7]

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Contributor-Level 10

L = 0.07H and R = 12Ω
Vs = 220V, 50Hz
XL = 2πfL = 2 (22/7) (50) (0.07) = 22Ω
Z = √ (R² + XL²) = √ (12² + 22²) = √ (144+484) = √628 ≈ 25Ω
i = V/Z = 220/25 = 8.8A
tan φ = XL/R = 22/12 = 11/6
φ = tan? ¹ (11/6)

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4 months ago

An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance R that must be put in series with the bulb so that the bulb delivers the same power is _______ Ω.

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Class 12th Electromagnetic Induction

Contributor-Level 9

Given: Power in R_B = 200 W, R_B = 50 Ω, V_source = 200 V
I² R_B = 200
I² (50) = 200
I² = 4 ⇒ I = 2A
Also, I = V_source / (R + R_B)
2 = 200 / (R + 50)
2 (R + 50) = 200
R + 50 = 100
R = 50Ω

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4 months ago

An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 kΩ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 µs is x/100 mA. Then x is equal to _______. (Take e?¹ = 0.37)

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Class 12th Physics Wave Optics

Contributor-Level 9

I = I? e? /?
= (20/10000) e^- (1*10? / 10*10? ³)
= 2 * 10? ³ e? ¹
The provided solution calculates as:
= 2 * 10? ³ e? ¹
= 2e? ¹ mA
= 2 * 0.37mA
= 0.74 mA = 74/100 mA
(Note: There seems to be a calculation discrepancy in the source image, the steps shown lead to I = 2e? ¹ mA)

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4 months ago

In Young's double slit experiment, if the source of light changes from orange to blue then:

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Class 12th Physics Electric Charge and Field

Contributor-Level 10

β = λD/d or β ∝ λ
Also, we know that λ_blue < _orange
So, β_orange > β_blue
So, dist b/w consecutive fringes will decrease.

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4 months ago

A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance 2 m when each is carrying same charge q'. If the free charged particle is displaced from its equilibrium position through distance 'x' (x << 1m). The particle executes SHM. Its angular frequency of oscillation will be ______ * 10? rad/s if q' = 1 C.

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