Physics

A parallel plate capacitor has plate area 100m² and plate separation of 10m. The space between the plates is filled up to a thickness 5m with material of dielectric constant of 10. The resultant capacitance of the system is 'x' pF.
The value of ε? = 8.85 * 10?¹² F.m?¹
The value of 'x' to the nearest integer is ----------.

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Contributor-Level 9

1/C = 5/ (ε? * 100) + 5/ (10ε? * 100)
⇒ C = (ε? * 1000) / 55 = (8.85 * 10? ¹² * 1000) / 55 = 1.61 * 10? ¹? F = 161 pF

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4 months ago

Sea-water at a frequency f = 9 * 10² Hz, has permittivity ε = 80ε₀ and resistivity ρ = 0.25Ωm. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V₀sin(2πft). Then the conduction current density become 10ˣ times the displacement current density after time t = 1/800 s. The value of x is ______. [Given: 1/(4πε₀) = 9 * 10⁹ Nm²C⁻²]

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Physics Physics Electromagnetic Waves

Contributor-Level 10

Conduction current density, J? = E/ρ = (V/d) / ρ = (V? sin (2πft) / (pd)
Displacement current density, J? = ε (dE/dt) = (ε/d) (dV/dt) = (2πfε/d) * V? cos (2πft)
The ratio of their magnitudes is:
J? / J? = tan (2πft) / (2πfε? ρ) = tan (2π * 900) / (2π * 9 * 10? * 80ε? * 0.25) = 10?

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4 months ago

The electric field intensity produced by the radiation coming from a 100W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60 W at the same distance is √(x/5) E. Where the value of x = ______.

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Contributor-Level 10

Electric field squared is proportional to the power P of the bulb (E² ∝ P).
(E' / E)² = (60 / 100) ⇒ E' = E * √* (3/5)*

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4 months ago

An npn transistor operates as a common emitter amplifier with a power gain of 10?. The input circuit resistance is 100Ω and the output load resistance is 10kΩ. The common emitter current gain 'β' will be -----------. (Round off to the Nearest Integer)

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Contributor-Level 9

Power gain = (i_c² R_c) / (i_b² R_B) = (i_c/i_b)² (R_c/R_B) = (10²)² (10? /10³) = 10?
i_c/i_b = 100 ⇒ β = i_c/i_b = 100

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4 months ago

An AC source rated 220 V, 50Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :

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Contributor-Level 9

T = 2π√ (l/g) ⇒ g = 4π²l/T²
Percentage error: Δg/g = Δl/l + 2 (ΔT/T) = (0.1/10.0) + 2 (0.005/0.5) = 0.03
Percentage error = (Δg/g) * 100 = 3%
ω = 2πf = 100π rad/s
i_rms = i? /√2
While current changes from its maximum to its rms value, its phase changes by π/4 rad.
t = (π/4)/ω = π/ (4 * 100π) = 2.5 * 10? ³ s = 2.5ms.

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4 months ago

Physics Class 11th

A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that boy can exert on the rope so that the piece of wood does not move from its place is _____ N. (Round off to the Nearest Integer) [Take g = 10 ms?²]

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Physics Units and Measurement

Contributor-Level 10

Assuming the rope in the boy's hand is vertical and using a Free Body Diagram (FBD):
f? = T
R + T = 90 ⇒ R = 90 - T
For the piece of wood not to move, f? ≤ µR:
T ≤ 0.5 (90 - T) ⇒ T ≤ 30N

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4 months ago

The time period of a simple pendulum is given by T = 2π√(l/g). The measured value of the length of pendulum is 10cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is,

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Physics Physics Ray Optics and Optical Instruments

Contributor-Level 9

l = 10.0 ± 0.1cm.
T = (100 ± 1) / 200 = 0.5 ± 0.005

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4 months ago

The image of an object placed in air formed by a convex refracting surface is at a distance of 10 m behind the surface. The image is real and is at 2/3 of the distance of the object from the surface. The wavelength of light inside the surface is 2/3 times the wavelength in air. The radius of the curved surface is x/13 m. The value of 'x' is _____.

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Physics Physics Alternating Current

Contributor-Level 10

Given the refractive index μ = λ? / λ = 3/2 and v = 10m, the object distance is u = - (3/2)v = -15m.
Using the lens maker's formula:
μ/v - 1/u = (μ - 1)/R
(3/2)/10 - 1/ (-15) = (3/2 - 1)/R
This gives the radius of curvature R:
R = -30/13 m

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4 months ago

In a series LCR resonance circuit, if we change the resistance only, from a lower to higher value :

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Contributor-Level 9

Resonance frequency is independent of R.
Quality factor = ωL/R ⇒ Quality factor decreases with increase in R.
Bandwidth of resonance circuit = R/L ⇒ increases with increase in R.

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4 months ago

Physics Class 11th

The disc of mass M with uniform surface mass density σ is shown in the figure. The centre of mass of the quarter disc (the shaded area) is at the position (xa/3π, xa/3π) where x is---------. (Round off to the Nearest integer) [a is an area as shown in the figure]

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