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New answer posted

5 months ago

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A
alok kumar singh

Contributor-Level 10

ni² = ne * nh
(1.5*10¹? )² = ne * (4.5*10²²)
ne = (2.25 * 10³²)/ (4.5 * 10²²) = 0.5 * 10¹? = 5 * 10? / m³

New answer posted

5 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Mean free path:
λ ∝ 1/ (nd²)
λA/λB = (d_B²/d_A²) = (5²/10²) = 25/100 = 25 * 10? ²

New answer posted

5 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Modulation Index
μ = Am/Ac = 20/20 = 1

New answer posted

5 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

K = hc/λ - W?
K = (20*10? ²? )/ (500*10? ) - 1.25 * 1.6*10? ¹?
K = 4*10? ¹? - 2*10? ¹? = 2*10? ¹? J
R = √ (2mK)/ (eB)
B = √ (2mK)/ (eR) = √ (2 * 9*10? ³¹ * 2*10? ¹? ) / (1.6*10? ¹? * 30*10? ²)
B = √ (36*10? ) / (48*10? ²¹) = (6*10? ²? ) / (4.8*10? ²? ) = 1.25 * 10? T
B = 125 * 10? T

New answer posted

5 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Power dissipated is same
P? = P?
P_R = P_RLC
(V²/R) = (V²/Z)cos (φ) = (V²/Z) (R/Z) = V²R/Z²
R = R²/Z² => Z² = R²
R² + (ωL - 1/ωC)² = R²
ωL = 1/ωC
ω = 1/√ (LC) = 1/√ (0.1 * 40 * 10? ) = 1/√ (4 * 10? ) = 1/ (2 * 10? ³) = 500 rad/s

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 = 1/λ = 30*10? ³ sec.
C = 200µF, R
Q/A = constant [Given]
A = A? e? & Q = Q? e? /?
Q/A = (Q? /A? ) e^ (-t/RC + λt) = constant
For this to be constant, the exponent must be zero.
-t/RC + λt = 0
1/RC = λ
R = 1/ (λC) = /C = 30*10? ³/200*10? = 150 Ω

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

r = 0.5Å = 0.5*10? ¹? m
v = 2.2*10? m/s
I =?
t = 2πr/v
I = e/t = ev/2πr = (1.6*10? ¹? * 2.2*10? )/ (2 * 22/7 * 0.5*10? ¹? )
I ≈ 1.12*10? ³ A = 1.12 mA = 112 * 10? ² mA

New answer posted

5 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

We have to find the point where the gravitational field must be zero.
EG = 0
GM/x² = G (9M)/ (8R-x)²
1/x² = 9/ (8R-x)²
8R - x = 3x => x = 2R
Potential at A (surface of first planet), VA = -GM/R - G (9M)/7R = -16GM/7R
Potential at point x, Vx = -GM/x - G (9M)/ (8R-x) = -GM/2R - G (9M)/6R = -2GM/R
ΔV = Vx - VA = -2GM/R - (-16GM/7R) = (-14+16)GM/7R = 2GM/7R
Using conservation of mechanical energy
ΔKE = ΔU = mΔV
½mv² = m (2GM/7R)
v² = 4GM/7R
v = √ (4GM/7R) => a = 4

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

M = 9.85*10? ² A/m²

I = 5*10? kg-m²

T = 5s/10osc = 0.5 sec

B =?

T = 2π√ (I/MB)

B = 4π²I/ (MT²) = (4*9.85*5*10? )/ (9.85*10? ² * 0.5²)

B = (20*10? )/ (10? ² * 0.25) = 800*10? = 0.8*10? ³ T = 8mT

New answer posted

5 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Amplitude of sidebands = µAc/2
Here µ = Am/Ac = 5/15 = 1/3
Amplitude = (1/3) (15)/2 = 2.5 V
a/10 = 2.5 => a = 25
b/10 = 2.5 => b = 25
a/b = 1

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