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New answer posted

a year ago

0 Follower 32 Views

V
Vishal Baghel

Contributor-Level 10

As b > a
The magnetic field inside the wire (rR) is B = µ? I/ (2πr).
For wire with radius a, B increases linearly to r=a, then decreases. For wire with radius b, B increases linearly to r=b, then decreases. Since a⇒ B? > B?
B? = µ? I / 2πa
B? = µ? I / 2πb
(Note: The question is likely asking for the graph representation, which is option A based on the formulas.)

New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

At terminal speed
Mg = Fv = 6πηRv
⇒ V = mg / 6πηR
V = (4/3)πR³ρg / 6πηR
⇒ V = (2/9) * (ρR²g/η)
= (2/9) * (1000 * (0.2 * 10? ³)² * 10) / (1.8 * 10? )
= 4.94 m/s

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

T² = (4π²r³)/GM
⇒ M = (4π²r³)/ (G T²)
⇒ M = 6 * 10¹¹ * (9 * 10? )³ / (450 * 60)²
= 6.48 x 10²³ kg
(Note: There is a calculation error in the source image. Using the formula and given values: M = (6e11 * (9e6)^3) / (27000)^2 = 6e23 kg)

New answer posted

a year ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

A = A? (1/2)?
A/A? = 1/8 = (1/2)³ ⇒ n = 3
t = n T½
30 = 3 T½
T½ = 10 years

New answer posted

a year ago

0 Follower 4 Views

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

a year ago

0 Follower 6 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

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

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

New answer posted

a year ago

0 Follower 11 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

a year ago

0 Follower 5 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

a year ago

0 Follower 5 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

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