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New answer posted

4 months ago

0 Follower 18 Views

A
alok kumar singh

Contributor-Level 10

ΔE = have = 13.6Z² (1/n? ² - 1/n? ²)eV.
hv = 13.6 (1²) (1/n² - 1/ (n+1)²) = 13.6 (n+1)²-n²)/ (n² (n+1)²)
v = (13.6/h) * (2n+1)/ (n² (n+1)²). For n>>1, v ≈ (13.6/h) (2n/n? ) ∝ 1/n³.

New answer posted

4 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

m = I (abk + abj)
|m| = Iab√2
Direction ⇒ (j+k)/√2

New answer posted

4 months ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

R = 100Ω. tanφ = (X_L-X_C)/R. tan45° = X_L/R (since it's an inductor). X_L=R=100Ω.
Lω=100 ⇒ L (2πf)=100 ⇒ L = 100/ (2π1000) ≈ 1.59 * 10? ² H. The closest option is (A).

New answer posted

4 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

4 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

σ4πr² + σ4πR² = Q ⇒ σ = Q/ (4π (R²+r²)
V_c = kq? /r + kq? /R = k (σ4πr²)/r + k (σ4πR²)/R = kσ4π (r+R)
= K (Q/ (R²+r²) (R+r)

New answer posted

4 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

On increasing the temperature, random velocity of molecules increases, therefore mean collision time between the molecules decreases. But the mean free path remains constant as it is product of velocity and time.? (b) and (c) are correct.

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

l = 10 * pitch = 10 * (2πmvcosθ/qB)
= 20π * 1.67*10? ²? * 4*10? * (1/2) / (1.6*10? ¹? * 0.3) = 0.44 m

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

λ = h/p = h/mv
λp/λe = (m? v? )/ (m? v? ) ⇒ 1.878 * 10? = (9.1*10? ³¹)/ (m? * 5)
m? = 9.1*10? ³¹ / (5 * 1.878 * 10? ) = 0.97 * 10? ²? kg

New answer posted

4 months ago

0 Follower 25 Views

A
alok kumar singh

Contributor-Level 10

Δl = lαΔT ⇒ Δl/l = αΔT = 0.02%
Δρ = -ργΔT
|Δρ/ρ| = γΔT = 3αΔT = 3 (0.02%) = 0.06%

New answer posted

4 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

 2θ = 60° ⇒ θ = 30°
h = 2Tcosθ / (ρsg) = 2 (0.05)cos30° / (667) (0.15 * 10? ³) (10) = (√3 * 100)/ (667 * 3) ≈ 173.2/2000 m = 8.66 cm

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