Class 12th

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New answer posted

4 months ago

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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 12 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 question posted

4 months ago

0 Follower 4 Views

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

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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

New answer posted

4 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

New question posted

4 months ago

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