Class 12th

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New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

Δ E 1 = 12400 e V ? 1085 ? = 11.43 e V

Δ E 2 = 12400 304 e V = 40.80 e V

Δ E 1 + Δ E 2 = 54.4 e V 1 n 1 2 - 1 n 2 2 52.23 = 54.4 1 n 1 2 - 1 n 2 2

If n 1 = 1 n 2 = 5

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

S3O9 has no

S - S linkage

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2 months ago

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A
alok kumar singh

Contributor-Level 10

Process of active methylene group  acidic strength and the side chain should be e- withdrawing.

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

1 20 = ( 1.55 - 1 ) 1 R 1 - 1 R 2

1 f = ( 1.50 - 1 ) 1 R 1 - 1 R 2

Divide (1) by (2), f 20 = 0.55 0.50

f = 22 c m

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

(I? /2)cos²θ = (18/100)I?
⇒ cos²θ = 9/25
⇒ cosθ = 3/5 ⇒ θ = 53°

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

    Potential difference across R?
(R? / (R? +R? ) * V is greater than zenor voltage
⇒ i? = V_z/R? = 20/1000 A = 20 mA
Current through R? , i? = (40-V_z)/R? = 20/500 = 40 mA


Current through zener diode = i? - i? = 20 mA

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

Neighboring group participation by a sulphide involved.

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

at t = 0, I = 0; E = L (dI/dt) ⇒ dI/dt = E/L = 1.5/0.1 = 15 A/s

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

V_common = (C? V? + C? V? )/ (C? + C? )? 2 = (3? F)*12)/ (3? F)+ (3? F)*K)? 2 = 12/ (K+1)? K = 5

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

β = (λD/d) ⇒ Δβ = (ΔDλ/d) ⇒ 4 * 10? = (λ/10? ³) * 8 * 10? ² ⇒ λ = 5 * 10? m = 500 nm

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