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4 months ago

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Vishal Baghel

Contributor-Level 10

z = √ [R² + (X? - X? )²] = √ [6² + (4-10)²] = 6√2 Ω

Power factor = cosφ = R/z = 6/ (6√2) = 1/√2

New answer posted

4 months ago

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Vishal Baghel

Contributor-Level 10

With the help of definition of e, we can write
e = v? /v? = 2v/u ⇒ u = 2v/e . (2)

Putting the value of e in equation (1), we have
m? (2v/e) = (m? - m? )v ⇒ 2m? = em? - em? ⇒ m? /m? = (2+e)/e = 1 + 2/e > 2

New answer posted

4 months ago

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Vishal Baghel

Contributor-Level 10

Kindly go through the answers

S? * S?

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Vishal Baghel

Contributor-Level 10

I? = I? + I? ⇒ I? /I? = 1 + I? /I? ⇒ 1/α = 1 + 1/β = (β+1)/β ⇒ α = β/ (1+β)

⇒ 1/β = 1/α - 1 = (1-α)/α ⇒ β = α/ (1-α)

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4 months ago

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Vishal Baghel

Contributor-Level 10

g? = g - ω²R ⇒ 0 = g - ω²R ⇒ ω = √ (g/R)

⇒ T = 2π/ω = 2π√ (R/g) = 2 * 3.14 * √ (6400 * 10³)/10)

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4 months ago

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Vishal Baghel

Contributor-Level 10

(A) sin (ωt) + cos (ωt) = √2 sin (ωt + π/4) ⇒ T = 2π/ω
(B) sin² (ωt) = 1/2 - (1/2)cos (2ωt) ⇒ T = 2π/ (2ω) = π/ω
(C) 3cos (π/4 - 2ωt) ⇒ T = 2π/ (2ω) = π/ω
(D) cos (ωt) + cos (2ωt) + cos (3ωt)
Time period of cos (ωt) = 2π/ω
Time period of cos (2ωt) = 2π/ (2ω)
Time period of cos (3ωt) = 2π/ (3ω)
Time period of combined function = 2π/ω

New answer posted

4 months ago

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Vishal Baghel

Contributor-Level 10

Always possible as it is associated only with β? decay


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Vishal Baghel

Contributor-Level 10

A = Area swept ⇒ dA/dt = (1/2)r² (dθ/dt) = (1/2) (Mr²ω)/M = L / (2M)

New answer posted

4 months ago

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Vishal Baghel

Contributor-Level 10

θ? > I = θ ⇒ sinθ? > sinθ ⇒ 1/μ > sinθ

⇒ μ < 1/sin

Note: If we assume θ = 45° ⇒ μ < 1.414, then red colour light ray will come out from face PR of prism.

New answer posted

4 months ago

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Vishal Baghel

Contributor-Level 10

PV? = C ⇒ V? (dP/dV) + P (γV^ (γ-1) = 0 ⇒ dP/dV = -γ (P/V) ⇒ dP/P = -γ (dV/V)

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