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New answer posted

7 months ago

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V
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)

New answer posted

7 months ago

0 Follower 8 Views

V
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

7 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Always possible as it is associated only with β? decay


New answer posted

7 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

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

New answer posted

7 months ago

0 Follower 15 Views

V
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

7 months ago

0 Follower 3 Views

V
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)

New answer posted

7 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

U = mV (r) = -Cm/r

F = -dU/dr = -Cm/r² ⇒ The force which provides required centripetal force

⇒ mv²/r = Cm/r² ⇒ r = C/v²

New answer posted

7 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

I = (V/R) (1 - e^ (-t/τ) where τ = L/R

E = (1/2)LI² = (1/2)L (V²/R²) (1 - e^ (-t/τ)² ⇒ Energy stored in inductor

According to Question, we can write
(1/4) * (1/2) (LV²/R²) = (1/2) (LV²/R²) (1 - e^ (-t/τ)²
⇒ 1/4 = (1 - e^ (-t/τ)² ⇒ 1/2 = 1 - e^ (-t/τ) ⇒ e^ (-t/τ) = 1/2

⇒ t/τ = ln (2) ⇒ t = τln (2) = (L/R)ln (2)

New answer posted

7 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

R = mv / (qB) = √ (2mK) / (qB) ⇒ K = (R²q²B²) / (2m)

⇒ K? /Kα = (R? ²q? ²B² / 2m? ) * (2mα / (Rα²qα²B²) = (mα/m? ) * (R? /Rα)² * (q? /qα)²

⇒ K? /Kα = (4) * (2)² * (1/2)² = 4:1

New answer posted

7 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

R = (2μ sinθ) / (1.22λ) . (1)

According to de-Broglie's hypothesis, we can write
λ = h / (mv) . (2)

With the help of equations (1) and (2), we can write
R = (2μmv sinθ) / (1.22h) ⇒ R? /R? = m? /m? = 1837 ⇒ R? = 1837R?

  • Concept involved: Resolving Power of Microscope
  • Topic: Optics
  • Difficulty level: Moderate
  • Point of Error: Formula

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