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New answer posted

4 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

|Δp| = 2mu sinθ = 2 * 5 * 10? ³ * 5√2 * (1/√2) = 5 * 10? ² kg m s? ¹

⇒ x = 5

New answer posted

4 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

V = (4/3)πr³

⇒ ΔV/V = 3 (Δr/r) ⇒ Relative error

% error in volume = (ΔV/V) * 100 = 3 (Δr/r) * 100 = 3 * (0.85/7.50) * 100 = 33.999 ≈ 34

New answer posted

4 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

R_eq = (R? )/ (R? +R? ) ⇒ p (l)/ (2A) = [ (p? l/A) (p? l/A)] / [ (p? l/A) + (p? l/A)]

⇒ ρ/2 = (p? )/ (p? +p? ) ⇒ ρ = (2p? p? )/ (p? +p? ) = (2 * 6 * 3)/9 = 4 Ωcm

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

v? ²/2 + P? /ρ + gh? = v? ²/2 + P? /ρ + gh?

⇒ v? ²/2 + (P? + ρgh + mg/A)/ρ + g * 0 = v? ²/2 + P? /ρ + g * 0

⇒ v? ²/2 + (ρgh + mg/A)/ρ = v? ²/2 . (1)

Using Continuity equation, we can write
Av? = av? ⇒ v? = (a/A)v?

Putting the value of v? in equation (1), we have
(a²/A²)v? ²/2 + (ρgh + mg/A)/ρ = v? ²/2 ⇒ (A² - a²)/A² * v? ²/2 = (ρgh + mg/A)/ρ

⇒ v? = √ [ (A²/ (A²-a²) * 2 (ρgh + mg/A)/ρ] = 3.09 m/s ≈ 3 m/s

New answer posted

4 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

12/x = 6/ (72-x) ⇒ x = 144 - 2x ⇒ x = 144/3 = 48cm

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

d_lm = Distance covered by transmitting antenna + Distance covered by receiving antenna

⇒ d_lm = √ (2Rh_transmitting) + √ (2Rh_receiving)
⇒ d_lm = √ (2 * 64 * 10? * 20) + √ (2 * 64 * 10? * 5) = 16000 + 8000 = 24000m

When h_receiving = 0 then
d_2m = √ (2 * 64 * 10? * 20) = 16000m

% increment = (8000/16000) * 100 = 50

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

β = I? /I? = (2 * 10? ³)/ (10 * 10? ) = 200

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

v_rms = √ (3RT/M) and v_av = √ (8RT/πM)

⇒ v_rms / v_av = √ [ (3RT/M) / (8RT/πM)] = √ (3π/8)

New answer posted

4 months ago

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Vishal Baghel

Contributor-Level 10

μ_min = (I tanθ)/ (I + mR²)
= [ (mR²/2)tanθ] / [mR²/2 + mR²] = (tanθ)/3 = (tan 60°)/3 = √3/3 = 1.732/3 = 0.5773

Since given coefficient of static friction is less than μ_min, so body will perform rolling with slipping and kinetic friction will act
F? = μN = μmg cosθ = (0.4) * mg cos 60° = mg/5

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

As we know that direction of propagation of electromagnetic wave is perpendicular to plane containing mutually perpendicular electric field and magnetic field, so option D will be correct answer.

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