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10 months agoContributor-Level 10
σ4πr² + σ4πR² = Q ⇒ σ = Q/ (4π (R²+r²)
V_c = kq? /r + kq? /R = k (σ4πr²)/r + k (σ4πR²)/R = kσ4π (r+R)
= K (Q/ (R²+r²) (R+r)
New answer posted
10 months agoContributor-Level 10
On increasing the temperature, random velocity of molecules increases, therefore mean collision time between the molecules decreases. But the mean free path remains constant as it is product of velocity and time.? (b) and (c) are correct.
New answer posted
10 months agoContributor-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
10 months agoContributor-Level 10
λ = h/p = h/mv
λp/λe = (m? v? )/ (m? v? ) ⇒ 1.878 * 10? = (9.1*10? ³¹)/ (m? * 5)
m? = 9.1*10? ³¹ / (5 * 1.878 * 10? ) = 0.97 * 10? ²? kg
New answer posted
10 months agoContributor-Level 10
Δl = lαΔT ⇒ Δl/l = αΔT = 0.02%
Δρ = -ργΔT
|Δρ/ρ| = γΔT = 3αΔT = 3 (0.02%) = 0.06%
New answer posted
10 months agoContributor-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
10 months agoContributor-Level 10
The excess pressure inside a soap bubble is given by ΔP = 4T/R, where T is the surface tension and R is the radius.
The pressures are given as P? = 1.01 atm and P? = 1.02 atm. Let the atmospheric pressure be P? = 1 atm.
ΔP? = P? - P? = 1.01 - 1 = 0.01 atm = 4T/R?
ΔP? = P? - P? = 1.02 - 1 = 0.02 atm = 4T/R?
Dividing the two equations: (ΔP? /ΔP? ) = (R? /R? )
0.01 / 0.02 = R? /R? ⇒ R? /R? = 2
The ratio of their volumes is V? /V? = ( (4/3)πR? ³ ) / ( (4/3)πR? ³ ) = (R? /R? )³ = 2³ = 8.
The ratio is 8:1.
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