Class 12th
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New answer posted
a year agoContributor-Level 10
As we know that magnetic force acting on a charge particle will be
F = q (v * B)
W = F.dl
Since force and displacement will be always perpendicular so work done is always zero.
New answer posted
a year agoContributor-Level 10
Since speed of light is constant for all colour so red colour and blue colour have different frequencies and different wavelengths.
New answer posted
a year agoContributor-Level 10
N? /N? = e?
For 33% decay, N? /N? = 0.67 ≈ 2/3.
2/3 = e? ⇒ t? = (1/λ)ln (3/2)
For 67% decay, N? /N? = 0.33 ≈ 1/3.
1/3 = e? ⇒ t? = (1/λ)ln (3)
Δt = t? - t? = (1/λ) [ln (3) - ln (3/2)] = (1/λ)ln (2) = T? /? = 20 min
New answer posted
a year agoContributor-Level 10
Since B, v and length are perpendicular
ε = Bvl
emf will induce only in wire CD
ε = B (d)v? (d) = B? (d/a)v? d = B? v? d²/a
New answer posted
a year agoContributor-Level 10
E_K - E_L = hc/λ_Kα
E_K - E_L = (4.14*10? ¹? * 3*10? )/0.071*10? = 17500 eV = 17.5 keV.
E_L = E_K - 17.5 = 27.5 - 17.5 = 10 keV.
New answer posted
a year agoContributor-Level 10
E = -dφ/dt = - (20t + 20) mV.
At t=5s, E = - (100+20) = -120mV.
I = |E|/R = 120mV/2Ω = 60mA.
New answer posted
a year agoContributor-Level 10
The graph of? vs I is a parabola-like curve with a minimum at i=e. So (B) is the correct representation.
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