Class 12th

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New answer posted

a year ago

0 Follower 1 View

A
alok kumar singh

Contributor-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 ago

0 Follower 2 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 9 Views

A
alok kumar singh

Contributor-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 ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-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 ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

The graph of? vs I is a parabola-like curve with a minimum at i=e. So (B) is the correct representation.

New answer posted

a year ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

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