Class 12th
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New answer posted
7 months agoContributor-Level 10
Enet = Eo/k
Enet = E_free - E_bound = qf/Aε? - qb/Aε?
Eo = qf/Aε?
So, (qf-qb)/Aε? = qf/ (kAε? )
qf - qb = qf/k
qb = qf (1 - 1/k)
New answer posted
7 months agoContributor-Level 10
Given 2q? = q? and v? /v? = 2/3, m? = m?
r = mv/qB
r? /r? = (m? /m? ) * (v? /v? ) * (q? /q? )
r? /r? = (1) * (2/3) * (2) = 4/3
New answer posted
7 months agoContributor-Level 10
μ = sin (A + δm)/2) / sin (A/2)
Since, δm = A
μ = sin (A) / sin (A/2) = (2sin (A/2)cos (A/2) / sin (A/2)
μ = 2cos (A/2
New answer posted
7 months agoContributor-Level 10
I = I? cosωt
Current is changing from its maximum value to rms value (I? /√2)
I? /√2 = I? cosωt
cosωt = 1/√2
ωt = π/4
2π * 50t = π/4
t = 1/400 s = 2.5 ms
New answer posted
7 months agoContributor-Level 10
tan 37° = E? /E?
E? = k (2p? /r³) axial direction
E? = k (p? /r³) equatorial direction
tan 37° = (k p? /r³) / (k 2p? /r³) = p? / (2p? ) = 3/4
p? /p? = 2/3
New answer posted
7 months agoContributor-Level 10
For electron,
λe = h/p ⇒ (KE)e = ½mv² = p²/ (2m) = (h²/λ²)/ (2m)
For photon,
λp = h/p ⇒ (K.E.)p = pc = hc/λ
KEe / KEp = (h²/ (2mλ²) / (hc/λ) = h/ (2mcλ)
But for electron p = mv = h/λ so h/λ = mv
KEe / KEp = mv / (2mc) = v/2c
New answer posted
7 months agoContributor-Level 10
ε? = 250 * Potential Gradient
ε? + ε? = 400 * Potential Gradient
(ε? )/ (ε? + ε? ) = 250/400 = 5/8
By solving above
ε? /ε? = 5/3
New answer posted
7 months agoContributor-Level 10
By shell's law
(sin θ)/ (sin θ') = 4/3 . (i)
For TIR on second surface
sin θ' > sin θc
(θ' + θ' = 90)
sin (90 - θ') > sin θc
cos θ' > sin θc
1 - sin²θ' > sin²θc (By equation (i) )
1 - (3/4 sin θ)² > sin²θc
[sin θc = 3/4]
1 - (9/16)sin²θ > (9/16)
1 - 9/16 > (9/16)sin²θ
7/16 > (9/16)sin²θ
sinθ < 7/3
New answer posted
7 months agoContributor-Level 10
Yes, Institute of Hotel Management Studies, Kotdwar accept Class 12th marks. Candidates seeking admission to all programmes can enrol for admission with Class 12 marks. The college offers Certificate and Diploma courses. Institute of Hotel Management Studies, Kotdwar admissions are based on merit.
New answer posted
7 months agoContributor-Level 9
λ = h/mv = h/√2mK
For same K:
λ ∝ 1/√m
λ? : λ? : λ? = 1/√m? : 1/√m? : 1/√4m?
As m? > m? ,
λ? > λ? > λ?
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