Class 12th

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New answer posted

10 months ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

By Einstein's equation of photoelectric effect,
hc/λ = hc/λ? + eVs
Now hc/λ = hc/λ? + e*4.8 . (i)
hc/ (2λ) = hc/λ? + e*1.6 . (ii)
Multiply (ii) by 2:
hc/λ = 2hc/λ? + e*3.2 . (iii)
Equating (i) and (iii):
hc/λ? + 4.8e = 2hc/λ? + 3.2e
1.6e = hc/λ?
Substitute this into (i)
hc/λ = 1.6e + 4.8e = 6.4e
Substitute this into (ii)
(6.4e)/2 = hc/λ? + 1.6e
3.2e = hc/λ? + 1.6e
1.6e = hc/λ?
From (i): hc/λ = hc/λ? + 3 (hc/λ? ) = 4hc/λ?
λ? = 4λ

New answer posted

10 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

Maximum value of emf is measured when pointer is at B with I? = 0.
So, I_AB = 6 / (20 + 0.1 * 1000) = 6/120 = 1/20 A
V_AB = E = I_AB * R_AB
= (1/20) * 100
= 5V

New answer posted

10 months ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

Reflected wave will be 3.1cos [1.8z + (5.4*10? )t] î N/C
K = 2π/λ
⇒ 1.8 = 2π/λ
λ = 2π/1.8 = π/0.9 = 10π/9 = (10/9) * 3.14
= 3.48m

New answer posted

10 months ago

0 Follower 32 Views

A
alok kumar singh

Contributor-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

10 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-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

10 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-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

10 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-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

10 months ago

0 Follower 21 Views

A
alok kumar singh

Contributor-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

10 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-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

10 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-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

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