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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 3 Views

A
alok kumar singh

Contributor-Level 10

According to question
|x| = |y| and
|x - y| = n|x + y|
x² + y² - 2x·y = n² (x² + y² + 2x·y)
(1 - n²) (x² + y²) = (1 + n²)2x·y
(1 - n²) (x² + y²) = (1 + n²)2xy cosθ
(1 - n²) (2x²) = (1 + n²) (2x²)cosθ
cos θ = (1 - n²)/ (1 + n²)
θ = cos? ¹ (1 - n²)/ (1 + n²) = cos? ¹ (- (n²-1)/ (n²+1)

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 9 Views

R
Raj Pandey

Contributor-Level 9

Displacement in 4? second
= S? - S?
= (1/2)g [2n - 1]
= (1/2)g [2*4 - 1]
= (1/2) * 9.8 * 7
= 34.3m
As this distance matches with data given in question for position of next drop. So drops are falling at the rate of 1 drop/second.

New answer posted

10 months ago

0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

Body is dropped from height 75m with initial velocity (upward) 10m/s
So, s = ut + ½ at²
-75 = 10t - ½ gt²
5t² - 10t - 75 = 0
By solving t = 5 sec.
In 5sec. balloon covered
h = 10 * 5 = 50m
Now height of balloon = 75 + 50 = 125m

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

New answer posted

10 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-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

10 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-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?

New answer posted

10 months ago

0 Follower 17 Views

R
Raj Pandey

Contributor-Level 9

μ? sin 30° = μ? sin θ

⇒ 2.42 * (1/2) = (1) sinθ
sinθ = 1.21 > 1
Refraction is not possible, total internal reflection

 

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