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New answer posted

5 months ago

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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

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

ME = PE + KE = 8J
At x? , PE=4J, so KE = 8-4=4J. (A is correct)
At x>x? , PE=6J, so KE=2J, which is constant. (B is correct)
At x8J. This is not possible. Particle cannot reach here. So it is not a correct statement. (C is incorrect)
At x? , PE=0J, so KE=8J, which is maximum. (D is correct)

New answer posted

5 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

5 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

Tsinθ = qE
Tcosθ = mg
tanθ = qE/mg
E = V/d_eff. V = V? - (-V? ) = V? +V?
C? =kε? A/t, C? =ε? A/ (d-t).
Capacitors are in series. Q=C_eqV. E inside dielectric = σ/ (kε? ) = Q/ (Akε? ).
E in air = σ/ε? = Q/Aε?
Let's follow the solution.
Q = (C? / (C? +C? ) (V? +V? ).
E = E_air = Q/ (Aε? )
tanθ = qQ/ (Aε? mg) = q/ (Aε? mg) * (C? / (C? +C? ) (V? +V? )
Substitute C? and C?
This is getting complicated. The solution directly uses an expression for E. E = Q/C? Aε?

New answer posted

5 months ago

0 Follower 1 View

R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

5 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

P = Fv = (ma)v = m (dv/dt)v
P dt = mv dv
∫P dt = ∫mv dv
Pt = ½mv²
v = √ (2Pt/m)
dx/dt = √ (2P/m) t¹/²
x = √ (2P/m) ∫t¹/² dt = √ (2P/m) * (2/3)t³/²
Squaring to match options:
x² = (8P/9m) t³
This does not match. Let's re-examine the options.
Position x is proportional to t³/².
x = (8P/9m)¹/² t³/²

New answer posted

5 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

ω = 2πf = 1/√ (LC)
L = 1/ (4π²f²C) = 1/ (4π² * 60² * 0.1*10? ) = 70.3 mH

New answer posted

5 months ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

E? (due to A and G) = 2 * (kq/l²)cos (45) = √2 kq/l² (downwards)
E? (due to B and F) = 2 * (kq/l²)cos (45) = √2 kq/l² (towards left)
E? (due to C and H) = 0
E? (due to D and E) = 0
Resultant E = √ (E? ²+E? ²) = √ (2 (kq/l²)²+2 (kq/l²)²) = 2kq/l²
(Solution in the image seems to be different.)

New answer posted

5 months ago

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V
Vishal Baghel

Contributor-Level 10

T = 2π√ (m/k)
0.2 = 2π√ (0.5/k)
k = (0.5) (2π/0.2)² = 50π² ≈ 500
x = A sin (ωt) = 5 sin (2π/T * t)
At t=T/4, x = 5 sin (π/2) = 5cm
PE = ½kx² = ½ * 500 * (0.05)² = 250 * 0.0025 = 0.625 J

New answer posted

5 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

Energy of electron = 3 eV
It forms H atom in n=3 state. Energy released E = 3 - (-13.6/9) = 3 + 1.51 = 4.51 eV.
Photon Energy = 4.51 eV
Threshold energy = hc/λ = 12400eVÅ / 4000Å = 3.1 eV.
kE_max = 4.51 - 3.1 = 1.41 eV

 

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