Class 12th

Resonance frequency is independent of R.
Quality factor = ωL/R ⇒ Quality factor decreases with increase in R.
Bandwidth of resonance circuit = R/L ⇒ increases with increase in R.

Let the charge on C? be q µC. For the capacitor network:
q/C? = (C? * 10 - q) / C? ⇒ q/8 = (2 * 10 - q) / 2 ⇒ q = 16

λ? /λ? = (m? v? ) / (m? v? ) ⇒ m? = m? (v? /v? ) (λ? /λ? ) = m? (1/4) (1/2) = (1/8)m?

The system of equations has no solution if the determinant of the coefficient matrix is zero.
Δ = |k 1|
|1 k 1|
|1 k|
Δ = k (k² - 1) - 1 (k - 1) + 1 (1 - k) = 0
Δ = k³ - k - k + 1 + 1 - k = 0
⇒ k³ - 3k + 2 = 0 ⇒ (k - 1)² (k + 2) = 0
∴ k = -2, 1
If k = 1 then all the equations are identical (infinite solutions). Hence k = -2 for no solution.

Let V? = 10V, V? = xV, V? = 0V, and V? = yV.
Applying Kirchhoff's current law at node B:
(x - 10)/100 + (x - y)/15 + (x - 0)/10 = 0 ⇒ 53x - 20y = 30 . (1)
Applying Kirchhoff's current law at node D:
(y - 10)/60 + (y - x)/15 + (y - 0)/5 = 0 ⇒ 17y - 4x = 10 . (2)
Solving equations (1) and (2), we get:
x = 0.865 and y = 0.792
The current i? is:
i? = (x - y) / 15 = 4.87 mA

dy/dx = xy - 1 + x - y = x (y + 1) - (y + 1) = (x - 1) (y + 1)
⇒ ∫ dy / (1 + y) = ∫ (x - 1) dx ⇒ ln (1 + y) = x²/2 - x + C
For y (0) = 0, c = 0. ∴ ln (1 + y) = x²/2 - x
1 + y = e^ (x²/2 - x)
y (1) = e^ (1/2 - 1) - 1 = e^ (-1/2) - 1

In external magnetic field, a magnetic force acts on every small part of the loop in direction perpendicular to the wire. Thus, loop assumes a shape (circular) in which it covers maximum area

E ∝ m ⇒ E / 13.6 = 207 ⇒ E = 207 * 13.6eV = 2815.2eV

The Balmer series lies in the visible region of the electromagnetic spectrum.

From Radioactive Decay Law,
dN/dt = λ? N + λ? N = λ_eff N
⇒ λ_eff = λ? + λ? ⇒ ln (2)/T = ln (2)/T? + ln (2)/T? ⇒ T = (T? ) / (T? + T? )
(where T, T? , and T? are half-lives)