Class 12th

T = 2π√ (l/g) ⇒ g = 4π²l/T²
Percentage error: Δg/g = Δl/l + 2 (ΔT/T) = (0.1/10.0) + 2 (0.005/0.5) = 0.03
Percentage error = (Δg/g) * 100 = 3%
ω = 2πf = 100π rad/s
i_rms = i? /√2
While current changes from its maximum to its rms value, its phase changes by π/4 rad.
t = (π/4)/ω = π/ (4 * 100π) = 2.5 * 10? ³ s = 2.5ms.

Let the equation of the plane passing through (1, 2, 3) be:
a (x - 1) + b (y - 2) + c (z - 3) = 0
The plane contains the y-axis, which has direction ratios (0, 1, 0).
Therefore, the normal to the plane must be perpendicular to the y-axis.
a (0) + b (1) + c (0) = 0 ⇒ b = 0
The equation becomes: a (x - 1) + c (z - 3) = 0
ax + cz = a + 3c
The plane also passes through the origin (0,0,0) since it contains the y-axis.
a (0) + c (0) = a + 3c ⇒ a + 3c = 0 ⇒ a = -3c
Substitute a = -3c into the plane equation:
-3c (x - 1) + c (z - 3) = 0
-3 (x - 1) + (z - 3) = 0
-3x + 3 + z - 3 = 0
-3x + z = 0 ⇒ 3x - z = 0

Given the refractive index μ = λ? / λ = 3/2 and v = 10m, the object distance is u = - (3/2)v = -15m.
Using the lens maker's formula:
μ/v - 1/u = (μ - 1)/R
(3/2)/10 - 1/ (-15) = (3/2 - 1)/R
This gives the radius of curvature R:
R = -30/13 m

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