Class 12th

Given r * a = r * b, which means r * a - r * b = 0 ⇒ r * (a - b) = 0.
This implies that vector r is parallel to vector (a - b).
So, r = λ (a - b) for some scalar λ.
a - b = (2i - 3j + 4k) - (7i + j - 6k) = -5i - 4j + 10k.
So, r = λ (-5i - 4j + 10k).
We are also given r ⋅ (i + 2j + k) = -3.
λ (-5i - 4j + 10k) ⋅ (i + 2j + k) = -3
λ (-51 - 42 + 10*1) = -3
λ (-5 - 8 + 10) = -3
λ (-3) = -3 ⇒ λ = 1.
So, r = 1 * (-5i - 4j + 10k) = -5i - 4j + 10k.
We need to find r ⋅ (2i - 3j + k).
(-5i - 4j + 10k) ⋅ (2i - 3j + k) = (-5) (2) + (-4) (-3) + (10) (1)
= -10 + 12 + 10 = 12.

We need to evaluate lim (x→0? ) (cos? ¹ (x - [x]²) ⋅ sin? ¹ (x - [x]²) / (x - x²).
As x → 0? , the greatest integer [x] = 0.
So the expression becomes:
lim (x→0? ) (cos? ¹ (x) ⋅ sin? ¹ (x) / (x (1 - x²)
= lim (x→0? ) cos? ¹ (x) * lim (x→0? ) (sin? ¹ (x) / x) * lim (x→0? ) (1 / (1 - x²)
We know lim (x→0) (sin? ¹ (x) / x) = 1.
lim (x→0? ) cos? ¹ (x) = cos? ¹ (0) = π/2.
lim (x→0? ) (1 / (1 - x²) = 1 / (1 - 0) = 1.
So the limit is (π/2) * 1 * 1 = π/2.

The flux φ is calculated as:
φ = (2/5) * 4 * 10³ * 0.4 = 640 Nm²C? ¹

From the given relations:
mv²/r = |dU/dr| = 4|U? |r³ . (1)
mvr = nh / 2π . (2)
By combining equations (1) and (2), we can derive the radius r:
r = ( (nh)² / (4π√* (U? )*) )¹/³ ⇒ r ∝ n²/³

1/C = 5/ (ε? * 100) + 5/ (10ε? * 100)
⇒ C = (ε? * 1000) / 55 = (8.85 * 10? ¹² * 1000) / 55 = 1.61 * 10? ¹? F = 161 pF

Conduction current density, J? = E/ρ = (V/d) / ρ = (V? sin (2πft) / (pd)
Displacement current density, J? = ε (dE/dt) = (ε/d) (dV/dt) = (2πfε/d) * V? cos (2πft)
The ratio of their magnitudes is:
J? / J? = tan (2πft) / (2πfε? ρ) = tan (2π * 900) / (2π * 9 * 10? * 80ε? * 0.25) = 10?

Electric field squared is proportional to the power P of the bulb (E² ∝ P).
(E' / E)² = (60 / 100) ⇒ E' = E * √* (3/5)*

Power gain = (i_c² R_c) / (i_b² R_B) = (i_c/i_b)² (R_c/R_B) = (10²)² (10? /10³) = 10?
i_c/i_b = 100 ⇒ β = i_c/i_b = 100

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