Class 11th

cotθ = (1+cos2θ)/sin2θ
cot (π/24) = (1+cos (π/12)/sin (π/12)
cos (π/12) = cos (15°) = cos (45-30) = (√3+1)/2√2
sin (π/12) = sin (15°) = sin (45-30) = (√3-1)/2√2
cot (π/24) = (1+ (√3+1)/2√2)/ (√3-1)/2√2) = (2√2+√3+1)/ (√3-1)
= (2√2+√3+1) (√3+1)/2 = (2√6+2√2+3+√3+√3+1)/2
= √6 + √2 + √3 + 2

Energy per second = 1000 J / 10 s = 100 J/s
Energy of one photon E = hc/λ = (6.626*10? ³? * 3*10? ) / (400*10? ) = 4.965 * 10? ¹? J
Number of electrons ejected = Total energy / Energy per photon = 100 / (4.965 * 10? ¹? ) = 20.14 * 10¹? ≈ 2 * 10²?

Truth table analysis shows that (P ∨ Q) ∧ (¬P) is equivalent to Q ∧ ¬P.
Then (Q ∧ ¬P) ⇒ Q. This is a tautology.
The provided solution seems to have an error.
Let's check the options. (P ∨ Q) is a tautology. (P ∧ ¬Q) is a contradiction.
~ (P ⇒ Q) ⇔ P ∧ ¬Q is true.

A + B? 2C
Initial: 1, 1
At eq: 1-x, 1+2x
K = [C]²/ ( [A] [B]) = (1+2x)²/ (1-x)² = 100
(1+2x)/ (1-x) = 10
1+2x = 10-10x => 12x = 9 => x = 3/4
[C] = 1+2x = 1+2 (3/4) = 1+1.5 = 2.5M = 25 * 10? ¹M

sinθ + cosθ = 1/2
16 (sin (2θ) + cos (4θ) + sin (6θ)
= 16 [2sin (4θ)cos (2θ) + cos (4θ)]
= 16 [4sin (2θ)cos² (2θ) + 2cos² (2θ) - 1] . (i)
Now, sinθ + cosθ = 1/2, squaring on both sides, we get
1 + sin (2θ) = 1/4
sin (2θ) = -3/4
cos² (2θ) = 1 - sin² (2θ) = 1 - 9/16 = 7/16
From equation (i)
16 [4 (-3/4) (7/16) + 2 (7/16) - 1]
16 [-21/16 + 14/16 - 16/16] = 16 [-23/16] = -23

C? H? + 13/2 O? → 4CO? + 5H? O
1 mole C? H? (58 g) produces 5 mole H? O (90 g)
∴ 90 g H? O obtained from 58 g C? H?
∴ 72g H? O obtained from (58/90) * 72g = 46.4 g
= 464 * 10? ¹g

ΔH? = ΔH? + ΔH?
= 2.8 + 98.2 = 101 kJ/mole

S? : |z - 3 - 2i|² = 8
|z - (3 + 2i)| = 2√2
(x - 3)² + (y - 2)² = (2√2)²
S? : Re (z) ≥ 5
x ≥ 5
S? : |z - z? | ≥ 8
|2iy| ≥ 8
2|y| ≥ 8
|y| ≥ 4
y ≥ 4 or y ≤ -4
From the graph of the circle (S? ) and the regions (S? and S? ), we can see that there is one point of intersection at (5, 4).
∴ n (S? ∩ S? ∩ S? ) = 1