Class 11th

( (1+i)/ (1-i) )^ (m/2) = ( (1+i)/ (i-1) )^ (n/3) = 1
⇒ ( (1+i)²/2 )^ (m/2) = ( (1+i)²/ (-2) )^ (n/3) = 1
⇒ (i)^ (m/2) = (-i)^ (n/3) = 1
⇒ m/2 = 4k? and n/3 = 4k?
⇒ m = 8k? and n = 12k?
Least value of m = 8 and n = 12
∴ GCD = 4
∴ GCD = 4

The height of capillary rise is given by the formula:
h = (2T cosθ) / (rρg)
Given: h = 15 cm = 0.15 m, r = 0.015 cm = 1.5 * 10? m, ρ = 900 kg/m ³, g = 10 m/s², and θ ≈ 0° (so cosθ ≈ 1).
We need to find the surface tension, T.
T = (h r ρ g) / (2 cosθ)
T = (0.15 * 1.5 * 10? * 900 * 10) / 2
T = 0.10125 N/m
The answer is required in milliNewton m? ¹, so we multiply by 1000.
T = 101.25 mN/m.

(0.16)^log? (1/3 + 1/3² + . to ∞)
= (4/25)^log? (1/2)
= ( (5/2)? ² )^log? /? (1/2) = (5/2)^ (-2 log? /? (1/2)
= (5/2)^ (log? /? ( (1/2)? ² ) = 4

By the principle of conservation of angular momentum, the total angular momentum of the system (platform + person) remains constant.
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
Initial state: Person is on the rim.
I_initial = I_platform + I_person_rim = (1/2)M_platform R² + M_person R²
Final state: Person is at the center.
I_final = I_platform + I_person_center = (1/2)M_platform R² + 0
[ (1/2) (200)R² + (80)R²] * 5 rpm = [ (1/2) (200)R²] * ω_final
(100R² + 80R²) * 5 = (100R²) * ω_final
180R² * 5 = 100R² * ω_final
ω_final = (180 * 5) / 100 = 9 rpm.

lim_ (x→0) (1/x? ) {1 - cos (x²/2) - cos (x²/4) + cos (x²/2)cos (x²/4)} = 2?
⇒ lim_ (x→0) ( (1 - cos (x²/2) (1 - cos (x²/4) / x? ) = 2?
⇒ 2? = 2? ⇒ k = 8

∴ center lies on x + y = 2 and in 1st quadrant center = (a, 2-a) where a > 0 and 2-a > 0 ⇒ 0 < a < 2
∴ circle touches x = 3 and y = 2
⇒ |3-a| = |2 - (2-a)| = radius
⇒ |3-a| = |a| ⇒ a = 3/2
∴ radius = a
⇒ Diameter = 2a = 3.

After burning, heat exchange occurs between helium and atmospheric. Hence, irreversible, isothermal process.

S = (2 . ¹P? - 3 . ²P? + 4 . ³P? upto 51 terms) + (1! - 2! + 3! - . upto 51 terms)
∴ [? ? P_ (n-1) = n!]
= (2! - 3! + 4! + 52!) + (1! - 2! + 3! - 4! + . . + (51)!)
= 1! + 52!

LHL : lim_ (x→0? ) |1-x-x|/|λ-x-1| = 1/|λ-1|
RHL: lim_ (x→0? ) |1-x+x|/|λ-x+0| = 1/|λ|
For existence of limit
LHL = RHL
⇒ 1/|λ-1| = 1/|λ| ⇒ λ = 1/2
∴ L = 1/|λ| = 2

The mass of the cylinder M = ρV = ρ (πR²L), where ρ is the density of the material.
We can express R² in terms of L: R² = M / (πρL).
The moment of inertia is I = M (R²/4 + L²/12).
Substitute R² into the equation for I:
I (L) = M * [ (M / (4πρL) + (L²/12) ]
To find the minimum possible I, we differentiate I with respect to L and set the derivative to zero.
dI/dL = M * [ -M/ (4πρL²) + 2L/12 ] = 0
M/ (4πρL²) = 2L/12 = L/6
M/ (πρ) = (4/6)L³ = (2/3)L³
Since R² = M/ (πρL), we have R²L = M/ (πρ).
Substitute M/ (πρ) into the differentiated equation:
R²L = (2/3)L³