Class 12th

Contributor-Level 10

dy/dx + (tanx)y = sinx. This is a linear differential equation.
Integrating Factor (I.F.) = e^ (∫tanx dx) = e^ (ln|secx|) = secx.
The solution is y * I.F. = ∫ (sinx * I.F.) dx + C.
y * secx = ∫ (sinx * secx) dx = ∫tanx dx = ln|secx| + C.
Given y (0) = 0.
0 * sec (0) = ln|sec (0)| + C => 0 = ln (1) + C => C = 0.
So, y * secx = ln (secx).
y = cosx * ln (secx).
At x = π/4:
y = cos (π/4) * ln (sec (π/4) = (1/√2) * ln (√2) = (1/√2) * (1/2)ln (2) = ln (2) / (2√2).

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a year ago

Class 12th Maths

Let A denote the event that a 6-digit integer formed by 0,1,2,3,4,5,6 without repetitions, is divisible by 3. Then probability of event A is equal to :

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Contributor-Level 10

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a year ago

Class 12th Maths

Let A = {2,3,4,5,.,30} and '~' be an equivalence relation on A*A, defined by (a,b) ~ (c,d), if and only if ad = bc. Then the number of ordered pairs which satisfy this equivalence relation with ordered pair (4, 3) is equal to :

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Class 12th Ncert Solutions Chemistry Class 12th

Contributor-Level 10

The equation of the circle is x²+y²+ax+2ay+c=0, with a<0.
x-intercept = 2√ (g² - c) = 2√ (a/2)² - c) = 2√2. So, a²/4 - c = 2 => a² = 8 + 4c - (i)
y-intercept = 2√ (f² - c) = 2√ (a² - c) = 2√5. So, a² - c = 5 => a² = 5 + c - (ii)
Equating (i) and (ii): 8 + 4c = 5 + c => 3c = -3 => c = -1.
Substituting c in (ii): a² = 5 - 1 = 4. Since a < 0, a = -2.
The equation of the circle is x² + y² - 2x - 4y - 1 = 0.
Completing the square: (x-1)² + (y-2)² = 1+4+1 = 6.
The center is (1,2) and radius is √6.
The tangent is perpendicular to the line x + 2y = 0 (slope -1/2).
So, the slope of the tangent is 2.
Equation of the tangent: (y-2) = 2 (x-1)

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a year ago

When light of wavelength 248 nm falls on a metal of threshold energy 3.0eV, the de-Broglie wavelength of emitted electrons is__________ Angstrom.

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Class 12th Chemistry Chemical Kinetics

Contributor-Level 9

K.E = φ - φ?
φ? = 3 eV = 3 * 1.6 * 10? ¹? J = 4.8 * 10? ¹? J
φ = hc/λ = (6.63 * 10? ³? * 3 * 10? ) / (248 * 10? ) J = 8 * 10? ¹? J
K.E = 8 * 10? ¹? - 4.8 * 10? ¹? = 3.2 * 10? ¹? J
Now using, λ = h / √ (2 K.E m)
λ = (6.63 * 10? ³? ) / √ (2 * 3.2 * 10? ¹? * 9.1 * 10? ³¹) m
λ = (6.63 * 10? ³? ) / (7.63 * 10? ²? ) m = 0.87 * 10? m = 8.7 Å
So, the nearest integer is 9.

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a year ago

The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0x10^-3 /s and the activation energy E_a = 11.488 kJ/mol, the rate constant at 200 K is ___________ x10-5 /s. (Round off to the nearest Integer).

[Given R = 8.314 J/molK]

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Contributor-Level 9

T? = 300 K; K? = 1 * 10? ³ s? ¹
T? = 200 K; K? =?
E_a = 11.488 kJ/mol
Using Arrhenius equation:
log (K? /K? ) = (E_a / 2.303R) * [ (T? - T? ) / (T? )]
log (K? / 10? ³) = (11.488 * 10³ / (2.303 * 8.314) * [ (-100) / (6 * 10? )]
log (K? / 10? ³) = -1
K? / 10? ³ = 10? ¹
K? = 10? s? ¹ or K? = 10 * 10? s? ¹

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a year ago

The equivalents of ethylene diamine required to replace the neutral ligands from the coordination sphere of the trans-complex of CoCl3.4NH3 is _________.

(Round off to the Nearest Integer).

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Contributor-Level 9

The complex CoCl? ·4NH? is represented as [CoCl? (NH? )? ]Cl.
Here, 4 NH? are neutral monodentate ligands. So, 2 equivalents of ethylene diamine replace 4NH? since ethylene diamine is a didentate ligand.

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New answer posted

a year ago

Class 12th Maths

Let α ∈ R be such that the function f(x) = { [cos⁻¹(1-{x}²)sin⁻¹(1-{x})] / ({x}-{x}³), x≠0; α, x=0 } is continuous at x=0, where {x} = x-[x], [x] is the greatest integer less than or equal to x. Then :

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Contributor-Level 10

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a year ago

AB2 is 10% dissociated in water to A²⁺ and B^-. The boiling point of a 10.0 molal aqueous solution of AB2 is_________ °C. (Round off to the nearest Integer).

[Given : Molal elevation constant of water Kb = 0.5 K kg mol^-1, boiling point of pure water = 100°C]

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Contributor-Level 9

m = 10 molal
K_b = 0.5 K kg mol? ¹
Using: ΔT_b = I K_b m
and α = (i - 1) / (n - 1)
n for AB? is 3; α = 0.1
0.1 = (i - 1) / (3 - 1) ⇒ I = 1.2
ΔT_b = 1.2 * 0.5 * 10 = 6 °C
So, boiling point of solution = 100 + 6 = 106 °C

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New answer posted

a year ago

A 6.50 molal solution of KOH (aq) has a density of 1.89 g cm^-3. The molarity of the solution is________ mol dm^-3. (Round off to the nearest Integer).

[Atomic masses : K = 39.0u, O = 16.0u, H= 1.0u]

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Contributor-Level 9

A 6.5 molal solution means 6.5 moles of KOH is in 1 kg (1000 g) of solvent (H? O).
Moles of solute, n_B = 6.5
Mass of solute, W_B = 6.5 * 56 = 364 g
Mass of solvent, W_A = 1000 g
Mass of solution = 1364 g
Volume of solution = 1364 / 1.89 mL
Now, molarity = [6.5 / (1364 / 1.89)] * 1000 M = 9 M

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New answer posted

a year ago

Two salts A2X and MX have the same value of solubility product of 4.0 x 10^-12. The ratio of their molar solubilities i.eS A2X/S MX = ____________.

(Round off to the Nearest Integer).

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