Class 11th

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New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the answers

S? * S?

New answer posted

4 months ago

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R
Raj Pandey

Contributor-Level 9

Bohr's theory accounts for the line spectrum of single electron species but Li? has two electrons. Bohr's theory fails to explain splitting of spectral lines in presence of magnetic field i.e. Zeeman effect.

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

To find the sum Σ[r=0 to 6] (?C?)².
This is the coefficient of x? in the expansion of (1+x)?(x+1)? = (1+x)¹².
By the binomial theorem, (1+x)¹² = Σ[k=0 to 12] ¹²C? x?.
The coefficient of x? is ¹²C?.
¹²C? = (12 * 11 * 10 * 9 * 8 * 7) / (6 * 5 * 4 * 3 * 2 * 1) = 11 * 2 * 3 * 2 * 7 = 924.

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

g? = g - ω²R ⇒ 0 = g - ω²R ⇒ ω = √ (g/R)

⇒ T = 2π/ω = 2π√ (R/g) = 2 * 3.14 * √ (6400 * 10³)/10)

New answer posted

4 months ago

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alok kumar singh

Contributor-Level 10

I = ∫[0 to 10] [sin(2πx)] / e^(x-[x]) dx.
The period of the integrand involves [sin(2πx)] which depends on the sign of sin(2πx) and {x} = x - [x], which has a period of 1.
Let f(x) = [sin(2πx)] / e^{x}.
The integral is ∫[0 to 10] f(x) dx = 10 * ∫[0 to 1] f(x) dx due to the periodicity of {x} and the integer period of sin(2πx).
In the interval (0, 1/2), sin(2πx) is between 0 and 1, so [sin(2πx)] = 0.
In the interval (1/2, 1), sin(2πx) is between -1 and 0, so [sin(2πx)] = -1.

At x=0, 1/2, 1, the value is 0.
So, ∫[0 to 1] f(x) dx = ∫[0 to 1/2] 0 dx + ∫[1/2 to 1] -1/e^x dx
= 0 + [-e^(-x) * (-1)] from 1/2 to 1 = [e^(-x)] from 1

...more

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4 months ago

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Vishal Baghel

Contributor-Level 10

(A) sin (ωt) + cos (ωt) = √2 sin (ωt + π/4) ⇒ T = 2π/ω
(B) sin² (ωt) = 1/2 - (1/2)cos (2ωt) ⇒ T = 2π/ (2ω) = π/ω
(C) 3cos (π/4 - 2ωt) ⇒ T = 2π/ (2ω) = π/ω
(D) cos (ωt) + cos (2ωt) + cos (3ωt)
Time period of cos (ωt) = 2π/ω
Time period of cos (2ωt) = 2π/ (2ω)
Time period of cos (3ωt) = 2π/ (3ω)
Time period of combined function = 2π/ω

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

A = Area swept ⇒ dA/dt = (1/2)r² (dθ/dt) = (1/2) (Mr²ω)/M = L / (2M)

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

Parabola: y² = 4x - 20 = 4(x - 5). Vertex at (5,0).
Line: The text seems to derive the tangent equation y = x - 4. This is not a tangent to the given parabola. The standard tangent to y²=4aX is Y=mX+a/m. Here X=x-5, a=1. So y = m(x-5)+1/m.
The other curve is an ellipse: x²/a² + y²/b² = 1.
The text says x²/2 + (x-4)²/b² = 1. This assumes a² = 2.
x²/2 + (x²-8x+16)/b² = 1
x²(1/2 + 1/b²) - (8/b²)x + (16/b² - 1) = 0.
For tangency, the discriminant (D) of this quadratic equation must be zero.
D = (8/b²)² - 4(1/2 + 1/b²)(16/b² - 1) = 0.
64/b? - 4(8/b² - 1/2 + 16/b? - 1/b²) = 0.
16/b? - (7/b² - 1/2 + 16/b?) = 0.
-7/b² + 1/2 = 0

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New answer posted

4 months ago

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Vishal Baghel

Contributor-Level 10

PV? = C ⇒ V? (dP/dV) + P (γV^ (γ-1) = 0 ⇒ dP/dV = -γ (P/V) ⇒ dP/P = -γ (dV/V)

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

U = mV (r) = -Cm/r

F = -dU/dr = -Cm/r² ⇒ The force which provides required centripetal force

⇒ mv²/r = Cm/r² ⇒ r = C/v²

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