Class 11th

Match List I with List II.

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Contributor-Level 10

[C] = Q/V = Q/ (W/Q) = Q²/W = (A²T²)/ (M¹L²T? ²) = M? ¹L? ²T? A²
[ε? ] = C/ (4πr²) * (Fr²/q²) = M? ¹L? ³T? A²
[µ? ] = 4πrF/I²l = M¹L¹T? ²A? ²
[E] = F/q = (MLT? ²)/ (AT) = MLT? ³A? ¹

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2 months ago

Two identical particles of mass 1 kg each go round a circle of radius R, under the action of their mutual gravitational attraction. The angular speed of each particle is:

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Class 11th Mechanical Properties of Fluids

Contributor-Level 10

F = Gm²/ (2R)² = mω²R Given m = 1kg
⇒ ω² = Gm/ (4R³)
⇒ ω = (1/2)√ (G/R³)

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2 months ago

A raindrop with radius R = 0.2 mm falls from a cloud at a height h = 2000 m above the ground. Assume that the drop is spherical through its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is :
[Density of water ρw = 1000 kg m⁻³, and Density of air ρa = 1.2 kg m⁻³, g = 10 m/s², Coefficient of viscosity of air η = 1.8 * 10⁻⁵ Nsm⁻²]

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Contributor-Level 10

At terminal speed
Mg = Fv = 6πηRv
⇒ V = mg / 6πηR
V = (4/3)πR³ρg / 6πηR
⇒ V = (2/9) * (ρR²g/η)
= (2/9) * (1000 * (0.2 * 10? ³)² * 10) / (1.8 * 10? )
= 4.94 m/s

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2 months ago

The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of 9.0 * 10³ km. Find the mass of Mars.
{Given 4π²/G = 6 * 10¹¹ N⁻²kg²}

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Contributor-Level 10

T² = (4π²r³)/GM
⇒ M = (4π²r³)/ (G T²)
⇒ M = 6 * 10¹¹ * (9 * 10? )³ / (450 * 60)²
= 6.48 x 10²³ kg
(Note: There is a calculation error in the source image. Using the formula and given values: M = (6e11 * (9e6)^3) / (27000)^2 = 6e23 kg)

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2 months ago

bDKJVbKJDVnsdkjvnskJDVnSKJDVn $\overset{?}{a}$ fjejkne

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New answer posted

2 months ago

The equation of a circle is Re(z²) + 2(Im(z))² + 2Re(z) = 0, where z = x + iy. A line which passes through the center of the given circle and the vertex of the parabola, x² - 6x - y + 13 = 0, has y-intercept equal to……….

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alok kumar singh

Contributor-Level 10

Re (z²) = x²-y². 2 (Im (z)² = 2y². 2Re (z) = 2x.
x²-y²+2y²+2x=0 ⇒ x²+y²+2x=0.
(x+1)²+y²=1. Center (-1,0).
Parabola: x²-6x+9 = y-13+9 ⇒ (x-3)²=y-4. Vertex (3,4).
Line through (-1,0) and (3,4): slope = (4-0)/ (3- (-1) = 1.
y-0 = 1 (x+1) ⇒ y=x+1.
y-intercept is 1.

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2 months ago

Let n ∈ N and [x] denote the greatest integer less than or equal to x. If the sum of (n+1) terms ⁿC₀, 3.ⁿC₁, 5.ⁿC₂, 7.ⁿC₃,. is equal to 2¹⁰⁰.101, then 2[ (n-1)/2 ] is equal to

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alok kumar singh

Contributor-Level 10

Sum S = Σ (2r+1)? C? = 2Σr? C? + Σ? C? = 2n2? ¹ + 2? = n2? + 2? = (n+1)2?
(n+1)2? = 101 * 2¹? ⇒ n=100.
2 [ (n-1)/2] = 2 [ (99)/2] = 2 [49.5] = 2*49 = 98.

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2 months ago

If log₃2, log₃(2ˣ - 5), log₃(2ˣ - 7/2) are in an arithmetic progression, then the value of x is equal to ________.

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Contributor-Level 10

2log? (2? -5) = log?2 + log? (2? -7/2).
(2? -5)² = 2 (2? -7/2) = 2*2? -7.
Let t=2? (t-5)² = 2t-7.
t²-10t+25=2t-7 ⇒ t²-12t+32=0 ⇒ (t-4) (t-8)=0.
t=4 or t=8.
2? =4 ⇒ x=2. log? (4-5) undefined.
2? =8 ⇒ x=3. log? (8-5)=log?3=1. log? (8-3.5)=log?4.5.
2 (1) = log?2+log?4.5 = log?9=2. Correct.
x=3.

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2 months ago

If the co-efficients of x⁷ and x⁸ in the expansion of (2 + x/3)ⁿ are equal, then the value of n is equal to…….

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alok kumar singh

Contributor-Level 10

Coeff of x? in (2+x/3)? is? C? 2? (1/3)?
Coeff of x? in (2+x/3)? is? C? 2? (1/3)?
? C? 2? / 3? =? C? 2? / 3?
(n!/ (7! (n-7)!) * 2 = (n!/ (8! (n-8)!) * (1/3).
2 / (n-7) = 1 / (8*3).
48 = n-7 ⇒ n=55.

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2 months ago

Class 11th Statistics

If the mean and variance of the following data: 6, 10, 7, 13, a, 12, b, 12 are 9 and 37/4 respectively, then (a-b)² is equal to:

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