Class 11th

Contributor-Level 10

ρdrω²r = ρgdh
ω²∫? rdr = g∫? dh
ω²R²/2 = gh
h = ω²R²/2g = 25ω²/2g

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10 months ago

Class 11th Physics

A bead of mass m stays at point P(a,b) on a wire bent in the shape of a parabola y = Cx² and rotating with angular speed ω (see figure). The value of ω is (neglect friction)

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Contributor-Level 10

mω²acosθ = mgsinθ
ω = √ (gtanθ/a)
y = 4cx²

tanθ = dy/dx = 8xC
(tanθ)? , b = 8aC
ω = √ (g*8aC/a) = 2√ (2gC)

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10 months ago

If |x| < 1, |y| < 1 and x y, then the sum to infinity of the following series (x+y) + (x+xy+y) + (x+xy+xy+y) + . is

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Contributor-Level 10

Let t? denotes r+1th term of (? x? +? x? )¹?
t? = ¹? C? (? x? )¹? (? x? )? = ¹? C? ¹? x? ¹?
If t? is independent of x
90-15r=0? r=6
This differs from the solution.
Let's follow the solution's powers.
(10-r)/9 - r/6 = 0? r=4
maximum value of t? is 10K (given)
? ¹? C? is maximum
By AM? GM (for positive numbers)
(? ³/2+? ³/2+? ²/2+? ²/2)/4? (? /16)¹/?
? ? ? 16
So, 10K = ¹? C?16
? K=336

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10 months ago

The number of integral values of k for which the line, 3x+4y=k intersects the circle, x²+y²-2x-4y+4=0 at two distinct points is.

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Contributor-Level 10

Circle x²+y²-2x-4y+4=0
⇒ (x-1)²+ (y-2)²=1
Centre: (1,2) radius=1
line 3x+4y-k=0 intersects the circle at two distinct points.
⇒ distance of centre from the line < radius
⇒ |3*1+4*2-k|/√ (3²+4²) < 1
⇒ |11-k|<5
⇒ 6⇒ k∈ {7,8,9, .,15} since k∈I
Number of K is 9.

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10 months ago

Class 11th Physics

The mass density of a spherical galaxy varies as K/r over a large distance 'r' from its center. In that region, a small star is in a circular orbit of radius R. Then the period of revolution, T depends on R as:

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Contributor-Level 10

M = ∫ ρdV
M = ∫? (k/r) 4πr²dr
M = 4πkR?²/2 = 2πkR?²
F? = GMm/R?² = 2ω?²R

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10 months ago

If lim(x→1) (x+x²+x³+.+xⁿ-n)/(x-1) = 820, (n∈N) then the value of n is equal to.

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Contributor-Level 10

lim (x→1) (x+x²+.+x? -n)/ (x-1) = 820
⇒ lim (x→1) (x-1)/ (x-1)+ (x²-1)/ (x-1)+.+ (x? -1)/ (x-1) = 820
⇒ 1+2+.+n = 820
⇒ n (n+1)=2*820
⇒ n (n+1)=40*41
Since n∈N, so n=40

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10 months ago

Class 11th Physics

A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a force F at its centre 'O' perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is:

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Contributor-Level 10

FR > mgcosθR
F > mgcosθ
F > mg √ (R²- (R-a)²)/R ⇒ Mg√ (1- (R-a)²/R²)

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10 months ago

If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is.

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Class 11th Units and Measurement

Contributor-Level 10

MOTHER
1? E
2? H
3? M
4? O
5? R
6? T
So position of word MOTHER in dictionary
2*5!+2*4!+3*3!+2!+1
=240+48+18+2+1
=309

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10 months ago

The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of cylinder the zero of the vernier scale is between 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is (VSD is vernier scale division)

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Contributor-Level 10

Zero error = 0 + 7 * 0.1 = 0.070
Vernier reading = (3.1 + 4 * 0.01) – 0.07 = 3.07

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New answer posted

10 months ago

If |x| < 1, |y| < 1 and x y, then the sum to infinity of the following series (x+y) + (x+xy+y) + (x+xy+xy+y) + . is

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