Class 11th

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New answer posted

7 months ago

0 Follower 14 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

7 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

| Class | No. of students | Number of possible cases |
| :-: | :-: | :- |
| 10 | 5 | (I) 2 | (II) 3 | (III) 2 |
| 11 | 6 | (I) 2 | (II) 2 | (III) 3 |
| 12 | 8 | (I) 6 | (II) 5 | (III) 5 |
Total cases =? C? *? C? *? C? +? C? *? C? *? C? +? C? *? C? *? C?
= 23,800 = 100K
∴ K = 238

New answer posted

7 months ago

0 Follower 1 View

A
Aadit Singh Uppal

Contributor-Level 10

This happens due to the fact that they possess strong C-C and H-H bonds which are non-polar sigma bonds and cannot be broken easily. Also, they lack polar functional groups and are saturated hydrocarbons. These properties combined make them less reactive overall.

New answer posted

7 months ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

F = -dU/dr = - [-12A/r¹³ + 6B/r? ]
F=0 ⇒ r= (2A/B)¹/?
U (at r= (2B/A)¹/? ) = -A²/4B

New answer posted

7 months ago

0 Follower 16 Views

A
alok kumar singh

Contributor-Level 10

Equation of tangent to y²=4 (x+1) is y=m (x+1)+1/m.
Equation of tangent to y²=8 (x+2) is y=m' (x+2)+2/m'.
m'=-1/m.
Solving for intersection point, x+3=0.

New answer posted

7 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

L.C. = 0.5 mm / 50 = 10? ² mm = 10? m = 10µm

New answer posted

7 months ago

0 Follower 7 Views

R
Raj Pandey

Contributor-Level 9

Σ (1/a) (1-rb/a)? ¹ = (1/a)Σ (1+rb/a+r²b²/a²+.)
≈ (1/a)Σ (1+rb/a) = n/a + (b/a²)n (n+1)/2
Compare coeffs: α=1/a, β=b/2a². γ=b²/3a³. This differs from solution.

New answer posted

7 months ago

0 Follower 14 Views

R
Raj Pandey

Contributor-Level 9

sinx+sin4x + sin2x+sin3x = 0
2sin (5x/2)cos (3x/2) + 2sin (5x/2)cos (x/2) = 0
2sin (5x/2) [cos (3x/2)+cos (x/2)] = 0
4sin (5x/2)cosxcos (x/2)=0.
sin (5x/2)=0 ⇒ 5x/2=kπ ⇒ x=2kπ/5. x=0, 2π/5, 4π/5, 6π/5, 8π/5, 2π.
cosx=0 ⇒ x=π/2, 3π/2.
cos (x/2)=0 ⇒ x=π.
Sum = 9π.

New answer posted

7 months ago

0 Follower 17 Views

R
Raj Pandey

Contributor-Level 9

gogog (3n+1)=gog (3n+2)=g (3n+3)=3n+1. So gogog=I.
If fog=f, then f must map range of g to values consistent with f.
There exists a one-one function f: N→N such that fog=f. e.g. f (x)=x.

New answer posted

7 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

S? = 3n/2 [2a+ (3n-1)d]. S? = 2n/2 [2a+ (2n-1)d].
S? =3S? ⇒ 3n/2 [2a+ (3n-1)d] = 3 (2n/2) [2a+ (2n-1)d].
2a+ (3n-1)d = 2 [2a+ (2n-1)d] ⇒ 2a+ (n-1)d=0.
S? /S? = (4n/2 [2a+ (4n-1)d]) / (2n/2 [2a+ (2n-1)d]) = 2 [- (n-1)d+ (4n-1)d]/ [- (n-1)d+ (2n-1)d] = 2 (3n)/ (n)=6.

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