Class 11th

f? = 1/ (2l) √ (T? /µ)
f? = 1/ (2l) √ (T? /µ)
f? /f? = 450/300 = √ (T? /T? )
⇒ T? /T? = 9/4 = 2.25

On moving Left to Right along a period.
Atomic Radius → decreases.
Electronegativity → Increases.
Electron gain enthalpy → Increases.
Ionisation Enthalpy → Increases

Let three terms of G.P. are a/r, a, ar product=27
⇒ a³=27 ⇒ a=3
S=3/r+3r+3
For r>0
(3/r+3r)/2 ≥ √9 = 3
⇒ 3/r+3r≥6
For r<0, 3/r+3r-6
From (1) and (2)
S∈ (-∞, -3]∪ [9, ∞)

R { (x, y):x, y∈z, x²+3y²≤8}
For domain of R? ¹
Collection of all integral of 's
For x=0, 3y²≤8
⇒ y∈ {-1,0,1}

α and β are roots of 5x²+6x-2=0
⇒ 5α²+6α-2=0
⇒ 5α? ²+6α? ¹-2α? =0
(By multiplying α? )
Similarly 5β? ²+6β? ¹-2β? =0
By adding (1) and (2)
5S? +6S? -2S? =0
For n=4
5S? +6S? =2S?

Let p denotes statement
p: I reach the station in time.
q: I will catch the train.
Contrapositive of p→q is q→p
q→p: I will not catch the train, then I do not reach the station in time.

σ²=variance
µ=mean
σ² = Σ (x? -µ)²/n
µ=17
⇒ Σ (ax+b)/17 = 17
⇒ 9a+b=17
σ²=216
⇒ Σ (ax+b-17)²/17 = 216
⇒ Σa² (x-9)²/17 = 216
⇒ a² (81-18*9+3*35) = 216
⇒ a² (24) = 216 ⇒ a²=9 ⇒ a=3 (a>0)
⇒ From (1), b=-10
So, a+b=-7

|x|/2 + |y|/3 = 1
x²/4 + y²/9 = 1
Area of Ellipse = πab = 6π
Required area = π*2*3 - (Area of quadrilateral)
= 6π - 1/2*6*4
= 6π-12
= 6 (π-2)

For ideal gas
PM = dRT
d = [PM]/R * 1/T
So graph between dV ST is not straight line

In this acid base Titrating there is no use of Bunsen burner and measuring cylinder other laboratory equipments will be required for getting the end point of titration.