Class 11th

Velocity of block in equilibrium, in first case,

$v=A\omega =A.\sqrt[]{\frac{k}{M}}$            

Velocity of block in equilibrium, is second case,

$v\text{'}=A\text{'}\omega \text{'}=A\text{'}\sqrt[]{\frac{k}{M+m}}$                  

From conservation of momentum,

Mv = (M + m) v'

$\Rightarrow MA\sqrt[]{\frac{k}{M}}=\left(M+m\right)A\text{'}\sqrt[]{\frac{k}{M+m}}\Rightarrow A\text{'}=A\sqrt[]{\frac{M}{M+m}}$          

(1 + x)? = C? + C? x + C? x² + . C? x?
Put x = 1
2? = C? + C? + C? + C? + … C?
Put x = ω
(1 + ω)? = C? + C? ω + C? ω² + . C? (ω)?
Put x = ω²
(1 + ω²)? = C? + C? ω² + C? ω² + … C? (ω²)?
∴ 1 + ω + ω² = 0
Now add all three equation (s)
2? + (1 + ω)? + (1 + ω²)? = 3 [C? + C? + C? + …]
C? + C? + C? + C? + …
= (2? + (1+ω)? + (1+ω²)? ) / 3
= (2? + (1+ω)? + (-ω)? ) / 3 = (2? - (-1)? ) / 3

Since, $\frac{x^2}{\alpha kT}$ should be dimensionless.

So, dimension of  $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$

Dimension of  $\alpha {\beta }^2$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$

Negation of (p → q) ∨ (p ∨ q)
~ [ (p → q) ∨ (p ∨ q)]
≡ ~ (p → q) ∧ ~ (p ∨ q)
[∴ ~ (p → q) ≡ p ∧ ~q]
≡ (p ∧ ~q) ∧ (~p ∧ ~q)
= F (contradiction)

by conservation of mechanical energy

              K.E_i + P.E_i = K.E_f + P.E_f

              $\frac{1}{2}*0.5v^2+0=\frac{1}{2}*0.5{\left(\frac{v}{2}\right)}^2+\frac{1}{2}k{x}^2$  

              $\Rightarrow \frac{1}{2}*0.5\left[v^2-\frac{v^2}{4}\right]=\frac{1}{2}k{x}^2$  

              $\Rightarrow \frac{1.5}{4}12*12=\frac{k*9}{100}$  

              $\frac{1.5*36}{9}*100=k$  

              k = 600 N/m¹

According to question
aα² + bα + c = 0 . (i)
−aβ² + bβ + c = 0 . (ii)
a/2 γ² + bγ + c = 0 . (iii)
bα + c = -aα² . (iv)
bβ + c = aβ² . (v)
bγ + c = -a/2 γ² . (vi)
By observing (v) and (vi), we get β > γ
By observing (iv) and (vi), we get α > γ
So α < <

By using condition of tangency,
c² = a² (m² + 1)
⇒ c² = 5 [ (2)² + 1]
⇒ c² = 25
⇒ c = ±5

F.B.D. of hanging length

F.B.D. of chain lying on the table

f = T = $\mu N$  

$\Rightarrow \lambda xg=\mu \lambda \left(L-x\right)g$  

$x=0.5\left(6-x\right)$  

$\Rightarrow x=3-0.5x$    

  $1.5x=3$  

x = 2m

  $\omega =\frac{2\pi }{T}\Rightarrow \frac{{\omega }_1}{{\omega }_2}=\frac{T_2}{T_1}=8$

There are 7 letters permutation.
The total number of 4 letters
Out of 2A's, 2K's and 3 different letters O, L, T.
= Coefficient of x? in 4! (1 + x + x²/2!)² (1 + x)³
= Coefficient of x? in 4! (1 + x + x²/2)² (1 + x)³
= Coefficient of x? in 4! [ (1 + x)? + (1 + x)? x² + (1 + x)³ + x? /4]
= 4! [? C? +? C? + ³C? /4] = 4! (5 + 6 + 1/4)
= 24 [11 + 1/4]
= 264 + 6 = 270