Class 11th

Pressure * time $=\frac{Ft}{A}=\left[ML{T}^{-2}\right]\left[T\right]\left[L^{-2}\right]=\left[M{L}^{-1}{T}^{-1}\right]$  

 Coefficient of viscosity, $\eta =\left[M{L}^{-1}{T}^{-1}\right]$  

  $F_{viscous}=-\eta A\frac{\Delta v}{\Delta y}$  

According to Newton's law of motion, we can write

ma = mg – N = mg -  $\frac{mg}{4}=\frac{3mg}{4}$                              

$\Rightarrow a=\frac{3g}{4}$

According to conservation of energy, we can write

Gain in kinetic energy = Loss in potential energy

$\Rightarrow$  K_f – K_in = U_in - U_f

$\Rightarrow$ K - = mgy – mg (y – y₀) = mgy₀

According to Newton's laws of motion, we can write

4a = 4g – T . (1), and

40a = T - f_k = T $\mu$ - (40g) ……………… (2)

Adding equations (1) and (2), we can write

44a = 40 – 0.02 * (400) = 32

$\Rightarrow a=\frac{32}{44}=\frac{8}{11}m/s^2$

 According to question, we can write

Total moles of gas = n = n_Oxygen + n_Oxygen = $\frac{16}{2}+\frac{128}{32}=12moles$  

Volume of gas = 12 * 22.4 litre = 268.8 litre = 2.688 * 10⁵ cm³ $?27*10^{4}c{m}^3$  

Slope of AH = $\frac{a+2}{1}$ slope of BC = $-\frac{1}{p}\therefore p=a+2$ $\therefore C\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$  

slope of HC =  $\frac{16p-p^2-31}{16p-32}$  

slope of BC * slope of HC = -1 Þ p = 3 or 5

hence p = 3 is only possible value.

According to question, we can write

$\omega =\pi =\sqrt[]{\frac{g}{\mathcal{l}}}\Rightarrow \mathcal{l}=\frac{g}{{\pi }^2}=\frac{9.8}{{\left(3.14\right)}^2}=0.99395=99.4cm$

Coefficient of x in ${\left(1+x\right)}^p{\left(1-x\right)}^q={-}^p{C_0}^q{C}_1{+}^p{C_1}^q{C}_0=-3\Rightarrow p-q=3$  

              Coefficient of x² in ${\left(1+x\right)}^p{\left(1-x\right)}^q{=}^p{C_0}^q{C}_2{-}^p{C_1}^q{C}_1{-}^p{C_2}^q{C}_0=-5$  

$\Rightarrow \frac{q\left(q-1\right)}{2}-pq+\frac{p\left(p-1\right)}{2}=-5\Rightarrow \frac{q\left(q-1\right)}{2}-\left(q-3\right)q+\frac{\left(q-3\right)\left(q-4\right)}{2}=-5\Rightarrow q=11,p=8$  

              Coefficient of x³ in ${\left(1+x\right)}^8{\left(1-x\right)}^{11}={-}^{11}{C}_3{+}^8{C_1}^{11}{C}_2{-}^8{C_2}^{11}{C}_1{+}^8{C}_3=23$  

$x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$  

$\Rightarrow x^7\left(x-1\right)-x^5\left(x-1\right)+3x^3\left(x-1\right)-x\left(x^2-1\right)+2x\left(1-x\right)+\left(x-1\right)=0$  

$\Rightarrow \left(x-1\right)\left(x^2-1\right)\left(x^5+3x-1\right)=0\therefore x=\pm 1$  are roots of above equation and x⁵ + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or -1.

 3 real roots.

Kindly go through the solution