Class 11th

Since velocity does not change, so acceleration will be zero.

mg = F_B + F_v $\Rightarrow \frac{4\pi }{3}{r}^3\rho g=\frac{4\pi }{3}{r}^3\sigma g+6\pi \eta rv$  

$\Rightarrow v=\frac{2r^2\left(\rho -\sigma \right)g}{9\eta }=\frac{2*0.1*0.1*10^{-6}*\left(10^4-10^3\right)*10}{9*1.0*10^{-5}}$  

$\Rightarrow h=\frac{400}{2g}=20m$  

Least  count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10^-2 cm.

According to question, we can write

$R=\frac{u^{2}sin2\theta }{g}\Rightarrow u^2=g{R}_{max},where\theta =45°$

$\Rightarrow H_{max}=\frac{u^2}{2g}=\frac{g{R}_{max}}{2g}=\frac{R_{max}}{2}=\frac{100}{2}=50m$

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

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