Class 11th

NaOH + Na₂CO₃

(1) When Hph is added

m.e NaOH + $\frac{1}{2}m.eN{a}_{2}C{O}_3=m.e.HCl=17.5*\frac{1}{10}=1.75$  (m.e = milli equivalents)

(2) When MeOH is added after Hph

$\frac{1}{2}meN{a}_{2}C{O}_3=m.eHCl=1.5*\frac{1}{10}=0.15$

$\begin{array}{l}\therefore m.eNaOH=1.75-0.15=1.6\\ m.eN{a}_{2}C{O}_3=0.15*2=0.3\end{array}$  

$\frac{W_{N{a}_{2}C{O}_3}}{E_{N{a}_{2}C{O}_3}}*1000=0.3$   

$Weight\mathrm{\%}ofN{a}_{2}C{O}_3=\left(\frac{\frac{0.3*53}{1000}}{0.4}\right)*100=\frac{0.3*53}{10*0.4}=\frac{15.9}{4}=3.975\mathrm{\%}\approx 4\mathrm{\%}.$               

$\begin{array}{l}\left(+6\right)\left(+2\right)\left(+3\right)\left(+6\right)\\ Cr{O_4}^{-2}+\underset{}{\underset{?}{S_2{O_3}^{-2}\to Cr{{\left(OH\right)}_4}^{-}+S{O_4}^{-2}}}\end{array}$  

O.N.C = 4

n factor of S₂O₃^-- = 8

n factor of CrO₄^-- = 3

$m.eCr{O}_4^{-2}=m.e.S_2{O}_3^{-2}$        

$V*\left(0.154*3\right)=40*\left(0.25*8\right)\therefore v=\frac{80}{0.462}$       

= 173.16 ml $\approx$ 173 ml

Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially.

After that, the slope of v-t curve is constant and positive.

After some time, velocity becomes constant and acceleration is zero.

After that, the slope of v-t curve is constant and negative.

Thermal strain $=$ Longitudinal strain $=\alpha \Delta T$

$\Rightarrow$ Longitudinal strain, $\delta =10^{-5}*10^2=10^{-3}$

$\Rightarrow$ Compressive stress $=\delta *$  Young's Modulus

$=10^{-3}*0.5*10^{11}$

$=0.5*10^8$

$\Rightarrow$ Compressive force $=0.5*10^8*10^{-3}=0.5*10^5$

$=5*10^4*\frac{10}{10}$

$=50*10^3 N$

From principle of homogeneity

$\left[F\right]=\left[\alpha t^2\right]=\left[\beta t\right]$

$\left[\alpha \right]=\frac{\left[F\right]}{\left[t^2\right]} and \left[\beta \right]=\frac{\left[F\right]}{\left[t\right]}$

$\therefore \left[\alpha \right]\left[t\right]=\left[\beta \right]$

$\therefore \frac{\alpha t}{\beta }= dimensionless$

Apply energy conservation,

$U_i+K_i=U_f+K_f$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}m{v}^2$

$\Rightarrow -\frac{GMm}{R}+K_i=-\frac{GMm}{3R}+\frac{1}{2}*m*\frac{GM}{3R}$

$\Rightarrow K_i=-\frac{1}{6}\frac{GMm}{R}+\frac{GMm}{R}$

$K_i=\frac{5}{6}\frac{GMm}{R}$

At same temperature, curve with higher volume corresponds to lower pressure.

$V_3>V_2>V_1$

$\Rightarrow P_1>P_2>P_3$

(We draw a straight line parallel to volume axis to get this)

$T^{\prime }=2\pi \sqrt[]{\frac{{?}^{\prime }}{g}}$ where $t^{\prime }=\frac{\mathcal{l}}{2}$

$T=2\pi \sqrt[]{\frac{t}{g}}$

$T^{\prime }=\frac{x}{2}T$

$2\pi \sqrt[]{\frac{t}{2g}}=\frac{x}{2}2\pi \sqrt[]{\frac{t}{g}}$

$\frac{1}{\sqrt[]{2}}=\frac{x}{2}\Rightarrow x=\sqrt[]{2}$

$T=m\widehat{?}{\omega }^2$

$T^{\prime }=m:{(2\omega )}^2$

$T^{\prime }=4T$

In the case for maximum elongation,

Stress = Elastic limit

${\delta }_{max}=\frac{{\sigma }_{elastic }*L}{ Young\text{'}s modulus }=\frac{8*10^8*1}{2*10^{11}}=4*10^{-3}$

$=4 mm$

i.e. maximum elongation is 4 mm