Class 11th

F. B. D of Beaker

By NLM2    

$\begin{array}{l}f=ma=m{\omega }^{2}R\\ \Rightarrow m{\omega }^{2}R\le \mu N\\ R\le \mu g/{\omega }^2\end{array}$

PT³ = k (constant)

$\Rightarrow \left(\frac{nRT}{V}\right)T^3=k$

$\Rightarrow T^4{V}^{-1}=k$

$\Rightarrow T^4=kV$

$\Rightarrow 4*\frac{\Delta T}{T}=\frac{\Delta V}{V}$ - (i) >

$\Rightarrow \Delta V=VY\Delta T$ - (ii) >

Comparing I > & ii >

$\Rightarrow Y=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$T$}\right.$

$P=\rho RT$

$\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{P_1}{P_2}*\frac{T_2}{T_1}=\frac{76}{45}*\frac{260}{300}$

$\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{M_1}{M_2}$

$\Rightarrow M_2=\frac{45*300*185}{76*266}=123.54kg$

$X_P\left(t\right)=\alpha t+\beta t^2$

$V_P\left(t\right)=\alpha +2\beta t$  - (i)

$X_Q\left(t\right)=ft-t^2$

$V_Q\left(t\right)=f-2t$ - (ii)
(i) = (ii)

$\alpha +2\beta t=f-2t\Rightarrow t\left(2\beta +2\right)=f-\alpha$

$t=\frac{f-\alpha }{2\left(\beta +1\right)}$

First angle, θ₁= θ, $R=\frac{u^{2}sin2\theta }{g}$      

Another angle (θ₂ = 90 - θ) for which range will be Same as that of θ₁ =θ

$R^1=\frac{u^{2}sin2\left(90-\theta \right)}{g}=\frac{u^2}{g}sin2\theta =R$

at ${\theta }_1=\theta ,h_1=\frac{u^{2}si{n}^2\theta }{2g}$

$\&{\theta }_2=90-\theta ,h_2=\frac{u^{2}si{n}^2{\theta }_2}{2g}=\frac{u^{2}co{s}^2\theta }{2g}$

$\Rightarrow h_1{h}_2=\frac{1}{4^2}{\left[\frac{u^{2}sin2\theta }{g}\right]}^2$

$h_1{h}_2=\frac{1}{4^2}{R}^2$

$R=4\sqrt[]{h_1{h}_2}$

So, Both statement is true & Reason is correct explanation for statement 1.

  $ta{n}^{-1}\frac{1}{2r^2}=ta{n}^{-1}\frac{2}{1+\left(4r^2-1\right)}$             

  $=ta{n}^{-1}\frac{\left(2r+1\right)-\left(2r-1\right)}{1+\left(2r+1\right)\left(2r-1\right)}$

${\displaystyle \sum _{r=1}^{50}\left[ta{n}^{-1}\left(2r+1\right)-ta{n}^{-1}\left(2r-1\right)\right]}$

$=ta{n}^{-1}101-ta{n}^{-1}1=ta{n}^{-1}\frac{100}{102}$

$\Rightarrow tanP=\frac{100}{102}=\frac{50}{51}$             

Acid = M (OH)2 Salt + H2O

M.E of Acid = me of Base

$\therefore 10*\left(0.1*nf\right)=\left(0.05*2\right)*30$

$\therefore n_f=3$

Thus basicity of acid = 3

$P\left(-2\sqrt[]{6},\sqrt[]{3}\right)lieson\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$              

$\Rightarrow \frac{24}{a^2}-\frac{3}{b^2}=1.........\left(i\right)$              

$b^2=\frac{a^2}{4}.........\left(ii\right)$

solving (i) & (ii) a² = 12 Þ b² = 3 hyperbola $\frac{x^2}{12}-\frac{y^2}{3}=1$  

Cuts conjugate axis at $R\left(0,R\sqrt[]{3}\right)$     

$\therefore QR=6\sqrt[]{3}$        

Let height of the wall = h than A(0, 0, 0) G (10, 10, h)

$\overrightarrow{AG}=10\widehat{i}+10\widehat{j}+h\widehat{k}$              

B(10, 0, 0) A (0, 10, h)

$\overrightarrow{BH}=-10\widehat{i}+10\widehat{j}+h\widehat{k}$              

  $cos\theta =\frac{1}{5}=\frac{\overrightarrow{BH}.\overrightarrow{AG}}{\left|\overrightarrow{BH}\right|\left|\overrightarrow{AG}\right|}$             

$=\frac{-100+100+h^2}{\left(200+h^2\right)}$              

$5h^2=200+h^2\Rightarrow h^2=50$              

$h=5\sqrt[]{2}$

${\left(\sqrt[]{3}+i\right)}^{100}=2^{100}{\left(\frac{\sqrt[]{3}}{2}+i\frac{1}{2}\right)}^{100}$

$=2^{100}{\left(cos\frac{\pi }{6}+isin\frac{\pi }{6}\right)}^{100}$

$=2^{99}\left(-1+i\sqrt[]{3}\right)=2^{99}\left(p+iq\right)$

$\therefore p=-1\&q=\sqrt[]{3}$              

$\therefore$ Required equation $x^2-\left(\sqrt[]{3}-1\right)x-\sqrt[]{3}=0$