Class 11th

Kindly consider the following figure

A tangent to y² = 4x is x – ty + t² = 0

$\frac{3+t^2}{\sqrt[]{1+t^2}}=3$

(3 + t²)² = 9 (1 + t²)

$9+t^4+6t^2=9+9t^2$

Point of contact $\left(3,2\sqrt[]{3}\right)=\left(a,b\right)$

$x-\sqrt[]{3}y+3=0$

$\frac{\sqrt[]{3}x+y-3\sqrt[]{3}=0]}{4x-6=0}$

$x=\frac{3}{2},y=\frac{\sqrt[]{3}}{2}+\sqrt[]{3}$

& $\left(\frac{3}{2},\frac{\sqrt[]{3}}{2}+2\right)=\left(c,d\right),2\left(a+c\right)=2\left(3+\frac{3}{2}\right)=9$

 $\underset{x\to 0}{lim}\frac{ax-\left(e^{4x}-1\right)}{ax\left(e^{4x}-1\right)}=b,$ use of L' Hospital rule implies

$\underset{x\to 0}{lim}\frac{a-4e^{4x}}{a\left(e^{4x}-1\right)+ax\left(4e^{4x}\right)}$

$=\frac{a-4}{0}\Rightarrow a=4$

$\Rightarrow \underset{n\text{?}\to 0}{lim}\frac{4\left(-4.e^{4x}\right)}{4.4e^{4x}+16e^{4x}+16x.4e^{4x}}$

$=\frac{-16}{16+16}=-\frac{1}{2}=b$

a – 2b = 4 – (1) = 5

  $\frac{a+2+a}{3}=\frac{10}{3}$

2a + 2 = 0

2a = 8 -> a = 4       .(i)

and            $\frac{c+b+b}{3}=\frac{7}{3}$

2b + c = 7 .(ii)

Since         a, b, c are in A.P.

2b = a + c

From (i)   $\therefore$ 2b = 4 + c .(iii)

Solving (ii) and (iii)

4 + c + c = 7

2c = 3

$c=\frac{3}{2}$    

$\therefore 2b=4+\frac{3}{2}=\frac{11}{2}$     

$\therefore b=\frac{11}{4}$  

As per question

$\alpha +\beta =\frac{-b}{a}and\alpha \beta =\frac{1}{a}$       

$=\frac{121-192}{256}=\frac{-71}{256}$

A -> m_A = 1 kg, P_A = P

B ->  m_B = 2 kg,   P_B = P

$?kE=\frac{P^2}{2m}$

$\therefore \frac{k{E}_A}{k{E}_B}=\frac{\left(\frac{P^2}{2mA}\right)}{\left(\frac{P^2}{2m_B}\right)}=\frac{m_B}{m_A}=\frac{2}{1}$

$7.2*\frac{5}{18}=2m/s$ [V_A = V_B = 2m/s]

$f{\text{'}}_B=f_A\left(\frac{v+v_B}{v-v_A}\right)$

$\downarrow$     

App freq heard by car B

$f{\text{'}}_B=676\left[\frac{340+2}{340-2}\right]=676*\frac{342}{338}=684Hz$  

Similarly, $f{\text{'}}_A=f_B\left(\frac{v+v_B}{v-v_B}\right)=684Hz$

$\therefore$ Beat frequency heard by both = 684 – 676 = 8Hz

As we know that

$Y=\frac{FL}{A?L}$          

$?L=0.04m=\frac{FL}{AY}...............\left(i\right)$           

If length and diameter both are doubled

$?L\text{'}=\frac{F.2L}{4A.Y}=\frac{FL}{2Y}=0.02m=2cm$       

Let m_AB = m = 1 kg

AB = 0.4 m =  $\mathcal{l}$

d = OM = $\sqrt[]{0.16-0.04}=\sqrt[]{0.12}$

$d=2\sqrt[]{3}*10^{-1}$  

$I_{ABaboutO}=\frac{m{\mathcal{l}}^2}{12}+m{d}^2$           

$\therefore l_{hexagon,aboutO}=6\left[\frac{m{\mathcal{l}}^2}{12}+m{d}^2\right]=\frac{m{\mathcal{l}}^2}{2}+6m{d}^2$       

$=\frac{1*{\left(0.4\right)}^2}{2}+6*1*{\left(2\sqrt[]{3}*10^{-1}\right)}^2=\frac{0.16}{2}+6*4*3*10^{-2}$   

= 0.08 + 0.72 = 0.8 kgm² = 8 * 10^-1 kgm²

$T=27°C,P=1atm=10^{5}N/m^2,V_{rms}=200m/s$

As we know

$v_{rms}=\sqrt[]{\frac{3RT}{M}}\Rightarrow \frac{v_{rms}}{v{\text{'}}_{rms}}=\frac{\sqrt[]{\frac{3RT}{M}}}{\sqrt[]{\frac{3RT\text{'}}{M}}}$

$V{\text{'}}_{rms}=200*\frac{2}{\sqrt[]{3}}=\frac{x}{\sqrt[]{3}}$

x = 400

$x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

$=\frac{\sigma \pi a^2*a-\frac{\sigma \pi a^2}{4}*\frac{3a}{2}}{\sigma \pi a^2-\frac{\sigma \pi a^2}{4}}=\frac{\frac{5}{8}\sigma \pi a^3}{\frac{3}{4}\sigma \pi a^2}=\frac{5}{8}\sigma \pi a^2*\frac{4}{3}\sigma \pi a^2=\frac{5}{6}a$