Class 11th

$V_{rms}=\sqrt[]{\frac{3RT}{M}}$

& $v_P=\sqrt[]{\frac{2RT}{M}}$

$v_{rms}=\sqrt[]{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{v}_P$

Let, Rain drop moving with terminal velocity v_t in the air F_b (Bnoyancy force) = $\frac{4}{3}\pi r^3{\rho }_{air}g$  

$mg=\frac{4}{3}\pi r^3{\rho }_{air}g$

$F_v=\left(visconsforce\right)=6\pi \eta r{v}_T$

$F_b+F_v=mg$

$\Rightarrow \frac{4}{3}\pi r^3{\rho }_{air}g+6\pi \eta r{v}_T=\frac{4}{3}\pi r^3\rho g$

$v_T=\frac{2}{9}\frac{r^2}{\eta }\left(\rho -{\rho }_{air}\right)$

$v_T\propto r^2$

$g^1=\frac{GM}{{\left(R+h\right)}^2}$

$g^1=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

Given h = D = 2R

$g^1=\frac{g}{{\left(1+\frac{2R}{R}\right)}^2}$

$g^1=\frac{g}{9}$

$\overrightarrow{r}=2\widehat{i}+\widehat{j}+2\widehat{k}$

$\overrightarrow{\tau }=\overrightarrow{r}*\overrightarrow{F}$

$=\left(2\widehat{i}+\widehat{j}+2\widehat{k}\right)*\left(3\widehat{i}+4\widehat{j}-2\widehat{k}\right)=\left|\begin{array}{ccc}\hfill \widehat{i}\hfill & \hfill -\widehat{j}\hfill & \hfill \widehat{k}\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 3\hfill & \hfill 4\hfill & \hfill -2\hfill \end{array}\right|$

$\overrightarrow{\tau }=-10\widehat{i}+10\widehat{j}+5\widehat{k}$

$\left|\widehat{A}+\widehat{B}\right|=\sqrt[]{A^2+B^2+2ABcos\theta }$

$=\sqrt[]{1+1+2cos\theta }$

$=\sqrt[]{2\left(1+cos\theta \right)}$

= $\sqrt[]{2*2co{s}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\left|\widehat{A}+\widehat{B}\right|=2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(1)

$\left|\widehat{A}-\widehat{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }$

$=\sqrt[]{1+1-2cos\theta }$

$=2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$  -(2)

(2) $÷$ $$ (1)

$\frac{\left|\widehat{A}-\widehat{B}\right|}{\left|\widehat{A}+\widehat{B}\right|}=\frac{2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow \left|\widehat{A}-\widehat{B}\right|=\left|\widehat{A}+\widehat{B}\right|tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

2x=tan (π/9)+tan (7π/18)
=sin (π/9+7π/18) / cos (π/9)cos (7π/18)
=sin (π/2) / cos (π/9)cos (7π/18)
=1 / cos (π/9)cos (7π/18)
=1 / cos (π/9)sin (π/2−7π/18)
=1 / cos (π/9)sin (π/9)
⇒x=1 / 2cos (π/9)sin (π/9)
=1 / sin (2π/9)=cosec (2π/9)

Again 2y=tan (π/9)+tan (5π/18)
⇒2y=sin (π/9+5π/18) / cos (π/9)cos (5π/18)
=sin (7π/18) / sin (π/2−π/9)sin (π/2−5π/18)
=sin (7π/18) / sin (7π/18)sin (4π/18) = cosec (2π/9)
⇒|x−2y|=0

for reflexive (x, x)
x³ - 3x²x + 3x³ = 0
. reflexive
For symmetric
(x, y)? R
x³ - 3x²y - xy² + 3y³ = 0
? (x - 3y) (x² - y²) = 0
For (y, x)
(y - 3x) (y² - x²) = 0
? (3x - y) (x² - y²) = 0
Not symmetric

Equation of the ellipse
(x−3)²/a² + (y+4)²/b² = 1
a=2
ae=1⇒e=1/2
⇒b²=3
Equation of tangent
y+4=m (x−3)±√4m²+3
⇒mx−y=4+3m±√4m²+3
⇒3m±√4m²+3=0
⇒9m²=4m²+3
⇒5m²=3

A= {1,2,3,4,5.100}
B= {4,7,10,13,16,19.}
C= {2,4,6,8,10.}
B−C= {7,13,19, .97}
A∩ (B−C)= {7,13,19, .97}
sum of elements=832

Direction ratio of line (1, -1, -6)
Equation of line (x−3)/1 = (y+4)/-1 = (z+5)/-6 =k
x=k+3, y=−k−4, z=−6k−5
Solving with plane k=−2
⇒x=1, y=−2, z=7
⇒Distance=√ (3−1)²+6²+3²=√49=7