Class 11th

From conservation of Angular momentum about hinged point

$mv\mathcal{l}=\left(\frac{m{\mathcal{l}}^2}{3}+2m.{\mathcal{l}}^2\right)\omega$            

$\omega =\left(\frac{3}{7}\frac{v}{\mathcal{l}}\right)$            

Now from conservation of Energy.

$\frac{1}{2}*\frac{7}{3}m{\mathcal{l}}^2*{\omega }^2=mgl+2mg.2\mathcal{l}$            

$v=\sqrt[]{\frac{70}{3}g\mathcal{l}}$            

Let $2x$ $$ length of rod is immersed in water.

$$ ${\tau }_{\text{Hinge}}($  net $)=0$ $$

$\Rightarrow \mathrm{}\mathrm{m}\mathrm{g}?\frac{b}{2}\mathrm{s}\mathrm{i}\mathrm{n}?\theta -F_B(b-x)\mathrm{s}\mathrm{i}\mathrm{n}?\theta =0$

$\Rightarrow \mathrm{}mgb=2(b-x)F_B$

$$ $\therefore \mathrm{}{F}_B=\frac{mgb}{2(b-x)}=\frac{\left(Abd\right)b}{2(b-x)}g$  , where $d=$ $$ density of material of rod

${}_{}$  $F_B=$ Buoyant force $\mathrm{}\frac{}{}\frac{}{}$

Equate $F_B,\frac{A{b}^{2}d}{2(b-x)}=2Ax\rho$ ${}_{}\frac{{}^{}}{}$

$\Rightarrow \mathrm{}{b}^2=4x(b-x)\frac{9}{5}$

$\mathrm{}{}^{}\frac{{}^{}}{}\frac{}{}$  $\Rightarrow \mathrm{}{x}^2-bx+\frac{5b^2}{36}=0\Rightarrow x=\frac{b}{6}\therefore$ immersed $=\frac{b}{3}$

Let the acceleration of string = a

For 2 kg 20 – T₁ = 2a ………… (i)

For monkey T₂ – 80 – T₁ = 8 (2 – a)  …. (ii)

For 10 kg T₂ – 100 = 10a    ……. (iii)

$\therefore a=0.8m/s^2$

Acceleration of monkey w.r. to ground

= 2 – 0.8 = 1.2 m/s²

  $s=ut+\frac{1}{2}a{t}^2$        

$2.4=0+\frac{1}{2}*1.2*t^2$            

t = 2 sec

$\mathrm{}g=\frac{G{M}_{\text{in}}}{r^2}$  , where $M_{\text{in}}={\int }_0^r?\rho 4\pi x^2\cdot dx$

$\begin{array}{rr}& M_{\text{in}}=4\pi {\rho }_0{\int }_0^r??\left(1-\frac{x^3}{R^3}\right)x^{2}dx=4\pi {\rho }_0\left(\frac{r^3}{3}-\frac{r^6}{6R^3}\right)\\ & \therefore \mathrm{}g=\frac{4\pi G{\rho }_0}{6}\left(2r-\frac{r^4}{R^3}\right)\end{array}$ ${}_{}{}_{}^{}{}^{}$

For $\mathrm{g}$ $$ to be maximum or minimum or constant $\frac{dg}{dr}=0\Rightarrow 2-\frac{4r^3}{R^3}=0\Rightarrow r=\frac{R}{2^{1/3}}$ $\frac{}{}\frac{{}^{}}{{}^{}}\frac{}{{}^{}}$

Since plane is smooth hence motion is only translational

$a=gsin\theta =10*\frac{1}{2}=5m/s^2$              

$s=ut+\frac{1}{2}a{t}^2$            

$1.6=0+\frac{1}{2}*5*t^2$            

$t^2=\sqrt[]{\frac{3.2}{5}}=\sqrt[]{0.64}=0.8sec.$            

PQ is focal chord.
Quadrilateral PTQR is square.

Area = (PQ * TR) / 2 = (4 * 4) / 2 = 8

$y\left(x,t_0\right)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{x}{b}\right)$

$y(x,0)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{x+v{t}_0}{b}\right)$

$y(x,t)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left(\frac{x+v{t}_0-vt}{b}\right)$

$\therefore y(x,t)=A\mathrm{s}\mathrm{i}\mathrm{n}?\left. [\frac{x-v\left(t-t_0\right)}{b}\right]$

lim (x→0) (ae²? - bcosx + c) / (x sinx)
= lim (x→0) (ae²? - bcosx + c) / x² * lim (x→0) x/sinx
For limit to exist
a - b + c = 0
2a + b = 0
(4a-b)/2 = 1
c = 2, b = 2, a = -1
a+b+c = 3

Family of circle through (2, 2) and (9, 9), (x-2) (x-9) + (y-2) (y-9) + λ (y-x) = 0
Touches y=0 (x-axis)
(x-2) (x-9) + 18 - λx = 0
x² - 11x - λx + 36 = 0
x² - (11+λ)x + 36 = 0 has repeated roots
D = 0 ⇒ λ = 1, λ = -23
x = 6 or x = -6
Difference = 12

All even + 2 odd 1 even
¹? C? + ¹? C? * ¹? C?
(10*9*8)/6 + (10*9)/2 * 10
120 + 450 = 570