Class 11th

Equation from the given graph:

$\begin{array}{l}{v}^2=2x+20\\ v^2=2x+20\\ \Rightarrow 2V\frac{dV}{dt}=2V\end{array}$

=>a = 1m/s²

$?\frac{3}{1^2*2^2}+\frac{5}{2^2*3^2}+\frac{7}{3^2*4^2}+........$

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

$S_{10}={\displaystyle \sum _{r=1}^{10}{t}_r=}{\displaystyle \sum _{r=1}^{10}\left(\frac{1}{r^2}-\frac{1}{{\left(r+1\right)}^2}\right)=1-\frac{1}{11^2}=\frac{120}{121}}$

Range R = $\frac{u^{2}sin2\theta }{g}$

For 42° and 48° Range will be same

$H_{max}\alpha$  maximum q

So maximum height will be for 48°

Work done is equal to change in K.E.

$So{W}_1+W_2=\frac{1}{2}M{\left(0.8\sqrt[]{gh}\right)}^2-0$ W₁ ^-> work done by mg

$mgh+W_2=\frac{1}{2}m*0.64gh$  W₂ -> work done by air friction

$W_2=0.32mgh-mgh=-0.68mgh$              

W₂ = -0.68 mgh

Time period T = 2π $\sqrt[]{\frac{l}{g_{eff}}}$

$\Rightarrow T_i=T=2\pi \sqrt[]{\frac{l}{g}}$    

$\Rightarrow T_f=T\text{'}=2\pi \sqrt[]{\frac{4l/3}{g\left(1-\frac{\rho }{\sigma }\right)}}=2\pi \sqrt[]{\frac{16l}{9g}}$

$\Rightarrow T\text{'}=\frac{4}{3}T$

$\overrightarrow{x}=\left|\overrightarrow{P}+\overrightarrow{Q}\right|=\sqrt[]{P^2+Q^2}$          

=> $\overrightarrow{y}=\left|\overrightarrow{P}-\overrightarrow{Q}\right|=\sqrt[]{P^2+Q^2}$

$\left|\overrightarrow{x}+\overrightarrow{y}\right|=\sqrt[]{x^2+y^2+2xy}$

$\Rightarrow \sqrt[]{3\left(P^2+Q^2\right)}=\sqrt[]{{\left(\sqrt[]{P^2+Q^2}\right)}^2+{\left(\sqrt[]{P^2+Q^2}\right)}^2+2\left(P^2+Q^2\right)cos{\theta }_1}$

$\Rightarrow {\theta }_1=60°$

=>Using same formula : θ₂ = 90°

Since the line $\frac{x}{5}cos\theta +\frac{y}{12}sin\theta =1$ is the equation of tangent to the ellipse $\frac{x^2}{5^2}+\frac{y^2}{12^2}=1$ at (5 cos q, 12 sin q). Hence option (A) is correct option.

Let $?\frac{a}{r}$ , a, ar are in G.P. $\therefore \frac{a}{r},$ 2a, ar are in A.P.

$2a-\frac{a}{r}=ar-2a\Rightarrow 2-\frac{1}{r}=r-2\Rightarrow r^2-4r+1=0\Rightarrow r=2+\sqrt[]{3}$

$A/q,a{r}^2=3r^2\Rightarrow a=3$

$\begin{array}{l}\therefore d=ar-2a=3\sqrt[]{3}\\ \therefore r^2-d=7+\sqrt[]{3}\end{array}$

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a₁ then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = $\frac{\sqrt[]{3}}{9}g$  

Now for 8 kg,

$8gsin30°+8acos30°=m{a}_1$  

$a_1=g*\frac{1}{2}+\frac{\sqrt[]{3}}{9}g*\frac{\sqrt[]{3}}{2}$  

$=\frac{2}{3}g$