Class 11th

q = 18° Þ 2q + 3q = 90° Þ sin 3q = 1 – sin 2q Þ cos q Þ cosq (4 sin² q + 2sinq - 1) = 0

$?cos\theta \ne 0\therefore 4si{n}^2\theta +2sin\theta -1=0\Rightarrow cose{c}^{2}18°-2cosec18°-4=0$

? x² – 2x – 4 = 0

${\left|2\overrightarrow{a}+3\overrightarrow{b}\right|}^2={\left|3\overrightarrow{a}+\overrightarrow{b}\right|}^2\Rightarrow 4{\left|\overrightarrow{a}\right|}^2+9{\left|\overrightarrow{b}\right|}^2+12\overrightarrow{a}.\overrightarrow{b}=9{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+6\overrightarrow{a}.\overrightarrow{b}$

$\Rightarrow -5{\left|\overrightarrow{a}\right|}^2+8{\left|\overrightarrow{b}\right|}^2+6\overrightarrow{a}.\overrightarrow{b}=0.........\left(i\right)$

$cos60°=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{1}{2}\therefore \overrightarrow{a}.\overrightarrow{b}=4\left|\overrightarrow{b}\right|and\left|\overrightarrow{a}\right|=8$

from (i) we get $\left|\overrightarrow{b}\right|=5$

Object is moving in upward direction with constant velocity so in upward motion (+2N) and for downward motion (-2N) So option (1) is correct representation.

Adiabatic equation : $P_1{V}^{\gamma }=P_2{V}^{\gamma }\Rightarrow \left(200\right){\left(1200\right)}^{\gamma }=P{\left(300\right)}^{\gamma }$

$\Rightarrow P=\left(200\right){\left(4\right)}^{\frac{3}{2}}=1600KPa$

$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{240-480}{1.5-1}=-480J$  

Maximum energy = 10 J

$\frac{1}{2}K{x}^2=10$              

K = 5

Given T_pendulum = T_spring

$2\pi \sqrt[]{\frac{\mathcal{l}}{g}}=2\pi \sqrt[]{\frac{m}{K}}$             

$\sqrt[]{\frac{4}{g}}=\sqrt[]{\frac{5}{5}}$           

g = 4m/s²

From question :

9MSD = 10VSD

LC = 1MSD – 1VSD = 1MSD - $\frac{9}{10}MSD=0.1MSD=0.1*1mm=0.1mm$

Reading = 10mm + 8 * 0.1 mm = 10.8 mm

Final reading = 10.8 mm – 0.04 mm = 10.4 mm = 104 * 10^-2 cm

So radius = 52 * 10^-2 cm

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

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

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

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