Class 11th

Let ball starts its motion with horizontal velocity v₀, so with the help of conservation of mechanical energy, we can write

$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{x}^2\Rightarrow v_0=x\sqrt[]{\frac{k}{m}}=0.05*\sqrt[]{\frac{100}{0.1}}=0.5*\sqrt[]{10}m/s$              

t = Time required to fall the ball =    $\sqrt[]{\frac{2h}{g}}=2*\sqrt[]{\frac{1}{10}}s$

^{$=d=v_0*t=0.5*\sqrt[]{10}*2*\sqrt[]{\frac{1}{10}}=1m$}              

Heat absorbed in cyclic process = Work done = 100? Joule

The equation of wave at any time t will be $y=\frac{1}{1+{\left(x?vt\right)}^2},$ so v * 1 = 2 -> v = 2m/s

In the frame of vehicle, vehicle is in equilibrium under the influence of pseudo force F_P

  $F_P=\frac{m{v}^2}{R}$            

N = mg cos 30° + F_P sin 30°

$\Rightarrow N=mgcos30°+\frac{m{v}^2}{R}sin30°.\left(i\right)$              

, and

$f_S=\frac{m{v}^2}{R}cos30°-mgsin30°$              

By doing (1) * cos 30° - (2) * sin 30°, we have

$\Rightarrow N=\frac{800*10}{0.87-0.2*0.5}=\frac{800*10}{0.77}=10.2*10^{3}kgm/s^2$

As we know that root mean square speed is given as

$v=\sqrt[]{\frac{3RT}{M}},so$

$v_A>v_B>v_C\Rightarrow \frac{1}{v_A}<\frac{1}{v_B}<\frac{1}{v_C}as{m}_A<m_B<m_C$

$\overrightarrow{A}.\overrightarrow{B}=\left|\overrightarrow{A}*\overrightarrow{B}\right|\Rightarrow ABcos\theta =ABsin\theta \Rightarrow \theta =45°$             

$\left|\overrightarrow{A}-\overrightarrow{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }=\sqrt[]{A^2+B^2-2AB*\frac{1}{\sqrt[]{2}}}=\sqrt[]{A^2+B^2-\sqrt[]{2}AB}$                

Since process is isochoric, so

$Q=n{C}_v\Delta T=n\left(\frac{f}{2}R\right)\Delta T=4\left(\frac{5}{2}*R\right)\left(50\right)=500R$

Let the true dip is $\delta$ $$ and apparent dip is $\delta \text{'}$ $$ , so

$tan\delta \text{'}=\frac{B_V}{B_{H}cos\theta }=\frac{tan\delta }{cos\theta }\Rightarrow tan\delta =tan\delta \text{'}cos\theta =tan45°cos30°=1*\frac{\sqrt[]{3}}{2}$

$\Rightarrow \delta =ta{n}^{-1}\left(\frac{\sqrt[]{3}}{2}\right)$

2ⁿ = 4096 -> n = 12

¹²C₆ = 924 (the greatest coefficient)

Since binary mass system performs circular motion about is common centre of mass, so

$m_A{\omega }_A^2{r}_A=\frac{G{m}_B{m}_A}{{\left(r_A+r_B\right)}^2}=\frac{G{m}_B{m}_A}{r^2}$

$m_A{\omega }_A^2*\frac{m_B}{\left(m_A+m_B\right)}r=\frac{G{m}_B{m}_A}{r^2}$

${\omega }_A=\sqrt[]{\frac{G\left(m_A+m_B\right)}{r^3}}$

Similarly we can show that

${\omega }_B=\sqrt[]{\frac{G\left(m_A+m_B\right)}{r^3}}$

Hence their angular velocity will be same, time period will be same, i.e. TA = TB