Class 11th

  ${\tau }_A=I\alpha$

$KxR=\frac{3}{2}M{R}^2\alpha$            

$KR\theta R=\frac{3}{2}M{R}^2\alpha$            

$\alpha =\frac{2K}{3M}$            

$\therefore T=2\pi \sqrt[]{\frac{3M}{2K}}$            

ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²

Vertical component of velocity just after collision $=\frac{u}{2\sqrt[]{2}}$

$k_i=\frac{m{u}^2}{2}{k}_f=\frac{1}{2}m\frac{u^2}{2}+\frac{1}{2}m\frac{u^2}{8}=\frac{5m{u}^2}{16}$ $\frac{}{\text{exception 25:}}$

Fraction  $=\frac{k_i-k_f}{k_i}=1-\frac{\frac{5m{u}^2}{16}}{\frac{m{u}^2}{2}}=1-\frac{5}{8}=\frac{3}{8}$

Heat lost $=$ =  Heat gained

$\Rightarrow \mathrm{}m*540+m*s_{\omega }*(100-5)=10*80+(10+74+10)*s_{\omega }*5$

$\Rightarrow \mathrm{}m(540+95)=800+94*5$

$\Rightarrow \mathrm{}m635=1270$

$\therefore \mathrm{}m=2gm$

For gas $\mathrm{A}:P_A($ ${}_{}$ initial $)={\left(\frac{v_f}{v_i}\right)}^{\gamma }P=(2{)}^{3/2}P$ ${\left(\frac{{}_{}}{{}_{}}\right)}^{}{}^{}$

$P_A=2\sqrt[]{2}P$

For gas B: $P_B($ ${}_{}$ initial $)=P$ $$

For gas $\mathrm{C}:P_C($  initial $)=\frac{2V_0}{V_0}P=2P\mathrm{}\therefore$  Ratio $=2\sqrt[]{2}:1:2$

  $\eta =\frac{W.D.bygas}{Heatabsorb}$

$\eta =\frac{Heat\left(Net\right)}{Heat\left(Absorb\right)}$

$\eta =\frac{Areaunderthecurve}{HeatAbsorb}$

$\eta =\frac{\frac{1}{2}*S_0{T}_0}{\frac{1}{2}\left(3T_0\right)S_0}=\frac{1}{3}$

To reach at point A particle must cross the peak point.

Loss in KE = gain in PE

$\frac{1}{2}*\frac{100}{1000}*v^2=\left(5-0\right)$            

v² = 100

v = 10 m/s

  $\frac{1}{2}\mu v_r^2=\frac{1}{2}k{x}^2$  

  $x={V_{rer}}^2\sqrt[]{\frac{\mu }{k}}$          

$=2*\sqrt[]{\frac{8/6}{10}}=\sqrt[]{\frac{8}{15}}$            

From conservation of momentum

2 * 4 + 4 * 2 = 2v₁ + 4v₂

8 = v₁ + 2v₂                           ….(1)

From conservation of energy

  $\frac{1}{2}*2*4^2+\frac{1}{2}*4*2^2=\frac{1}{2}*2*v_1^2+\frac{1}{2}*4*v_2^2$           ….(ii)

On solving   $v_1=\frac{4}{3}m/s,v_2=\frac{10}{3}m/s$