Class 11th

  ${\tau }_A=0$

2mg *30 = Mg *10

$\Rightarrow 2*0.01g*30=Mg*10$          

M = 6 * 10^-2kg

COME

$k\cdot E_i+P\cdot E_i=k\cdot E_f+P\cdot E_f$

$\Rightarrow 0+mg\left(h+\frac{h}{2}\right)=0+\frac{1}{2}k{\left(\frac{h}{2}\right)}^2$ $\Rightarrow \frac{3h}{2}mg=\frac{1}{2}k\frac{h^2}{4}$

$\Rightarrow 3mgh=\frac{k}{4}{h}^2\Rightarrow 12\frac{mg}{h}=k\Rightarrow k=\frac{12*0.1*10}{0.1}$               

k = 120Nm^-1

For first ball

 s = ut +   $\frac{1}{2}a{t}^2$

$\Rightarrow -h=u*6-\frac{1}{2}*10*6*6$

$\Rightarrow -h=6u-180$      

h = 180 - 6u                     -(1)

for second boy

$h=ut\frac{1}{2}a{t}^2$

$h=u*1.5+\frac{1}{2}*10*1.5*1.5$             

h = 1.5u + 11.25               -(2)

from (1) & (2)

180 - 6u = 1.5u + 11.25

7.5u = 180 - 11.25

$u=\frac{168.75}{7.5}=22.5$    

h = 180 - 6u

= 180 - 6 * 22.5

45m

for third ball

$h=ut+\frac{1}{2}a{t}^2$

$h=0*t+\frac{1}{2}*10t^2$     

$\Rightarrow 45=5t^2$          

t² = 9

t = 3 sec

Is minimum at the highest position of the circular path.

 $T^2\alpha R^3$

$\begin{array}{l}\Rightarrow \frac{{T_1}^2}{{T_2}^2}=\frac{{R_1}^3}{{R_2}^3}\\ \Rightarrow \frac{{T_1}^2}{{T_2}^2}=\frac{R^3}{{\left(3R\right)}^3}\\ T_2=3\sqrt[]{3}years\end{array}$

Velocity gradient = $\frac{dv}{dx}=\frac{L{T}^{-1}}{L}=T^{-1}$

Decay constant $\left(\lambda \right)=\frac{0.693}{T_{\frac{1}{2}}}=T^{-1}$

$V_{rms}\alpha \sqrt[]{T}|T_i=127°C=400k.$

$V_{rm{s}^1}=2V_{rms},m=0.056kg=56g$   

T_f = 4T_i = 4 * 400 = 1600 k

$Q=n{c}_v\Delta T=\left(\frac{m}{M}\right)C_v\Delta T$        

=  $\left(\frac{56}{28}\right)\frac{5}{2}R\Delta T$

=  $2*\frac{5}{2}*2*1200$

Q = 12 * 10³ cal

= 12 k cal

At steady state condition -> Heat conducted through slab AB = Heat conducted though slab BC

$\frac{k_{1}A}{AB}\left(100-80\right)=\frac{kA}{B{C}_R}\left(80-0\right)$

-> $\frac{k_1}{16}*20=\frac{k}{8}*80$ $\frac{{}_{}}{}\frac{}{}$

k1 = 8 k

0.25 = 1 - $\frac{T_2}{T_1}$ $\frac{{}_{}}{{}_{}}$

0.25 = 1 - $\frac{\left(27+273\right)}{T_1}\Rightarrow \frac{300}{T_1}=1-0.25$ $\frac{}{{}_{}}\frac{}{{}_{}}$

$\Rightarrow \frac{300}{T_1}=0.75$

$T_1=\frac{300*100}{75}$

T1 = 400 k

Now, efficiency increases by 100%

${\eta }_2=100\mathrm{\%}{\eta }_1+{\eta }_1$

$=2{\eta }_1$

= 0.25 * 2

= 0.50

0.50 = 1 - $\frac{T_2\text{'}}{T_1}$ $\frac{{}_{}}{{}_{}}$

$\Rightarrow \frac{T_2\text{'}}{T_1}=0.50$

$T_1\text{'}=\frac{300}{0.5}$

$T_1\text{'}=600k$

$$ $T_1\text{'}-T_1=$  (600 – 400) = 200 k or 200° C

$\pi =\frac{n}{v}RT$

$\pi =\frac{W}{M*v}RT$

$5.03*10^{-3}=\frac{2.5}{M*0.5}*0.083*300$

$M=\frac{2.5*0.083*300}{5.03*0.5}*10^{3}g/mol$

Molar mass of protein, M = 24751 g/mol

 Molar mass of glycine = 75 g/mol

So; number of glycine units =   $\frac{24751}{75}=330$