Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

$\overrightarrow{F}=10\widehat{i}+5\widehat{j}$

$\overrightarrow{a}=\frac{10}{m}\widehat{i}+\frac{5}{m}\widehat{j}$

$=\frac{10}{0.1kg}\widehat{i}+\frac{5}{0.1}\widehat{j}$

$\overrightarrow{a}=100\widehat{i}+50\widehat{j}$

$a_x=100,a_y=50$

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

$=0*2+\frac{1}{2}*100*2*2$

Sx = 200 m

$\frac{a}{b}=\frac{200}{100}=2$

Given, L = 2H

l = 2 sin (t²)

$u={\displaystyle \underset{0}{\overset{2}{\int }}Lidi=\frac{L}{2}{\left[i^2\right]}_0^2=\frac{L}{2}\left[4-0\right]}$

$u=\frac{2}{2}*4=4J$

$\equiv Req=2.5\Omega$

$l=\frac{v}{R_{eq}}=\frac{5}{2.5}=2A$

$H=I^{2}Rt=2^2*R*15$

$\Rightarrow 300=60R$

$R=5\Omega$

Now for 3A, time = 10 sec

$H\text{'}=l^{2}Rt=3^2*5*10=450J$

ceq = $\frac{24*8}{32}=6\mu F$

$H=\frac{\lambda }{4}$

$\Rightarrow 20=\frac{\lambda }{4}$

$\lambda =80cm$

Now first overtone for open organ pipe

$L_2=4\frac{\lambda }{4}=\lambda =80cm$

$Q_1=\underset{water}{20°C}\to 100°steam$

$Q_1=mc\Delta T+mL$

$=m\left[c\Delta T+L\right]$

$=31000=31*10^{3}cal$

$Q_2=\frac{293}{373}*31*10^3$

 

$\tau =\eta \frac{v}{h}$  

              Given

              $\tau =10^{-3}N/m^2$  

              (shear stress)

              h =?

              v = 36 km/hr = 10 m/sec

              $10^{-3}=\frac{10^{-2}*10}{h}$

h = 100 m

by conservation of mechanical energy

              K.E_i + P.E_i = K.E_f + P.E_f

              $\frac{1}{2}*0.5v^2+0=\frac{1}{2}*0.5{\left(\frac{v}{2}\right)}^2+\frac{1}{2}k{x}^2$  

              $\Rightarrow \frac{1}{2}*0.5\left[v^2-\frac{v^2}{4}\right]=\frac{1}{2}k{x}^2$  

              $\Rightarrow \frac{1.5}{4}12*12=\frac{k*9}{100}$  

              $\frac{1.5*36}{9}*100=k$  

              k = 600 N/m¹

F.B.D. of hanging length

F.B.D. of chain lying on the table

f = T = $\mu N$  

$\Rightarrow \lambda xg=\mu \lambda \left(L-x\right)g$  

$x=0.5\left(6-x\right)$  

$\Rightarrow x=3-0.5x$    

  $1.5x=3$  

x = 2m