physics ncert solutions class 11th

For first resonance,

$\mathcal{l}+0.3d=\frac{v}{4f}$

$\Rightarrow \mathcal{l}+0.3*6=\frac{336*100}{4*504}\Rightarrow \mathcal{l}=14.8cm$

Translational kinetic energy will be equal to rotational kinetic energy corresponds to each degree of freedom.

At same temperature, curve with higher volume corresponds to lower pressure.

$V_3>V_2>V_1$

$\Rightarrow P_1>P_2>P_3$

(We draw a straight line parallel to volume axis to get this)

Energy difference $\Delta E=\frac{hc}{\lambda }$

$\therefore \lambda \propto \frac{1}{\Delta E}$

${(\Delta E)}_{6-2}>{(\Delta E)}_{5-2}>{(\Delta E)}_{4-2}>{(\Delta E)}_{3-2}$

${\lambda }_{6-2}<{\lambda }_{5-2}<{\lambda }_{4-2}<{\lambda }_{3-2}$

$A-III,\; B-IV,\; C-II,\; D-I$

K.E. energy of electron = eV

Translational K.E. of N₂ =   $\frac{3}{2}KT$

So eV = $\frac{3}{2}KT$  

$1.6*10^{-19}*0.1=\frac{3}{2}*1.38*10^{-23}*T$                

T = 773 – 273 = 500°C

Cross product will represent a vector perpendicular to both the vector.

Dir. Of Motion of P : $=\left(\widehat{i}+\widehat{j}\right)*\left(\widehat{j}*\widehat{k}\right)$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

= $\widehat{k}-\widehat{j}+\widehat{i}$ $\stackrel{}{}\stackrel{}{}$

Dir. Of Motion of Q = $\left(\widehat{i}+\widehat{j}\right)*\left(-\widehat{i}+\widehat{j}\right)$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$=2\widehat{k}$

$cos\theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{ab}=\frac{2}{2\sqrt[]{3}}=\frac{1}{\sqrt[]{3}}\Rightarrow \theta =co{s}^{-1}\left(\frac{1}{\sqrt[]{3}}\right)$

Consider translational motion of a body

mg sinq - f_S = ma             .(i)

Consider rotational motion of body about its centre of mass

$f_{S}R=l\alpha \Rightarrow \alpha =\frac{f_{S}R}{l}$ .(ii)

Using the condition of rolling without slipping at contact, we have

$\alpha R=a\Rightarrow \frac{f_S{R}^2}{l}=\frac{mgsin\theta -f_S}{m}\Rightarrow f_S=\frac{lmgsin\theta }{l+m{R}^2}$

Using the condition of static friction, we have

$f_S\le \mu N\Rightarrow \frac{lmgsin\theta }{l+m{R}^2}\le \mu mgcos\theta \Rightarrow \mu \ge \frac{ltan\theta }{l+m{R}^2}...........\left(iii\right)$

$\alpha =\frac{mgRsin\theta }{l+m{R}^2}\Rightarrow a=\alpha R=\frac{mg{R}^{2}sin\theta }{l+m{R}^2}=\frac{g{R}^{2}sin\theta }{k^2+R^2}...........\left(iv\right)$

$t=\sqrt[]{\frac{2\mathcal{l}}{a}}\Rightarrow$ Time required to reach the ground

$\Rightarrow \frac{v_{ring}}{v_{cylinder}}=\sqrt[]{\frac{2g\mathcal{l}{R}^{2}sin\theta }{k_{ring}^2+R^2}}*\sqrt[]{\frac{k_{cylinder}^2}{2g\mathcal{l}{R}^{2}sin\theta }}=\sqrt[]{\frac{\frac{R^2}{2}+R^2}{R^2+R^2}}=\frac{\sqrt[]{3}}{2}$

As we know that root mean square speed is given as

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

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

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

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

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

According to ideal gas law we know that

PV = nRT