Class 11th

In the case for maximum elongation,

Stress = Elastic limit

${\delta }_{max}=\frac{{\sigma }_{elastic }*L}{ Young\text{'}s modulus }=\frac{8*10^8*1}{2*10^{11}}=4*10^{-3}$

$=4 mm$

i.e. maximum elongation is 4 mm

Excess force  $=T*2\pi R$

$=\frac{7}{100}*2*3.14*\frac{4.5}{100}$

$=197.82*10^{-4}$

$=19.8*10^{-3} N$

$=19.8mN$

(A) Ethane                   one $\left(C-C\right)\sigma$ bond

(B) Ethene                         one $\left(C-C\right)\sigma$ and one $\left(C-C\right)\pi$ bond

(C) $C_2$                                                       two $\left(C-C\right)\pi$ bonds

(D) Ethyne $H-C\equiv C-H$                       two $\left(C-C\right)\pi$ bonds and one $\left(C-C\right)\sigma$ bond

$V.C=MSD-VSD$     … (i)

given :  $\left(N+1\right)VSD=NMSD$

$VSD=\left(\frac{N}{N+1}\right)MSD$     … (ii)

From (1) and (2)

$V.C=\left(MSD\right)-\frac{N}{N+1}\left(MSD\right)$

$=MSD\left(1-\frac{N}{N+1}\right)=\frac{MSD}{N+1}$

$=\frac{0.01}{N+1}=\frac{1}{100\left(N+1\right)}$

$E_n=-R_H\left(\frac{Z^2}{n^2}\right)J$

For ${}^{+}\left(n=1\right)$ ,

$E_n=-x=-R_H\left(\frac{2^2}{1^2}\right)=-4R_H$

$\therefore R_H=\frac{x}{4}$

For ${}^{3+}\left(n=2\right)$ ,

$E_n=-R_H\left(\frac{z^2}{n^2}\right)J$

$=-\frac{x}{4}*\left(\frac{4*4}{2*2}\right)=-xJ$

A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.

(A) Isothermal process $\Rightarrow$ Temperature is constant throughout the process

(B) Isochoric process $\Rightarrow$ Volume is constant throughout the process

(C) Isobaric process $\Rightarrow$ Pressure is constant throughout the process

(D) Adiabatic process $\Rightarrow$ No exchange of heat $\left(q\right)$ between system and surrounding

(1) 4 mol of $=4{}_A$ atoms

(2) 4 u of $=\frac{4u}{4u}=1$ atom

(3) 4 g of Helium $=\frac{4}{4}$ mole $=1$ mole $=N_A$ He atom

(4) $2.2710982$ of He at STP $=\frac{2.271}{22.710982}$ mole

$=0.1$

$=0.1{}_A$

$2?+C,K_C=4*10^{-3}$

At a given time $t,Q_c$ is to be calculated and been compared with $K_c$ .

$Q_C=\frac{\left[B\right]\left[C\right]}{{[A]}^2}=\frac{\left(2*10^{-3}\right)\left(2*10^{-3}\right)}{{\left(2*10^{-3}\right)}^2}$

$Q_C=1$

Capacitive Reactance $X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2*3.14*50*10*10^{-6}}$

$=\frac{1000}{3.14}$

$V_{rms}=210 V$

$i_{rms}=\frac{V_{rms}}{X_C}=\frac{210}{X_C}$

$Peak current =\sqrt[]{2}{i}_{ms}=\sqrt[]{2}*\frac{210}{1000}*3.14=0.932$

$?0.93 A$