Class 11th

A simple pendulum shows simple harmonic motion only for small angular displacements. Otherwise, its motion is oscillatory but not purely SHM.

Tension in the string and gravitational force are common forces. These forces are resolved into tangential and radial components.

$2C\left(s\right)+2H_2\left(g\right)\text{?}\text{?}\to C_2{H}_6\left(g\right)\Delta H=\Delta H_f\left(C_2{H}_6\right)$  

$\Delta H_f\left(C_2{H}_6\right)=\left[2*\left(-394\right)\right]+\left[3*\left(-286\right)\right]-\left[1*\left(-1560\right)\right]KJ/mole$  

= (-788 – 858 + 1560) KJ/mole = -86 kJ/mole

For small displacements,  sinθ≈θ (in radians), the small-angle approximation simplifies pendulum motion equations and makes it to approximate simple harmonic motion.

Bond order of $C_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-4}{2}=3$  

Bond order of  $N_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-6}{2}=2$   

Bond order of  $O_2^{2-}=\frac{N_B-N_A}{2}=\frac{10-8}{2}=1$  

N_B = No. of electrons in bonding molecular orbitals.

N_A = No. of electron is Anti bonding molecular orbitals

Energy of 1 photon = $\frac{hc}{\lambda }$  

Energy of 1 mole of photon =  $N_A*\frac{hc}{\lambda }$  

$=\frac{6.022*10^{23}*6.63*10^{-34}*3*10^8}{300*10^{-9}}$  

= 0.399 * 10⁶ J = 399 KJ

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

Number of moles of H = 2 * no. of moles of H₂O = $\left(\frac{270}{18}*2\right)$ moles

Mass of C =  $\frac{330}{44}*12$ gm = 90 gm

Mass of H =  $\frac{270}{18}*2*1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}*100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}*100\mathrm{\%}=25\mathrm{\%}\end{array}$  

$F=\left(M_1+M_2\right)a$

$a=\frac{10}{2+3}=2 m{s}^{-2}$

$F^{\prime }=M_2\left(2\right)=3*2 N=6 N$

$g^{\prime }=\frac{G{M}^{\prime }}{R^{\text{'}2}}=\frac{GM}{10{\left(\frac{R}{2}\right)}^2}$

$=\frac{4}{10}\frac{GM}{R^2}=0.4*9.8$

$=3.92 m s^{-2}$

$x=5sin\left(\pi t+\frac{\pi }{3}\right)m$

Amplitude  $=5 m$

$=5 m$

$\omega =\pi =\frac{2\pi }{T}$

$T=\frac{2\pi }{\pi }=2 s$