Class 11th

$?\frac{P_i=P_f}{m*40}=\frac{m}{2}v\text{'}+\frac{m}{2}*60$

v’ = 20 m/s

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$\frac{\Delta KE}{KE}=\frac{\frac{1}{2}m\left(2000\right)-\frac{1}{2}m\left(1600\right)}{\frac{1}{2}m\left(1600\right)}$

$\therefore x=1$

Sound waves and water waves are mechanical waves, which always require a medium to travel. Light, however, does not require a medium to propagate, as it is electromagnetic.  Light can travel through vacuum, while sound and water waves cannot travel through vacuum.   

$x=Asin\omega t$   (Eq. of a particle executing SHM)

When KE = PE

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$\frac{1}{2}m{\omega }^2\left(A^2-x^2\right)=\frac{1}{2}k{x}^2\left[?k=m{\omega }^2\right]$

A2 – x2 = x2

 $So,\frac{A}{\sqrt[]{2}}=Asin\omega t$

$\frac{1}{\sqrt[]{2}}=sin\omega t$

$\therefore t=\frac{T}{8}$

mvr = $\frac{nh}{2\pi }\left(angularmomentum\right)$  

$KE=\frac{n^2{h}^2}{8{\pi }^{2}m{r}^2}$ (Bohr's kinetic energy)

$=\frac{4h^2}{8{\pi }^{2}m{\left(4a_0\right)}^2}\left(n=2\right)$

$K.E.=\frac{1}{32{\pi }^2}\frac{h^2}{m{a}_0^2}$

Comparing with $\frac{h^2}{xm{a}_0^2}$

x = 32π² = 315.50

10x = 3155

(1) Standard enthalpy of formation for alkali metal bromides becomes more negative on descending down the group.

(2) Standard enthalpy of formation for LiF is most negative among alkali metal fluorides.

(3) In case of Csl, lattice energy is less but Cs⁺ having less hydration energy due to which it is less soluble in water.

(4) For alkali metal fluorides, the solubility in water increases from Li to Cs. LiF is least soluble in water.

$P=-B\left(\frac{\Delta v}{v}\right)$

or  $B=\frac{-P}{\left(\frac{\Delta v}{v}\right)}$

$B=\frac{-\rho gh}{\left(\frac{-0.005V_0}{V_0}\right)}\Rightarrow h=\frac{908*10^8*0.005}{10^3*9.8}$

$\Rightarrow h=500m$

$\mu =9*10^{-4}kg/m$

T = 900 N

$f_0=500Hz\to$ resonance frequency

$f_0^{\text{'}}=550Hz\to$ Next higher frequency

$v=\sqrt[]{\frac{T}{\mu }}=\sqrt[]{\frac{900N}{9*10^{-4}}}=1000m/s$

$\therefore 500Hz=\frac{10*1000}{2*\mathcal{l}}$

$\mathcal{l}=10m$

For $W_1,$  

$\frac{\left(m+10+20\right)*10}{8*10^{-7}}\le 1.25*10^9$              

$\Rightarrow m\le 70kg$              

For W₂,

$\frac{\left(m+10\right)*10}{4*10^{-7}}\le 1.25*10^9$              

-> $m\le 40kg$               

Thus, $m\le 40kg$  

The Bond dissociation energy of D₂ is greater than H₂, therefore D₂ reacts slower than H₂.

$Unitof\frac{a{n}^2}{V^2}=atm$

$Unitofa=\frac{atm*unitof{V}^2}{unitof{n}^2}$

$=\frac{atm*{\left(d{m}^3\right)}^2}{mo{l}^2}$

= atm dm⁶ mol^-2