physics ncert solutions class 11th

1 litre, T = 300K, P = 2 atm, KE = 2*10? J/molecule, no of molecule =?
No. of molecules = (no of moles) * NA = nNA
Also, n = PV/RT = PV/ (NAkT)
KE = (3/2)kT = 2*10? J [Given]
kT = (4/3)*10?
P = 2 atm = 2 * 1.013 * 10? N/m²
vol = 1 lit = 10? ³ m³
No. of molecules = PV/kT = (2*1.013*10? * 10? ³)/ (4/3)*10? ) ≈ 1.5 * 10¹¹

According to question
|x| = |y| and
|x - y| = n|x + y|
x² + y² - 2x·y = n² (x² + y² + 2x·y)
(1 - n²) (x² + y²) = (1 + n²)2x·y
(1 - n²) (x² + y²) = (1 + n²)2xy cosθ
(1 - n²) (2x²) = (1 + n²) (2x²)cosθ
cos θ = (1 - n²)/ (1 + n²)
θ = cos? ¹ (1 - n²)/ (1 + n²) = cos? ¹ (- (n²-1)/ (n²+1)

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

$I_{AB}=\frac{2}{5}m{a}^2*2+\left(\frac{2}{5}m{a}^2+m{b}^2\right)*2=\frac{8m{a}^2}{5}+2m{b}^2$

$7.2*\frac{5}{18}=2m/s$ [V_A = V_B = 2m/s]

$f{\text{'}}_B=f_A\left(\frac{v+v_B}{v-v_A}\right)$

$\downarrow$     

App freq heard by car B

$f{\text{'}}_B=676\left[\frac{340+2}{340-2}\right]=676*\frac{342}{338}=684Hz$  

Similarly, $f{\text{'}}_A=f_B\left(\frac{v+v_B}{v-v_B}\right)=684Hz$

$\therefore$ Beat frequency heard by both = 684 – 676 = 8Hz

$T=27°C,P=1atm=10^{5}N/m^2,V_{rms}=200m/s$

As we know

$v_{rms}=\sqrt[]{\frac{3RT}{M}}\Rightarrow \frac{v_{rms}}{v{\text{'}}_{rms}}=\frac{\sqrt[]{\frac{3RT}{M}}}{\sqrt[]{\frac{3RT\text{'}}{M}}}$

$V{\text{'}}_{rms}=200*\frac{2}{\sqrt[]{3}}=\frac{x}{\sqrt[]{3}}$

x = 400

Travelling wave is represented by $f\left(t\pm \frac{x}{v}\right)$ $\frac{}{}$

Thus, y = A sin (15x – 2t) represents a travelling wave

According to KTG, the gas exerts pressure because its molecule :

suffer change in momentum when impinge on the walls of container.

Let v is velocity at the highest position.

$T_{max}=5T_{min}\Rightarrow mg+\frac{m\left(v^2+4g\mathcal{l}\right)}{\mathcal{l}}=5\left(\frac{m{v}^2}{\mathcal{l}}-mg\right)\Rightarrow 4.\frac{v^2}{\mathcal{l}}=10g$

$\Rightarrow v=\sqrt[]{\frac{5}{2}g\mathcal{l}}=\sqrt[]{\frac{5}{2}*10*1}=5m/s$

$\Delta U+\Delta KE=0$

$\Rightarrow \frac{f}{2}nR\Delta T=\frac{1}{2}m{v}^2\Rightarrow \Delta T=\frac{m}{n}\frac{v^2}{fR}=4*10^{-3}*\frac{30^2}{3R}=\frac{3.6}{3R}K.$