Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

r? = 10αt²î + 5β (t-5)?
v? = dr? /dt = 20αtî + 5β?
As L? = m (r? * v? )
So, at t=0, L=0
given L is same at t=t as at t=0
⇒ r? * v? = 0
⇒ (10αt²î + 5β (t-5)? ) * (20αtî + 5β? ) = 0
⇒ 50αβt² (î*? ) + 100αβt (t-5) (? *î) = 0
⇒ 50αβt² k? - 100αβt (t-5) k? = 0
⇒ 50t² - 100t (t-5) = 0
⇒ 50t² - 100t² + 500t = 0
⇒ -50t² + 500t = 0
⇒ 50t (10 - t) = 0
⇒ t = 10 second

Assume the length of train be I and its acceleration be a.

$v^2=u^2+2al\Rightarrow al=\frac{v^2-u^2}{2}$

Velocity when middle point crosses the post,

$V_m=\sqrt[]{u^2+2a\frac{\mathcal{l}}{2}}.=\sqrt[]{u^2+\frac{v^2-u^2}{2}}=\sqrt[]{\frac{u^2+v^2}{2}}$

$T=2\pi \sqrt[]{\frac{r^3}{G{M}_e}}$

$T_B-T_A=\frac{2\pi }{\sqrt[]{G{M}_e}}\left(r_B^{3/2}-r_A^{3/2}\right)=\frac{2*3.14}{\sqrt[]{6.67*10^{-11}*6*10^{24}}}\left[{\left(8*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(7*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$v=\frac{p}{m}\Rightarrow v_p:v_d:v_{\alpha }=1:\frac{1}{2}:\frac{1}{4}=4:2:1$

$F_{mag}=qvB\Rightarrow F_P:F_d:F_{\alpha }=1*4:1*2:2*1=2:1:1$

[h] = ML²T^-1

[E] = ML²T^-2

[V] = ML²T^-2C^-1

[P] = MLT^-1

For first resonance,

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

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

Magnetic field on the axis of a circular coil at distance x from centre, B = $\frac{{\mu }_{0}Ni{r}^2}{2{\left(r^2+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ $\frac{{}_{}{}^{}}{{{}^{}{}^{}}^{\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.}}$

$\frac{{\left(r^2+{\left(0.2\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left(r^2+{\left(0.05\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=8$ $\frac{r^2+{\left(0.2\right)}^2}{r^2+{\left(0.05\right)}^2}=4\Rightarrow r^2=0.01\Rightarrow r=0.1m.$

Number of half lives of Y = 3

Number of half lives of X = 6 [As half life of X is half of that of Y].

$\frac{N_1}{2^6}=\frac{N_2}{2^3}\Rightarrow \frac{N_1}{N_2}=8$

$T=2\pi \sqrt[]{\frac{\mathcal{l}}{g}}$

$\Rightarrow 2=2\pi \sqrt[]{\frac{2}{g}}$

$\Rightarrow g=2{\pi }^{2}m/s^2$

At constant pressure,

$dU=n{C}_{V}dT=n.\frac{5}{2}RdT$

$dQ=n{C}_{P}dT=n.\frac{7}{2}RdT$

dW = nRdT

dU : dQ : dW = 5 : 7 : 2