Physics

According to de-Broglie's hypothesis

$\lambda =\frac{h}{mv}\Rightarrow p=mv=\frac{h}{\lambda }\Rightarrow K=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2}$

Cut-off wavelength of emitted X-ray = $\frac{hc}{K}=hc*\frac{2m{\lambda }^2}{h^2}=\frac{2mc{\lambda }^2}{h}$

Let Mass α  $\left[t^a{\upsilon }^b{\mathcal{l}}^c\right]\Rightarrow \left[M^1{L}^0{T}^0\right]\equiv T^a*{\left[L{T}^{-1}\right]}^b*{\left[M{L}^2{T}^{-1}\right]}^c$

=> a – b – c = 0    c = 1, and b + 2c = 0 Þ b = -2c = -2

=> a = b + c = 1 – 2 = -1

$Y=\frac{Mg{L}^3}{4b{d}^3\delta }$

For significant error in Y

$=\frac{\Delta M}{M}+\frac{3\Delta L}{L}+\frac{\Delta b}{b}+3\frac{\Delta d}{d}+\frac{\Delta \delta }{\delta }$              

$=\frac{1*10^{-3}}{2}+\frac{3*10^{-3}}{1}+\frac{10^{-2}}{4}+3*\frac{0.01*10^{-1}}{0.4}+\frac{10^{-2}}{5}$             

= 0.0155

$p=\sqrt[]{2mK}\Rightarrow \frac{p_2}{p_1}=\sqrt[]{\frac{K_2}{K_1}}\Rightarrow \frac{p_2}{p}=\sqrt[]{\frac{4K_1}{K_1}}=2\Rightarrow p_2=2p$

The percentage change in its momentum = $\frac{\Delta p}{p}*100=\frac{p}{p}*100=100\mathrm{\%}$

Direction of E in the direction of y axes so flux is only due to top and bottom surface for bottom surface y = 0 E = 0

and for top surface y = 0.5 m      So

$E=150*{\left(0.5\right)}^2=\frac{150}{4}$

$fluxflowing?=EA=\frac{150}{4}*{\left(0.5\right)}^2$              

$=\frac{150}{16}$             

Gausses law $?=\frac{q}{{\epsilon }_0}$  

  $\frac{150}{16}=\frac{q}{{\epsilon }_0}$            

$=8.3*10^{-11}C$              

  $\omega =100\pi ,R=100\Omega ,X_L=L\omega =0.5*10^{-3}*100\pi =5\pi *10^{-2}\Omega ,$ and

  $X_C=\frac{1}{C\omega }=\frac{1}{0.1*10^{-12}*100\pi }=\frac{10^{11}}{\pi }\Omega$            

Since X_C is approximately infinite, so the phase angle between current and supplied voltage and the nature of the circuit is $\approx 90°,$ predominantly capacitive circuit.

In given circuit inductor behave as a simple wire so resultant circuit will be

R_ef = 2 + 1 = 3 $\Omega$  

V = IR

$l=\frac{30}{3}=10A$  

$l=\frac{V_{0}B\mathcal{l}}{R}=\frac{V_0*5*20*10^{-2}}{4+1}$         

  $2*10^{-3}=V_0*20*10^{-2}$

$V_0=\frac{2*10^{-3}}{20*10^{-2}}=\frac{2}{2}*10^{-2}=1*10^{-2}m/s$

V₀ = 1 cm/s

Object is moving in upward direction with constant velocity so in upward motion (+2N) and for downward motion (-2N) So option (1) is correct representation.

Maximum energy = 10 J

$\frac{1}{2}K{x}^2=10$              

K = 5

Given T_pendulum = T_spring

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

$\sqrt[]{\frac{4}{g}}=\sqrt[]{\frac{5}{5}}$           

g = 4m/s²