Physics

W = 10 kg m/s²

A = 100 cm²

^{$\mathcal{l}=20cm$}

Y = 2 * 10¹¹ N/m²

^{$Y=\frac{F\mathcal{l}}{A\Delta \mathcal{l}}\Rightarrow \Delta \mathcal{l}=\frac{F\mathcal{l}}{AY}$}

For elemental mass of length, dx

Change in its length

$\Delta \mathcal{l}=5*10^{-10}m$

$\to \left[W\right]=\left[\tau \theta \right]=\left[M{L}^2{T}^{-2}\right]\to correct$

$\to \left[h\right]=\left[L\right]=\left[\frac{2scos\theta }{\rho rg}\right]\to correct$

$\left[v\right]=\left[L^3\right]\left[\frac{\pi p{a}^4}{8\eta L}\right]=\left[L^3{T}^{-1}\right]\to$ Incorrect

$\to J=\epsilon \frac{\partial E}{\partial t}\to$ correct

$\left[A\stackrel{\downarrow }{L}{}^{-2}\right]\left[A\stackrel{\downarrow }{L}{}^{-2}\right]$

$Z=\sqrt[]{R^2+{\left(X_C-X_L\right)}^2}$

$Z=\sqrt[]{R^2+{\left(R-R\right)}^2}\left[?X_L=X_C=R\to Given\right]$

Z = R

Transverse waves can travel through solids or any medium that can withstand shearing strain. Fluids, including liquids and gases, cannot sustain shear force, so such types of mechanical waves cannot travel through them. 

Longitudinal waves can travel through all elastic media that can sustain compressive strain. That means, these mechanical waves can travel through solids, liquids, and gases.  

Diode, in forward biased condition only, will allow current to flow through it.

Pot. different across resistor is

$\Delta V=\left(10sin\omega t-3\right)volt$

But in reverse biased condition of diode,

$\Delta V=0$     (across diode)

$dM=\sqrt[]{2R{h}_T}+\sqrt[]{2R{h}_R}$

$?h_T+h_R=160$

[Given]

So, hR = 160 - hT

$d_M=\sqrt[]{2R{h}_T}+\sqrt[]{2R\left(160-h_T\right)}$

$\therefore d_M=\sqrt[]{2*6400000*80}+\sqrt[]{2*6400000*80}$

$=2*4*8*10^3$

dM = 64 km

E = (50 N/s) sin (t – x/c)

Energy 5.5 * 10^-12 J., vol v =?

$u=\frac{1}{2}{\epsilon }_0{E}^2\to$ energy density

$?u=\frac{U}{v}\Rightarrow U=uv$

= $\frac{1}{2}{\epsilon }_0{E}^2.v$

$v=\frac{5.5*10^{-12}*2}{8.55*2500*10^{-12}}*10^{6}c{m}^3$

= 497

$\cong 500c{m}^3$

(a) $Calamine:ZnC{O}_3$  

(b) Malachite : $CuC{O}_3.Cu{\left(OH\right)}_2$  

(c) Siderite : FeCO₃

(d) Sphalerite : ZnS

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

v’ = 20 m/s

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

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

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

$\therefore x=1$