Physics

First angle, 1 = $R=\frac{u^{2}sin2?}{g}$ $\frac{{}^{}}{}$

Another angle (2 = 90 - ) for which range will be

Same as that of 1 =

$R^1=\frac{u^{2}sin2\left(90??\right)}{g}=\frac{u^2}{g}sin2?=R$

at ${?}_1=?,\text{?}\text{?}\text{?}\text{?}\text{?}{h}_1=\frac{u^{2}si{n}^2?}{2g}$ ${}_{}{}_{}\frac{{}^{}{}^{}}{}$

$\&\text{?}\text{?}\text{?}{?}_2=90??,h_2=\frac{u^{2}si{n}^2{?}_2}{2g}=\frac{u^{2}co{s}^2?}{2g}$

$?h_1{h}_2=\frac{1}{4^2}{\left[\frac{u^{2}sin2?}{g}\right]}^2$

$h_1{h}_2=\frac{1}{4^2}{R}^2$

$R=4\sqrt[]{h_1{h}_2}$

So, Both statement is true & Reason is correct explanation for statement 1.

In galvanometer

$\Rightarrow \left(l-l_g\right)S=l_{g}G$

$\frac{I_g}{l}=\frac{S}{S+G}$

$\Rightarrow \frac{1}{3}=\frac{S}{S+G}\Rightarrow S+G=3S\Rightarrow G=2S$

${\mu }_B=\frac{C}{v_B}$

$v_B=\frac{3*10^8}{1.47}=20.408*10^7$

$v_A=2.6*10^7+20.408*10^7=23*10^7$

${\mu }_B=\frac{C}{v_B}\&{\mu }_A=\frac{C}{v_A}$

= 1.127

$\approx 1.13$

$\frac{\Delta z}{z}=\frac{2\Delta A}{A}+\frac{3\Delta B}{B}+\frac{4\Delta C}{C}$

Optical fibre frequency range is 1 THz to 1000 THz.

Very small change in minority charge carriers produces high value of reverse bias current.

  $v\propto \frac{z}{n}$ $\frac{}{}$

$\frac{}{}$  $v=k\frac{z}{n}$ [K constant]

for 3rd orbit of He+

${}_{}$ $v_{H{e}^{+}}=\frac{k*2}{3}=\frac{2k}{3}$  - (1)

$$ $v_H=\frac{k*1}{3}=\frac{k}{3}$  - (2)

$\frac{v_{H{e}^{+}}}{v_H}=2:1$

Y ->And gate

Velocity of light in medium 2

$=\frac{1}{\sqrt[]{{\mu }_m{\in }_m}}$

$v_2=\frac{1}{\sqrt[]{{\mu }_0{\mu }_r{\in }_0{\in }_r}}$

$=\frac{1}{\sqrt[]{1*{\mu }_0{\in }_0*4}}$

$v_2=\frac{1}{2\sqrt[]{{\mu }_0{\in }_0}}$

By snell's law for total internal Reflection

${\mu }_{2}sin{\theta }_C>{\mu }_{1}sin90°$

${\theta }_c>30°$