Physics

msx (16 – 10) = msy (20 – 16)

$\Rightarrow \frac{s_x}{s_y}=\frac{2}{3}$

$m{s}_y\left(26-20\right)=m{s}_z\left(30-26\right)$

$\Rightarrow \frac{s_y}{s_z}=\frac{2}{3}$

Tension in wire, $T=\frac{2*3*5*10}{3+5}=\frac{75}{2}N$

$\frac{T}{\pi r_{min}^2}=\frac{24}{\pi }*10^2$

$\Rightarrow r_{min}^2=\frac{75}{2*24}*10^{-2}=\frac{25}{16}*10^{-2}$
$\Rightarrow r_{min}=\frac{5}{4}*10^{-1}m=12.5cm$

Intersity a number of photons kinetic Energy a f

$R=\frac{\sqrt[]{2Em}}{B.q}$

$\Rightarrow R*\frac{\sqrt[]{m}}{q}$

$\Rightarrow \frac{R_1}{R_2}=\frac{\sqrt[]{4}}{2}*\frac{3}{\sqrt[]{16}}=\frac{3}{4}$

$\Rightarrow R_2=\frac{4R_1}{3}$

$sin\theta =\frac{d}{R}\Rightarrow \theta \alpha \frac{1}{2}\Rightarrow {\theta }_2<{\theta }_1$

$\lambda >h\Rightarrow \lambda >400m$

$Q_1=C_{1}V=2V\mu C$

$Q_2=Q_3=\frac{C_2{C}_3}{C_2+C_3}V=\frac{6*12}{6+12}V=4V\mu C$

$Q_1:Q_2:Q_3=1:2:2$

Let us consider an elementary ring of radius r and thickness dr in which current is flowing.

So, No. of turns in this elementary ring

$dN=\left(\frac{N}{b-a}\right)dr$

$\therefore {\left(dB\right)}_{atcentre}=\frac{{\mu }_{0}ldN}{2r}$

$B=\frac{{\mu }_{0}lN}{2\left(b-a\right)}\mathcal{l}n\frac{b}{a}$

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)

Using F = MA = $m\frac{V}{T}$ $\frac{}{}$

m = FTV¹

Load of mass will be equally distributed among the four colours so force on each columns will be 125 * 10³ N.

Cross section area of the column = $\pi \left[{\left(1\right)}^2-{\left(0.5\right)}^2\right]=2.355m^2$

Using young's modulus : $\epsilon =\frac{\sigma }{Y}=\frac{F}{AY}=\frac{125*10^3}{2.355*2*10^{11}}=2.65*10^{-7}$