Physics

$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{t{|}_{t_1|2}}$

$\Rightarrow N=10^{10}*{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{10^{10}}{\sqrt[]{2}}?7*10^9$

Given

6a² = 24

a2 = 4

a = 2m

$\Delta v=v\left(\gamma \right)\Delta T$

$=3\alpha v\Delta T$

$=3*5*10^{-4}*\left(a^3\right)*10$

$=3*5*10^{-4}*8*10$

$\begin{array}{l}=120*10^3\\ =1.2*10^{5}c{m}^3\end{array}$

$V_{AB}=\frac{5}{5}*4=4V$

$\Rightarrow Q=CeqV=2x=8\mu c$

Emf = $\frac{-d?}{dt}$ $\frac{}{}$

When magnet passes through the centre rogion of loop, no emf is induced and the polarity of emf will be negative while entering the loop.

F. B. D of Beaker

By NLM2    

$\begin{array}{l}f=ma=m{\omega }^{2}R\\ \Rightarrow m{\omega }^{2}R\le \mu N\\ R\le \mu g/{\omega }^2\end{array}$

PT³ = k (constant)

$\Rightarrow \left(\frac{nRT}{V}\right)T^3=k$

$\Rightarrow T^4{V}^{-1}=k$

$\Rightarrow T^4=kV$

$\Rightarrow 4*\frac{\Delta T}{T}=\frac{\Delta V}{V}$ - (i) >

$\Rightarrow \Delta V=VY\Delta T$ - (ii) >

Comparing I > & ii >

$\Rightarrow Y=\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$T$}\right.$

dA = 2r dr

$dq=\sigma *2\pi rdr$

$\Rightarrow E={\displaystyle \underset{0}{\overset{R}{\int }}\frac{kdqz}{{\left(z^2+r^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

$\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{z}{\sqrt[]{R^2+z^2}}\right]$

$P=\rho RT$

$\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{P_1}{P_2}*\frac{T_2}{T_1}=\frac{76}{45}*\frac{260}{300}$

$\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{M_1}{M_2}$

$\Rightarrow M_2=\frac{45*300*185}{76*266}=123.54kg$

$X_P\left(t\right)=\alpha t+\beta t^2$

$V_P\left(t\right)=\alpha +2\beta t$  - (i)

$X_Q\left(t\right)=ft-t^2$

$V_Q\left(t\right)=f-2t$ - (ii)
(i) = (ii)

$\alpha +2\beta t=f-2t\Rightarrow t\left(2\beta +2\right)=f-\alpha$

$t=\frac{f-\alpha }{2\left(\beta +1\right)}$

First angle, θ₁= θ, $R=\frac{u^{2}sin2\theta }{g}$      

Another angle (θ₂ = 90 - θ) for which range will be Same as that of θ₁ =θ

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

at ${\theta }_1=\theta ,h_1=\frac{u^{2}si{n}^2\theta }{2g}$

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

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

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

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

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