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$l_1=\frac{50}{2000}=25mA,andl=\frac{V_i-V_Z}{1000}=\frac{100-50}{1000}=50mA$

$\Rightarrow l_z=l-l_1=25mA$

The sentence before 3^rd states a problem related to ChatGPT and the given statement provides a solution.

$h\nu =3.4-1.51=1.89eV$

As we know that radius of circular path in magnetic field is given as

$r=\frac{\sqrt[]{2mK}}{qB}\Rightarrow K=\frac{r^2{q}^2{B}^2}{2m}$              

$\Rightarrow K=\frac{{\left(7\times 10^{-3}\right)}^2\times \left(1.6\times 10^{-19}\right)\times {\left(5\times 10^{-4}\right)}^2}{2\times 9.1\times 10^{-31}}eV=107.7\times 10^{-2}eV$  = 1.08 eV

$\Rightarrow ?=h\nu -K=1.89-1.08=0.81eV$              

The given sentence concludes the statement given before the 4^th blank. Hence, it meaningfully fits in the 4^th blank only.

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Here, 1, 3 and 4 seem to be talking about diary writing and methods that can be adopted to do the same. 2 seems odd here.

Method-I : Using Kirchhoff’s Law for loop $\overrightarrow{BDABB},$  we can write

$-140+20l+6l_1=0\Rightarrow 10l+3l_1=70.......\left(i\right)$              

Using Kirchhoff’s Law for loop $\overrightarrow{ACBBA},$  

we can write

$5\left(l-l_1\right)+90-6l_1=0$              

->-5l + 11l₁ = 90 .(ii)

Adding equation (1) with twice the equation (2), we have

$25l_1=250\Rightarrow l_1=\frac{250}{25}=10A$              

Method-II : Using concept of equivalent cell, we can write

$V_{AB}=\frac{\frac{140}{20}+\frac{0}{6}+\frac{90}{5}}{\frac{1}{20}+\frac{1}{6}+\frac{1}{5}}=\frac{2100+0+5400}{15+50+60}$

$=\frac{7500}{125}=60Volt$

$\Rightarrow l_1=\frac{60}{6}=10A$              

Other than 4^th, all the sentences talk about problems associated with water usage.

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As we know that equivalent half life of radioactive material decays by simultaneous emissions, is given as

$\frac{1}{T}=\frac{1}{T_1}+\frac{1}{T_2}=\frac{1}{700}+\frac{1}{1400}=\frac{3}{1400}\Rightarrow T=\frac{1400}{3}years$

$\Rightarrow t=\frac{\mathcal{l}n\left(3\right)}{\lambda }=\frac{\mathcal{l}n\left(3\right)}{\mathcal{l}n\left(2\right)}T=\frac{1.1}{0.693}\times \frac{1400}{3}\approx 740.74years$

The second sentence talks about the English language while the rest talk about the power of words.

Using conservation linear momentum, we can write

$Mv=\frac{h}{\lambda }=\frac{h\nu }{c}\Rightarrow v=\frac{h\nu }{Mc}$                                                           

Loss of internal energy = Gain in Kinetic energy + energy of gamma ray

= $\frac{1}{2}M{v}^2+h\nu =\frac{1}{2}M{\left(\frac{h\nu }{Mc}\right)}^2+h\nu =h\nu \left[1+\frac{h\nu }{2M{c}^2}\right]$  

The first option states all the facts correctly and talks about all the relevant points. Other options are irrelevant. The paragraph given does not explicitly say so.

As we know that root mean square speed is given as

$v=\sqrt[]{\frac{3RT}{M}},so$

$v_A>v_B>v_C\Rightarrow \frac{1}{v_A}<\frac{1}{v_B}<\frac{1}{v_C}as{m}_A<m_B<m_C$

Option A brings all the points together and mainly focuses on the central idea of the passage without citing any examples.

$\overrightarrow{A}.\overrightarrow{B}=\left|\overrightarrow{A}\times \overrightarrow{B}\right|\Rightarrow ABcos\theta =ABsin\theta \Rightarrow \theta =45°$             

$\left|\overrightarrow{A}-\overrightarrow{B}\right|=\sqrt[]{A^2+B^2-2ABcos\theta }=\sqrt[]{A^2+B^2-2AB\times \frac{1}{\sqrt[]{2}}}=\sqrt[]{A^2+B^2-\sqrt[]{2}AB}$                

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Following the flow of information, it can be easily stated that 4 can never start the paragraph. 2341 sequence is more logical here.

As we know that radius of circular path in magnetic field is given as $r=\frac{mv}{qB}=\frac{\sqrt[]{2mK}}{qB},so$  

  $r=\frac{mv}{qB}=\frac{\sqrt[]{2mK}}{qB},so$

$\frac{r_d}{r_{\alpha }}=\sqrt[]{\frac{m_d}{m_{\alpha }}}\times \frac{q_{\alpha }}{q_d}=\frac{1}{\sqrt[]{2}}\times 2=\sqrt[]{2}$               

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