The temperature of the cold junction of the thermocouple is 0 °C and the temperature of the hot junction is T °C. The emf is E = 16T- 0.04T² µV. The inversion temperature T_i is:

1.  300⁰ C

2.  200⁰ C

3.  500⁰ C

4.  400⁰ C

A thin bar magnet of length \(2L\) is bent at the mid-point so that the angle between them is \(60^\circ\). The new length of the magnet is:
1. \( \frac{\mathrm{L}}{2 \sqrt{3}} \)
2. \( \frac{\sqrt{3} \mathrm{~L}}{2} \)
3. \(\mathrm{~L} \)
4. \(\frac{2 \mathrm{~L}}{3}\)

1.  0.7 V

2.  1.2 V

3.  0.8 V

4.  0.9 V

Given that the reduced temperature, the reduced pressure, the reduced volume, Thus, it can be said that the reduced equation of state may be given as:

1.   

2.  

3.  

4.  

The suitable reaction steps to carry out the following transformation is:

1.  $\underset{\left(\mathrm{ii}\right) {\mathrm{H}}_2{\mathrm{O}}_2 . \mathrm{NaOH}}{\overset{\left(\mathrm{i}\right) {\mathrm{BH}}_3. \mathrm{THF}}{\to }} \stackrel{ {\mathrm{HlO}}_4 }{\to }$

2.  $\underset{{\left({\mathrm{CH}}_3\right)}_3 \mathrm{C}. \mathrm{COOH}. {\mathrm{OH}}^{-}}{\overset{\mathrm{Os} {\mathrm{O}}_4}{\to }} \underset{{\mathrm{H}}_2\mathrm{O}}{\overset{{\mathrm{K}}_3{\mathrm{Cr}}_2{\mathrm{O}}_7 . {\mathrm{H}}_2{\mathrm{SO}}_4}{\to }}$

3.  $\underset{\left(\mathrm{ii}\right) {\mathrm{H}}_2{\mathrm{O}}_2 . \mathrm{NaOH}}{\overset{\left(\mathrm{i}\right) {\mathrm{BH}}_3. \mathrm{THF}}{\to }} \underset{{\mathrm{CH}}_2{\mathrm{Cl}}_2}{\overset{ \mathrm{PCC} }{\to }}$

4.  $\stackrel{{\mathrm{H}}_2\mathrm{O} {\mathrm{H}}_2{\mathrm{SO}}_4 \left(\mathrm{dil}.\right)}{\to } \underset{{\mathrm{CH}}_2{\mathrm{Cl}}_2}{\overset{\mathrm{PCC}}{\to }}$