A) \[\text{3}.\text{97}\times \text{1}{{0}^{\text{3}}}\text{min}\]
B) 0
C) \[9.13\times {{10}^{3}}N/{{m}^{2}}\]
D) \[\text{5}.\text{25}\times \text{1}{{0}^{\text{3}}}\text{min}\]
Correct Answer: A
Solution :
The equation of any chord which is normal, is \[\frac{1}{3}=1-\frac{{{T}_{L}}-50}{{{T}_{H}}}\] \[\frac{{{T}_{L}}}{{{T}_{H}}}=\frac{4}{5}\] \[{{T}_{H}}=\frac{5}{4}{{T}_{L}}\] Let the equation of parabola is \[{{T}_{H}}\]. The straight lines joining the origin to the intersection of these two equations is given by \[\frac{1}{3}=1-\frac{{{T}_{L}}-50}{\frac{5}{4}{{T}_{L}}}\] Since, it is at right angle. \[\frac{4({{T}_{L}}-50)}{5{{T}_{L}}}=\frac{2}{3}\]Coefficient of \[{{T}_{L}}-50=\frac{2}{3}\times \frac{5}{4}{{T}_{L}}\]Coefficient of \[{{T}_{L}}-\frac{5}{6}{{T}_{L}}=50\] \[{{T}_{L}}=50\times 6=300K\] \[X=\overline{A.B}\] \[\overline{A.B}=\overline{A}+\overline{B}\] \[X=\overline{A}+\overline{B}\] \[Y=\overline{X}=\overline{A}+\overline{B}\] \[\overline{\overline{A}+\overline{B}}=\overline{\overline{A}}.\overline{\overline{B}}\] \[=A.B\] \[[\because \overline{\overline{A}}=A]\] \[Y=A.B\] \[3.08\times {{10}^{16}}m\]
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