A) \[6.47{}^\circ \]
B) \[4.67{}^\circ \]
C) \[9{}^\circ \]
D) \[5{}^\circ \]
Correct Answer: B
Solution :
\[\therefore \]\[{{\delta }_{1}}-{{\delta }_{2}}=0\] or, \[({{\mu }_{1}}-1){{A}_{1}}-({{\mu }_{2}}-1){{A}_{2}}=0\] or, \[(1.5-1){{7}^{0}}=(1.75-1){{A}_{2}}\] or, \[\frac{7}{2}=\frac{3}{4}{{A}_{2}}\]or \[{{A}_{2}}=\frac{7}{2}\times \frac{4}{3}=\frac{14}{3}={{4.67}^{0}}\] Hence, the correction option is [b].
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