A) 2.25 g
B) 0.6g
C) 0.9g
D) 0.12g
Correct Answer: A
Solution :
\[T\propto \frac{1}{\sqrt{H}}\] \[\begin{align} & \frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{H}_{2}}}{{{H}_{1}}}} \\ & \frac{5}{10}=\sqrt{\frac{{{H}_{2}}}{3}} \\ & {{H}_{2}}=0.75 \\ \end{align}\] Correction in field = 3 - 0.75 = 2.25
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