A) 6h
B) \[\frac{1}{6}h\]
C) h
D) Zero
Correct Answer: A
Solution :
\[h=\frac{2T\cos \theta }{rdg}\]\\[h\propto \frac{1}{g}\] As \[{{g}_{m}}=\frac{{{g}_{e}}}{6}\]\ \[{{h}_{m}}=6{{h}_{e}}\]
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