A) 4 times
B) 6 times
C) 8 times
D) 2 times
Correct Answer: D
Solution :
\[{{n}_{2}}={{n}_{1}}{{\left[ \frac{{{M}_{2}}}{{{M}_{1}}} \right]}^{1}}{{\left[ \frac{{{L}_{2}}}{{{T}_{1}}} \right]}^{2}}{{\left[ \frac{{{T}_{2}}}{{{T}_{1}}} \right]}^{-2}}\] Given: \[{{M}_{2}}=2{{M}_{1}},{{L}_{2}}=2{{L}_{1}},{{T}_{2}}=2{{T}_{1}}\] \[\therefore \] \[{{n}_{2}}={{n}_{1}}{{[2]}^{1}}{{[2]}^{2}}{{[2]}^{-2}}\] \[=2{{n}_{1}}\]
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