A) 4a
B) 2 a
C) \[\frac{a}{4}\]
D) \[\frac{a}{2}\]
Correct Answer: A
Solution :
\[\frac{m{{v}^{2}}}{R}=ma\] Or \[a=\frac{{{v}^{2}}}{R}\] Or \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{v_{1}^{2}}{v_{2}^{2}}\] Here, \[{{v}_{1}}=v,{{v}_{2}}=2v,{{a}_{1}}=a\] \[\therefore \] \[\frac{a}{{{a}_{2}}}=\frac{{{v}^{2}}}{{{(2v)}^{2}}}=\frac{1}{4}\] or \[{{a}_{2}}=4a\]
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