A) \[\frac{{{v}^{2}}}{r}\]
B) \[a\]
C) \[\sqrt{{{a}^{2}}+{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}}\]
D) \[\sqrt{u+\frac{{{v}^{2}}}{r}}\]
Correct Answer: C
Solution :
Radial acceleration\[{{a}_{r}}=\frac{{{v}^{2}}}{r}\] Tangential acceleration \[{{a}_{t}}=a\] \[\therefore \]Resultant acceleration \[a=\sqrt{a_{r}^{2}+a_{t}^{2}+2{{a}_{r}}{{a}_{t}}\cos \theta }\] But here \[\theta =90{}^\circ \] \[\therefore \] \[\cos \theta =\cos 90{}^\circ =0\] and \[a=\sqrt{a_{r}^{2}+a_{t}^{2}}=\sqrt{{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}+{{a}^{2}}}\]
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