A) \[0.1 m/{{s}^{2}}\]
B) \[0.15 m/{{s}^{2}}\]
C) \[0.18 m/{{s}^{2}}\]
D) \[0.2 m/{{s}^{2}}\]
Correct Answer: A
Solution :
[a] Given: Mass of particle, \[\operatorname{M}=10g=\frac{10}{1000}kg\] radius of circle \[R=6.4\text{ }cm\] Kinetic energy E of particle \[=8\times 1{{0}^{-4}}J\] acceleration \[{{a}_{t}}=?\] \[\frac{1}{2}m{{v}^{2}}=E\Rightarrow \frac{1}{2}\left( \frac{10}{1000} \right){{v}^{2}}=8\times {{10}^{-4}}\] \[\Rightarrow {{v}^{2}} = 16\times {{10}^{-2}}\Rightarrow v=4\times {{10}^{-1}} = 0.4 m/s\] Now, using \[{{\operatorname{v}}^{2}}={{u}^{2}}+2{{a}_{t}}S~~~~~\left( s=4\pi R \right)\] \[{{\left( 0.4 \right)}^{2}}={{0}^{2}}+2{{a}_{t}}\left( 4\times \frac{22}{7}\times \frac{6.4}{100} \right)\] \[\Rightarrow {{a}_{t}}={{\left( 0.4 \right)}^{2}}\times \frac{7\times 100}{8\times 22\times 6.4}=0.1m/{{s}^{2}}\]
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