A) 4:1
B) 1:1
C) 2:1
D) 1:2
Correct Answer: C
Solution :
When charges are placed in an electric field, the force acting on it is \[F=qE\] ...(1) where, q is the charge on the particle, E is the electric field intensity. Also, when a body of mass m, moves with velocity v, it possesses kinetic energy given by \[K=\frac{1}{2}m{{v}^{2}}\] ...(2) From equation of motion, we have \[v=u+at\] Since particle moves from rest,\[u=0\] \[v=at\] ...(3) Putting v in Eq. (2), we get \[K=\frac{1}{2}m{{(at)}^{2}}\] From Newton's law\[F=ma=qE,\]we have \[K=\frac{1}{2}m{{\left( \frac{qE}{m}t \right)}^{2}}\] \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{\frac{1}{2}{{m}_{1}}{{\left( \frac{qE}{{{m}_{1}}}t \right)}^{2}}}{\frac{1}{2}{{m}_{2}}{{\left( \frac{qE}{{{m}_{2}}}t \right)}^{2}}}\] \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{{{m}_{2}}}{{{m}_{1}}}\] Given,\[{{m}_{2}}=2m\]and\[{{m}_{1}}=m\] \[\therefore \] \[\frac{{{K}_{1}}}{{{K}_{2}}}=\frac{2m}{m}=\frac{2}{1}\]
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