A) 3 : 4
B) 1 : 3
C) 9 : 4
D) None of these
Correct Answer: C
Solution :
Using \[{{Z}^{2}}=k\left( \frac{q}{m} \right)y;\] where \[k=\frac{{{B}^{2}}LD}{E}\]. For parabolas to coincide in the two photographs, the \[\frac{k\,q}{m}\] should be same for the two cases. Thus, \[\frac{B_{1}^{2}\ LDe}{{{E}_{1}}{{m}_{1}}}=\frac{B_{2}^{2}LD\,(2e)}{{{E}_{2}}{{m}_{2}}}\] Þ \[\frac{{{m}_{1}}}{{{m}_{2}}}={{\left( \frac{{{B}_{1}}}{{{B}_{2}}} \right)}^{2}}\times \left( \frac{{{E}_{2}}}{{{E}_{1}}} \right)\times \frac{1}{2}\]\[=\frac{9}{4}\times \frac{2}{1}\times \frac{1}{2}=\frac{9}{4}\]
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