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Question: Answered & Verified by Expert
Two masses M1 and M2 carry positive charges Q1 and Q2, respectively. They are dropped to the floor in a laboratory set up from the same height, where there is a constant electric field vertically upwards. M1 hits the floor before M2. Then,
PhysicsElectrostaticsKVPYKVPY 2019 (SA)
Options:
  • A Q1>Q2
  • B Q1<Q2
  • C M1Q1>M2Q2
  • D M1Q2>M2Q1
Solution:
1897 Upvotes Verified Answer
The correct answer is: M1Q2>M2Q1
\(\mathrm{Q}_{1}=\frac{\mathrm{M}_{1} \mathrm{~g}-\mathrm{Q}_{1} \mathrm{E}}{\mathrm{M}_{1}}, \mathrm{Q}_{2}=\frac{\mathrm{M}_{2} \mathrm{~g}-\mathrm{Q}_{2} \mathrm{E}}{\mathrm{M}_{2}}\)
as \(\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\)
So, \(\mathrm{h}=\mathrm{a} \times \mathrm{t}_{1}+\frac{1}{2}\left(\mathrm{~g}-\frac{\mathrm{Q}_{1} \mathrm{E}}{\mathrm{M}_{1}}\right) \mathrm{t}_{1}^{2} \ldots \ldots \ldots(1)\)
\(\mathrm{h}=\mathrm{a} \times \mathrm{t}_{2}+\frac{1}{2}\left(\mathrm{~g}-\frac{\mathrm{Q}_{2} \mathrm{E}}{\mathrm{M}_{2}}\right) \mathrm{t}_{2}^{2} \ldots \ldots . .(2)\)
given \(\mathrm{t}_{1}<\mathrm{t}_{2} \Rightarrow \mathrm{t}_{1}^{2}<\mathrm{t}_{2}^{2}\)
\(\mathrm{So}, \frac{2 \mathrm{~h}}{\mathrm{~g}-\frac{\mathrm{Q}_{1} \mathrm{E}}{\mathrm{M}_{1}}}<\frac{2 \mathrm{~h}}{\mathrm{~g}-\frac{\mathrm{Q}_{2} \mathrm{E}}{\mathrm{M}_{2}}}\)
\(\mathrm{~g}-\frac{\mathrm{Q}_{1} \mathrm{E}}{\mathrm{M}_{1}}>\mathrm{g}-\frac{\mathrm{Q}_{2} \mathrm{E}}{\mathrm{M}_{2}}\)
\(\frac{\mathrm{Q}_{1} \mathrm{E}}{\mathrm{M}_{1}}<\frac{\mathrm{Q}_{2} \mathrm{E}}{\mathrm{M}_{2}}\)
\(\Rightarrow \mathrm{Q}_{2} \mathrm{M}_{1}>\mathrm{Q}_{1} \mathrm{M}_{2}\)

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