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MHT CET 2021 (22 Sep, Shift 2) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from MHT CET 2021 (22 Sep, Shift 2)

Q25MHT CET 2021 · 22 SEP · SHIFT 2Easy

Motion of Charges in an Electric Field

Two particles AA and BB having same mass have charge +q+q and +4q+4q respectively. When they are allowed to fall from rest through same electric potential difference, the ratio of their speeds 'VAV_A' to 'VBV_B' will become

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Answer: (a)

  1. 12mv2=qV⇒v∝q\frac{1}{2}mv^2 = qV \Rightarrow v \propto \sqrt{q} for the same mm and VV.
  2. VAVB=q4q=12\frac{V_A}{V_B} = \sqrt{\frac{q}{4q}} = \frac{1}{2}
  3. So the answer is (a) 1:21:2.
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