Q136 / 180 · COMEDK 2026 (9 May Shift 2) · Physics
Bodies P, Q, R, S are labelled as having charges $Q_P = 0.5 \times 10^{-19}C$, $Q_Q = 0.7 \times 10^{-19}C$, $Q_R = 2.1 \times 10^{-19}C$, $Q_S = 4.8 \times 10^{-19}C$ respectively. Select the body having the correct charge.
[Given electronic charge $= e = 1.6 \times 10^{-19}C$]
[Given electronic charge $= e = 1.6 \times 10^{-19}C$]
- A$Q_Q = 0.7 \times 10^{-19}C$
- B$Q_P = 0.5 \times 10^{-19}C$
- Correct answerC$Q_S = 4.8 \times 10^{-19}C$
- D$Q_R = 2.1 \times 10^{-19}C$
Solution
According to the principle of quantization of charge, the charge on a body must be an integral multiple of the elementary charge $e = 1.6 \times 10^{-19} C$.
$Q = ne$
For body P, $n = \dfrac{0.5 \times 10^{-19}}{1.6 \times 10^{-19}} = \dfrac{5}{16}$ (not an integer)
For body Q, $n = \dfrac{0.7 \times 10^{-19}}{1.6 \times 10^{-19}} = \dfrac{7}{16}$ (not an integer)
For body R, $n = \dfrac{2.1 \times 10^{-19}}{1.6 \times 10^{-19}} = \dfrac{21}{16}$ (not an integer)
For body S, $n = \dfrac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}} = 3$ (an integer)
Since $n$ is an integer only for body S, the charge $Q_S = 4.8 \times 10^{-19} C$ is possible.
Answer: $Q_S = 4.8 \times 10^{-19}C$
$Q = ne$
For body P, $n = \dfrac{0.5 \times 10^{-19}}{1.6 \times 10^{-19}} = \dfrac{5}{16}$ (not an integer)
For body Q, $n = \dfrac{0.7 \times 10^{-19}}{1.6 \times 10^{-19}} = \dfrac{7}{16}$ (not an integer)
For body R, $n = \dfrac{2.1 \times 10^{-19}}{1.6 \times 10^{-19}} = \dfrac{21}{16}$ (not an integer)
For body S, $n = \dfrac{4.8 \times 10^{-19}}{1.6 \times 10^{-19}} = 3$ (an integer)
Since $n$ is an integer only for body S, the charge $Q_S = 4.8 \times 10^{-19} C$ is possible.
Answer: $Q_S = 4.8 \times 10^{-19}C$
Hence, the correct answer is option: 'C'.