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JEE Main 2025 (23 Jan, Shift 1) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from JEE Main 2025 (23 Jan, Shift 1)

Q18JEE Main 2025 · 23 JAN · SHIFT 1Easy

Electric Charge & Coulomb's Law

A positive ion AA and a negative ion BB has charges 6.67×10−19 C6.67 \times 10^{-19}\text{ C} and 9.6×10−10 C9.6 \times 10^{-10}\text{ C}, and masses 19.2×10−27 kg19.2 \times 10^{-27}\text{ kg} and 9×10−27 kg9 \times 10^{-27}\text{ kg} respectively. At an instant, the ions are separated by a certain distance rr. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is P×10−13P \times 10^{-13}, where the value of P is _________.

(Take 14πε0=9×109 N m2 C−2\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\text{ N m}^2\text{ C}^{-2} and universal gravitational constant as 6.67×10−11 N m2 kg−26.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2})

Numerical answer type. Work it out, then open the answer.

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Answer: b

  1. FeFg=kqAqBGmAmB\dfrac{F_e}{F_g} = \dfrac{kq_Aq_B}{Gm_Am_B} (rr cancels).
  2. =9×109×6.67×10−19×9.6×10−106.67×10−11×19.2×10−27×9×10−27= \dfrac{9\times10^9\times6.67\times10^{-19}\times9.6\times10^{-10}}{6.67\times10^{-11}\times19.2\times10^{-27}\times9\times10^{-27}}
  3. =9×9.6×10−20172.8×10−65=5×1044= \dfrac{9\times9.6\times10^{-20}}{172.8\times10^{-65}} = 5\times10^{44}
  4. With the data as printed the ratio is 5×10445\times10^{44}, which cannot be written as P×10−13P\times10^{-13} with a sensible PP.
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