Verified Solution • 2025–26 Session moderate

QJEEM25Phff7d: JEE Main 2025 Physics - Class 12 JEEM25Phff7d Solution: A positive ion and a negative ion has charges and , and masses…

A positive ion AA and a negative ion BB has charges 6.67×10−19C6.67 \times 10^{-19} \mathrm{C} and 9.6×10−10C9.6 \times 10^{-10} \mathrm{C} , and masses 19.2×10−27 kg19.2 \times 10^{-27} \mathrm{~kg} and 9×10−27 kg9 \times 10^{-27} \mathrm{~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×10−13\mathrm{P} \times 10^{-13} , where the value of P is _________.

(Take 14πε0=9×109Nm2C−1\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-1} and universal gravitational constant as 6.67×10−11Nm2 kg−26.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2} )

Step-by-Step Verified Solution

The ratio of the magnitudes of electrostatic force ( FeF_e ) to gravitational force ( FgF_g ) is given by:
FeFg=kqAqBGmAmB\frac{F_e}{F_g} = \frac{k q_A q_B}{G m_A m_B}
Plugging in the provided values:
Ratio=(9×109)(6.67×10−19)(9.6×10−10)(6.67×10−11)(19.2×10−27)(9×10−27)\text{Ratio} = \frac{(9 \times 10^9)(6.67 \times 10^{-19})(9.6 \times 10^{-10})}{(6.67 \times 10^{-11})(19.2 \times 10^{-27})(9 \times 10^{-27})}
Simplifying the expression:
Ratio=9.6×10−2019.2×10−65\text{Ratio} = \frac{9.6 \times 10^{-20}}{19.2 \times 10^{-65}} Ratio=0.5×1045=5×1044\text{Ratio} = 0.5 \times 10^{45} = 5 \times 10^{44}
Comparing this result to the form P×10−13P \times 10^{-13} mentioned in the question:
P×10−13=5×1044P \times 10^{-13} = 5 \times 10^{44} P=5×1057P = 5 \times 10^{57}
Note: In this question, the ratio value should have been P×1045P \times 10^{45} , but the exam had P×10−13P \times 10^{-13} . That's why it was given as a bonus.

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