A positive ion \(A\) and a negative ion \(B\) has charges \(6.67 \times 10^{-19} \mathrm{C}\) and \(9.6 \times 10^{-10} \mathrm{C}\), and masses \(19.2 \times 10^{-27} \mathrm{~kg}\) and \(9 \times 10^{-27} \mathrm{~kg}\) respectively. At an instant, the ions are separated by a certain distance r. At that instant the ratio of the magnitudes of electrostatic force to gravitational force is \(P \times 10^{45}\), where the value of 10\(P\) is _______.
(Take \(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-1}\) and universal gravitational constant as \(6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\) )
Assume that charge may not be an integral multiple of electrons.

Explanation:
Let us calculate the ratio of forces step-by-step using Coulomb's Law and Newton's Law of Gravitation from the Physics curriculum, applying direct numerical cancellation:

Step 1: Write the formula for the ratio of forces
The electrostatic force (\(F_e\)) and gravitational force (\(F_g\)) between two sharing bodies separated by distance \(r\) are given by:
\(F_e = \frac{k \cdot |q_A| \cdot |q_B|}{r^2}\) and \(F_g = \frac{G \cdot m_A \cdot m_B}{r^2}\)

Taking the ratio eliminates the distance parameter (\(r^2\)) completely:
\(\frac{F_e}{F_g} = \frac{k \cdot |q_A| \cdot |q_B|}{G \cdot m_A \cdot m_B}\)

Step 2: Substitute the values and apply the smart cancellation trick
Substitute the given values into the equation:
\(\frac{F_e}{F_g} = \frac{(9 \times 10^9) \times (6.67 \times 10^{-19}) \times (9.6 \times 10^{-10})}{(6.67 \times 10^{-11}) \times (19.2 \times 10^{-27}) \times (9 \times 10^{-27})}\)

Separating the constant coefficients from the powers of 10:
\(\frac{F_e}{F_g} = \left[ \frac{9 \times 6.67 \times 9.6}{6.67 \times 19.2 \times 9} \right] \times \left[ \frac{10^9 \times 10^{-19} \times 10^{-10}}{10^{-11} \times 10^{-27} \times 10^{-27}} \right]\)

Now, cancel out the identical numbers directly without multiplying them:
- The \(9\) in the numerator cancels out with the \(9\) in the denominator.
- The \(6.67\) in the numerator cancels out with the \(6.67\) in the denominator.
- We are left with \(\frac{9.6}{19.2}\). Since \(19.2 = 2 \times 9.6\), this reduces beautifully to \(\frac{1}{2} = 0.5\).

Step 3: Simplify the powers of 10
- Numerator power = \(9 - 19 - 10 = -20\)
- Denominator power = \(-11 - 27 - 27 = -65\)
- Net power = \(-20 - (-65) = 45\)

Combining the components gives:
\(\frac{F_e}{F_g} = 0.5 \times 10^{45}\)

Step 4: Find the final value of \(10P\)
Comparing our result with the expression given in the question (\(P \times 10^{45}\)):
\(P = 0.5\)

Therefore:
\(10P = 10 \times 0.5 = 5\)

Hence, Option A is the correct choice.