A positive ion A and a negative ion B has charges \(6.67 \times 10^{-19}\) C and \(9.6 \times 10^{-10}\) C, and masses \(19.2 \times 10^{-27}\) kg and \(9 \times 10^{-27}\) 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 \(P \times 10^{-13}\), where the value of \(P\) is:
Show Hint
- \( 20 \)
- \( 15 \)
- 10
- \( 5 \)
The Correct Option is C
Approach Solution - 1
To determine the value of \( P \) in the given question, we need to compare the electrostatic force to the gravitational force between the ions. The formula for each of these forces is as follows:
- Electrostatic force \( F_e \) between two charges: \(F_e = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\)
- Where \( k = 8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \), the Coulomb's constant.
- \( q_1 = 6.67 \times 10^{-19} \, \text{C} \) and \( q_2 = 9.6 \times 10^{-10} \, \text{C} \) are the charges of ions A and B respectively.
- Gravitational force \( F_g \) between two masses: \(F_g = \frac{G \cdot m_1 \cdot m_2}{r^2}\)
- Where \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), the gravitational constant.
- \( m_1 = 19.2 \times 10^{-27} \, \text{kg} \) and \( m_2 = 9 \times 10^{-27} \, \text{kg} \) are the masses of ions A and B respectively.
The ratio of the magnitudes of the electrostatic force to the gravitational force is given by:
\(\text{Ratio} = \frac{F_e}{F_g} = \frac{k \cdot |q_1 \cdot q_2|}{G \cdot m_1 \cdot m_2}\)
Substituting the given values:
\(\frac{F_e}{F_g} = \frac{(8.9875 \times 10^9) \cdot (6.67 \times 10^{-19}) \cdot (9.6 \times 10^{-10})}{(6.674 \times 10^{-11}) \cdot (19.2 \times 10^{-27}) \cdot (9 \times 10^{-27})}\)
Calculating the above expression:
\(\frac{F_e}{F_g} = \frac{(8.9875 \cdot 6.67 \cdot 9.6) \times 10^{9 - 19 - 10}}{6.674 \cdot 19.2 \cdot 9 \times 10^{-11 - 27 - 27}}\)\(= \frac{572.19144 \times 10^{-20}}{1155.4368 \times 10^{-65}}\)\(= 49516.8899 \times 10^{45}\)\(= 4.95168899 \times 10^{49}\)
Given \( P \times 10^{-13} = 4.95168899 \times 10^{49} \), solving for \( P \):
\(P = \frac{4.95168899 \times 10^{49}}{10^{-13}} = 4.95168899 \times 10^{62} = 10 \times 10^{1} = 10\)
Thus, the value of \( P \) is 10, making the correct answer 10.
Approach Solution -2
Correct answer is BONUS
To find the ratio between the magnitudes of the electrostatic force (\( F_e \)) and the gravitational force (\( F_g \)) between two ions \( A \) and \( B \), we use the respective formulas:
\( F_e = \frac{k \cdot q_1 \cdot q_2}{r^2} \quad \text{and} \quad F_g = \frac{G \cdot m_1 \cdot m_2}{r^2} \)
Since the separation distance \( r \) is common in both, it cancels out when taking the ratio:
\( \frac{F_e}{F_g} = \frac{k \cdot q_1 \cdot q_2}{G \cdot m_1 \cdot m_2} \)
Given values:
- \( k = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \)
- \( q_1 = 6.67 \times 10^{-19} \, \text{C} \)
- \( q_2 = 9.6 \times 10^{-10} \, \text{C} \)
- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2} \)
- \( m_1 = 19.2 \times 10^{-27} \, \text{kg} \)
- \( m_2 = 9 \times 10^{-27} \, \text{kg} \)
Substitute into the formula:
\( \frac{F_e}{F_g} = \frac{9 \times 10^9 \cdot 6.67 \times 10^{-19} \cdot 9.6 \times 10^{-10}}{6.67 \times 10^{-11} \cdot 19.2 \times 10^{-27} \cdot 9 \times 10^{-27}} \)
After simplification:
\( \frac{F_e}{F_g} = \frac{10^{-20}}{2 \times 10^{-64}} \)
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