- 1(current)
Q. No.Consider an \(HCl\) molecule, where one of its electrons is located exactly midway between the two nuclei, at a given instant. Ignore all the other electrons and any quantum effects.
Let \(\vec F_H\) be the electrostatic force exerted by the electron on the \(H\)-nucleus and \(\vec F_{Cl}\) be the force exerted by the electron on the \(Cl\) nucleus. Then:
Let \(\vec F_H\) be the electrostatic force exerted by the electron on the \(H\)-nucleus and \(\vec F_{Cl}\) be the force exerted by the electron on the \(Cl\) nucleus. Then:
| 1. | \(|\vec F_{Cl}|=|\vec F_H|\)and \(\vec F_{Cl}\) is opposite to \(\vec F_H\) |
| 2. | \(|\vec F_{Cl}|=|\vec F_H|\)and \(\vec F_{Cl}\) & \(\vec F_H\) are in the same direction |
| 3. | \(|\vec F_{Cl}|>|\vec F_H\) and \(\vec F_{Cl}\) is opposite to \(\vec F_H\) |
| 4. | \(|\vec F_{Cl}|<|\vec F_H|\) and \(\vec F_{Cl}\) is opposite to \(\vec F_H\) |
Subtopic: Â Coulomb's Law |
Level 3: 35%-60%
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Consider electrostatic and gravitational forces among the following: electron-electron\((ee)\), electron-proton \((ep)\) & proton-proton\((pp)\). All the distances between the particles are the same. Let \(F^{gr}\) 'denote' gravitational force and \(F^{el}\) 'denote' electrostatic force and the subscripts denote the particle pairs. We consider only the magnitudes of the forces. Then:
| (a) | \(F_{e p}^{e l}=F_{p p}^{e l}=F_{e e}^{e l}\) |
| (b) | \(F_{p p}^{e l}>F_{p p}^{g r}\) |
| (c) | \(F_{e p}^{g r}<F_{e p}^{el}\) |
| (d) | \(F_{e p}^{g r}=F_{p p}^{g r}=F_{ee}^{gr}\) |
Which of the above is/are true?
1. (a) only
2. (a) and (b) only
3. (a), (b), and (c) only
4. (a) and (d) only
Subtopic: Â Coulomb's Law |
 66%
Level 2: 60%+
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Q. No.
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