Charge and Mass Invariance – Rankers Physics

Properties of Charges: Practice Problem & Solution

Assertion (A): Both the charge and mass are invariant. Reason (R): Charge is always associated with mass.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Properties of Charges. Understanding the underlying formula is key to arriving at the correct answer below:

Assertion (A) is true, considering 'mass' as rest mass, which is invariant, similar to charge. Reason (R) is true: All charged particles possess mass; a massless particle cannot carry charge.

(R) is not the correct explanation for (A), as invariance of charge and rest mass are fundamental properties, not directly explained by their association.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Two charged spherical conductors of radii \( R_1 \) and \( R_2 \) (\( R_1 > R_2 \)) have equal surface charge densities placed at large distance from each other. If they are connected by a conducting wire, then
2. A point charge is placed at origin. Assuming potential to be zero at infinity, potential difference at two point A and B is found to be 10 V i.e., \(V_A - V_B = 10 \text{V}\). Now if the reference at infinity is changed to 10 V, then \(V_A - V_B\) will be
3. The electric field in a region is directed outward and is proportional to the distance \(r\) from the origin. If we take a spherical volume of radius \(r\) taking centre at the origin, then the charge contained inside the volume is proportional to
4. Two charged spherical conductors of radius \(R_1\) and \(R_2\) are connected by a wire. Then the ratio of final surface charge densities of the spheres \(\sigma_1 / \sigma_2\) is
5. Assertion (A): If electric field in x-y plane is given by \( \vec{E} = y\hat{i} + x\hat{j} \) then equipotential curve is given by \( xy = \text{ constant} \). Reason (R): Electric field may not be perpendicular to equipotential surface/curve/line.