Question

Two particles $A$ and $B$ having same mass have charge $+q$ and $+4q$ respectively. When they are allowed to fall from rest through same electric potential difference, the ratio of their speeds $V_A$ to $V_B$ will become

• A. $1 : 2$
• B. $2 : 1$
• C. $1 : 4$
• D. $4 : 1$

Hint:

Since kinetic energy scales linearly with charge ($K.E. \propto q$) for a constant voltage, particle $B$ will gain exactly 4 times the kinetic energy of particle $A$. Because velocity scales with the square root of kinetic energy ($v \propto \sqrt{K.E.}$), particle $B$ will move at $\sqrt{4} = 2$ times the speed of particle $A$, giving the ratio $1 : 2$ instantly!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Two particles of equal mass with charges +q and +4q accelerate from rest through the same potential V; find speed ratio V_A : V_B.

Step 2: Key Formula or Approach:
Work-energy theorem: qV = ½mv² → v = √(2qV/m). Since m and V are same, v ∝ √q.

Step 3: Detailed Explanation:
V_A / V_B = √(q / 4q) = √(1/4) = 1/2. Ratio is 1 : 2.

Step 4: Final Answer:
V_A : V_B = 1 : 2, matching option (A).