Question

A particle of mass \(3\,\text{g}\) and charge \(60\,\mu C\) is released from rest in a uniform electric field of intensity \(10^5\,\text{N C}^{-1}\). If the value of kinetic energy attained by the particle after moving through a distance of \(2\,\text{cm}\) is \(m \times 10^{-2}\,\text{J}\), then the value of \(m\) is:

• A. \(12\)
• B. \(6\)
• C. \(5\)
• D. \(4\)

Hint:

In a uniform electric field, work done is \(qEd\) which directly converts into kinetic energy. No need to use mass here.

Answer 1

The Correct Option is A

Step 1: Use work-energy theorem.
The work done by electric field is converted into kinetic energy:
\[ W = qEd \]
\[ \text{K.E.} = qEd \]

Step 2: Convert given quantities into SI units.
\[ q = 60\,\mu C = 60 \times 10^{-6}\,C \]
\[ E = 10^5\,\text{N/C} \]
\[ d = 2\,\text{cm} = 2 \times 10^{-2}\,\text{m} \]

Step 3: Substitute values.
\[ \text{K.E.} = (60 \times 10^{-6}) \times (10^5) \times (2 \times 10^{-2}) \]

Step 4: Simplify step by step.
\[ = 60 \times 10^{-6} \times 10^5 \times 2 \times 10^{-2} \]
\[ = 60 \times 2 \times 10^{-6+5-2} \]
\[ = 120 \times 10^{-3} \]
\[ = 0.12\,\text{J} \]

Step 5: Express in required form.
\[ 0.12 = 12 \times 10^{-2} \]

Step 6: Identify value of \(m\).
\[ m = 12 \]

Step 7: Final answer.
\[ \boxed{12} \]