KCET · Physics · Electrostatics
A mass of \( 1 \mathrm{Kg} \) carrying a charge of \( 2 \mathrm{C} \) is accelerated through a potential of \( 1 \mathrm{~V} \). The velocity acquired by it is
- A \( \sqrt{2} m s^{-1} \)
- B \( 2 m s^{-1} \)
- C \( \frac{1}{\sqrt{2}} m s^{-1} \)
- D \( \frac{1}{2} m s^{-1} \)
Answer & Solution
Correct Answer
(B) \( 2 m s^{-1} \)
Step-by-step Solution
Detailed explanation
Given, a mass is accelerated through a potential. Therefore,
\(\frac{1}{2} m v^{2}=q V\)
Here, \(\mathrm{m}=1 \mathrm{~kg}\); \(\mathrm{q}=2 \mathrm{C} ; \mathrm{V}=1\)
\(\therefore \frac{1}{2} \times 1 \times V^{2}=2 \times 1\)
\(\Rightarrow V^{2}=4 \Rightarrow V=2\)
Thus, velocity acquired by mass is \(2 \mathrm{~ms}^{-1}\)
\(\frac{1}{2} m v^{2}=q V\)
Here, \(\mathrm{m}=1 \mathrm{~kg}\); \(\mathrm{q}=2 \mathrm{C} ; \mathrm{V}=1\)
\(\therefore \frac{1}{2} \times 1 \times V^{2}=2 \times 1\)
\(\Rightarrow V^{2}=4 \Rightarrow V=2\)
Thus, velocity acquired by mass is \(2 \mathrm{~ms}^{-1}\)
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