KCET · Physics · Nuclear Physics
The angle between velocity and acceleration of a particle describing uniform circular motion is
- A \( 45^{\circ} \)
- B \( 60^{\circ} \)
- C \( 90^{\circ} \)
- D \( 180^{\circ} \)
Answer & Solution
Correct Answer
(C) \( 90^{\circ} \)
Step-by-step Solution
Detailed explanation
As the lift is freely falling, thus its acceleration in down ward direction \(\mathrm{a}=\mathrm{g}\)
Let the mass of the body be \(m\) and force acting on it due to spring be \(F\).
Writing equation of motion for the body: \(m a=m g-F\)
Or \(\mathrm{mg}=\mathrm{mg}-\mathrm{F}\)
\(\Longrightarrow \mathrm{F}=0\)
Thus no force is acting on the body due to spring, so the reading of the spring balance is zero.
Let the mass of the body be \(m\) and force acting on it due to spring be \(F\).
Writing equation of motion for the body: \(m a=m g-F\)
Or \(\mathrm{mg}=\mathrm{mg}-\mathrm{F}\)
\(\Longrightarrow \mathrm{F}=0\)
Thus no force is acting on the body due to spring, so the reading of the spring balance is zero.
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