JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2023
- Hard
- MCQ
A small block of mass \(100\,g\) is tied to a spring of spring constant \(7.5\,N / m\) and length \(20\, cm\). The other end of spring is fixed at a particular point \(A\). If the block moves in a circular path on a smooth horizontal surface with constant angular velocity \(5\,rad / s\) about point \(A\), then tension in the spring is \(.........\,N\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(0.75\)
Step-by-step Solution
Detailed explanation
Let extension in length of spring be \(x\).
Radius of circle \(r=0.2+ x\)
\(Kx = m \omega^2 r\)
\(7.5 x =\left(\frac{1}{10}\right)\left(5^2\right)(0.2+ x )\)
\(\Rightarrow \frac{15}{2} x =\frac{5}{2}\left( x +\frac{1}{5}\right)\)
\(\Rightarrow x =\frac{1}{10}\)
\(\therefore \text { Tension in spring }= kx =7.5 \times \frac{1}{10}=0.75\,N\)
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