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JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion

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 \(.........\,N\)

  1. A \(1.5\)
  2. B \(0.75\)
  3. C \(0.25\)
  4. D \(0.50\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(0.75\)

Step-by-step Solution

Detailed explanation

Let extension in length of spring be \(x\). Radius of circle \(r=0.2+ x\) \(Kx = m \omega^2 r\) \(7.5 x =\left(\frac{1}{10}\right)\left(5^2\right)(0.2+ x )\) \(\Rightarrow \frac{15}{2} x =\frac{5}{2}\left( x +\frac{1}{5}\right)\) \(\Rightarrow x =\frac{1}{10}\)…
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