Question:

A body of mass $100\text{ gram}$ is tied to a spring of spring constant $8\text{ N/m}$, while the other end of a spring is fixed. If the body moves in a circular path on smooth horizontal surface with constant angular speed $8\text{rad/s}$ then the ratio of extension in the spring to its natured length will be}

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For spring-based circular motion, do not forget that radius becomes: \[ r=l+x \] not just the extension \(x\).
Updated On: May 14, 2026
  • $1 : 1$
  • $8 : 1$
  • $2 : 1$
  • $4 : 1$
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The Correct Option is D

Solution and Explanation

Concept:
The spring force provides centripetal force. If natural length is \(l\) and extension is \(x\), then radius of circular path is: \[ r=l+x \] Also, \[ kx = m\omega^2 r \] ip

Step 1:
Write the given values.
\[ m=100\text{ g}=0.1\text{ kg} \] \[ k=8\text{ N/m} \] \[ \omega=8\text{ rad/s} \] ip

Step 2:
Apply centripetal force condition.
\[ kx = m\omega^2(l+x) \] Substitute values: \[ 8x = 0.1\times 64 (l+x) \] \[ 8x = 6.4(l+x) \] \[ 8x = 6.4l + 6.4x \] \[ 1.6x = 6.4l \] \[ x=4l \] ip

Step 3:
Find the ratio of extension to natural length.
\[ x:l = 4:1 \] ip Hence, the correct answer is:
\[ \boxed{(D)\ 4:1} \]
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