JEE Main 2022 · 27 Jun · Shift 2 · Physics Q34

← JEE Main 2022 · 27 Jun · Shift 2

PhysicsSingle correct+4 / -1Question 34 of 90

One end of a massless spring of spring constant $k$ and natural length $l_0$ is fixed while the other end is connected to a small object of mass $m$ lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity $\omega$ about an axis passing trough fixed end, then the elongation of the spring will be :

\dfrac{k-m\omega^2l_0}{m\omega^2}

\dfrac{m\omega^2l_0}{k+m\omega^2}

\dfrac{m\omega^2l_0}{k-m\omega^2}

\dfrac{k+m\omega^2l_0}{m\omega^2}

Answer and worked solution

Official answer: C

Spring supplies the centripetal force. $kx=m\omega^2(l_0+x)$ , so $x=\dfrac{m\omega^2l_0}{k-m\omega^2}$ .

(C)

Topics:Circular motion dynamics Spring potential energy

← Previous question Next question →

JEE Main 2022 · 27 Jun · Shift 2, Physics question 34