A small block of mass 100,g is tied to a spring of spring constant 7.5,N / m and length 20, cm . oth

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4-1.Newton's Laws of Motion

hard

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 $.........\,N$

A$1.5$

B$0.75$

C$0.25$

D$0.50$

(JEE MAIN-2023)

Solution

Let extension in length of spring be $x$.
Radius of circle $r=0.2+ x$
$Kx = m \omega^2 r$
$7.5 x =\left(\frac{1}{10}\right)\left(5^2\right)(0.2+ x )$
$\Rightarrow \frac{15}{2} x =\frac{5}{2}\left( x +\frac{1}{5}\right)$
$\Rightarrow x =\frac{1}{10}$
$\therefore \text { Tension in spring }= kx =7.5 \times \frac{1}{10}=0.75\,N$

Standard 11

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