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A small block of mass 100g 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

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1.5 N

0.75 N

0.25 N

0.50 N

answer is B.

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Detailed Solution

$Kx = m (ℓ_{0} + x) ω^{2}$

$\begin{array}{l} \Rightarrow (7.5) (x) = \frac{100}{1000} (20 + x) (25) \\ \Rightarrow \frac{15 x}{2} = \frac{5}{2} (20 + x) \\ \Rightarrow \frac{15 x}{2} = 50 + \frac{5 x}{2} \end{array}$

$x = 10 cm$

$Tension = K x$

= 7.5 (0.1)

= 0.75 N

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