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Question

One end of a massless spring of spring constant 100 N/M and natural length 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. if the mass is made to rotate at an angular velocity of 2 rad/s. Find the elongation of the spring.



A
1 cm
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B
4 cm
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C
7 cm
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D
10 cm
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Solution

The correct option is A 1 cm
We know that,
$$kl=mv^2/r=m\omega^2 r=m\omega^2(l_0+l)$$
Here $$\omega$$ is the angular velocity $$l_0$$ is the natural length $$(0.5)m$$ and $$(l_0+l)$$
Thus $$(k-m\omega^2)l=m\omega^2l_0\\l=\cfrac{m\omega^2l_0}{k-m\omega^2}$$
Putting value
$$l=\cfrac{0.5\times4\times0.5}{100-0.5\times4}\quad m=\cfrac{1}{100m}=1cm$$



Physics

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