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One end of a massless spring of spring constant 100 N/m is fixed at the centre of a frictionless horizontal table and the other end is connected to a body of mass 7 kg lying on the table at rest. The spring has a natural length of 2 m. If the spring remains horizontal and mass is made to rotate at an angular speed of 1 rad/s, then find the elongation in the spring, assuming the elongation to be small compared to the natural length of the spring.

A
14 cm
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B
3.5 cm
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C
35 cm
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D
1.4 cm
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Solution

The correct option is A 14 cm
When the body is rotating in a horizontal circular path, then the spring undergoes elongation to provide the required centripetal force for the body.

Applying the equation of circular motion along radial direction
Fnet=mrω2 .....(1)

Now let us assume the extension in spring as x metres.
Spring force Fspring=kx

kx=mrω2 ....(2)

Substituting m= 7 kg, ω= 1 rad/s
r= natural length of spring =2 m in equation (2)

x=mrω2k=7×2×(1)2100=0.14 m= 14 cm

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