Laws of Motion

JEE Main level, single correct, medium

113 of 210

Question 113 of 210

A small block of mass 100g100g is tied to a spring of spring constant 7.5Nm−17.5N{m}^{-1} and length 20cm.20cm. The other end of spring is fixed at a particular point A.A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5rads−15rad{s}^{-1}about point A,A, then tension in the spring is

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Answer: (A) 0.75N0.75N

Centripetal force will be provided by the spring force. Let the elongation in the spring be xx, then we can write
⇒kx=mω2(r+x)\Rightarrow kx=m{\omega }^{2}(r+x)
⇒7.5x=2.5(0.2+x)\Rightarrow 7.5x=2.5(0.2+x)
⇒x=0.55=0.1\Rightarrow x=\dfrac{0.5}{5}=0.1
Therefore, required tension will beT=kx=0.75NT=kx=0.75N.

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