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A spring balance has a scale which ranges from 0 to 25 kg and the length of the scale is 0.25 m. It is taken to an unknown planet X where the acceleration due to gravity is 11.5ms^(-1). Suppose a body a mass M kg is suspended in this spring and made to oscillate with a period of 0.50 s. Compute the gravitational force acting on the body.

Answer» <html><body><p></p>Solution :Let us first calculate the stiffness constant of the spring balance by <a href="https://interviewquestions.tuteehub.com/tag/using-7379753" style="font-weight:bold;" target="_blank" title="Click to know more about USING">USING</a> equation `(m)/(k)=(l)/(g)` <br/> `k=(mg)/(l)=(25xx11.5)/(0.25)=1150Nm^(-1)` <br/> The time period of oscillations is given by `T=2pisqrt((M)/(k))`, where M is the mass of the <a href="https://interviewquestions.tuteehub.com/tag/body-900196" style="font-weight:bold;" target="_blank" title="Click to know more about BODY">BODY</a>. <br/> Since, M is unknown, rearranging, we get <br/> `M=(<a href="https://interviewquestions.tuteehub.com/tag/kt-1063108" style="font-weight:bold;" target="_blank" title="Click to know more about KT">KT</a>^(2))/(4pi^(2))=((1150)(0.5)^(2))/(4pi^(2))=7.3kg` `M=(kT^(2))/(4pi^(2))=((1150)(0.5)^(2))/(4pi^(2))=7.3kg` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/gravitational-476409" style="font-weight:bold;" target="_blank" title="Click to know more about GRAVITATIONAL">GRAVITATIONAL</a> force <a href="https://interviewquestions.tuteehub.com/tag/acting-269" style="font-weight:bold;" target="_blank" title="Click to know more about ACTING">ACTING</a> on the body is <br/> `W=Mg=7.3xx11.5=83.95N~~84N`</body></html>


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