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A spherical steel ball released at the topp of a long column of glycerine of length l falls through a distance l//2 with accelerated motion and the remaining distance l//2 with uniform velocity V. Let t_(1) and t_(2) denote the times taken to cover the first and second half and W_(1) and W_(2) the work done agaist gravity in the two halves, then compare times and work done.

Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/average-13416" style="font-weight:bold;" target="_blank" title="Click to know more about AVERAGE">AVERAGE</a> <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> in the first half of the distance `ltv`, while in the second half the average velocity is v. There fore, `t_(1s)gtt_(2)`. The workdone against gravity in both halves is `<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> l//2` <br/> `:.t_(1s)gt_(2)""w_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)=w_(2)`</body></html>


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