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A block of mass 1 kg is fastended to one end of a wire of cross-sectional area 2 mm^(2) and is rotated in a verticalcircle of radius 20 cm. The speed of the block at the bottom of the circle is 3.5 ms^(-1). Find the elongation of the wire when the block is at the top of the circle.

Answer» <html><body><p></p>Solution : i) Tension at the <a href="https://interviewquestions.tuteehub.com/tag/bottom-401202" style="font-weight:bold;" target="_blank" title="Click to know more about BOTTOM">BOTTOM</a> of the circle,<br/>`T = (mv^2)/(r ) + <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>= (1 xx (3.5)^2)/(0.2) + 1 xx 9.8` <br/>` = 61.25 + 9.8 = 71.05 N`<br/>This tension in the <a href="https://interviewquestions.tuteehub.com/tag/string-11290" style="font-weight:bold;" target="_blank" title="Click to know more about STRING">STRING</a> is equal to the force, F i.e., F=71.05N, L =r=0.2m. <br/>The increase in length,<br/>` e = (FL)/(AY) = (71.05 xx 0.2)/(2 xx 10^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>) xx 2 xx 10^11)` <br/>` = 3.553 xx 10^(-5) m` <br/>ii) Tension at the top of the circle,T = Tension at the bottom - 6 mg<br/>`= 71.05 - <a href="https://interviewquestions.tuteehub.com/tag/6xx-1913072" style="font-weight:bold;" target="_blank" title="Click to know more about 6XX">6XX</a> 1 xx 9.8 = 71.05 - 58.8 = 12.25 N.`<br/>F= 12.25 N, L = 0.2 m.<br/>The increase in length ` = (FL)/(AY)` <br/> ` = (12.25 xx 0.2)/(2 xx 10^(-6) xx 2 xx 10^11)` <br/>` = 0.6125 xx 10^(-5) m`</body></html>


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