When load of $5\,kg$ is hung on a wire then extension of $3\,m$ takes place, then work done will be ....... $joule$
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(a) $W = \frac{1}{2}Fl = \frac{1}{2} \times Mg \times l = \frac{1}{2} \times 5 \times 10 \times 3 = 75\;Joule$
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