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A conductor of radius r, length l, and resistivity p and resistance R. It is melted down and formed into a new conductor, also cylindrical, with one-fourth the length of the original conductor. The resistance of the new conductor is

a. R/16 b. R/4 c. R d. 4R e. 16R
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Added Mon, 03 Aug '15
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Answer is a)R/16

dimension of the conductor before melting is l, r

reistivity is p

R=(p*l)/(pie*r2)

after reforming length is reduced to L=l/4

volume in both the cases will be same

ie pie*r2*l=pie*R2L

due to this radius will become R=2r

now new reistance is given by Rx=(p*L)/(pie*R2)

ie Rx=(p*l/4)/(pie*r2*4)

after simplification RX=((p*l)/(pie*r2))/16

ieRx=R/16
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