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The resistance of a wire is20Omega. What will be new resistance, if it is stretched uniformly 8 times its original length? |
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Answer» Solution :`R_1 =20 Omega , R_2=?` LET the original length `(l_1)` be l. The new length `L_2 =8L_1(i.e)l_2=8l` The original resistance. `R_1 =rho(l_1)/(A_1)` The new resistance `R_2=rho(l_2)/(A_2)=(rho(8l))/(A_2)` Though the wire is stretched its volume is unchanged Initial volume =Final volume `A_1l_1=A_2I_2,A_1l=A_28l` `(A_1)/(A_2)=(8l)/l=8` By dividing equation `R_2` by equation `R_1` we get `(R_2)/(R_1) =(rho(8l))/(A_2)xx(A_1)/(rhol)RARR (R_2)/(R_1)=(A_1)/(A_2)XX8` Substituting the value of `(A_1)/(A_2)` we get `(R_2)/(R_1)=8xx 8 =64 rArr R_2=64 xx20 =1280 Omega` HENCE, stretching the length of the wire has increased its resistance. |
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