A cylindrical wire has a resistance R and resistivity rho If its length and diameter are BOTH cut
in half, what will be its resistance? 4R 2R R R/4 R/2 If its length and diameter are BOTH cut
in half, what will be its resistivity? 2rho rho rho/4 rho/2 4rho
Solution
a)R=4L/d2 this is the original resistance
now length and diameter are half means
l=l/2 ,d=d/2
Rnew =4(L/2)/((d/2)2)
Rnew/R =4(L/2)/((d/2)2)/4L/d2
Rnew/R =(1/2)/(1/4)/0.25 =2
Rnew =2R
so the answer is B
b)resistivity is a constant,so it remains same
so the answer is B
in half, what will be its resistance? 4R 2R R R/4 R/2 If its length and diameter are BOTH cut
in half, what will be its resistivity? 2rho rho rho/4 rho/2 4rho
Solution
a)R=4L/d2 this is the original resistance
now length and diameter are half means
l=l/2 ,d=d/2
Rnew =4(L/2)/((d/2)2)
Rnew/R =4(L/2)/((d/2)2)/4L/d2
Rnew/R =(1/2)/(1/4)/0.25 =2
Rnew =2R
so the answer is B
b)resistivity is a constant,so it remains same
so the answer is B
