Question

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x from the left end and obeys the formula rho(x)=a+bx2, where a and b are constants. At the left end, the resistivity is 2.25×10−8Ω⋅m, while at the right end it is 8.50×10−8Ω⋅m. What is the resistance of the rod?

Answer 1

Answer:

Resistance = 3.35*$10^{-4}$ Ω

Explanation:

Since resistance R = ρ$\frac{L}{A}$

whereas $\rho(x) = a + bx^2$

resistivity is given for two ends. At the left end resistivity is $2.25* 10^{-8}$ whereas x at the left end will be 0 as distance is zero. Thus

$2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a$

At the right end x will be equal to the length of the rod, so $x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25} = b\\b = 2.77 *10^{-8}$

Thus resistance will be R = ρ$\frac{L}{A}$

where A = π $r^2$

so,

$R = \frac{8.50*10^{-8} * 1.50}{3.14*(1.10*10^{-2})^2} \\R=3.35 * 10 ^{-4}$