Question

The external diameter of of a 5m long hollow tube is 10cm and the thickness ofits wall is 5mm. If the specific resistance of copper 1.7 x 10-5 ohm-meter, then determine its resistance.

Answer 1

Answer:

The resistance will be $2.17\times10^{-5}\Omega$

Explanation:

Given that,

Diameter d = 5 m

Length l = 10 cm

Thickness t = 5 mm

Specific resistance of copper $\rho= 1.7\times10^{-5}\ ohm-meter$

We know that,

The resistance of the tube is defined as:

$R = \dfrac{\rho l}{A}$     ....(I)

Area is defined as:

$A=\pi(r^2_{1}-r^2_{2})$   ...(II)

$r_{2}=r_{1}-t$

$r_{2}=2.5-0.005$

$r_{2}=2.495\m$

Put the value of $r_{2}$ in equation (II)

$A= 3.14\times((2.5)^2-(2.495)^2)$

$A = 0.0784\ m^2$

Now put the value of area in equation (I)

$R = \dfrac{1.7\times10^{-5}\times 10\times10^{-2}}{ 0.0784}$

$R = 0.00002168$

$R = 2.17\times10^{-5}\Omega$

Hence, The resistance will be $2.17\times10^{-5}\Omega$.

Answer 2

Answer:

The answer will be 0.057 ohms

Explanation:

Refer to this image please!