Question

Figure-ashows a rod of resistive material. The resistance per unit length of the rod increases in the positive direction of the xaxis. At any position xalong the rod, the resistancedRof a narrow (differential) section of width dxis given by dR = 5.00 xdx, whereis in ohms and xis in meters. Figure-bshows such a narrow section. You are to slice off a length of the rod between x = 0 and some positionx = Land then connect that length to a battery with potential difference V = 5.0V(Figure-c). You want the current in the length to transfer energy to thermal energy at the rate of 200 W. At what position x = Lshould you cut the rod?

Short answer

The position at which you should cut the rod is 0.224 m .

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The resistance of the narrow section is dR = 5.00 xdx .
• The width of the narrow section is dx .
• The slice of the length between x = 0 and x = L .
• The potential difference is V = 5.0 V .
• The power is P = 200 W .

Understanding the concept of the resistance

By using equation (i), we can find the resistance and by using this value of resistance in the given statement condition, we can find the position you should cut the rod.

Formula:

The electric power generated due to the potential difference,

$P=\frac{V^2}{R}$ … (i)

Calculation of the position at which the rod is to be cut

Using the given data in equation (i), we can get the value of the resistance as follows:

$R=\frac{V^2}{P}$

Substitute the values in the above equation.

$$\begin{gathered} R=\frac{{\left(5.0V\right)}^2}{200W} \\ =0.125\Omega \end{gathered}$$

To meet the condition of the problem statement, we must therefore set the value as follows:

$$\begin{gathered} {\int }_0^{L}dR=R \\ {\int }_0^{L}5.00xdx=0.125\Omega \\ L=0.224m \end{gathered}$$

Hence, the value of the length at which the rod is to be cut is$0.224m$ .