Question

A uniform conducting wire of the length and radius ( = (10 + 0.1) cm and r = (1 0.01) cm
respectively, is cut to form a resistor. The maximum percentage error in the resistance is (assume
that the resistivity of the material of the wire is known very accurately)
1 %
O 2 %
O
3 %
4 %​

Answer 1

Answer:

1°\°

Explanation:

1. happy new year to you and your family and friends

Answer 2

Answer:

A high percentage of error in the resistance states $1 %$ %.

Explanation:

• The resistance of a wire depends on the cross-sectional area and the length of the wire.
• The resistivity depends on the material of the wire.
• The variation in resistance of a conductor is there with temperature change.

The formula for resistance is given as:

$R=\frac{\alpha *L}{A}$

where, $\alpha$ is the resistivity.

            $L$ is the length.

            $A$ is the cross-sectional area.

Step 1:

Resistance of wire without error $=\frac{\alpha *(10 cm)}{\pi *(1 cm)^{2} }$  $ohm$

                                                     $=\frac{10\alpha }{\pi }$  $ohm$

Resistance of wire with error  $=\frac{\alpha *(10 .1cm)}{\pi *(1.01 cm)^{2} }$  $ohm$

                                                 $=\frac{10.1\alpha }{1.0201\pi }$  $ohm$

Step 2:

Error is given by  $=\frac{10\alpha }{\pi }-\frac{10.1\alpha }{1.0201\pi }$  $ohm$

                            $=0.099*\frac{\alpha }{\pi }$  $ohm$

So, the error percentage is given by $=\frac{0.099*\frac{\alpha }{\pi }}{\frac{10\alpha }{\pi }} * 100$ %

                                                             $=\frac{0.099}{10} *100$  %

                                                             ≈  $1 %$ %

Hence, a high percentage of error in the resistance states $1 %$ %.