A lead (= 5 x 10 S/m) bar of square cross section has a hole bored along its length of 4 m so that its cross section becomes that of Figure 5.5. Find the resistance between the square ends. Solution: R= aS s = d² − xr² = 3² − x( ¹)² = 9 − −— cm². 4 R= 974 µ 5 × 10 (9 x/4) × 104 3 cr I cm 3 cm

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hi can you explane the calculation of S

Example:
A lead (= 5 x 10° S/m) bar of square cross section has a hole bored along its length of
4 m so that its cross section becomes that of Figure 5.5. Find the resistance between the
square ends.
Solution: R
aS
S - - -
=9 cm².
4
R=
= 974 µ
5x 10°(9-x/4) × 10*
3 cm
I cm
3 cm
Transcribed Image Text:Example: A lead (= 5 x 10° S/m) bar of square cross section has a hole bored along its length of 4 m so that its cross section becomes that of Figure 5.5. Find the resistance between the square ends. Solution: R aS S - - - =9 cm². 4 R= = 974 µ 5x 10°(9-x/4) × 10* 3 cm I cm 3 cm
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