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The resistance of a conductor R = ρl/A ,where ρ = resistivity of the conductor  in Ohm metres

l= length of the conductor in metres and A = area of cross section of the conductor in sq.metres  

Now given l = 10 metres ,  diameter of the conductor = 3 mm.Therefore area of the conductor=

A= 3.14*(3/2)^2* 10^-6 sq.metres = 7.065*10^-6 sq.metres

Also   ρ = 2.65*10^-8 Ohm metres

Therefore resistance of the conductor =  R =  2.65*(10^-8)*10/ (7.065*10^-6)

Simplifying we get R =  0.0375 Ohms or 37.5 milliOhms

Answer:  Option b

 

In the given circuit,there exists a short between terminals “c”  and “d” and so the circuit does not see any resistance beyond the short. Therefore  the resistances 2KΩ and 3KΩ connected beyond the short will be by passed ans should be ignored. The resistance as viewed from the input side( as shown in the sketch) will thus consist only of the resistance 2KΩ in series with the parallel combination of resistances 3KΩ and 6KΩ connected between the terminals a,b,c,d.

The equivalent resistance of the parallel combination of 3KΩ and 6KΩ connected between 

“a”,”b”,”c”    = 6*3/(6+3)= 2KΩ 

Therefore  Req =  Series resistance of 2KΩ (between input terminal and “a”) and the equivalent resistance of the above parallel combination  

That is Req =  2KΩ + 2KΩ  = 4KΩ

Answer: Option a

 

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