Q 1 :    

A wire of length 10 cm and radius 7×10-4 m connected across the right gap of a meter bridge. When a resistance of 4.5Ω is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm from the left end. If the resistivity of the wire is R×10-7Ωm, then value of R is       [2024]

  • 63

     

  • 70

     

  • 66

     

  • 35

     

(C)            For null point, 4.560=R'40

                 Also, R'=ρlA=ρlπr2

                 4.5×40=ρ×0.1π×7×10-8×60

                ρ=66×10-7Ω×m

 



Q 2 :    

Wheatstone bridge principle is used to measure the specific resistance (S1) of given wire, having length L, radius r. If X is the resistance of wire, then specific resistance is; S1=X(πr2L).

If the length of the wire gets doubled then the value of specific resistance will be:                  [2024]

  • S1/4

     

  • 2S1

     

  • S1/2

     

  • S1

     

(D)            As specific resistance does not depend on dimension of wire so, it will not change.