NEET Physics Current Electricity Questions Solved

BOARD

You are given two sets of potentiometer circuit to measure the emf E1 of a cell.

Set A: consists of a potentiometer wire of a material of resistivity ρ1, area of cross-section A1 and length l.

Set B: consists of a potentiometer of two composite wires of equal lengths l/2 each, of resistivity ρ1, ρ2 and area of cross-section A1, A2 respectively.

(i) Find the relation between resistivity of the two wires with respect to their area of cross section, if the current flowing in the two sets is same.

(ii) Compare the balancing length obtained in the two sets.                           (3) mark

12 (a)  I =εR+ρ1lA1

 

for Set A

 

 

I =εR+ρ1A1

 

 

for set B

 

 

 

Equating the above two expressions and simplifying = ρ1A1                       (1/2) mark

   

(b) Potential gradient of the potentiometer wire for set A, K= I

Potential drop across the potentiometer wire in set B

 

V = 

I×ρ1lA1=ρ2l2A2

 

V =  

12ρ1A1+ρ2A2l

(1/2)

K’ =12ρ1A1+ρ2A2

, using the condition obtained in part (i)

(1/2)

K’= Iρ1A1 , which is equal to K.

Therefore, balancing length obtained in the two sets is same.             (1/2)   mark

 

               

Difficulty Level:

Crack NEET with Online Course - Free Trial (Offer Valid Till September 24, 2019)