wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

When a solution of specific conductance 1.342 Sm1 was placed in a conductivity cell with parallel electrodes, the resistance was found to be 170.5 ohm. The area of the electrodes is 1.86×104 m2 . Calculate the distance between the electrodes.

A
0.0452 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
0.425 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4.25 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
42.5 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 0.0452 m
As per the equation:
G=1R (G is conductance ) ……… (1)
κ=G×lA ………….. (2)
From 1 and 2 we can also write: κ=1R×lA …………. (3)
Putting the values in equation 3:
1.342=1170.5×l1.86×104
l=1.342×170.5×1.86×104
On solving we get;
l=0.0425 m


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Resistivity and Conductivity
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon