Answer:
Option d is correct.
Explanation:
We know , resistance of a body is directly proportional to its length and inversely proportional to its area.
( Here,
is constant dependent on object material )
Writing
also :
( since they are of same material therefore,
is same.)
Now , if
.
Then 
Therefore, option d. is correct.
Hence, this is the required solution.
Conductivity is required for the electric current to flow.
So....
imagine the x and y axis or draw it on a piece of paper, a white board or even on your wall if you like. I'm kidding just draw it on a piece of paper
50m up and 20m down
the dog's displacement is 50 - 20 which is 30
your answer is 30m north
To solve this problem we will apply the concept of voltage given by Coulomb's laws. From there we will define the charges and the distance, and we will obtain the total value of the potential difference in the system.
The length of diagonal is given as

The distance of the center of the square from each of the corners is

The potential electric at the center due to each cornet charge is




The total electric potential at the center of the given square is


Al the charges are equal, and the distance are equal to a, then


Therefore the correct option is E.