Answer:
Option B is correct.
The leads should be attached to the opposite faces that measure 3 cm × 5 cm to obtain the maximum possible resistance.
Explanation:
The resistance of a material is given by
R = (ρL)/A
where ρ = resistivity of the material
A = Cross sectional Area through which current would flow
L = length of the material.
From the relation, it is evident that the resistance of a material is directly proportional to its length and inversely proportional to its cross sectional Area.
As the length of material increases, the resistance of the material also increases, and a decreasing length translates to a decreasing resistance too. (Direct Proportionality)
And as the cross sectional Area of the material increases, the resistance of the material decreases. Decrease in cross sectional Area of the material translates to an increase in resistance. (Inverse Proportionality)
To maximize resistance of a material, it would make sense to maximize the length and minimize the cross sectional Area of that material. (Since resistivity is assumed to be constant)
And with a dimension of (3 × 5 × 8), the longest length is 8 cm and the smallest cross sectional Area is (3 × 5).
So, the leads should be atrached to the face with area (3 cm × 5 cm), which is the smallest cross sectional Area and gives the largest length between the faces (8 cm).
This subsequently maximizes the resistance.
Hope this Helps!!!
