The surface area of a piece of gold is directly proportional to the square of its weight. By what ratio should the weight be adjusted so as to double the surface area.
surface area= k w^2
w=sqrt (area/k)
double surface area.
w'=w sqrt2
Thanks.I had a similar question in my recent exam. I followed a similar procedure...Made a silly error though at the end,forgot to square root..I ended up with w^2'=2w^2.
To find the ratio by which the weight should be adjusted, we need to understand the relationship between surface area and weight as described in the question. According to the problem, the surface area (A) of the gold is directly proportional to the square of its weight (W). Mathematically, we can represent this relationship as:
A ∝ W^2
Now, let's consider the ratio by which the weight should be adjusted to double the surface area. If we denote the initial weight as W1 and the final weight as W2, and the initial surface area as A1 and the doubled surface area as A2, we can set up the following relationship:
A1 : A2 = (W1^2) : (W2^2)
Since we want to find the ratio of weight adjustment, we can rewrite the equation as:
(W1^2) : (W2^2) = 1 : 2
To simplify the equation, we can take the square root of both sides:
W1 : W2 = 1 : √2
Hence, the ratio by which the weight should be adjusted to double the surface area is approximately 1 : √2.