A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of:

We have to compare the mechanical stress on the leg before and after the man becomes a giant. In mechanics, stress is defined first:

$$\sigma=\frac{F}{A}$$

where $$\sigma$$ is the stress, $$F$$ is the force acting on the area (here the weight of the body that the legs must support), and $$A$$ is the area over which that force is distributed (the cross-sectional area of the legs).

The problem states that all linear dimensions (height, width, depth, diameter of the leg, etc.) are enlarged by a common factor of $$9$$. Let us call this scaling factor $$k$$, so here $$k = 9$$.

Step 1: How does the volume change?

The volume of any three-dimensional object is proportional to the cube of a characteristic length. Because every linear dimension grows by $$k$$, the new volume $$V'$$ becomes

$$V' = k^3 V = 9^3 V = 729\,V.$$

Step 2: How does the mass (and hence weight) change?

The density is given to remain the same. Mass is density times volume, so if volume increases by the factor $$729$$, mass increases by exactly the same factor:

$$m' = 729\,m.$$

The weight is the gravitational force $$F = mg$$. Therefore, the new weight $$F'$$ is

$$F' = m' g = 729\,m g = 729\,F.$$

Step 3: How does the supporting area of the leg change?

The cross-section of the leg is a two-dimensional quantity, so it scales with the square of the linear factor:

$$A' = k^2 A = 9^2 A = 81\,A.$$

Step 4: Compute the new stress.

Using the stress formula on the enlarged man, we have

$$\sigma' = \frac{F'}{A'} = \frac{729\,F}{81\,A}.$$

We now separate the numerical factor:

$$\sigma' = \frac{729}{81}\,\frac{F}{A} = 9\,\sigma.$$

So the stress in the giant’s legs is nine times the original stress.

Hence, the correct answer is Option B.