Step 1: Understanding the Question:
We are given a flat triangular lamina that can rotate about four distinct axes lying in or perpendicular to its plane ($PR$, $QS$, $QR$, and $PQ$). We need to identify the specific axis about which the lamina exhibits its maximum moment of inertia.
Step 2: Key Formula or Approach:
The moment of inertia $I$ for a collection of continuous mass points is defined by:
$$I = \sum m_i r_i^2$$
where $m_i$ represents elemental masses and $r_i$ represents their perpendicular distances from the chosen axis of rotation. Consequently, the moment of inertia increases when the bulk mass distribution is located farther away from the operational axis.
Step 3: Detailed Explanation:
Let's analyze the geometry of the triangular lamina relative to each option:
An axis running through the bulk body, like the altitude $QS$ or the diagonal segments, keeps the surrounding mass relatively close to the line of rotation, leading to a smaller distribution radius ($r$).
Now consider the baseline edges $PQ$ and $QR$. In a standard triangular layout where $QR$ serves as the primary base opposite to the furthest vertex $P$, choosing $QR$ as the axis means that the mass extending toward vertex $P$ is positioned at the largest average perpendicular distance possible.
Because more mass elements are situated at greater perpendicular distances ($r_i$) relative to the base axis $QR$ compared to any other axis option, the mathematical summation $\sum m_i r_i^2$ reaches its absolute maximum value here.
Step 4: Final Answer:
The moment of inertia is maximum about the axis QR, which corresponds to option (C).