This question asks about the work done by the string on a stone when it's whirled in a circle. To understand this, let's break down the concept of work and the physics involved in circular motion.
In physics, work (\(W\)) is done when a force causes a displacement. Mathematically, work is calculated as the dot product of the force vector (\(\vec{F}\)) and the displacement vector (\(\vec{d}\)):
\(W = \vec{F} \cdot \vec{d} = Fd\cos\theta\)
Where:
Importantly, work is only done if the force has a component along the direction of displacement. If the force is perpendicular to the displacement (\(\theta = 90^\circ\)), the value of \(\cos(90^\circ) = 0\), and thus, no work is done.
Consider a stone tied to a string and being whirled in a horizontal circle:
Using the work formula \(W = Fd\cos\theta\):
Since the angle \(\theta\) between the tension force exerted by the string and the instantaneous displacement of the stone is \(90^\circ\), we have:
\(W = (\text{Tension}) \times (\text{Displacement}) \times \cos(90^\circ)\)
\(W = F \times d \times 0\)
\(W = 0\)
Because the force applied by the string (tension) is always perpendicular to the direction of the stone's movement (displacement) at every point on the circular path, the work done by the string on the stone is zero. The kinetic energy of the stone remains constant (assuming the speed doesn't change), but no energy is transferred *by the string* in the form of work.
Therefore, the correct option is that the work done is zero.
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