A 0.75-m-long piano wire (radius of 0.5 mm) has one end fixed and the other end attached to a rotatable tuning peg (having a radius of 2 mm). The tuning peg is rotated so that the wire is just taut, and then the peg is given three full turns (tightening the wire). If the Young's modulus is $2 \times 10 ^ { 11 }$ Pa, the tension in the piano wire is
A) 7900 N.
B) 6600 N.
C) 5600 N.
D) 4400 N.

Question

Quizplus · Accepted Answer

The tension in the wire can be calculated using the formula for the change in length due to twisting: $\Delta L = \frac{2 \pi R n}{L}$, where $R$ is the radius of the peg, $n$ is the number of turns, and $L$ is the original length of the wire. The tension $T$ is then given by $T = \frac{Y \cdot A \cdot \Delta L}{L}$, where $Y$ is Young's modulus and $A$ is the cross-sectional area of the wire. Substituting the given values, the tension is approximately 7900 N.

A 075-M-Long Piano Wire (Radius of 0 2×10112 \times 10 ^ { 11 }2×1011

A 075-M-Long Piano Wire (Radius of 0 $2 \times 10 ^ { 11 }$