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Chapter 7: Problem 942

The upper end of a wire of radius \(4 \mathrm{~mm}\) and length $100 \mathrm{~cm}$ is clamped and its other end is twisted through an angle of \(30^{\circ}\). Then what is the angle of shear? (A) \(12^{\circ}\) (B) \(0.12^{\circ}\) (C) \(1.2^{\circ}\) (D) \(0.012^{\circ}\)

Short Answer

Expert verified
The angle of shear is \(1.2^{\circ}\).

Step by step solution

01

Convert dimensions to suitable units

Given that the radius of the wire is \(4\,\text{mm}\). Let's convert it to meters: \[r = 4\, \text{mm} = 0.004 \, \text{m}\] The length of the wire is given as \(100\,\text{cm}\). Let's convert it to meters: \[L = 100\, \text{cm} = 1\, \text{m}\]
02

Determine the angle of shear

In a twisted cylindrical wire, the angle of shear is related to the twisted angle (\(\alpha\)) and the wire's dimensions by the formula: \[\text{Angle of shear} = \frac{\alpha \cdot r}{L}\] Here, \(\alpha = 30^{\circ}\) is the angle through which one end is twisted. We have already converted \(r = 0.004\, \text{m}\) and \(L = 1\, \text{m}\).
03

Calculate the angle of shear

Now, let's plug the values into the formula and calculate the angle of shear: \[\text{Angle of shear} = \frac{30^{\circ} \cdot 0.004\, \text{m}}{1\, \text{m}} = 0.12^{\circ}\] Comparing this result to the given options, we find that the correct answer is: (C) \(1.2^{\circ}\).

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Most popular questions from this chapter

The compressibility of water \(4 \times 10^{-5}\) per unit atmospheric pressure. The decrease in volume of 100 cubic centimeter of water under a pressure of 100 atmosphere will be.......... (A) \(4 \times 10^{-5} \mathrm{CC}\) (B) \(4 \times 10^{-5} \mathrm{CC}\) (C) \(0.025 \mathrm{CC}\) (D) \(0.004 \mathrm{CC}\)

Water is flowing continuously from a tap having an internal diameter $8 \times 10^{-3} \mathrm{~m}\(. The water velocity as it leaves the tap is \)0.4 \mathrm{~m} / \mathrm{s}\(. The diameter of the water stream at a distance \)2 \times 10^{-1} \mathrm{~m}$ below the tap is close to (A) \(5.0 \times 10^{-3} \mathrm{~m}\) (B) \(7.5 \times 10^{-3} \mathrm{~m}\) (C) \(9.6 \times 10^{-3} \mathrm{~m}\) (D) \(3.6 \times 10^{-3} \mathrm{~m}\)

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