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A wire can be broken by applying a load of \(20 \mathrm{~kg}\) wt. The force required to break the wire of twice the diameter is (A) \(20 \mathrm{~kg}\) wt (B) \(5 \mathrm{~kg} \mathrm{wt}\) (C) \(80 \mathrm{~kg} \mathrm{wt}\) (D) \(160 \mathrm{~kg} \mathrm{wt}\)

Short Answer

Expert verified
The force required to break the wire of twice the diameter is found by using the relationship between breaking strength and diameter. The breaking strength of a wire is proportional to the square of its diameter. By substituting the known values and solving for the unknown force, we find that the force required to break the wire of twice the diameter is 80 kg wt. So, the correct answer is (C) \(80\mathrm{~kg~wt}\).

Step by step solution

01

Understand the relationship between breaking strength and diameter

The breaking strength of a wire is proportional to the square of its diameter. This means that if the diameter of the wire increases, the breaking strength will increase by the square of that increase.
02

Calculate the ratio of the diameters

Let the diameter of the original wire be \(d_1\) and the diameter of the wire twice the size be \(d_2\). Since the diameter of the second wire is twice the diameter of the original wire, we can write: \(d_2 = 2d_1\)
03

Find the ratio of breaking strengths

Since breaking strength is proportional to the square of the diameter, we can write the relationship between the breaking strengths of the two wires as: \(\frac{S_2}{S_1} = \frac{(d_2)^2}{(d_1)^2}\) Where \(S_1\) is the breaking strength of the original wire, and \(S_2\) is the breaking strength of the wire with twice the diameter.
04

Substitute the values

We are given that the force required to break the original wire is 20 kg wt, which means \(S_1 = 20 \mathrm{~kg} \mathrm{wt}\). We also know that \(d_2 = 2d_1\). Substitute these into the previous equation: \[\frac{S_2}{20} = \frac{(2d_1)^2}{(d_1)^2}\]
05

Solve for the force required to break the wire of twice the diameter

Now, solve the equation for \(S_2\): \[S_2 = 20 \cdot \frac{(2d_1)^2}{(d_1)^2}\] \[S_2 = 20\cdot \frac{(2)^2}{1}\] \[S_2 = 20\cdot 4\] \[S_2 = 80\] The force required to break the wire of twice the diameter is \(80\mathrm{~kg~wt}\). So, the correct answer is (C) \(80\mathrm{~kg~wt}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Breaking strength of wire
Breaking strength is a crucial attribute in physics that describes the point at which a material will fail under tension. Imagine pulling both ends of a wire until it snaps - that's what we mean by breaking strength. In the case of wires, one fundamental factor influencing their breaking strength is the cross-sectional area, which in turn is dependent on the diameter of the wire.

For a cylindrical wire, the cross-sectional area is given by the area of a circle, which is computed using the formula \( A = \pi d^2 / 4 \) where \( d \) is the diameter of the wire, and \( \pi \) is a constant (approximately 3.14159). Since the breaking strength is directly proportional to the cross-sectional area, if you double the diameter, the area—and thereby the strength—quadruples (because \( (2d)^2 = 4d^2 \) ). This relationship is fundamental in ensuring materials are used appropriately in construction, manufacture, and engineering fields for safety and resilience.
Relationship between diameter and strength
In physics, it's essential to understand how changes in dimensions of a material affect its properties. Specifically, for a wire, the diameter directly influences its strength, and this is a practical consideration in numerous applications like bridges, electrical wires, and even medical devices.

The strength we are talking about here refers not just to resistance to breaking under tension, but also to other structural qualities like durability and resilience under various loads. As mentioned earlier, if you increase the diameter of a wire, it doesn't just become 'a little' stronger, it becomes significantly stronger, proportional to the square of the increase in diameter. This is a nonlinear relationship where a small increase in diameter can lead to a much larger increase in strength. It's something that designers and engineers must account for when selecting materials for specific applications to ensure that structures and devices are not only able but more than competent to handle the expected stresses.
Proportional relationships in physics
Physics is filled with proportional relationships that describe how two quantities vary in relation to one another. In the context of the breaking strength of a wire, the relationship is quadratic, meaning the strength increases exponentially with a linear increase in diameter. Proportional relationships are not always intuitive; for example, doubling the diameter leads to a fourfold increase in strength, not double, as one might initially think.

Understanding these relationships is vital for problem-solving in physics. It allows us to create equations and formulas that predict outcomes in various physical situations, ensuring we can design and build with confidence. Recognizing these patterns in proportionalities also helps in simplifying complex problems by reducing them to more manageable and calculable relationships, thereby aiding students and professionals alike in their calculations and design processes.

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Most popular questions from this chapter

One end of a uniform wire of length \(L\) and of weight \(W\) is attached rigidly to a point in the roof, and a weight \(W_{1}\) is suspended from its lower end. If \(S\) is the area of cross-section of the wire, the stress in the wire at a height ( \(3 L / 4\) ) from its lower end is (A) \(W_{1} / S\) (B) \(\left[W_{1}+(W / 4)\right] / S\) (C) \(\left[W_{1}+(3 W / 4)\right] / S\) (D) \(\left[W_{1}+W\right] / S\)

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