Question Video: Determining the Resistance of the Rod after Being Pulled Uniformly | Nagwa Question Video: Determining the Resistance of the Rod after Being Pulled Uniformly | Nagwa

# Question Video: Determining the Resistance of the Rod after Being Pulled Uniformly Physics • Third Year of Secondary School

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If a cylindrical metallic rod of cross-sectional area 3 cm² and resistance 10 Ω is pulled uniformly until its cross-sectional area becomes 1.5 cm², what will the resistance of the rod be?

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### Video Transcript

If a cylindrical metallic rod of cross-sectional area three square centimeters and resistance 10 ohms is pulled uniformly until its cross-sectional area becomes 1.5 square centimeters, what will the resistance of the rod be?

In this question, we are being asked how the resistance of a metallic rod will change if we reduce its cross-sectional area uniformly by pulling the rod. We can begin by drawing a diagram of the rod as it is initially. We know that it’s cylindrical. We’ve also been told that before we stretch the rod, the cross-sectional area is three square centimeters. We can label this as the area 𝐴 one. The rod will also have some length, which we haven’t been told. So let’s call this 𝑙 one. Finally, we’ve been given the resistance of the rod before stretching, which is 10 ohms. Let’s call this 𝑅 one.

Next, we’re told that the rod is uniformly pulled until its cross-sectional area changes. This means the rod is carefully stretched from both ends without any deformation or bending in the cross section. In other words, the cross section is still circular and the rod is still straight along its length. However, the new cross-sectional area is smaller than before. Let’s label this as 𝐴 two. We’ve been told that 𝐴 two is now 1.5 square centimeters.

Now, if we’ve stretched the rod, then its cross-sectional area will’ve decreased, but the length of the rod will have increased. It will’ve gotten longer. We can calculate how much longer the rod is compared to before. We can do this because even though the rod has been stretched, the volume of the rod will not have changed. The amount of metal making up the rod has not changed as we have not cut anything off or added anything.

So, we can say that the volume of the rod before stretching, 𝑉 one, must be equal to the volume of the rod after stretching, 𝑉 two. The volume, 𝑉, of a cylindrical rod like this is equal to the cross-sectional area, 𝐴, multiplied by the length, 𝑙. So, 𝑉 one is the same as 𝐴 one times 𝑙 one, and 𝑉 two is the same as 𝐴 two times 𝑙 two. Remember, these are equal. So, we can write 𝐴 one 𝑙 one is equal to 𝐴 two 𝑙 two.

We can now rearrange this to find that 𝑙 two is equal to 𝐴 one over 𝐴 two all multiplied by 𝑙 one. Substituting in our values for the initial and final cross-sectional areas, we find that 𝑙 two is equal to three square centimeters divided by 1.5 square centimeters all multiplied by 𝑙 one. Since the units of square centimeters cancel out, we find that 𝑙 two is two times larger than 𝑙 one. The stretching process halved the cross-sectional area of the wire but doubled the length.

So, how can we use this information to find the new resistance of the wire? Well, we can recall that the material that the wire is made out of will have a property known as resistivity. The resistivity, 𝜌, of a material is equal to the resistance, 𝑅, of a piece of that material multiplied by the cross-sectional area, 𝐴, of that piece divided by its length, 𝑙. We can apply this to our wire before and after it gets stretched. And importantly, because resistivity is a property of the material that the wire is made out of, it must be the same before and after the material is stretched. This is because the stretching process does not turn the metal that the wire is made out of into something else. In other words, 𝜌 one is equal to 𝜌 two.

Let’s now use this equation to find an expression for the resistivities of the wire before and after it gets stretched. 𝜌 one, the resistivity before stretching, is equal to the resistance before stretching 𝑅 one multiplied by the cross-sectional area 𝐴 one divided by the length 𝑙 one. And similarly, 𝜌 two is equal to 𝑅 two 𝐴 two over 𝑙 two. We know these two quantities must be equal. So 𝑅 one 𝐴 one over 𝑙 one is equal to 𝑅 two 𝐴 two over 𝑙 two.

Finally, we can substitute in all our known quantities. We know the resistance before stretching, 𝑅 one, is 10 ohms. We also know the areas before and after stretching. And lastly, even though we don’t know the length of the wire before or after stretching, we do know that 𝑙 two is equal to two 𝑙 one. We can substitute this in and then multiply both sides of the equation by 𝑙 one so that it cancels on both sides. In other words, it doesn’t matter that we didn’t know the length of the wire as long as we could calculate how the length changed after stretching.

Finally, we can rearrange the equation to solve for 𝑅 two, the resistance of the wire after stretching. If we evaluate this expression, we see that the units of square centimeters cancel out in the numerator and denominator. This is good because we’re left with a unit of ohms, which is exactly what we want since we’re calculating a resistance. We find that the resistance of the wire after stretching is 40 ohms, and this is the final answer to our question. The resistance of the wire after being pulled uniformly is 40 ohms.

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