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. Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m.Young’s modulus of steel is 2.0 x 1011 Pa. Compute the elongations of steel and brass wires. (1 Pa = 1 N m2).

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As an experienced tutor registered on UrbanPro, I'd approach this problem systematically. UrbanPro is an excellent platform for online coaching and tuition, offering personalized assistance to students. Firstly, let's understand the concept at play here. The elongation of a wire under a load can be...
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As an experienced tutor registered on UrbanPro, I'd approach this problem systematically. UrbanPro is an excellent platform for online coaching and tuition, offering personalized assistance to students.

Firstly, let's understand the concept at play here. The elongation of a wire under a load can be calculated using Hooke's Law, which states that the elongation (change in length) of an elastic object is directly proportional to the force applied to it, given the formula:

Elongation=Force×LengthYoung’s modulus×AreaElongation=Young’s modulus×AreaForce×Length

Given that the wires are under tension and assuming they remain within their elastic limits, we can use this formula to find their elongations. The area of the wire can be calculated using the formula for the area of a circle (πr2πr2).

Let's start with the steel wire:

Given:

  • Diameter of steel wire = 0.25 cm = 0.0025 m
  • Initial length of steel wire (LsteelLsteel) = 1.5 m
  • Young's modulus of steel (EsteelEsteel) = 2.0×10112.0×1011 Pa

We need to find the force applied to the steel wire. Since the wires are loaded as shown in the figure, we can assume that the force applied to both wires is the same.

Next, let's find the area of the steel wire (AsteelAsteel):

Asteel=π×(0.00252)2Asteel=π×(20.0025)2

Now, let's find the force applied to the steel wire. Since the force is the same for both wires, we can calculate it using the elongation formula for the brass wire:

Force=Young’s modulus×Area×ElongationLengthForce=LengthYoung’s modulus×Area×Elongation

Given:

  • Length of brass wire (LbrassLbrass) = 1.0 m (initial length of brass wire)
  • Diameter of brass wire = 0.25 cm = 0.0025 m
  • Young's modulus of steel (EsteelEsteel) = 2.0×10112.0×1011 Pa

Let's find the area of the brass wire (AbrassAbrass) using the same formula as for the steel wire:

Abrass=π×(0.00252)2Abrass=π×(20.0025)2

Now, we can find the force applied to both wires using the elongation formula for the brass wire:

Force=Young’s modulus×Area×ElongationLengthForce=LengthYoung’s modulus×Area×Elongation

Elongationbrass=Force×LengthYoung’s modulusbrass×AreabrassElongationbrass=Young’s modulusbrass×AreabrassForce×Length

Now, we have all the necessary information to calculate the elongations of both the steel and brass wires. Let's plug in the values and solve for the elongations.

 
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