Physics # A light rigid bar is suspended horizontally from two vertical wires, one of steel and one of brass, as shown in figure. Each wire is 2.00 m long. The diameter of the steel wire is 0.60 mm and the length of the bar AB is 0.20 m. When a mass of 10 kg is suspended from the centre of AB bar remains horizontal.i) What is the tension in each wire ? ii) Calculate the extension of the steel wire and the energy stored in it.iii) Calculate the diameter of the brass wire. ​ ➡ See the attachment free body diagram.

=》T₁ + T₂ = mg

• m is the mass of AB bar = 10 kg
• g is gravitational acceleration = 9.8 m/s²

=》 T₁ + T₂ = 10 × 9.8

=》 T₁ + T₂ = 98 N ------(1)

✅ Consider torque at point O (Centre/mid pt. of AB bar).

=》 (T₁ × r) - (T₂ × r) = 0

• r is the length of OA for T₁ & OB for T₂. But both r are same, because O is the midpoint of AB.

=》 r (T₁ - T₂) = 0

=》 T₁ - T₂ = 0

=》 T₁ = T₂ =

The tension in each wire is 49 N.

[NOTE ➡ If we take g = 10 m/s², then the value of tension will be different.]

• E be the Young's modulus of steel = 2.0 × 10¹¹ Pa
• Force = T₁ = 49 N
• r = radius of steel wire = = 0.30 mm = 0.0003 m
• l = 2.00 m.

The extension of the steel wire is 1.737 × 10⁻³ m.

The energy stored is 42.5 × 10⁻³ J

☆ Extension is same in both steel and brass wire. Because the resultant tension is same in both.

• = 1.0 × 10¹¹ Pa
• Force = T₂ = 49 N
• l = 2.00 m
• = 1.737 × 10⁻³ m

The diameter of the brass wire is 8.45 × 10⁻⁴ m.

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