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In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 × 10−3 cm2in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. Compute the elongation.
a)3.7 mm upper, 1.0 mm lower
b)3.4 mm upper, 1.0 mm lower
c)3.5 mm upper, 1.1 mm lower
d)3.1 mm upper, 1.0 mm lower
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
In constructing a large mobile, an artist hangs an aluminum sphere of ...
First see Tension Force at topmost it will be due to both massesT1= (10+6)*9.8 ;;;;;;
,Now tension at the point just below 6 kg is due to 10 kg only so T2=10*9.8;;;;;


T1/A/∆L1/L= Ysteel
Steel becoz elongation steel ki hogi Aluminium,Brass ki nh Not for Aluminium;;;;;
Steel ka remember 200*10^9 Pa is Y
similarly T2/A/∆L2/L=Ysteel again;where
∆L1 elongation of wire 1,,
∆L2elongation of wire 2
calculate them ,brass Aluminium just for confusion
Community Answer
In constructing a large mobile, an artist hangs an aluminum sphere of ...
Given:
Mass of aluminum sphere, m1 = 6.0 kg
Length of vertical steel wire, L = 0.50 m
Cross-sectional area of steel wire, A = 2.5 × 10−3 cm2
Mass of brass cube, m2 = 10.0 kg

To find: Elongation of the wires

Formula used:
Stress, σ = F/A
Strain, ε = ΔL/L
Young's modulus, Y = σ/ε
Force, F = mg

Explanation:
1. Calculation of stress in the wire holding the aluminum sphere:
The weight of the aluminum sphere can be calculated as:
F1 = m1g = 6.0 kg × 9.8 m/s2 = 58.8 N
The stress in the wire holding the aluminum sphere can be calculated as:
σ1 = F1/A = 58.8 N/(2.5 × 10−3 m2) = 2.352 × 107 N/m2

2. Calculation of strain in the wire holding the aluminum sphere:
The wire holding the aluminum sphere will undergo an elongation ΔL1 due to the weight of the sphere. The strain in the wire can be calculated as:
ε1 = ΔL1/L
Since the wire is under tension, the strain will be positive. Let's assume the elongation is in the upward direction, then:
ΔL1 = F1L/Y1
where Y1 is the Young's modulus of the wire holding the aluminum sphere. The Young's modulus of steel is 2 × 1011 N/m2. Therefore:
ΔL1 = (58.8 N)(0.5 m)/(2 × 1011 N/m2) = 1.47 × 10−6 m
ε1 = ΔL1/L = (1.47 × 10−6 m)/0.50 m = 2.94 × 10−6

3. Calculation of stress in the wire holding the brass cube:
The weight of the brass cube can be calculated as:
F2 = m2g = 10.0 kg × 9.8 m/s2 = 98 N
The stress in the wire holding the brass cube can be calculated as:
σ2 = F2/A = 98 N/(2.5 × 10−3 m2) = 3.92 × 107 N/m2

4. Calculation of strain in the wire holding the brass cube:
The wire holding the brass cube will undergo an elongation ΔL2 due to the weight of the cube. The strain in the wire can be calculated as:
ε2 = ΔL2/L
Since the wire is under tension, the strain will be positive. Let's assume the elongation is in the upward direction, then:
ΔL2 = F2L/Y2
where Y2 is the Young's modulus of the wire holding the brass cube. The Young's modulus of steel is 2 × 1011 N/m2. Therefore:
ΔL2 = (98 N)(0.5 m)/(2 × 1011 N/m2) = 2.45 × 10−6 m
ε2 = ΔL2/L = (2.45 × 10−6 m)/0.50 m = 4.9 × 10
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In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 × 10−3 cm2in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. Compute the elongation.a)3.7 mm upper, 1.0 mm lowerb)3.4 mm upper, 1.0 mm lowerc)3.5 mm upper, 1.1 mm lowerd)3.1 mm upper, 1.0 mm lowerCorrect answer is option 'D'. Can you explain this answer?
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In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 × 10−3 cm2in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. Compute the elongation.a)3.7 mm upper, 1.0 mm lowerb)3.4 mm upper, 1.0 mm lowerc)3.5 mm upper, 1.1 mm lowerd)3.1 mm upper, 1.0 mm lowerCorrect answer is option 'D'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 × 10−3 cm2in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. Compute the elongation.a)3.7 mm upper, 1.0 mm lowerb)3.4 mm upper, 1.0 mm lowerc)3.5 mm upper, 1.1 mm lowerd)3.1 mm upper, 1.0 mm lowerCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In constructing a large mobile, an artist hangs an aluminum sphere of mass 6.0 kg from a vertical steel wire 0.50 m long and 2.5 × 10−3 cm2in cross-sectional area. On the bottom of the sphere he attaches a similar steel wire, from which he hangs a brass cube of mass 10.0 kg. Compute the elongation.a)3.7 mm upper, 1.0 mm lowerb)3.4 mm upper, 1.0 mm lowerc)3.5 mm upper, 1.1 mm lowerd)3.1 mm upper, 1.0 mm lowerCorrect answer is option 'D'. Can you explain this answer?.
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