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A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?
- a)4.80 mm
- b)4.90 mm
- c)4.80 mm2.4.90 mm
- d)80 mm2.4.90 mm3.
Correct answer is option 'C'. Can you explain this answer?
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A 200-kg load is hung on a wire with a length of 4.00 m, a cross-secti...
Cm², and a Young's modulus of 2.00 x 10⁵ N/cm². What is the elongation of the wire under this load?
We can use the formula for the elongation (ΔL) of a wire under tension:
ΔL = (F * L) / (A * E)
Where:
F = force (in N)
L = length (in m)
A = cross-sectional area (in m²)
E = Young's modulus (in N/m²)
First, we need to convert the units of the area and Young's modulus:
A = 0.200 cm² = 0.0000200 m²
E = 2.00 x 10⁵ N/cm² = 2.00 x 10¹⁰ N/m²
Now we can plug in the values:
ΔL = (200 kg * 9.81 m/s² * 4.00 m) / (0.0000200 m² * 2.00 x 10¹⁰ N/m²)
ΔL = 0.0392 m
Therefore, the elongation of the wire under the 200-kg load is 0.0392 m (or 3.92 cm).
We can use the formula for the elongation (ΔL) of a wire under tension:
ΔL = (F * L) / (A * E)
Where:
F = force (in N)
L = length (in m)
A = cross-sectional area (in m²)
E = Young's modulus (in N/m²)
First, we need to convert the units of the area and Young's modulus:
A = 0.200 cm² = 0.0000200 m²
E = 2.00 x 10⁵ N/cm² = 2.00 x 10¹⁰ N/m²
Now we can plug in the values:
ΔL = (200 kg * 9.81 m/s² * 4.00 m) / (0.0000200 m² * 2.00 x 10¹⁰ N/m²)
ΔL = 0.0392 m
Therefore, the elongation of the wire under the 200-kg load is 0.0392 m (or 3.92 cm).
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A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?a)4.80 mmb)4.90 mmc)4.80 mm2.4.90 mmd)80 mm2.4.90 mm3.Correct answer is option 'C'. Can you explain this answer?
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A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?a)4.80 mmb)4.90 mmc)4.80 mm2.4.90 mmd)80 mm2.4.90 mm3.Correct answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?a)4.80 mmb)4.90 mmc)4.80 mm2.4.90 mmd)80 mm2.4.90 mm3.Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?a)4.80 mmb)4.90 mmc)4.80 mm2.4.90 mmd)80 mm2.4.90 mm3.Correct answer is option 'C'. Can you explain this answer?.
A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?a)4.80 mmb)4.90 mmc)4.80 mm2.4.90 mmd)80 mm2.4.90 mm3.Correct answer is option 'C'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?a)4.80 mmb)4.90 mmc)4.80 mm2.4.90 mmd)80 mm2.4.90 mm3.Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 200-kg load is hung on a wire with a length of 4.00 m, a cross-sectional area of 0.200 × 10−4 m2, and a Young’s modulus of 8.00 × 1010N/ m2. What is its increase in length?a)4.80 mmb)4.90 mmc)4.80 mm2.4.90 mmd)80 mm2.4.90 mm3.Correct answer is option 'C'. Can you explain this answer?.
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