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A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10−2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.
a) 1.0 cm
b) 1.1 cm
c) 0.9 cm
d) 1.2 cm 
Correct answer is option 'A'. Can you explain this answer?
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A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10−2...
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A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10−2...
To calculate the depression at the midpoint of the mild steel wire, we can use the formula for the depression of a stretched wire under the influence of a hanging mass. The formula is given by:

d = (F * L^2) / (4 * A * T)

Where:
d = depression at the midpoint
F = force applied (weight of the hanging mass)
L = length of the wire
A = cross-sectional area of the wire
T = tension in the wire

Let's plug in the given values:

Length of the wire, L = 1.0 m
Cross-sectional area of the wire, A = 0.50 × 10^(-2) cm^2 = 0.50 × 10^(-6) m^2
Force applied, F = 100 g = 0.1 kg (since 1 kg = 1000 g)
Tension in the wire, T = ?

Now, we need to find the tension in the wire. Since the wire is stretched horizontally between two pillars and the mass is suspended at the midpoint, the tension in the wire is equal at both ends and acts vertically upwards. It can be calculated using the formula:

T = (F / 2)

T = (0.1 kg) / 2 = 0.05 kg

Now, let's substitute the values into the depression formula:

d = (0.1 * 1.0^2) / (4 * 0.50 × 10^(-6) * 0.05)

d = (0.1) / (4 * 0.50 × 10^(-6) * 0.05)

d = (0.1) / (2 * 0.50 × 10^(-6) * 0.05)

d = (0.1) / (0.50 × 10^(-6) * 0.10)

d = (0.1) / (0.50 × 10^(-7))

d = (0.1) / (5 × 10^(-8))

d = 2 × 10^(6) m = 2 cm

Therefore, the depression at the midpoint of the mild steel wire is 2 cm.

However, in the given options, the correct answer is given as 1.0 cm. It seems there might be an error in the question or the options provided. Based on the calculations, the correct answer should be 2 cm.
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A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10−2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.a)1.0 cmb)1.1 cmc)0.9 cmd)1.2 cmCorrect answer is option 'A'. Can you explain this answer?
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A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10−2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.a)1.0 cmb)1.1 cmc)0.9 cmd)1.2 cmCorrect answer is option 'A'. 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 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10−2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.a)1.0 cmb)1.1 cmc)0.9 cmd)1.2 cmCorrect answer is option 'A'. 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 A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10−2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.a)1.0 cmb)1.1 cmc)0.9 cmd)1.2 cmCorrect answer is option 'A'. Can you explain this answer?.
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