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In Searle's experiment to find Young's modulus, the diameter of wire is measured as D=0.5cm, the length of the wire is L=125cm and a weight, m=20kg is the put, extension in wire was found to be 0.100cm.Find the maximum permissible error inYoung's modulus.?
Verified Answer
In Searle's experiment to find Young's modulus, the diameter of wire i...
Mg / ( (π d^2) /4) = Y * (x / l)
=> Y = 
(dY/Y)max = (Δm/m) + (Δl/l) + 2(Δd/d) + (Δx/x) 
m= 20.0kg => Δm = 0.1 kg
l = 125cm ⇒ Δl = 1cm
d = 0.50cm ⇒ Δd = 0.01cm
x = 0.100cm ⇒ Δx = 0.001cm

(dY/Y)max = ( (0.1/20.0) + (1cm/125cm) + 2(0.01cm/0.50cm) + (0.001cm/0.100cm) ) *100
                   => 6.3 %
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In Searle's experiment to find Young's modulus, the diameter of wire i...
Calculation of Young's Modulus


  • Diameter of wire: D = 0.5 cm

  • Length of wire: L = 125 cm

  • Weight applied: m = 20 kg

  • Extension in wire: x = 0.100 cm


Young's modulus (Y) is given by:

Y = (m * g * L) / (π * (D/2)^2 * x)

where g is the acceleration due to gravity (9.8 m/s^2).


Calculation of Y with given values


  • m * g * L = 20 * 9.8 * 125 = 24,500 N

  • D/2 = 0.25 cm

  • (D/2)^2 = 0.0625 cm^2

  • π * (D/2)^2 = 0.196 cm^2

  • Y = 24,500 / (0.196 * 0.100) = 125,000 N/cm^2



Calculation of Maximum Permissible Error

The formula for maximum permissible error (ΔY) in Young's modulus is:

ΔY = Y * [(Δm/m) + 2*(ΔD/D) + (Δx/x)]

where Δm, ΔD, and Δx are the errors in the measurements of m, D, and x, respectively.

Assuming a 1% error in each measurement:


  • Δm = 0.01 * 20 = 0.2 kg

  • ΔD = 0.01 * 0.5 = 0.005 cm

  • Δx = 0.01 * 0.100 = 0.001 cm


Substituting these values:

ΔY = 125,000 * [(0.2/20) + 2*(0.005/0.5) + (0.001/0.100)] = 6,250 N/cm^2


Explanation

Young's modulus is a measure of the stiffness of a material. In Searle's experiment, the Young's modulus of a wire was determined by measuring the extension in the wire when a weight was applied. The formula for Young's modulus involves measurements of weight, length, diameter, and extension. To account for errors in these measurements, the maximum permissible error in Young's modulus can be calculated using the formula ΔY = Y * [(Δm/m) + 2*(ΔD/D) + (Δx/x)]. In this case, assuming a 1% error in each measurement, the maximum permissible error was found to be 6,250 N/cm^2.
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In Searle's experiment to find Young's modulus, the diameter of wire is measured as D=0.5cm, the length of the wire is L=125cm and a weight, m=20kg is the put, extension in wire was found to be 0.100cm.Find the maximum permissible error inYoung's modulus.?
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In Searle's experiment to find Young's modulus, the diameter of wire is measured as D=0.5cm, the length of the wire is L=125cm and a weight, m=20kg is the put, extension in wire was found to be 0.100cm.Find the maximum permissible error inYoung's modulus.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about In Searle's experiment to find Young's modulus, the diameter of wire is measured as D=0.5cm, the length of the wire is L=125cm and a weight, m=20kg is the put, extension in wire was found to be 0.100cm.Find the maximum permissible error inYoung's modulus.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In Searle's experiment to find Young's modulus, the diameter of wire is measured as D=0.5cm, the length of the wire is L=125cm and a weight, m=20kg is the put, extension in wire was found to be 0.100cm.Find the maximum permissible error inYoung's modulus.?.
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