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This section contains 5 questions. The answer to each of the questions is a single digit integer ranging from 0 to 9. In Searle’s experiment, which is used to find Young’s Modulus of elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a scale of least count 0.001 cm) and length is L = 110 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer of least count 0.001 cm). The maximum possible error in the values of Young’s modulus is (x + 2) × 10 9 N/m 2 . Find the value of x. Screw gauge and meter scale are free from error.



This section contains 5 questions. The answer to each of the questions is a single digit integer ranging from 0 to 9.


 






In Searle’s experiment, which is used to find Young’s Modulus of elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a scale of least count 0.001 cm) and length is L = 110 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of X = 0.125 cm (measured by a micrometer of least count 0.001 cm). The maximum possible error in the values of Young’s modulus is (x + 2) × 109 N/m2. Find the value of x. Screw gauge and meter scale are free from error.



 

Grade:upto college level

1 Answers

Radhika Batra
247 Points
7 years ago

9

    Maximum percentage error in Y is given by

    Y = W/((?D^2)/4)×L/X

    (?Y/Y)_max=2(?D/D)+?X/X+?L/L

        = 2(0.001/0.05)+(0.001/0.125)+(0.1/110) = 0.0489

    So, maximum percentage error = 4.89%

    It is given that

    W = 50 N; D = 0.05 cm; = 0.05 × 10–2 m;

    X = 0.125 cm = 0.125 × 10–2m;

    L = 110 cm = 110 × 10–2 m

    Y = (50×4×110×?10?^(-2))/(3.14(0.05×?10?^(-2) )×(0.125×?10?^(-2) ) )

    = 2.24 × 1011 N/m2

    ? ?Y = 0.0489 × 2.24 × 1011 = 1.09 × 1010 N/m2

    = 11 × 109 N/m2 = (9 + 2) × 109 N/m2

    ? x = 9

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