21∂Y/Y = 4*10-3 + 1/1100 + 1/125
In searle's apparatus, diameter of the wire was measured 0.05 cm by screw gauge of least count 0.0001 cm.
length of wire was measured 110 cm by meter scale of least count .1 cm. An external load of 50N was applied. The extension in the length of the wire was 0.125 cm by micrometer of least count 0.001 cm.
Fidn the maximum possible error in measurement of young's modulus.
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6 Answers
1
E is the Young's modulus (modulus of elasticity)
F is the force applied to the object;
A0 is the original cross-sectional area through which the force is applied;
ΔL is the amount by which the length of the object changes;
L0 is the original length of the object.
the maximun error in calculation of each instrument is its lest count so ..
∂E/E = .1 /110 + 2 * 0.0001/0.05 + 0.001/0.125 = .0129 or 1.29 %
E = 50 * 1.10 / 0.00125 * Π(0.0005/2)2 = (2.24 ± .03 ) * 1011
21woh kya??
maine to dY/Y ka xpresion hi ddiya hai..........
usme bhi koi galti hai kya??????/
24method is 100% correct..........find the answer from last step.....