Mathematics in Physics

JEE Main level, single correct, medium

85 of 121

Question 85 of 121

A student determined Young's Modulus of elasticity using the formula Y=MgL34bd3δ.Y=\dfrac{Mg{L}^{3}}{4{bd}^{3}\delta }. The value of gg is taken to be 9.8ms−29.8m{s}^{-2} without any significant error, his observations are as following. Physical Quantity Least count of the Equipment used for measurement Observed Value Mass (M)(M) 1g1g 2kg2kg Length of bar (L)(L) 1mm1mm 1m1m Breadth of bar (b)(b) 0.1mm0.1mm 4cm4cm Thickness of bar (d)(d) 0.01mm0.01mm 0.4cm0.4cm Depression (δ)(\delta ) 0.01mm0.01mm 5mm5mm Then the fractional error in the measurement of YY is :

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Answer: (D) 0.01550.0155

Y=MgL34bd3δY=\dfrac{Mg{L}^{3}}{4{bd}^{3}\delta }
∆YY=∆MM+3∆l1+∆bb+3∆dd+∆δδ⇒∆YY=0.0005+3×0.001+0.0025+3×0.0025+0.002=0.0155\dfrac{∆Y}{Y}=\dfrac{∆M}{M}+3\dfrac{∆l}{1}+\dfrac{∆b}{b}+3\dfrac{∆d}{d}+\dfrac{∆\delta }{\delta } \Rightarrow \dfrac{∆Y}{Y}=0.0005+3\times 0.001+0.0025+3\times 0.0025+0.002=0.0155

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