Question:
A student determined Young's Modulus of elasticity using the formula \(Y = \frac{MgL^3}{4bd^3\delta}\). The value of g is taken to be 9.8 m/s2, without any significant error, his observation are as following.Physical Quantity Least count of the Equipment used for measurement Observed value Mass (M) 1 g 2 kg Length of bar (L) 1 mm 1 m Breadth of bar (b) 0.1 mm 4 cm Thickness of bar (d) 0.01 mm 0.4 cm Depression (\(\delta\)) 0.01 mm 5 mm
Then the fractional error in the measurement of Y is :
A student determined Young's Modulus of elasticity using the formula \(Y = \frac{MgL^3}{4bd^3\delta}\). The value of g is taken to be 9.8 m/s2, without any significant error, his observation are as following.
Then the fractional error in the measurement of Y is :
| Physical Quantity | Least count of the Equipment used for measurement | Observed value |
| Mass (M) | 1 g | 2 kg |
| Length of bar (L) | 1 mm | 1 m |
| Breadth of bar (b) | 0.1 mm | 4 cm |
| Thickness of bar (d) | 0.01 mm | 0.4 cm |
| Depression (\(\delta\)) | 0.01 mm | 5 mm |
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For a product-quotient formula, the fractional errors add with the powers as multipliers: \(\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + 3\frac{\Delta L}{L} + \frac{\Delta b}{b} + 3\frac{\Delta d}{d} + \frac{\Delta \delta}{\delta}\). Convert all quantities to the same unit first.
Updated On: Nov 4, 2026
- 0.0083
- 0.0155
- 0.155
- 0.083
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The Correct Option is B
Solution and Explanation
\(Y = \frac{MgL^3}{4bd^3\delta}\), so \(\frac{\Delta Y}{Y} = \frac{\Delta M}{M} + 3\frac{\Delta L}{L} + \frac{\Delta b}{b} + 3\frac{\Delta d}{d} + \frac{\Delta\delta}{\delta}\) (g, 4 treated as exact).
\(\frac{\Delta M}{M} = \frac{1\text{ g}}{2000\text{ g}} = 0.0005\)
\(3\frac{\Delta L}{L} = 3\times\frac{1\text{ mm}}{1000\text{ mm}} = 0.003\)
\(\frac{\Delta b}{b} = \frac{0.1\text{ mm}}{40\text{ mm}} = 0.0025\)
\(3\frac{\Delta d}{d} = 3\times\frac{0.01\text{ mm}}{4\text{ mm}} = 0.0075\)
\(\frac{\Delta \delta}{\delta} = \frac{0.01\text{ mm}}{5\text{ mm}} = 0.002\)
Sum \(= 0.0005 + 0.003 + 0.0025 + 0.0075 + 0.002 = 0.0155\).
Answer: (2) 0.0155
\(\frac{\Delta M}{M} = \frac{1\text{ g}}{2000\text{ g}} = 0.0005\)
\(3\frac{\Delta L}{L} = 3\times\frac{1\text{ mm}}{1000\text{ mm}} = 0.003\)
\(\frac{\Delta b}{b} = \frac{0.1\text{ mm}}{40\text{ mm}} = 0.0025\)
\(3\frac{\Delta d}{d} = 3\times\frac{0.01\text{ mm}}{4\text{ mm}} = 0.0075\)
\(\frac{\Delta \delta}{\delta} = \frac{0.01\text{ mm}}{5\text{ mm}} = 0.002\)
Sum \(= 0.0005 + 0.003 + 0.0025 + 0.0075 + 0.002 = 0.0155\).
Answer: (2) 0.0155
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