Y=4bd3δMgL3
YΔY=MΔM+31Δl+bΔb+3dΔd+δΔδ⇒YΔY=0.0005+3×0.001+0.0025+3×0.0025+0.002=0.0155
JEE Main 2021 — Physics Mechanics
A student determined Young's Modulus of elasticity using the formula Y=4bd3δMgL3. The value of g is taken to be 9.8ms−2 without any significant error, his observations are as following.
| Physical Quantity | Least count of the Equipment used for measurement | Observed Value |
| Mass (M) | 1g | 2kg |
| Length of bar (L) | 1mm | 1m |
| Breadth of bar (b) | 0.1mm | 4cm |
| Thickness of bar (d) | 0.01mm | 0.4cm |
| Depression (δ) | 0.01mm | 5mm |
Held on 1 Sept 2021 · Verified 6 Jul 2026.
0.155
0.0083
0.083
0.0155
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