JEE MAIN - Physics (2021 - 1st September Evening Shift - No. 8)
A student determined Young's Modulus of elasticity using the formula $$Y = {{Mg{L^3}} \over {4b{d^3}\delta }}$$. 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 |
Then the fractional error in the measurement of Y is :
0.0083
0.0155
0.155
0.083
Explanation
$$y = {{Mg{L^3}} \over {4b{d^3}\delta }}$$
$${{\Delta y} \over y} = {{\Delta M} \over M} + {{3\Delta L} \over L} + {{\Delta b} \over b} + {{3\Delta d} \over d} + {{\Delta \delta } \over \delta }$$
$${{\Delta y} \over y} = {{{{10}^{ - 3}}} \over 2} + {{3 \times {{10}^{ - 3}}} \over 1} + {{{{10}^{ - 2}}} \over 4} + {{3 \times {{10}^{ - 2}}} \over 4} + {{{{10}^{ - 2}}} \over 5}$$
$$ = {10^{ - 3}}[0.5 + 3 + 2.5 + 7.5 + 2] = 0.0155$$
Option (b)
$${{\Delta y} \over y} = {{\Delta M} \over M} + {{3\Delta L} \over L} + {{\Delta b} \over b} + {{3\Delta d} \over d} + {{\Delta \delta } \over \delta }$$
$${{\Delta y} \over y} = {{{{10}^{ - 3}}} \over 2} + {{3 \times {{10}^{ - 3}}} \over 1} + {{{{10}^{ - 2}}} \over 4} + {{3 \times {{10}^{ - 2}}} \over 4} + {{{{10}^{ - 2}}} \over 5}$$
$$ = {10^{ - 3}}[0.5 + 3 + 2.5 + 7.5 + 2] = 0.0155$$
Option (b)
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