JEE MAIN - Physics (2021 - 16th March Evening Shift - No. 11)
In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1 g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young's Modulus determined by this experiment?
0.14%
9%
1.4%
0.9%
Explanation
$${{\Delta Y} \over Y} = \left( {{{\Delta m} \over m}} \right) + \left( {{{\Delta g} \over g}} \right) + \left( {{{\Delta A} \over A}} \right) + \left( {{{\Delta l} \over l}} \right) + \left( {{{\Delta L} \over L}} \right)$$
$$ = \left( {{{1g} \over {1kg}}} \right) + 0 + 2\left( {{{\Delta r} \over r}} \right) + \left( {{{\Delta l} \over l}} \right) + \left( {{{\Delta L} \over L}} \right)$$
$$ = \left( {{{1g} \over {1kg}}} \right) + 2\left( {{{0.001cm} \over {0.2cm}}} \right) + \left( {{{0.001cm} \over {0.5cm}}} \right) + \left( {{{0.001m} \over {1m}}} \right)$$
$$ = \left( {{1 \over {1000}}} \right) + 2\left( {{{1 \times 10} \over {2 \times {{10}^3}}}} \right) + \left( {{1 \over 5} \times {{{{10}^2}} \over {{{10}^3}}}} \right) + \left( {{1 \over {{{10}^3}}}} \right)$$
$$ = {1 \over {1000}} + {1 \over {100}} + {2 \over {{{10}^3}}} + {1 \over {{{10}^3}}}$$
$$ = {{1 + 10 + 2 + 1} \over {1000}} = {{14} \over {1000}} \times 100\% $$
$$ = 1.4\% $$
$$ = \left( {{{1g} \over {1kg}}} \right) + 0 + 2\left( {{{\Delta r} \over r}} \right) + \left( {{{\Delta l} \over l}} \right) + \left( {{{\Delta L} \over L}} \right)$$
$$ = \left( {{{1g} \over {1kg}}} \right) + 2\left( {{{0.001cm} \over {0.2cm}}} \right) + \left( {{{0.001cm} \over {0.5cm}}} \right) + \left( {{{0.001m} \over {1m}}} \right)$$
$$ = \left( {{1 \over {1000}}} \right) + 2\left( {{{1 \times 10} \over {2 \times {{10}^3}}}} \right) + \left( {{1 \over 5} \times {{{{10}^2}} \over {{{10}^3}}}} \right) + \left( {{1 \over {{{10}^3}}}} \right)$$
$$ = {1 \over {1000}} + {1 \over {100}} + {2 \over {{{10}^3}}} + {1 \over {{{10}^3}}}$$
$$ = {{1 + 10 + 2 + 1} \over {1000}} = {{14} \over {1000}} \times 100\% $$
$$ = 1.4\% $$
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