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QJEEM21Ph81f3: JEE Main 2021 Physics - Class 11 JEEM21Ph81f3 Solution: In order to determine the Young's Modulus of a wire of radius…

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?

Multiple Choice Options

A 0.14%
B 9%
C 1.40%
D 0.90%

Step-by-Step Verified Solution

{{\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\%

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