Young's modulus is determined by the equation given by $Y =49000 \frac{ M }{ l } \frac{\text { dyne }}{ cm ^2}$ where $M$ is the mass and $l$ is the extension of wire used in the experiment. Now error in Young modulus $(Y)$ is estimated by taking data from $M-l$ plot in graph paper. The smallest scale divisions are $5 g$ and $0.02 cm$ along load axis and extension axis respectively. If the value of $M$ and $l$ are $500 g$ and $2 cm$ respectively then percentage error of $Y$ is :

[JEE Main 2024, 08 Apr (Shift 1)]

Question

Young's modulus is determined by the equation given by $Y =49000 \frac{ M }{ l } \frac{\text { dyne }}{ cm ^2}$ where $M$ is the mass and $l$ is the extension of wire used in the experiment. Now error in Young modulus $(Y)$ is estimated by taking data from $M-l$ plot in graph paper. The smallest scale divisions are $5 g$ and $0.02 cm$ along load axis and extension axis respectively. If the value of $M$ and $l$ are $500 g$ and $2 cm$ respectively then percentage error of $Y$ is :

[JEE Main 2024, 08 Apr (Shift 1)]

YESPYQ · Accepted Answer

Correct answer: d) $2\%$. Given Values Mass ($M$): $500\text{ g}$ Extension ($l$): $2\text{ cm}$ Least Count of $M$ ($\Delta M$): $5\text{ g}$ (smallest scale division along load axis) Least Count of $l$ ($\Delta l$): $0.02\text{ cm}$ (smallest scale division along extension axis) Calculation of Percentage Error $$\text{Percentage Error} = \left( \frac{\Delta M}{M} + \frac{\Delta l}{l} \right) \times 100\%$$ $$\text{Percentage Error} = \left( \frac{5}{500} + \frac{0.02}{2} \right) \times 100\%$$ $$\text{Percentage Error} = (0.01 + 0.01) \times 100\%$$ $$\text{Percentage Error} = 0.02 \times 100\% = 2\%$$ Correct Option: D (2%)

Young's modulus is determined by the equation given by \(Y =49000 \frac{ M }{ l } \frac{\text { dyne }}{ cm ^2}\) where …

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