Young's modulus is determined by the equation given by $Y=49000 \frac{\mathrm{~m}}{\ell} \frac{\text { dyne }}{\mathrm{cm}^{2}}$ where $M$ is the mass and $\ell$ is the extension of wire used in the experiment. Now error in Young modules(Y) is estimated by taking data from $\mathrm{M}-\ell$ 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 $\ell$ are 500 g and 2 cm respectively then percentage error of Y is :

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Units & Measurement

Young's modulus is determined by the equation given by

Y = 49000 \frac{m}{ℓ} \frac{dyne}{cm ^{2}}

where

M

is the mass and

ℓ

is the extension of wire used in the experiment. Now error in Young modules(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 :

Answer: (C)

Y = 49000 \frac{m}{ℓ}

, so

\frac{Δ Y}{Y} = \frac{Δ m}{m} + \frac{Δ ℓ}{ℓ}

. Using the smallest scale divisions as the least counts,

\frac{Δ Y}{Y} = \frac{5}{500} + \frac{0.02}{2} = 0.01 + 0.01 = 0.02 = 2%

Option (C).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option