JEE Main 2024 — 6 April, Shift 2 — Question 33

$1 \mathrm{MSD}=\frac{1 \mathrm{~cm}}{20}=0.05 \mathrm{~cm}$ $1 \mathrm{VSD}=\frac{49}{50} \mathrm{MSD}=\frac{49}{50} \times 0.05 \mathrm{~cm}=0.049 \mathrm{~cm}$

$\mathrm{LC}=1 \mathrm{MSD}-1 \mathrm{VSD}=0.001 \mathrm{~cm}$ For mark on paper, $\mathrm{L}_{1}=8.45 \mathrm{~cm}+26 \times 0.001 \mathrm{~cm}$

$=84.76 \mathrm{~mm}$

For mark on paper through slab, L2

$=7.12 \mathrm{~cm}+$

$41 \times 0.001 \mathrm{~cm}=71.61 \mathrm{~mm}$

For powder particle on top surface, $\mathrm{ZE}=4.05 \mathrm{~cm}$ $+1 \times 0.001 \mathrm{~cm}=40.51 \mathrm{~mm}$

$\therefore$ actual $\mathrm{L}_{1}=84.76-40.51=44.25 \mathrm{~mm}$

actual $\mathrm{L} 2=71.61-40.51=31.10 \mathrm{~mm}$ $\mathrm{L}_{2}=\frac{\mathrm{L}_{1}}{\mu}$

$\Rightarrow \mu=\frac{L_{1}}{L_{2}}=\frac{44.25}{31.10}=1.42$