The diameter of a cylinder is measured using a Vernier calipers with no zero error. If is found that the zero of the Vernier scale lies between 5.10cm and 5.15cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45cm. The 24division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of cylinder is

$$\begin{gathered} \mathrm{value} \mathrm{of} 1\mathrm{Vernier} \mathrm{Scale} \mathrm{Division}= 1 \mathrm{VSD}=\frac{2.45\mathrm{cm}}{50}=0.049\mathrm{cm} \\ \mathrm{value} \mathrm{of} 1\text{\hspace{0.17em}Main} \mathrm{Scale} \mathrm{Division}=1\mathrm{MSD}=5.15-5.10=0.05\mathrm{cm} \\ \mathrm{Least} \mathrm{Count}=\mathrm{L}.\mathrm{C}=\text{\hspace{0.17em}value of\hspace{0.17em}}1\mathrm{MSD}-\mathrm{value} \mathrm{of}1\mathrm{VSD}=0.05-0.049 \\ =0.001\mathrm{cm} \\ \mathrm{given} \mathrm{Vernier} \mathrm{Coincidence}= \mathrm{VC}=24 \\ \mathrm{Diameter} \mathrm{of} \mathrm{wire} =\mathrm{MSD}+\mathrm{VC}\times \mathrm{L}.\mathrm{C} \\ =5.10+24\times 0.001 \\ =5.10+0.024 \\ Diameter of wire =5.124cm \end{gathered}$$