Question

A student measured the length of a rod and wrote it as $3.50cm$. Which instrument did he use to measure it?

• A. A meter scale
• B. A vernier calliper where the $10$ divisions in vernier scale matches with $9$ division in main scale and main scale has $10$ divisions in $1cm$
• C. A screw gauge having $100$ divisions in the circular scale and pitch as $1mm$
• D. A screw gauge having $50$ divisions in the circular scale and pitch as $1mm$

Answer 1

Answer: (B)

A vernier calliper where the $10$ divisions in vernier scale matches with $9$ division in main scale and main scale has $10$ divisions in $1cm$

Given- * Meauswred length of the Rod $=3.50cm$ count of the instrement is $0.01cm.$ (exact $y$ )

Now for the velitics verenier calipers we have -

10MSD= 1 cm (MSD-Main Scale division) (VSD - Verenier Scale div) and

$\begin{array}{cc}\Rightarrow 1VSD& =0.9.MSD\\ & \text{IVSD}=0.09cm\end{array}$

Least Count = IMSD-1VSD $=0.1cm-0.09cm=0.01cm$ this is our required instrument.

Hence option B is correct