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In an experiment the angles are required to be measured using an instrument in which 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half a degree $\left(=0.5^{\circ}\right)$, then the least count of the instrument is
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one minute
According to the question, given that
29 main scale division $($ MSD $)=30$ vernier scale division (VSD)
$\begin{aligned}
& \Rightarrow 29 \mathrm{MSD}=30 \mathrm{VSD} \Rightarrow 1 \mathrm{VSD}=\frac{29}{30} \times 1 \mathrm{MSD} \\
& \Rightarrow \quad 1 \mathrm{VSD}=\frac{29}{30}\left(0.5^{\circ}\right) \quad\left[\because 1 \mathrm{MSD}=0.5^{\circ}\right] \\
& \Rightarrow \quad 1 \mathrm{VSD}=\left(\frac{29}{60}\right)^{\circ}
\end{aligned}$
$\therefore$ Least count of vernier calliperse is given as
$\begin{aligned}
\text { LC } & =1 \text { MSD }-1 \text { VSD } \\
& =0.5^{\circ}-\left(\frac{29}{60}\right)^{\circ}=\left(\frac{30-29}{60}\right)^{\circ}=\left(\frac{1}{60}\right)^{\circ} \\
& =1 \min \quad\left[\because 1 \min =\frac{1}{60} \text { degree }\right]
\end{aligned}$
29 main scale division $($ MSD $)=30$ vernier scale division (VSD)
$\begin{aligned}
& \Rightarrow 29 \mathrm{MSD}=30 \mathrm{VSD} \Rightarrow 1 \mathrm{VSD}=\frac{29}{30} \times 1 \mathrm{MSD} \\
& \Rightarrow \quad 1 \mathrm{VSD}=\frac{29}{30}\left(0.5^{\circ}\right) \quad\left[\because 1 \mathrm{MSD}=0.5^{\circ}\right] \\
& \Rightarrow \quad 1 \mathrm{VSD}=\left(\frac{29}{60}\right)^{\circ}
\end{aligned}$
$\therefore$ Least count of vernier calliperse is given as
$\begin{aligned}
\text { LC } & =1 \text { MSD }-1 \text { VSD } \\
& =0.5^{\circ}-\left(\frac{29}{60}\right)^{\circ}=\left(\frac{30-29}{60}\right)^{\circ}=\left(\frac{1}{60}\right)^{\circ} \\
& =1 \min \quad\left[\because 1 \min =\frac{1}{60} \text { degree }\right]
\end{aligned}$
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