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  • One main scale division of a vernier callipers is 'a' cm and nth division of the vernier scale coincide with division of the main scale. The least count of the callipe

One main scale division of a vernier callipers is 'a' cm and nth division of the vernier scale coincide with (n-1)^{th} division of the main scale. The least count of the callipers in mm is :
 
Option: 1 \frac{10 a}{n}
Option: 2 \left ( \frac{n-1}{10n} \right )a
Option: 3 \frac{10\; na}{(n-1)}  
Option: 4 \frac{10 a}{(n-1)}

Answers (1)

best_answer

 As nth division of the vernier scale coincide with (n-1)^{th} division of the main scale

So

\begin{array}{l} (\mathrm{n}-1) \mathrm{a}=\mathrm{n}\left(\mathrm{a}^{\prime}\right) \\ \\ \Rightarrow \mathrm{a}^{\prime}=\frac{(\mathrm{n}-1) \mathrm{a}}{\mathrm{n}} \end{array}

Where \mathrm{a}^{\prime} is the One main scale division of vernier calipers.

S0

\begin{aligned} \therefore \mathrm{L.C} &=1 \mathrm{MSD}-1 \mathrm{VSD} \\ &=\left(\mathrm{a}-\mathrm{a}^{\prime}\right) \mathrm{cm} \\ &=\mathrm{a}-\frac{(\mathrm{n}-1) \mathrm{a}}{\mathrm{n}} \\ &=\frac{\mathrm{na}-\mathrm{na}+\mathrm{a}}{\mathrm{n}}=\frac{\mathrm{a}}{\mathrm{n}} \mathrm{cm} \end{aligned}

So

LC=\frac{10a}{n}\ mm

Posted by

avinash.dongre

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