In an instrument, there are 25 divisions on the vernier scale which coincides with 24th division of the main scale. 1 cm on main scale is divided into 20 equal parts. The least count of the instrument is

0.002 cm

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0.05 cm

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0.001 cm

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0.02 cm

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Solution

The correct option is A $0.002 cm$
Its given that,
$24 M . S . D . = 25 V . S . D .$
$\Rightarrow 1 V . S . D . = \frac{24}{25} M . S . D .$

Least count is given by:
$L . C . = 1 M . S . D . - 1 V . S . D .$
$L . C . = 1 M . S . D . - \frac{24}{25} M . S . D .$
$L . C . = \frac{1}{25} M . S . D .$

As given, $1 M . S . D . = 1 / 20 c m$
$L . C . = \frac{1}{25} \times \frac{1}{20} c m = 0.002 c m$

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In an instrument, there are 25 divisions on the vernier scale which coincides with 24th division of the main scale. 1 cm on main scale is divided into 20 equal parts. The least count of the instrument is

The correct option is A 0.002 cmIts given that, 24 M.S.D.=25 V.S.D. ⇒1 V.S.D.=2425M.S.D. Least count is given by: L.C.=1 M.S.D.−1 V.S.D. L.C.=1 M.S.D.−2425M.S.D. L.C.=125M.S.D. As given, 1 M.S.D.=1/20 cm L.C.=125×120 cm=0.002 cm