A vernier calipers have 1 mm marks on the main scale- It has 20 equal divisions on the vernier scale which matches with 16 main scale divisions- For this vernier calipers- the least count is -0-02 mm0-05 mm0-1 mm0-2 mm

Question

Toppr Admin · Accepted Answer

Least count-xA0-L-C-1M-S-D-x2212-1V-S-xA0-DAs vernier scale division equals matches on main scale divison- therefore-xA0-L-C-1-x2212-1620-M-S-D-xA0- -xA0- -xA0- -xA0- -xA0-1-x2212-0-8-1mm-0-2mm

A vernier calipers have $1 m m$ marks on the main scale. It has $20$ equal divisions on the vernier scale which matches with $16$ main scale divisions. For this vernier calipers, the least count is :
$0.02 m m$
$0.05 m m$
$0.1 m m$
$0.2 m m$

Least count, $L . C = 1 M . S . D - 1 V . S . D$
As vernier scale division equals matches on main scale divison, therefore
$L . C = (1 - \frac{16}{20}) M . S . D$
$= (1 - 0.8) (1 m m) = 0.2 m m$

A vernier calipers have 1mm marks on the main scale. It has 20 equal divisions on the vernier scale which matches with 16 main scale divisions. For this vernier calipers, the least count is :0.02mm0.05mm0.1mm0.2mm

Least count, L.C=1M.S.D−1V.S. DAs vernier scale division equals matches on main scale divison, therefore L.C=(1−1620)M.S.D =(1−0.8)(1mm)=0.2mm

A vernier calipers have $1 m m$ marks on the main scale. It has $20$ equal divisions on the vernier scale which matches with $16$ main scale divisions. For this vernier calipers, the least count is :
$0.02 m m$
$0.05 m m$
$0.1 m m$
$0.2 m m$

Least count, $L . C = 1 M . S . D - 1 V . S . D$
As vernier scale division equals matches on main scale divison, therefore
$L . C = (1 - \frac{16}{20}) M . S . D$
$= (1 - 0.8) (1 m m) = 0.2 m m$