For a screw gauge least count of the main scale is 1 mm. Find the minimum number of divisions on the circular scale so that it can measure a diameter of 5 μm

Question

For a screw gauge least count of the main scale is 1 mm. Find the minimum number of divisions on the circular scale so that it can measure a diameter of 5 μm (A) 500 (B) 100 (C) 50 (D) 200

Tardigrade Admin · Accepted Answer

Least count of screw gauge =5 Î¼m L.C=(pitch/no . of div on circular scale) 5Î¼m=(1 mm/N) N=200

Q. For a screw gauge least count of the main scale is $1 mm .$ Find the minimum number of divisions on the circular scale so that it can measure a diameter of $5 μ m$

$500$

$100$

$50$

$200$

Least count of screw gauge $= 5 μ m$
$L . C = \frac{p i t c h}{n o . o f d i v o n c i rc u l a r sc a l e}$
$5 μ m = \frac{1 mm}{N}$
$N = 200$

Q. For a screw gauge least count of the main scale is 1mm. Find the minimum number of divisions on the circular scale so that it can measure a diameter of 5μm

500

100

50

200