The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μ m diameter of wire is :

Question

The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 μ m diameter of wire is : (A) 50 (B) 100 (C) 200 (D) 500

Tardigrade Admin · Accepted Answer

Least count = ( textPitch/ textNumber textof textdivisio texton textcircular textscale ) beginaligned L C &=(P/ N ) L C &=5 μ m =5 × 10-6 m . &=10-3 m . ⇒ N=(P/L C)=(10-3/5 × 10-6) ⇒ N=200 . endaligned

Q. The least count of the main scale of a screw gauge is $1 mm$ . The minimum number of divisions on its circular scale required to measure $5 μ m$ diameter of wire is :

50

100

200

500

Least count = $\frac{Pitch}{Number of divisio on circular scale}$
$L C L C \Rightarrow \Rightarrow = \frac{P}{N} = 5 μ m = 5 \times 1 0^{- 6} m . = 1 0^{- 3} m . N = \frac{P}{L C} = \frac{1 0 ^{- 3}}{5 \times 1 0 ^{- 6}} N = 200.$

Q. The least count of the main scale of a screw gauge is 1mm. The minimum number of divisions on its circular scale required to measure 5μm diameter of wire is :

50

100

200

500

Least count = NumberofdivisiooncircularscalePitch​ LCLC⇒⇒​=NP​=5μm=5×10−6m.=10−3m.N=LCP​=5×10−610−3​N=200.​

Q. The least count of the main scale of a screw gauge is $1 mm$ . The minimum number of divisions on its circular scale required to measure $5 μ m$ diameter of wire is :

Least count = $\frac{Pitch}{Number of divisio on circular scale}$
$L C L C \Rightarrow \Rightarrow = \frac{P}{N} = 5 μ m = 5 \times 1 0^{- 6} m . = 1 0^{- 3} m . N = \frac{P}{L C} = \frac{1 0 ^{- 3}}{5 \times 1 0 ^{- 6}} N = 200.$