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  • Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm. Reason R :</st

Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm. Reason R : Least Count = \frac{Pitch }{Total\: \: divisions\: \: on\: \: circular\: \: scale} In the light of the above statements, choose the most appropriate answer from the options given below :
Option: 1 A is not correct but R is correct.
Option: 2 Both A and R are correct and R is the correct explanation of A.
Option: 3 A is correct but R is not correct.
Option: 4 Both A and R are correct and R is NOT the correct explanation of A.

Answers (1)

best_answer

Distance travelled on the main scale in one rotation is equal to pitch.

$$ \begin{aligned} \therefore 5 \mathrm{~mm} &=5 \text { (pitch) } \\ \text { pitch } &=1 \mathrm{~mm} \rightarrow(1) \end{aligned}

\text{No. of CSD}=50\; \text{division}

\text{L.C}= \frac{\text{pitch}}{\text{No.of CSD}}

          \begin{aligned} =\frac{1 \mathrm{~mm}}{50} &=0.02 \times 10^{-3} \mathrm{~m} \\ &=0.002 \mathrm{~cm} \end{aligned}

The correct option is (1)

Posted by

vishal kumar

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