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Student A and Student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 \mathrm{~cm}. The absolute value of the difference between the final circular scale readings observed by the students \mathrm{A}$ and $\mathrm{B} is__________.
[Figure shows position of reference ' O ' when jaws of screw gauge are closed]
Given pitch =0.1 \mathrm{~cm}.
 

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pitch= 0\cdot 1\, cm
No.of circular divisions = 100
Least\, count = \frac{pitch}{No.of \, circular \, division}
                         = 10^{-3}\, cm
                         = 0\cdot 001 cm
Actual value of the radius of the wire = 0\cdot 322 cm
Let the main scale reading be X
\left ( Reading \right )_{A}= MSR+CSR
                         = X+\left ( CSD \right )\times LC
                        = X+\left ( 5 \right )\times 0\cdot 001
R_{A}=X+0\cdot 005
(i.e 5 divisions ahead of zero mark)
\left ( Reading \right )_{B}= R_{B}= MSR+CSR
                         = \left ( X-1 \right )+\left ( CSD \right )\times \left ( LC \right )
                         = \left ( X-1 \right )+\left ( 92 \right )\times 0\cdot 001
                         =X-0\cdot 008
(i.e 8 CSD behind zero mark )  
\Delta R= R_{A}-R_{B}= 0\cdot 013
Reading corresponds to 13 circular scale division
The answer is 13

Posted by

vishal kumar

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