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 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is [Figure shows position of reference 'O' when jaws of screw gauge are closed]

Question

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 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is [Figure shows position of reference 'O' when jaws of screw gauge are closed] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

For (A) Reading = MSR + CSR + Error 0.322=0.300+ CSR +5 × LC 0.322=0.300+C S R+0.005 CSR =0.017 For B Reading = MSR + CSR + Error 0.322=0.200+C S R+0.092 C S R=0.030 Difference =0.030-0.017=0.013 cm Division on circular scale =(0.013/0.001)=13

For (A) Reading =MSR+CSR+ Error 0.322=0.300+CSR+5×LC 0.322=0.300+CSR+0.005 CSR=0.017 For B Reading =MSR+CSR+ Error 0.322=0.200+CSR+0.092 CSR=0.030 Difference =0.030−0.017=0.013cm Division on circular scale =0.0010.013​=13

For $(A)$
Reading $= MSR + CSR +$ Error
$0.322 = 0.300 + CSR + 5 \times L C$
$0.322 = 0.300 + CSR + 0.005$
$CSR = 0.017$
For $B$
Reading $= MSR + CSR +$ Error
$0.322 = 0.200 + CSR + 0.092$
$CSR = 0.030$
Difference $= 0.030 - 0.017 = 0.013 c m$
Division on circular scale $= \frac{0.013}{0.001} = 13$