Question

To measure the diameter of a wire, a screw gauge is used. In a complete rotation, the spindle of the screw gauge advances by $ \dfrac{1}{2}mm $ and its circular scale has $ 50 $ deviations. The main scale is graduated to $ \dfrac{1}{2}mm $. If the wire is put between the jaws, $ 4 $ main scale divisions are clearly visible and $ 1 $ dimensions of circular scale co-inside with the reference line. The resistance of the wire is measured to be $ \left( {10\Omega \pm 1\% } \right) $. Length of the wire is measured to be $ 10cm $ using a scale of least count $ 1mm $. Maximum permissible error in resistivity measurement is:
(A) $ 1.5\% $
(B) $ 2\% $
(C) $ 2.9\% $
(D) $ 3\% $

Answer 1

Hint: To solve this question, we need to find out the measurement of the diameter from the information given. Then we have to use the formula of the resistance in terms of the length and the diameter of the wire to find out the required error in the resistivity measurement.

Formula used: The formula which is used in solving this question is given by
 $ R = \dfrac{{\rho l}}{A} $, here $ R $ is the resistance of a wire which has a length of $ l $, cross sectional area of $ A $, and a resistivity of $ \rho $.

Complete step by step answer
According to the question, the spindle of the screw gauge moves through a distance of $ \dfrac{1}{2}mm $ in one complete rotation. So the pitch of the screw gauge is $ \dfrac{1}{2}mm $.
Now, we know that the least count of a screw gauge is the pitch divided by the number of divisions on the circular scale. According to the question, the number of divisions on the circular scale is equal to $ 50 $. So the least count is given by
 $ L.C. = \dfrac{{1/2mm}}{{50}} = 0.01mm $
Now, when the wire is put between the jaws, $ 4 $ main scale divisions are clearly visible and the first division of the circular scale coincides with the reference line. So the MSR and CSR are given by
 $ MSR = 4 \times \dfrac{1}{2}mm = 2mm $ (least count of the main scale is $ \dfrac{1}{2}mm $ )
 $ CSR = 1 \times 0.01 = 0.01mm $
So the total diameter of the wire is given by
 $ D = MSR + CSR $
 $ \Rightarrow D = 2mm + 0.01mm = 2.01mm $ …………………...(1)
The maximum error in the diameter is equal to the least count of the screw gauge, that is,
 $ \Delta D = 0.01mm $ …………………...(2)
Now, the resistance of the wire is given to be $ \left( {10\Omega \pm 1\% } \right) $. This means that the absolute value of resistance is
 $ R = 10\Omega $…………………... (3)
Also the error in the measurement of resistance is given to be equal to $ 1\% $. This means that
 $ \Delta R = 0.01\Omega $ …………………...(4)
Also the length of the wire is measured to be equal to $ 10cm $ using a scale of least count $ 1mm $. This means that
 $ L = 10cm = 100mm\; $ …………………...(5)
 $ \Delta L = 1mm $ …………………...(6)
Now, we know that the resistance of a wire is given by
 $ R = \dfrac{{\rho l}}{A} $
So the resistivity is
 $ \rho = \dfrac{{RA}}{l} $
We know that the area of a wire is
 $ A = \dfrac{{\pi {D^2}}}{4} $
So the resistivity becomes
 $ \rho = \dfrac{{\pi {D^2}R}}{{4l}} $
Taking relative errors on both the sides, we get
 $ \dfrac{{\Delta \rho }}{\rho } = 2\dfrac{{\Delta D}}{D} + \dfrac{{\Delta R}}{R} + \dfrac{{\Delta l}}{l} $
Substituting (1), (2), (3), (4), (5) and (6) we have
 $ \dfrac{{\Delta \rho }}{\rho } = 2\dfrac{{0.01}}{{2.01}} + \dfrac{{0.01}}{{10}} + \dfrac{1}{{100}} $
On solving we get
 $ \dfrac{{\Delta \rho }}{\rho } = 0.02 = 2\% $
So the maximum error in the resistivity measurement is equal to $ 2\% $.
Hence, the correct answer is option B.

Note
In this question no information regarding the zero error of the screw gauge was given to us. Hence, we simply assumed it to be equal to zero while doing the calculations.