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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 12

  • question_answer
    Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of- 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is

    A) 3.32 mm 

    B) 3.73 mm

    C) 3.67 mm        

    D) 3.38 mm

    Correct Answer: D

    Solution :

    [d]: Pitch of screw gauge \[=\frac{1}{2}mm=0.5mm\] Least count of the screw gauge \[=\frac{0.5}{50}mm=0.01mm\] Zero error = -0.03 mm Zero correction = + 0.03 mm Main scale reading = 3 mm Circular scale reading = 35 Observed diameter of the wire \[=3\text{ }mm+35\times \left( 0.01 \right)mm=3.35\text{ }mm\] Corrected diameter of the wire = (3.35 + 0.03) mm = 3.38 mm


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