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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) 5.725 mm                     

    B) 5.950 mm

    C) 5.755 mm     

    D)               5.740 mm

    Correct Answer: C

    Solution :

     Least count \[=\,\,\frac{0.5}{100}\,\,mm\] Actual value \[=5.5\text{ }mm+\left( 48+3 \right)\text{ }{{10}^{-3}}mm\] \[=\text{ }5.755\text{ }mm\]


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