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JEE Main & Advanced JEE Main Paper (Held On 10 April 2015)

  • question_answer
    Diameter of a steel ball is measured using a Vernier callipers which lias divisions of cm on its main scale (MS) and 10 divisions of its vernier scale (VS) match 9 divisions on the main scale. Three such measurements for a ball are given as :
    S. No. MS (cm) VS divisions
    1. 0.5 8
    2. 0.5 4
    3. 0.5 6
    If the zero error is - 0.03 cm, then mean corrected diameter is : JEE Main Online Paper (Held On 10 April 2015)

    A)  0.56 cm

    B)  0.53 cm

    C)   0.52 cm

    D)  0.59 cm

    Correct Answer: D

    Solution :

                     \[\overline{x}=\frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3}\] \[=\frac{0.58+0.54+0.56}{3}\] \[\overline{x}=0.56\] so value = 0.56 + error = 0.56 + 0.03 = 0.59 cm


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