2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Morning)

  • question_answer
    The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure \[5\mu m\]diameter of a wire is [JEE Main Online Paper Held On 12-Jan-2019 Morning]

    A) 50                                            

    B) 100

    C) 200                              

    D)   500

    Correct Answer: C

    Solution :

    Least count = 1 mm                    \[{{N}_{d}}\](Number of divisions) \[=\frac{L.C.}{5\mu m}\] \[=\frac{1mm}{5\mu m}=\frac{{{10}^{3}}}{5}=200\]


You need to login to perform this action.
You will be redirected in 3 sec spinner