2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.25cm. There are 100 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. Assuming negligible zero error, the thickness of the wire is:                                                             [JEE Online 15-04-2018]

    A) \[0.0430cm\]               

    B) \[0.3150cm\]               

    C) \[0.4300cm\]               

    D) \[0.2150cm\]   

    Correct Answer: D

    Solution :

    In one rotation scale moves \[\frac{0.25}{5}=0.05cm\] Least count \[=0.05\times {{10}^{-2}}cm\] For 4 main scale division \[=4\times 0.05=0.2cm\] For circular scale division \[=30\times 0.05\times {{10}^{-2}}=1.5\times {{10}^{-2}}cm\] Thickness of wire\[=0.2+0.015=0.2150cm\]


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