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JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer
    The circular head of a screw gauge is divided into 200 divisions and move \[1\text{ }mm\] ahead in one revolution. If the same instrument has a zero error of- \[0.05\text{ }mm\] and the reading on the main scale in measuring diameter of a wire is \[6\text{ }mm\] and that on circular scale is 45. The diameter of the wire is

    A)  \[6.275\,mm\]               

    B)  \[6.375\,mm\]

    C)  \[5.75\,mm\]                

    D)  \[5.50\,mm\]

    Correct Answer: A

    Solution :

     Pitch = 1 mm                                        Number of divisions on circular scale = 200 \[L.C=\frac{Pitch}{Number\,ofdivisions\,on\,circular\,scale}\] \[=\frac{1\,mm}{200}=0.005\,mm=0.0005\,cm\] Diameter of the wire = (Main scale reading + Circular scale reading \[\times \] L.C.) - zero error \[=6\,mm+45\times 0.05-(-0.05)\] \[=6\,mm+0.225mm+0.05mm=6.275\,mm\]


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