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Manipal Medical Manipal Medical Solved Paper-2006

  • question_answer
    A steel scale measures the length of a copper wire as 80.0 cm, when both are at\[20{}^\circ C,\]the calibration temperature for the scale. What would the scale read for the length of the wire when both are at\[40{}^\circ C\]? Given, \[{{a}_{S}}=11\times {{10}^{-6}}per{}^\circ C\] and\[{{a}_{Cu}}=17\times {{10}^{-6}}per{}^\circ C\]:

    A)  80.0096 cm     

    B)  80.0272 cm

    C)  1 cm           

    D)  25.2 cm

    Correct Answer: A

    Solution :

     Using the relation \[{{l}_{t}}={{l}_{0}}(1+\alpha t)\] \[=1\times [1+11\times {{10}^{6}}\times (40{}^\circ -20{}^\circ )]\] \[=1.00022\text{ }cm\] Now, length of copper rod at \[40{}^\circ C\] \[l{{}_{t}}=l{{}_{0}}(1+\alpha t)\] \[=80[1+17\times {{10}^{6}}(40{}^\circ -20{}^\circ )]\] \[=80.0272\text{ }cm\] Now, number of cms observed on the scale \[=\frac{80.0272}{1.00022}=80.0096\]


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