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NEET Sample Paper NEET Sample Test Paper-30

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

    A)  \[90.0096\,cm\]          

    B)  \[80.0272\,cm\]

    C)  \[1\,cm\]                     

    D)  \[25.2\,cm\]

    Correct Answer: B

    Solution :

    From relation     \[{{l}_{t}}=10\,(1+\alpha t)\] Here,       \[\alpha \] steel \[=11\times {{10}^{-6}}pe{{r}^{o}}C.\] Putting the value in equation (i) we get             \[{{l}_{t}}=1\times [(1+11\times {{10}^{-6}})\times ({{40}^{o}}-{{20}^{o}})]\] \[\Rightarrow \]\[{{l}_{i}}=1.00022\,cm\] Now, length of copper rod at \[{{40}^{o}}C\]             \[{{i}_{t}}={{i}_{0}}.\,\,(1+\alpha t)\] Given a copper \[=17\times {{10}^{-6}}({{40}^{o}}-{{20}^{o}})\] \[\Rightarrow \]            \[{{i}_{1}}=80[1+17+{{10}^{-6}}(40-{{20}^{o}})]\] \[\Rightarrow \]            \[{{i}_{l}}=80.0272cm.\] Hence, number of cms observed on the scale             \[\frac{{{i}_{t}}}{{{l}_{t}}}=\frac{80.272}{1.00022}=80.0096.\]


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