CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    The radius of a sphere is measured as 5 cm with an error possibly as large as 0.02 cm. The error and percentage error in computing the surface area of the sphere are:

    A)  \[0.87\pi \,and\,0.2%\]                

    B)  \[0.8\pi \,and\,0.8%\]

    C)  \[0.4\pi \,and\,0.4%\]

    D)         \[\pi \,and\,1%\]

    E)  \[0.6\pi \,and\,0.6%\]

    Correct Answer: B

    Solution :

    Surface area\[=4\pi {{r}^{2}}\] \[\therefore \]Error in surface area\[=4\pi {{(r+\Delta r)}^{2}}-4\pi {{r}^{2}}\] where\[r=5,\text{ }\Delta r=0.02.\] \[\Rightarrow \]Error \[=4\pi [{{r}^{2}}+\Delta {{r}^{2}}+2r\Delta R-{{R}^{2}}]\]                 \[=0.8\,\pi \] % Error\[=\frac{0.8\pi }{100\pi }\times 100=0.8%\]


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