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CMC Medical CMC-Medical Ludhiana Solved Paper-2015

  • question_answer
    A steel plate efface area 4 \[c{{m}^{2}}\] and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. The lateral displacement of the upper surface with respect to the lower surface will be [Modulus of rigidity for steel\[=8.4\times {{10}^{10}}N{{m}^{-2}}\]]

    A)  \[2.1\times {{10}^{-8}}m\]                         

    B)  \[1.5\times {{10}^{-9}}m\]

    C)  \[3.7\times {{10}^{-7}}m\]                         

    D)  \[0.8\times {{10}^{-10}}m\]

    Correct Answer: B

    Solution :

                    Given, \[A=4\times {{10}^{-4}}{{m}^{2}}\] thickness, \[t=0.5\times {{10}^{-2}}\,m\] force,           \[F=10\text{ }N\] Now           \[E=\frac{F}{A\theta }\] Here, E = modulus of rigidity \[\Rightarrow \]               \[\theta =\frac{F}{AE}=\frac{10}{4\times {{10}^{-4}}\times 84\times {{10}^{10}}}\] Now, lateral displacement\[=\theta t\] \[=0.297\times {{10}^{-6}}\times 0.5\times {{10}^{-2}}\] \[=1.5\times {{10}^{-9}}m\]


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