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Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    A force vector applied on a body is given by \[\overset{\text{ }\!\!\hat{\ }\!\!\text{ }}{\mathop{\text{F}}}\,\text{=6}\overset{\text{ }\!\!\hat{\ }\!\!\text{ }}{\mathop{\text{i}}}\,\text{-8}\overset{\text{ }\!\!\hat{\ }\!\!\text{ }}{\mathop{\text{j}}}\,\text{+10}\overset{\text{ }\!\!\hat{\ }\!\!\text{ }}{\mathop{\text{k}}}\,\] and acquires an acceleration of 1 m/s2. Then the mass of the body is

    A)  10\[\sqrt{2}\]kg        

    B)  \[2\sqrt{10}\] kg

    C)  10kg          

    D)  20kg

    Correct Answer: A

    Solution :

     Force applied on a body \[\overrightarrow{F}=6\hat{i}-8\hat{j}+10\hat{k}\] \[\therefore \] \[|\overrightarrow{F}|=\sqrt{{{(6)}^{2}}+{{(8)}^{2}}+{{(10)}^{2}}}\] \[|\overrightarrow{F}|=\sqrt{36+64+100}\] \[|\overrightarrow{F}|=\sqrt{100+100}=\sqrt{200}\] \[|\overrightarrow{F}|=10\sqrt{2}N\] \[\therefore \] \[m=\frac{|\overrightarrow{F}|}{a}=\frac{10\sqrt{2}}{1}=10\sqrt{2}kg\]


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