RAJASTHAN PMT Rajasthan - PMT Solved Paper-2001

  • question_answer
    If a sphere is inserted in water, then it flows with \[\frac{1}{3}\text{rd}\] of it outside the water. When it is inserted in an unknown liquid  then it flows with \[\frac{3}{4}\text{th}\] of it outside, then density of unknown liquid is:      

    A)  \[10.9\,gm\,/c.c.\]    

    B)         \[\frac{9}{4}gm\,/c.c.\]

    C)  \[\frac{8}{3}gm\,/c.c.\]               

    D)         \[\frac{3}{8}gm\,/c.c.\]

    Correct Answer: C

    Solution :

    \[\frac{\text{Relative}\,\text{density}\,\text{of}\,\text{body}}{\text{Relative}\,\text{density}\,\text{of}\,\text{water}}\] \[\text{=}\frac{\text{volume}\,\text{of}\,\text{inserted}\,\text{part}\,\text{of}\,\text{body}}{\text{total}\,\text{volume}\,\text{of}\,\text{body}}\] \[\frac{d}{1}=\frac{2V}{3V}\]                 \[\Rightarrow \]               \[d=\frac{2}{3}\] Again using the same formula                                       \[\frac{1}{4}=\frac{2}{3d}\] \[\therefore \]  \[d=\frac{8}{3}gm/c.c.\]


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