Physics

# If weight of an object in air $W_a$ is $(10.0\pm 0.1)$g and its weight in water $W_L$ is $(5.00\pm 0.2)$g, then percentage error in its relative density is?$\left(Relative density =\dfrac{W_a}{W_a-W_L}\right)$

$7\%$

##### SOLUTION
Given ${ W }_{ a }=10.00\pm 0.1;{ W }_{ L }=5.00\pm 0.2$
Loss of weight in water$={ W }_{ a }-{ W }_{ L }$
=(10.00$\pm$0.1)-(5.00$\pm$0.2)
=(10-5)$\pm$(0.1+0.2)
=5$\pm$0.3
Now percentage error in relative density is given by$=\dfrac { { W }_{ a } }{ { W }_{ a }-{ W }_{ L } } =\frac { 10.00\pm 0.1 }{ 5\pm 0.3 } =(\frac { 10 }{ 5 } )\pm (\frac { 0.1 }{ 10 } +\frac { 0.3 }{ 5 } )\times 100\\ =2\pm (1+6)\\ =2\pm 7\\$
7 percent error

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Single Correct Medium Published on 18th 08, 2020
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