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?$$\left(Relative density =\dfrac{W_a}{W_a-W_L}\right)$$


ANSWER

$$7\%$$


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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