The relative density of the material of a bod | vedclass

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(A) Weight in air $W_a = (5.00 \pm 0.05) \ N$.Weight in water $W_w = (4.00 \pm 0.05) \ N$.Loss of weight in water $W_L = W_a - W_w = (5.00 - 4.00) \pm

Vedclass · Accepted Answer

$5.0 \pm 11\%$

Vedclass · Answer

(A) Weight in air $W_a = (5.00 \pm 0.05) \ N$.Weight in water $W_w = (4.00 \pm 0.05) \ N$.Loss of weight in water $W_L = W_a - W_w = (5.00 - 4.00) \pm (0.05 + 0.05) = (1.00 \pm 0.10) \ N$.Relative density $RD = \frac{W_a}{W_L} = \frac{5.00}{1.00} = 5.0$.The maximum permissible percentage error is given by $\frac{\Delta RD}{RD} 	imes 100 = \left( \frac{\Delta W_a}{W_a} + \frac{\Delta W_L}{W_L} ight) 	imes 100$.Substituting the values: $\left( \frac{0.05}{5.00} + \frac{0.10}{1.00} ight) 	imes 100 = (0.01 + 0.10) 	imes 100 = 1\% + 10\% = 11\%$.Thus,the relative density is $5.0 \pm 11\%$.

The relative density of the material of a body is found by weighing it first in air and then in water. If the weight in air is $(5.00 \pm 0.05) \ N$ and the weight in water is $(4.00 \pm 0.05) \ N$,then the relative density along with the maximum permissible percentage error is:

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