A) 0.003
B) 0.001
C) 0.004
D) 0.002
Correct Answer: D
Solution :
(0.002) Maximum error in measuring mass = 0.001 g, because least count is 0.001 g, similarly, maximum error in measuring volume is \[0.01\text{ }c{{m}^{3}}.\] \[\frac{\Delta \rho }{\rho }=\frac{\Delta M}{M}+\frac{\Delta V}{V}=\frac{0.001}{20.000}+\frac{0.01}{10.00}\] \[=(5\times {{10}^{-5}})+(1\times {{10}^{-3}})=1.05\times {{10}^{-3}}\] \[\Delta \rho =(1.05\times {{10}^{-3}})\times \rho \] \[=1.05\times {{10}^{-3}}\times \frac{20.000}{10.00}=0.002gc{{m}^{-3}}\]
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