The mass of a beaker is (10.1±0.1)g when empty and (17.3±0.1)g when filled with a liquid. The mass of the liquid with possible limits of accuracy is

m1=(10.1±0.1) gm2=(17.3±0.1) gm=m2-m1=17.3-10.1=7.2Δm=Δm2+Δm1=0.1+0.1=0.2∴m1±m2=[7.2±0.2] g

The mass of a beaker is (10.1±0.1)g when empty and (17.3±0.1)g when filled with a liquid. The mass of the liquid with possible limits of accuracy is

m1=(10.1±0.1) gm2=(17.3±0.1) gm=m2-m1=17.3-10.1=7.2Δm=Δm2+Δm1=0.1+0.1=0.2∴m1±m2=[7.2±0.2] g

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The mass of a beaker is $(10.1 \pm 0.1) g$ when empty and $(17.3 \pm 0.1) g$ when filled with a liquid. The mass of the liquid with possible limits of accuracy is

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$(7.2 \pm 0.2) g$

$(7.2 \pm 0.1) g$

$(7.2 \pm 0.3) g$

$(7.1 \pm 0.2) g$

answer is A.

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$\begin{array}{rcl} m_{1} & = & (10.1 \pm 0.1) g \\ m_{2} & = & (17.3 \pm 0.1) g \\ m & = & m_{2} - m_{1} = 17.3 - 10.1 = 7.2 \\ Δm & = & {Δm}_{2} + {Δm}_{1} = 0.1 + 0.1 = 0.2 \\ ∴ m_{1} \pm m_{2} & = & [7.2 \pm 0.2] g \end{array}$

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The mass of a beaker is (10.1±0.1)g when empty and (17.3±0.1)g when filled with a liquid. The mass of the liquid with possible limits of accuracy is

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The mass of a beaker is $(10.1 \pm 0.1) g$ when empty and $(17.3 \pm 0.1) g$ when filled with a liquid. The mass of the liquid with possible limits of accuracy is