First slide

Accuracy; precision of instruments and errors in measurements

Question

The volumes of two bodies are measured to be $V_{1} = (10.2 \pm 0.02) {cm}^{3}$ and $V_{2} = (6.4 \pm 0.01) {cm}^{3}$ . Calculate sum and difference in volumes with error limits.

Moderate

A

$(16.6 \pm 0.02) {cm}^{3} and (3.8 \pm 0.02) {cm}^{3}$
B

$(16.6 \pm 0.03) {cm}^{3} and (3.8 \pm 0.03) {cm}^{3}$
C

$(19 \pm 0.03) {cm}^{3} a n d (3.8 \pm 0.03) {cm}^{3}$
D

$(16.6 \pm 0.03) {cm}^{3} a n d (8.3 \pm 0.03) {cm}^{3}$

Solution

$V_{1} = (10.2 \pm 0.02) {cm}^{3}$
$V_{2} = (6.4 \pm 0.01) {cm}^{3}$
$Δ V = \pm (Δ V_{1} + Δ V_{2})$
$= \pm (0.02 + 0.01) {cm}^{3} = \pm 0.03 {cm}^{3}$
$V_{1} + V_{2} = (10.2 + 6.4) {cm}^{3} = 16.6 {cm}^{3}$
$V_{1} - V_{2} = (10.2 - 6.4) {cm}^{3} = 3.8 {cm}^{3}$
Hence, sum of volumes $= (16.6 \pm 0.03) {cm}^{3}$
and difference of volumes $= (3.8 \pm 0.03) {cm}^{3}$

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