Question

Two bodies of masses (15.6 +- 0.2)g and (23.9 +- 0.3)g are put in a bag of negligible mass what is the total mass of bag

Answer 1

Answer:

Two bodies of mass are (15.6+-0.2) and(23.9+-0.3)

M1+M2 =(15.6+23.9)=39.5

∆m1+∆m2=(+-0.2)+(0.5)=0.5

So total masses are 39.5+-0.5

Answer 2

Answer:

The total mass of bag = $(39.5 \pm 0.5 )gm$

Explanation:

Given:

mass, $m_1 = (15.6 \pm 0.2 ) gm$

$m_2 = (23.9 \pm 0.3 ) gm$

From the rules of propagation of errors, we know that when the quantities are added or subtracted their errors get added.

Therefore,

Mass of bag, $M=$ $m_1+m_2$

$M=m_1+m_2=15.6+23.9 = 39.5gm$

and, $\Delta M = \Delta m_1 + \Delta m_2$

$\Delta M = 0.2+0.3 = 0.5gm$

Therefore, mass of the bag is, $M = (39.5 \pm 0.5 )gm$

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