Answer

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**Hint:**We all have seen vernier calipers in our laboratory which are used to measure outer dimensions, height or width of microscale objects. While in our question we used calipers to measure the diameter of the brass metal bob. The main key of this question is to know about the significant figures and rounding off method.

**Complete step by step answer:**

We are provided with a brass metal bob with diameter $1.92 \times {10^{ - 2}}m$

Radius of the bob $r = \dfrac{d}{2}$

$r = \dfrac{{1.92 \times {{10}^{ - 2}}}}{2}$

Hence, $r = 0.86 \times {10^{ - 2}}m$

Now, finding volume of the brass metal bob $V = \dfrac{4}{3}\pi {r^3}$

$V = \dfrac{4}{3}\pi {\left( {0.86 \times {{10}^{ - 2}}} \right)^3}$

$\Rightarrow V = 3.7 \times {10^{ - 6}}{m^3}$

Now, the density of the brass bob is mass per unit volume of the bob.

$D = \dfrac{M}{V}$

$\Rightarrow D = \dfrac{{29.150 \times {{10}^{ - 3}}}}{{3.7 \times {{10}^{ - 6}}}} \\

\therefore D= 7.86 \times {10^3}kg{m^{ - 3}}$

Hence the density of brass bob is $7.86 \times {10^3}kg{m^{ - 3}}$.Now, we should know a little bit about the significant figures. Significant figures are the number of digits that contributes to the accuracy of a value. In our answer we had three significant figures. And we are asked to round it off too. So, the round off should be $7.9 \times {10^3}kg{m^{ - 3}}$.

**Hence, the answer of the question should be $7.9 \times {10^3}kg{m^3}$.**

**Note:**Rules for significant figures:

-All non-zero numbers are significant. E.g.:67

-Zero between non-zero numbers are significant. Eg:9087

-Trailing zero to the right are significant while to the left are not significant. Eg. 660 is significant and 097 is not significant.

Rules for rounding off is:

-If the number is followed by $5$, $6$, $7$, $8$, $9$ then round the number up.

-If the number you are rounding is followed by $1$, $2$, $3$ or $4$ then round the number down.

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