Question

$5.74gm$ of a substance a volume of $1.2c{m^3}$ .Calculate its density with due regard for significant figures.
(A) $4.8gc{m^{ - 3}}$
(B) $2.3gc{m^{ - 3}}$
(C) $2.4gc{m^{ - 3}}$
(D) $3.8gc{m^{ - 3}}$

Answer 1

Hint : Use the definition of density and use the concept of significant figures to find the density of the substance up to significant figures. The density of a substance of mass $m$ occupying volume $V$ is given by, $\rho = \dfrac{m}{V}$ .

Complete Step By Step Answer:
We know that the substance of mass $m$ occupying volume $V$ has a density $\rho = \dfrac{m}{V}$ .
Now, we have to find the density of it with regard of significant figures,
Let's first calculate the density. Putting $m = 5.74gm$ and $V = 1.2c{m^3}$ we get,
density $\rho = \dfrac{{5.74}}{{1.2}}gc{m^{ - 3}}$ .
it becomes, $\rho = 4.78333gc{m^{ - 3}}$
Now, here we have $m = 5.74gm$ contains up to second decimal place and $V = 1.2c{m^3}$ contains only up to first decimal place. Since, the volume contains the least digits after decimal point, hence we have to get the density up to two significant figures or up to first decimal place.
Now, to round off we have to look at the number $4.78$ . Since, the second decimal place contains a number which is greater than $5$ . So, the first decimal place will increase by one point. So, Rounded off figure of density will be, $\rho = 4.8gc{m^{ - 3}}$ .
So, the density of the substance will be $4.8gc{m^{ - 3}}$ with regard for significant figures.
Hence, option ( A) is correct.

Note :
If The approximation is done of a number is done say up to ${n^{th}}$ digit, we have to check the ${(n + 1)^{th}}$ if it is greater ,less or equal to $5$ . If ${(n + 1)^{th}}digit \geqslant 5$ we increase the ${n^{th}}$ digit by $1$ and if ${(n + 1)^{th}}digit < 5$ we keep the ${n^{th}}$ digit as it is and get the approximate value of the number up to ${n^{th}}$ digit .