Each side of a metallic cube of mass 5.580 te | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Density $ho = \frac{	ext{Mass}}{	ext{Volume}}$.Mass $m = 5.580 	ext{ kg}$.Side length $a = 9.0 	ext{ cm} = 9.0 	imes 10^{-2} 	ext{ m}$.Vol

Vedclass · Accepted Answer

$7.7$

Vedclass · Answer

(B) Density $ho = \frac{	ext{Mass}}{	ext{Volume}}$.Mass $m = 5.580 	ext{ kg}$.Side length $a = 9.0 	ext{ cm} = 9.0 	imes 10^{-2} 	ext{ m}$.Volume $V = a^3 = (9.0 	imes 10^{-2} 	ext{ m})^3 = 729 	imes 10^{-6} 	ext{ m}^3 = 7.29 	imes 10^{-4} 	ext{ m}^3$.Density $ho = \frac{5.580 	ext{ kg}}{7.29 	imes 10^{-4} 	ext{ m}^3} \approx 7654.32 	ext{ kg m}^{-3} = 7.65432 	imes 10^3 	ext{ kg m}^{-3}$.Since the side length $9.0 	ext{ cm}$ has only two significant figures, the final result must be rounded to two significant figures.Rounding $7.65432$ to two significant figures gives $7.7$.Therefore, $X = 7.7$.

Each side of a metallic cube of mass $5.580 \text{ kg}$ is measured to be $9.0 \text{ cm}$. Keeping the significant figures in view, the density of the material of the cube can be best expressed as $X \times 10^3 \text{ kg m}^{-3}$, where the value of $X$ is:

Similar Questions

What is the value of the number $96378$ when rounded off to two significant figures?

If $97.52$ is divided by $2.54$,the correct result in terms of significant figures is

Three measurements are made as $18.425 \, cm$,$7.21 \, cm$,and $5.0 \, cm$. The addition should be written as ........ $cm$.

Taking into account the significant figures,what is the value of $9.99\, m - 0.0099\, m$ (in $, m$)?

Value of round off of numbers $2.745$ and $2.735$ up to $3$ significant digits will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module