The dimensions of an aluminum block are 3.0 , | vedclass

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Question

(A) Volume of the block in cubic inches = $3.0 	imes 4.0 	imes 5.0 = 60.0 \, 	ext{inch}^3$. Conversion factor: $1 \, 	ext{inch} = 2.54 \, 	ext{cm

Vedclass · Accepted Answer

$2.65 	imes 10^4$

Vedclass · Answer

(A) Volume of the block in cubic inches = $3.0 	imes 4.0 	imes 5.0 = 60.0 \, 	ext{inch}^3$. Conversion factor: $1 \, 	ext{inch} = 2.54 \, 	ext{cm}$,so $1 \, 	ext{inch}^3 = (2.54)^3 \, 	ext{cm}^3 = 16.387 \, 	ext{cm}^3$. Volume in $	ext{cm}^3 = 60.0 	imes 16.387 = 983.22 \, 	ext{cm}^3$. Mass = $	ext{Density} 	imes 	ext{Volume} = 2.7 \, 	ext{g/cm}^3 	imes 983.22 \, 	ext{cm}^3 = 2654.694 \, 	ext{g}$. Rounding to two significant figures (based on the given dimensions),the mass is approximately $2.7 	imes 10^3 \, 	ext{g}$. However,calculating $60 	imes (2.54)^3 	imes 2.7$ gives $2654.69$,which is $2.65 	imes 10^3 \, 	ext{g}$. Given the options,the calculation $2.65 	imes 10^4$ appears to be a typo in the provided options' exponent,but $2.65$ is the correct numerical coefficient.

List-$A$	List-$B$
$(1)$ Electric current	$(A)$ Kilogram
$(2)$ Luminous intensity	$(B)$ Mole
$(3)$ Amount of substance	$(C)$ Ampere
$(4)$ Mass	$(D)$ Candela

Column-$A$	Column-$B$
$1$. Femtometer	$(P)$. $10^{-3} \ \mu m$
$2$. Picometer	$(Q)$. $10^{-3} \ mm$
$3$. Nanometer	$(R)$. $10^{-3} \ pm$
$4$. Micrometer	$(S)$. $10^{-3} \ cm$
	$(W)$. $10^{-3} \ nm$

The dimensions of an aluminum block are $3.0 \, \text{inch} \times 4.0 \, \text{inch} \times 5.0 \, \text{inch}$. What is the mass of this block in grams? The density of aluminum is $2.7 \, \text{g/cm}^3$ and $1 \, \text{inch} = 2.54 \, \text{cm}$.

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