A substance of mass 49.53 ~g occupies 1.5 ~cm^3 of volume. The density of the substance (in g .cm^-3 ) with correct number of significant figures is

Correct answer: (A) 33.02. Density =Mass/Volume Given, mass =49.53 ~g ; volume =1.5 ~cm^2 Therefore, Density =49.53/1.5 gcm^-3=33.02 gcm^-1

Q121 / 180 · KCET 2017 · Physics

A substance of mass $ 49.53 \mathrm{~g} $ occupies $ 1.5 \mathrm{~cm}^{3} $ of volume. The density of the substance (in $ \mathrm{g} $
$ \left.\mathrm{cm}^{-3}\right) $ with correct number of significant figures is

Correct answer
A
$ 33.02 $
B
$ 3.300 $
C
$ 3.3 $
D
$ 3.30 $

Solution

Density $=\frac{\text { Mass }}{\text { Volume }}$ Given, mass $=49.53 \mathrm{~g} ;$ volume $=1.5 \mathrm{~cm}^{2}$ Therefore, Density $=\frac{49.53}{1.5} \mathrm{gcm}^{-3}=33.02 \mathrm{gcm}^{-1}$

Hence, the correct answer is option: 'A'.

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A substance of mass \( 49.53 \mathrm{~g} \) occupies \( 1.5 \mathrm{~cm}^{3} \) of volume. The density of the substance (in \( \mathrm{g} \)\( \left.\mathrm{cm}^{-3}\right) \) with correct number of significant figures is

Solution

Relevant Questions

A substance of mass \( 49.53 \mathrm{~g} \) occupies \( 1.5 \mathrm{~cm}^{3} \) of volume. The density of the substance (in \( \mathrm{g} \)
\( \left.\mathrm{cm}^{-3}\right) \) with correct number of significant figures is