Tag: fundamental and derived units

Questions Related to fundamental and derived units

Among the following identify derived quantities?

physics measurements and units some examples of derived units fundamental and derived quantities fundamental and derived units

  1. Speed

  2. Temperature

  3. Volume

  4. Mass

Show answer
Correct Option: A,C
Explanation:

(A) Speed $\displaystyle = \frac{distance}{time}$ - a derived quantity of length and time.
(B) Temperature - fundamental quantity.
(C) Volume (cube of length) - a derived quantity of length.
(D) Mass - fundamental quantity.

Which of the following is larger unit of density ?

physics measurements and units some examples of derived units fundamental and derived quantities fundamental and derived units

  1. $gm^{-3}$

  2. $gkm^{-3}$

  3. $kgm^{-3}$

  4. $gcm^{3}$

Show answer
Correct Option: D
Explanation:
m is the mass of an object divided by its volume. Density often has units of grams per cubic centimeter (g/cm3). Remember, grams is a mass and cubic centimeters is a volume (the same volume as 1 milliliter).

The density of a cuboid of mass 200 g with dimensions 2 cm $\times$ 4 cm $\times$ 5 cm is

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 1000 kg m$^{-3}$

  2. 3000 kg m$^{-3}$

  3. 5000 kg m$^{-3}$

  4. 2000 kg m$^{-3}$

Show answer
Correct Option: C
Explanation:
The density of a body is defined as the amount of substance contained per unit volume.
Density $=\dfrac{mass}{volume}$

$\displaystyle d = \frac{0.2 kg}{40  cm^3} = \frac{0.2 \times 10^6}{40} kg  m^{-3}$

$=5000  kg  m^{-3}$

A wire of length 50 cm has a mass of 20 gm. It its radius is halved by stretching, its new mass per unit length will be

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 0.4 g cm$^{-1}$

  2. 0.2 kg m$^{-1}$

  3. 0.1 g cm$^{-1}$

  4. 0.2 g cm$^{-1}$

Show answer
Correct Option: C
Explanation:

On stretching, volume remains constant. Hence on decreasing the radius by half, the length of the wire becomes quadrupled. 
Mass per unit length $\displaystyle = \frac{20}{4 \times 50 cm}$
$=0.1  g  cm^{-1}$

Density of water is :

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 1 g cm$^{-3}$

  2. 100 kg m$^{-3}$

  3. 1000 kg m$^{-3}$

  4. None

Show answer
Correct Option: A,C
Explanation:

 Density of water = 1 g $ cm^{-3 }$ = 1000 kg $ m^{-3}$

so option (A) and option (C) are correct

If 5 litres of kerosene has a mass of 5 kg, then what is the density of kerosene?

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 500 kg/$m^3$

  2. 1000 kg/$m^3$

  3. 100 kg/$m^3$

  4. 50 kg/$m^3$

Show answer
Correct Option: B
Explanation:

Density of kerosene $= \displaystyle \frac{mass}{volume}$
$\displaystyle = \frac{5  kg}{5  litres}= \frac{5kg}{5 \times 10^{-3}m^3} = 1000 kg  m^{-3}$

Two spheres are having the ratio of their densities 1:3 and masses in 3:4 respectively. Find the ratio of their volumes. i.e $\dfrac{V _1}{V _2}$

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. 1:3

  2. 3:1

  3. 4:9

  4. 9:4

Show answer
Correct Option: D
Explanation:

$\displaystyle \frac{d _1}{d _2} = \frac{1}{3} ; \frac{m _1}{m _2} = \frac{3}{4}$
$d = m/v$
$\therefore \displaystyle \frac{V _1}{V _2}  = \frac{m _1}{m _2} \times \frac{d _2}{d _1} = \frac{3}{4} \times \frac{3}{1} = 9:4$

Density of air at room temperature is

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $120 \ {g\ cm}^{-3}$

  2. $120 \ {g\ mm}^{-3}$

  3. $12 \ {g\ m}^{-3}$

  4. $1.3 \ {kg\ m}^{-3}$

Show answer
Correct Option: D
Explanation:

The density of air at NTP is $1.293 kg m^{-3}$.

So after rounding, it is $1.3 kg m^{-3}$.

If the mass of a body is 12.1 g and its density is 2.2 g/cc, its volume is.

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $5.5 cm^3$

  2. $8$ cc

  3. $11$ cc

  4. $55$ cc

Show answer
Correct Option: A
Explanation:

Given:

mass $(m)=12.1g$
density $(d)=2.2g/cc$
Let the volume be $v$
As $density=\dfrac{mass}{volume}$
$d=\dfrac{m}{v}$
$v=\dfrac{m}{d}$
$v=\dfrac{12.1}{2.2}=5.5cc$

The density of water is equal to :

chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

  1. $10^{-3}$ $kgm^{-3}$

  2. $10^{-2}$ $kgm^{-3}$

  3. $10^{2}$ $kgm^{-3}$

  4. $10^{3}$ $kgm^{-3}$

Show answer
Correct Option: D
Explanation:

density of water in S.I. unit= ${10}^3$ kg/${m}^3$