Courses
Courses for Kids
Free study material
Offline Centres
More

A metallic sphere weighs $210g$ in air, $180g$ water, and $120g$ an unknown liquid. Find the density of metal and liquid.
\[\left( a \right)\]$3,7$
\[\left( b \right)\]$7,3$
\[\left( c \right)\]$4,6$
\[\left( d \right)\]$6,4$

seo-qna
Last updated date: 29th Feb 2024
Total views: 20.4k
Views today: 1.20k
IVSAT 2024
Answer
VerifiedVerified
20.4k+ views
Hint: The quantitative relation of density (mass of a unit volume) of a substance to the density of given reference material is the Relative density or specific gravity. Relative density typically suggests that denseness concerning water. It’s outlined as a quantitative relation of the density of an explicit substance with that of water.

Formula used:
The relative density of metal $ = \dfrac{{{\text{Weight in air}}}}{{Change{\text{ in weight of water}}}}$

Complete step by step solution: In the above question there is a sphere whose weights get vary as the medium gets changed in an unknown liquid. Therefore in this question, we are going to find the density in both liquid and metal.
As we know,
The relative density of metal $ = \dfrac{{{\text{Weight in air}}}}{{Change{\text{ in weight of water}}}}$
In terms of the statement, we can say that relative density is equal to the weight in the air upon the change in weight of the water.
Therefore putting the values, we get
$ \Rightarrow \dfrac{{210}}{{210 - 180}}$
$ \Rightarrow 7$
Therefore the density of the metal will be equal to $7g/c{m^2}$
Now since
The upthrust in the liquid will be equal to the change in the weight of the liquid.
Therefore it can be written as,
$ \Rightarrow \left( {{V_{solid}}} \right)\left( {{\rho _{liquid}}} \right)g$
Or it can be written as
$ \Rightarrow \vartriangle \omega \propto {\rho _{liquid}}$
Therefore, mathematically it will be expressed as
$ \Rightarrow \dfrac{{\vartriangle {\omega _l}}}{{\vartriangle {\omega _\omega }}} = \dfrac{{{\rho _l}}}{{{\rho _\omega }}}$
Therefore,
$ \Rightarrow {\rho _l} = \dfrac{{\vartriangle {\omega _l}}}{{\vartriangle {\omega _\omega }}}{\rho _\omega }$
Now we will put the values of it, we get
$ \Rightarrow \left( {\dfrac{{210 - 120}}{{210 - 180}}} \right)\left( 1 \right)gm/c{m^3}$
$ \Rightarrow 3g/c{m^3}$

Therefore the answer will be an option $b$

Notes we can measure denseness (properly called specific gravity) employing a measuring system if the substance is a liquid, like beer or wine. If employing a density bottle, we get SG instead of pure density, though we'll convert into density by using the density of water at the actual temperature taken throughout the measurement. One cannot use a cheap instrument, to my information, to measure density directly.