Question

a stone weights 392.4 n in air and 196.2N in water specific gravity of stone is​

Answer 1

Answer:

w= wt of stone in air             w°= wt of stone in water

specific gravity of stone =   w/(w-w°)

                                        =  392.4/(392.4-196.2)

                                         = 392.4/196.2

                                         = 2

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Answer 2

Answer:

Specific gravity of stone is​ equal to 2.

Explanation:

Given, the weight of stone in air = 392.4N

The weight of stone in water = 196.2N

Density = mass/volume

mass = density×Volume

In air, $392.4=Vdg$                                                      ...............(1)

Here, d is density of the stone.

In water, $196.2=Vdg$                                                 ...............(2)

Density of water d=1000kgm⁻³ and g=10 put in eq.(2);

$196.2=1000*10*V$

$V=1.962*10^{-2}m^{3}$

Put the value of V in equation (1):

$392.4=(1.962*10^{-2})(10)*d$

$d=2000$

Specific gravity or relative density can be measured from the ratio of density of substance to the density of reference material.

$Specific\;\; gravity=\frac{Density \; \; of \; \; stone}{Density \; \; of \; \; water}$

specific gravity of stone $=\frac{2000}{1000}=2$

Therefore, the specific gravity of stone is 2 when the weight of stone is 392.4N in air and 196.2N in water.