Question

PART ONE
An object weighing 301 N in air is immersed
in water after being tied to a string connected
to a balance. The scale now reads 270 N .
Immersed in oil, the object appears to weigh
276 N .
Find the density of the object.
Answer in units of kg/m^3

PART TWO
Find the density of the oil.
Answer in units of kg/m^3

Answer 1

Answer:

Density of the object is $9708.52$ $kg/m^3$

Density of the oil is $806.46$ $kg/m^3$

Explanation:

Given Information:

$F_{air} =301N\\F_{water} =270N\\F_{oil} =276N$

Part one: Find the density of the object

To find the volume of an object we use

$V=\frac{F_{B} }{D*g}$

Where $F_{B}$ is the Buoyant force and $D$ is the density of the object

The Buoyant force is given by

$F_{B}=F_{air}-F_{water}$

$F_{B}=301-270=31N$

The density of water is $1000$ $kg/m^3$

Therefore Volume will be

$V=31/1000*9.8=0.0031632$ $m^3$

Mass of the object is given by

$m=F_{air} /g$

$m=301/9.8=30.71$ $kg$

Density of the object is given by

$D=m/V$

$D=30.71/0.0031632=9708.52$ $kg/m^{3}$

Part two: Find the density of the oil

The Buoyant force is given by

$F_{B} =F_{air} -F_{oil}$

$F_{B}=301-276=25N$

$V=\frac{F_{B} }{D*g}$

Rearranging above equation yields

$D=\frac{F_{B} }{V*g}$

$D=\frac{25}{0.0031632*9.8}$

$D=806.46$ $kg/m^{3}$

Answer 2

1. The density of the object is equal to 9596.87 $kg/m^3$.

2. The density of the oil is equal to 797.19 $kg/m^3$.

Given the following data:

• Weight of object in air = 301 N

• Weight of object in water = 270 N

• Weight of object in oil = 276 N

Density of water = 1000 $kg/m^3$

Acceleration due to gravity = 9.8 $m/s^2$

1. To find the density of the object:

First of all, we would determine the buoyant force acting on the object by using the formula:

$F_b = F_a - F_w$

Where:

• $F_b$ is the buoyant force.

• $F_a$ is the weight of object in air.

• $F_w$ is the weight of object in water.

Substituting the given parameters into the formula, we have;

$F_b = 301-270\\\\F_b =31\;Newton$

Also, we would find the mass of the object when in air:

$M_a = \frac{F_a}{g} \\\\M_a = \frac{301}{9.8} \\\\M_a =30.71 \;kg$

Next, we would determine the volume of the object by using the formula:

$Volume = \frac{F_b}{D_wg} \\\\Volume = \frac{31 }{1000 \times 9.8} \\\\Volume = \frac{31}{9800} \\\\Volume = 0.0032\;m^3$

For the object's density:

$Density = \frac{Mass}{Volume} \\\\Density = \frac{30.71}{0.0032} \\\\Density = 9,596.87 \;kg/m^3$

2. To find the density of the oil:

First of all, we would determine the buoyant force acting on the object by using the formula:

$F_b = F_a - F_o$

Where:

• $F_b$ is the buoyant force.

• $F_a$ is the weight of object in air.

• $F_o$ is the weight of object in oil.

$F_b = 301-276\\\\F_b =25\;Newton$

For the oil's density:

$Density = \frac{F_b}{vg} \\\\Density = \frac{25}{0.0032 \times 9.8} \\\\Density = \frac{25}{0.03136} \\\\Density = 797.19 \; kg/m^3$

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