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A metal cylinder measures 3.0 cm in diameter, is 12.0 cm high, and has a mass of 750 g. The cylinder hangs from a string attached to a scale that measures weight in Newtons (N). The cylinder is lowered into a cylindrical beaker of water until it is submerged completely. The diameter of the beaker is 10.0 cm. (a) By how much does the height of the water in the beaker rise when the cylinder is submerged? (b) What weight does the scale read after the cylinder is submerged?

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lublana

Answer with Explanation:

We are given that

Diameter of cylinder,[tex]d_1=[/tex]3 cm

Radius,[tex]r_1=\frac{d_1}{2}=\frac{3}{2}=1.5cm=1.5\times 10^{-2} m[/tex]

1m=100 cm

Height,h=12 cm=[tex]\frac{12}{100}=0.12 m[/tex]

Mass,m=750 g=[tex]\frac{750}{1000}=0.75kg[/tex]

1kg=1000 g

Diameter of beaker,[tex]d_2=10 cm[/tex]

Radius of beaker,[tex]r_2=\frac{d_2}{2}=\frac{10}{2}=5cm[/tex]

A.[tex]\pi r^2_2h_2=\pi r^2_1 h_1[/tex]

[tex]h_2=\frac{r^2_1}{r^2_2}h[/tex]

[tex]h_2=\frac{(1.5)^2}{(5)^2}\times 12=1.08 cm[/tex]

Hence, the height of the water in the beaker rises when the cylinder is submerged=1.08 cm

B.Weight read on the scale after the cylinder is submerged

[tex]w=mg-\rho A_1h_1g[/tex]

Where [tex]g=9.8 m/s^2[/tex]

[tex]A_1=\pi r^2_1[/tex]

Density of water=[tex]\rho=1000kg/m^3[/tex]

[tex]w=0.75\times 9.8-1000\times \pi(1.5\times 10^{-2})^2\times 0.12\times 9.8[/tex]

[tex]w=6.52 N[/tex]

ajeigbeibraheem

The height at which the water rose in the beaker after which the cylinder gets submerged is 1.08 cm. The new weight of the scale after the cylinder is being submerged is 6.52 N.

Suppose we denote h to be the height at which the water rose in the beaker.

Then, when the cylinder gets submerged in the beaker, there is a replacement of volume similar to the amount of water.

∴

πr²h = πR²H

Given that:

  • the radius of the beaker (r) = 10 cm/2 = 5 cm
  • the radius of the cylinder (R) = 3.0 cm/2 = 1.5 cm
  • the height of the cylinder (H) = 12 cm

Then;

Making (h) the subject of the formula from the above equation, we have:

[tex]\mathbf{h = \dfrac{R^2 H}{r^2}}[/tex]

[tex]\mathbf{h = \dfrac{1.5^2 \times 12}{5^2}}[/tex]

[tex]\mathbf{h = \dfrac{2.25 \times 12}{25}}[/tex]

h = 1.08 cm

The initial weight loss refers to the weight of the water column substituted by the cylinder.

However, the new weight of the cylinder after being submerged is computed as:

= mg - ρ(πR²H)g

where;

  • mass = 750 g
  • density of water ρ = 1 g/c  

= mg - ρ(πR²H)g

= (750× 9.8) - 1(π× 1.5²× 12)9.8

= 7350 - 831.265

= 6.52 N

Therefore, we can conclude that the height at which the water rose in the beaker after which the cylinder gets submerged is 1.08 cm. The new weight of the scale after the cylinder is being submerged is 6.52 N.

Learn more about the volume of the cylinder here:

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