The diameter of a solid metal sphere is measured using a micrometer screw gauge.

Question 26

The diameter of a solid metal sphere is measured using a micrometer screw gauge. The diagram shows an enlargement of the shaft of the micrometer screw gauge when taking the measurement.

The mass of the sphere is 0.450 g.

What is the density of the metal used to make the sphere?

A 965 kg m^-3                    B 1340 kg m^-3                 C 7720 kg m^-3                                 D 10 700 kg m^-3

Reference: Past Exam Paper – June 2016 Paper 11 Q15

Solution:

Answer: C.

From the micrometer reading,

Diameter of sphere = 4.50 + 0.31 = 4.81 mm

Volume of sphere = 4/3 πr³ = 4/3 π(d/2)³ = 4πd³ / 24

Volume of sphere = πd³ / 6 

Density = mass / volume

The quantities need to be in SI units.

Density of metal used = (0.450×10^-3) / [π × (4.81×10^-3)³ / 6]

Density = 7723 ≈ 7720 kg m^-3 

State what is meant by the specific acoustic impedance of a medium.

Question 11

(a) (i) State what is meant by the specific acoustic impedance of a medium. [2]

(ii) The intensity reflection coefficient á is given by the expression

α = (Z2 − Z1)² / (Z2 + Z1)² .

Explain the meanings of the symbols in this expression.

α :

Z2 and Z1: [2]

(b) A parallel beam of ultrasound has intensity I0 as it enters a muscle.

The beam passes through a thickness of 3.4 cm of muscle before being incident on the

boundary with a bone, as shown in Fig. 4.1.

Fig. 4.1

The intensity of the ultrasound beam as it passes into the bone is IT.

Some data for muscle and bone are given in Fig. 4.2.

Fig. 4.2

Calculate the ratio IT / I0. [5]

[Total: 9]

Reference: Past Exam Paper – June 2016 Paper 41 & 43 Q4

Solution:

(a) (i) The specific acoustic impedance of a medium is the product of the speed (of the ultrasound wave) in the medium and the density of the medium.

(ii)

α: ratio of reflected intensity and/to incident intensity

Z1 and Z2: (specific) acoustic impedances of media (on each side of boundary)

(b)

{As the ultrasound beam moves through the muscle, its intensity decreases as the ultrasound is absorbed by the muscle. The intensity of ultrasound transmitted in the muscle is obtained by}

in muscle: IM = I0 e^-μx

{where x is the thickness of the muscle (in m)}

IM = I0 exp(–23 × 3.4×10^-2)

IM / I0 = 0.457              

{This (IM / I0) is the fraction of the incident intensity that has been transmitted in the muscle (as it moved in the muscle).

At the boundary, some of this fraction (IM / I0) that has been transmitted will be reflected while some will be transmitted.

The fraction of the intensity that has been REFLECTED can be obtained by}

at boundary: α = (6.3 – 1.7)² / (6.3 + 1.7)²

α = 0.33

{This (α) is the fraction of the intensity coming to the boundary that has been reflected. To obtain the fraction that has been transmitted at the boundary:

Fraction reflected (α) + Fraction transmitted (IT /IM) = 1

Fraction transmitted (IT /IM) = 1 – fraction reflected (α)}

IT /IM = [(1 – α) =] 0.67

{This (IT /IM) is the fraction of the beam that has been transmitted in the muscle. It is NOT the fraction of the incident intensity I₀. Therefore, we need to take the fraction of the intensity that has been transmitted (IM / I0).

IT / I0 = (IM / I0) × (IT /IM)}

IT / I0 = 0.457 × 0.67

IT / I0 = 0.31