A rigid cross-shaped structure having four arms PO, SO, QO and RO, each 1.00 m long, is pivoted at O. Forces act on the ends of the arms and on the midpoints of the arms as shown.

Question 44

A rigid cross-shaped structure having four arms PO, SO, QO and RO, each 1.00 m long, is pivoted at O. Forces act on the ends of the arms and on the midpoints of the arms as shown.

What is the magnitude of the resultant moment on the structure about O?

A 45 N m                     B 90 N m                     C 120 N m                  D 190 N m

Reference: Past Exam Paper – November 2016 Paper 12 Q17

Solution:

Answer: C.

In this question, we need to carefully identify the forces that cause a clockwise moment and those that cause an anticlockwise moment.

Consider RS.

Consider the two 70 N forces. Both of them cause anticlockwise moments. They are each 1.00 m from the pivot.

As for the two 20 N forces, the one on arm OS cause a clockwise moment while that one arm RO cause an anticlockwise moment. But since they are at equal distances from the pivot, they cancel out each other. Thus, they can be neglected.

The 30 N force on arm OS is at 0.50 m from the pivot and also causes an anticlockwise moment.

Consider PQ.

The two 50 N forces cause clockwise moments. They are at 0.50 m from the pivot.

The 30 N force on arm PO is at 0.50 m from the pivot and causes an anticlockwise moment.

Sum of clockwise moments = (50 × 0.5) + (50 × 0.5) = 50 Nm

Sum of anticlockwise moments = (70 × 1) + (70 × 1) + (30 × 0.5) + (30 × 0.5)

Sum of anticlockwise moments = 170 Nm

Resultant moment = 170 – 50 = 120 Nm

Explain briefly the main principles of the use of magnetic resonance to obtain diagnostic information about internal body structures.

Question 15

Explain briefly the main principles of the use of magnetic resonance to obtain diagnostic information about internal body structures. [8]

Reference: Past Exam Paper – June 2011 Paper 42 & 43 Q10

Solution:

A strong / large (uniform) magnetic field is applied which causes the nuclei to precess / rotate about the field direction.

A (r.f) radio frequency pulse is applied at the Larmor frequency which causes resonance / nuclei absorb energy.

On relaxation / de-excitation, the nuclei emit r.f pulse which is detected and processed.     

Superimposing a non-uniform field on the uniform field allows the position of the resonating nuclei to be determined and for the location of detection to be changed.

A thin rectangular slice of aluminium has sides of length 65 mm, 50 mm and 0.10 mm, as shown in Fig. 9.1.

Question 14

A thin rectangular slice of aluminium has sides of length 65 mm, 50 mm and 0.10 mm, as shown in Fig. 9.1.

Fig. 9.1 (not to scale)

Some of the corners of the slice are labelled.

A current I of 3.8 A is normal to face RSXY of the slice.

In aluminium, the number of free electrons per unit volume is 6.0 × 1028 m−3.

A uniform magnetic field of magnetic flux density B equal to 0.13 T is normal to face QRYZ of the aluminium slice in the direction from Q to P.

A Hall voltage VH is developed across the slice and is given by the expression

VH = BI / ntq

.

(a) Use Fig. 9.1 to state the magnitude of the distance t. [1]

(b) Calculate the magnitude of the Hall voltage VH. [2]

[Total: 3]

Reference: Past Exam Paper – June 2016 Paper 41 & 43 Q9

Solution:

(a)

0.10 mm

{In the equation, t is the thickness of materials through which the current passes.}

(b)

{V_H = BI / ntq

Thickness t should be in metres.

Charge of electrons: q = 1.60×10^-19}

VH = (0.13 × 3.8) / (6.0×10²⁸ × 0.10×10^-3 × 1.60×10^-19)

VH = 5.1 × 10^-7 V