Matter

10 methods · 15 questions

★ Signature method unlocked: the technique this chapter rewards most often, with 3 worked questions.

The full chapter (10 methods · 15 worked questions) comes with Platform Access. See pricing

Read the units in the options before computing anything: they tell you which system to work in, and on a question that offers several expressions they decide the answer on their own. With no calculator, every conversion avoided is time saved and one slip fewer.

Trigger: options carrying units, or a question that lists several quotients and asks which of them give a stated quantity.

Instances:
(i) options in $g cm^{- 3}$ : stay in grams and centimetres from start to finish, never convert to SI and back;
(ii) options in pascals: convert first, since $ρ g h$ fed with $g cm^{- 3}$ and centimetres does not produce pascals;
(iii) mixed-looking units such as $N cm^{- 2}$ with $cm^{2}$ are consistent as given, so use them as given and only bring in $g$ at the last step;
(iv) a list of expressions: reduce each to base units and keep only those whose dimensions match the quantity asked, because the examiner usually arranges for every option to carry the same number.

Linked questions (3)

NSAA 2016 Section 1 Part B (ESAT Physics) · Q30

A cubic block of side

10 cm

has a hole of square cross-section, of side

5.0 cm

, running straight through it from one face to the opposite face. The weight of the block is

30 N

.

What is the density of the material from which the block is made?

(The gravitational field strength

g

10 N kg^{- 1}

A 3D perspective diagram of a cube of side 10 cm, with a straight hole of square cross-section (side 5.0 cm) cut through it from the top face to the opposite (bottom) face, shown with dashed lines marking the hidden edges of the hole inside the block. — illustrative

A $0.30 g cm^{- 3}$
B $0.40 g cm^{- 3}$
C $0.60 g cm^{- 3}$
D $1.2 g cm^{- 3}$
E $3.0 g cm^{- 3}$
F $4.0 g cm^{- 3}$
G $6.0 g cm^{- 3}$
H $12 g cm^{- 3}$

Worked solution

Answer:F

The overall cube has volume

10 \times 10 \times 10 = 1000 cm^{3}

. The square hole has cross-section

5.0 \times 5.0 = 25 cm^{2}

and runs all the way through one

10 cm

dimension of the cube, so it removes a volume of

25 \times 10 = 250 cm^{3}

.

The volume of material actually present is

1000 - 250 = 750 cm^{3}

.

The mass of the block is

m = \frac{W}{g} = \frac{30}{10} = 3.0 kg = 3000 g

.

Density

= \frac{m}{V} = \frac{3000}{750} = 4.0 g cm^{- 3}

.

The answer is F.

NSAA 2017 Section 1 Part B (ESAT Physics) · Q32

Consider the following four calculations:

1.

(200 kg m s^{- 1}) \div (5 m s^{- 1})

(200 J) \div (5 m)

(200 kg m^{- 3}) \div (5 m^{3})

(200 N) \div (5 m s^{- 2})

Which two calculations result in the answer

40 kg

A1 and 2
B1 and 3
C1 and 4
D2 and 3
E2 and 4
F3 and 4

Worked solution

Answer:C

Every calculation gives the numerical value

200 \div 5 = 40

, so the question is really about which two also have the units of kilograms.

1.

\frac{kg m s ^{- 1}}{m s ^{- 1}} = kg

: matches,

40 kg

(momentum divided by velocity is mass).

2.

\frac{J}{m} = N

(energy divided by distance is force), not kg: rejected despite the matching number.

3.

\frac{kg m ^{- 3}}{m ^{3}} = kg m^{- 6}

, not kg: rejected.

4.

\frac{N}{m s ^{- 2}} = \frac{kg m s ^{- 2}}{m s ^{- 2}} = kg

: matches,

40 kg

(force divided by acceleration is mass, by Newton's second law).

So calculations 1 and 4 give

40 kg

.

The answer is C.

NSAA 2021 Section 1 Part B (ESAT Physics) · Q39

Liquid X has density

0.80 g cm^{- 3}

and liquid Y has density

1.0 g cm^{- 3}

80 cm^{3}

of liquid X and

100 cm^{3}

of liquid Y are poured into a cylindrical container and allowed to settle. The two liquids do not mix or react.

The internal cross-sectional area of the container is

20 cm^{2}

. The base of the container rests on a horizontal surface.

What is the pressure due to the liquids at a height of

4.0 cm

above the interior of the base of the container?

(gravitational field strength

= 10 N kg^{- 1}

)

A $10 Pa$
B $40 Pa$
C $42 Pa$
D $50 Pa$
E $100 Pa$
F $400 Pa$
G $420 Pa$
H $500 Pa$

Worked solution

Answer:G

Liquid Y is denser, so it settles at the bottom, with liquid X floating on top.

Height of the liquid Y layer:

h_{Y} = \frac{V _{Y}}{A} = \frac{100 cm ^{3}}{20 cm ^{2}} = 5.0 cm .

Height of the liquid X layer:

h_{X} = \frac{V _{X}}{A} = \frac{80 cm ^{3}}{20 cm ^{2}} = 4.0 cm .

So liquid Y occupies from

0

5.0 cm

above the base, and liquid X occupies from

5.0 cm

9.0 cm

.

At a height of

4.0 cm

(still within the Y layer, since

4.0 < 5.0

), the liquid pressure comes from everything ABOVE that point: the remaining

5.0 - 4.0 = 1.0 cm

of liquid Y, plus the full

4.0 cm

of liquid X above it.

Converting to SI units (

ρ_{Y} = 1000 kg m^{- 3}

ρ_{X} = 800 kg m^{- 3}

, heights in metres):

P = ρ_{Y} g (0.010 m) + ρ_{X} g (0.040 m)

P = (1000) (10) (0.010) + (800) (10) (0.040) = 100 Pa + 320 Pa = 420 Pa .

The answer is G.