| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dimensional Analysis |
| Type | Derive dimensions from formula |
| Difficulty | Moderate -0.8 This is a straightforward dimensional analysis question requiring only systematic application of standard techniques: rearranging formulas to find dimensions, unit conversions using scale factors, checking dimensional consistency, and solving simultaneous equations from dimensional analysis. All parts are routine textbook exercises with no novel problem-solving required, making it easier than average despite being multi-part. |
| Spec | 6.01a Dimensions: M, L, T notation6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions6.01e Formulate models: dimensional arguments |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([g] = LT^{-2}\) | B1 | |
| \([S] = \left[\frac{mg}{2a}\right] = \frac{M(LT^{-2})}{L} = MT^{-2}\) | M1, E1 | Obtaining dimensions of \(S\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.073 \times 1000 \times 60^2\) | M1M1 | For \(\times 1000\) and \(\times 60^2\); Give M1 for \(\frac{1}{1000} \times \frac{1}{60^2}\); 3 600 000 implies M2; \(2.8\times10^{-7}\) or \(2.0\times10^{-8}\) implies M1 |
| \(= 262800\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\rho] = ML^{-3}\) | B1 | |
| \([\text{RHS}] = \frac{MT^{-2}}{(ML^{-3})(LT^{-2})(L)} = L\) | M1 | Obtaining dimensions of RHS |
| \([\text{LHS}] = L\), so it is dimensionally consistent | E1 | Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = \frac{2S}{\rho g h} = \frac{2\times0.073}{1000\times9.8\times0.25}\) | M1 | Correct explicit numerical expression for \(r\). May have 25 for \(h\) |
| \(= 5.96\times10^{-5}\) m (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(L = (MT^{-2})^\alpha (ML^{-3})^\beta (LT^{-2})^\gamma\) | ||
| \(\alpha+\beta=0,\quad -3\beta+\gamma=1,\quad -2\alpha-2\gamma=0\) | M2 | For 3 equations (give M1 for 2 equations); CAO |
| \(\alpha=\frac{1}{2},\quad \beta=-\frac{1}{2},\quad \gamma=-\frac{1}{2}\) | A2 | Give A1 (dep M2) for two correct; If \([\rho]\) and/or \([g]\) wrong, give A1 for at least two correct values FT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(d = kS^{\frac{1}{2}}\rho^{-\frac{1}{2}}g^{-\frac{1}{2}}\) | ||
| \(d=0.0044,\ S=0.073,\ \rho=1000,\ g=9.8 \Rightarrow k=1.612\) | M1 | Obtaining a value for \(k\); Or \(\left(\frac{0.487}{0.073}\right)^{\frac{1}{2}}\) or \(\left(\frac{13500}{1000}\right)^{-\frac{1}{2}}\) |
| \(d_M = 1.612\times0.487^{\frac{1}{2}}\times13500^{-\frac{1}{2}}\times9.8^{-\frac{1}{2}}\) | M1 | Obtaining expression for \(d_M\) |
| Depth for mercury is \(0.00309\) m (3 sf) | A1 | CAO or 0.31 cm; But A0 for 0.31 m |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[g] = LT^{-2}$ | B1 | |
| $[S] = \left[\frac{mg}{2a}\right] = \frac{M(LT^{-2})}{L} = MT^{-2}$ | M1, E1 | Obtaining dimensions of $S$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.073 \times 1000 \times 60^2$ | M1M1 | For $\times 1000$ and $\times 60^2$; Give M1 for $\frac{1}{1000} \times \frac{1}{60^2}$; 3 600 000 implies M2; $2.8\times10^{-7}$ or $2.0\times10^{-8}$ implies M1 |
| $= 262800$ | A1 | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\rho] = ML^{-3}$ | B1 | |
| $[\text{RHS}] = \frac{MT^{-2}}{(ML^{-3})(LT^{-2})(L)} = L$ | M1 | Obtaining dimensions of RHS |
| $[\text{LHS}] = L$, so it is dimensionally consistent | E1 | Correctly shown |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = \frac{2S}{\rho g h} = \frac{2\times0.073}{1000\times9.8\times0.25}$ | M1 | Correct explicit numerical expression for $r$. May have 25 for $h$ |
| $= 5.96\times10^{-5}$ m (3 sf) | A1 | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $L = (MT^{-2})^\alpha (ML^{-3})^\beta (LT^{-2})^\gamma$ | | |
| $\alpha+\beta=0,\quad -3\beta+\gamma=1,\quad -2\alpha-2\gamma=0$ | M2 | For 3 equations (give M1 for 2 equations); CAO |
| $\alpha=\frac{1}{2},\quad \beta=-\frac{1}{2},\quad \gamma=-\frac{1}{2}$ | A2 | Give A1 (dep M2) for two correct; If $[\rho]$ and/or $[g]$ wrong, give A1 for at least two correct values FT |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $d = kS^{\frac{1}{2}}\rho^{-\frac{1}{2}}g^{-\frac{1}{2}}$ | | |
| $d=0.0044,\ S=0.073,\ \rho=1000,\ g=9.8 \Rightarrow k=1.612$ | M1 | Obtaining a value for $k$; Or $\left(\frac{0.487}{0.073}\right)^{\frac{1}{2}}$ or $\left(\frac{13500}{1000}\right)^{-\frac{1}{2}}$ |
| $d_M = 1.612\times0.487^{\frac{1}{2}}\times13500^{-\frac{1}{2}}\times9.8^{-\frac{1}{2}}$ | M1 | Obtaining expression for $d_M$ |
| Depth for mercury is $0.00309$ m (3 sf) | A1 | CAO or 0.31 cm; But A0 for 0.31 m |
---1 The surface tension of a liquid enables a metal needle to be at rest on the surface of the liquid. The greatest mass $m$ of a needle of length $a$ which can be supported in this way by a liquid of surface tension $S$ is given by
$$m = \frac { 2 S a } { g }$$
where $g$ is the acceleration due to gravity.\\
(i) Show that the dimensions of surface tension are $\mathrm { MT } ^ { - 2 }$.
The surface tension of water is 0.073 when expressed in SI units (based on kilograms, metres and seconds).\\
(ii) Find the surface tension of water when expressed in a system of units based on grams, centimetres and minutes.
Liquid will rise up a capillary tube to a height $h$ given by $h = \frac { 2 S } { \rho g r }$, where $\rho$ is the density of the liquid and\\
$r$ is the radius of the capillary tube. $r$ is the radius of the capillary tube.\\
(iii) Show that the equation $h = \frac { 2 S } { \rho g r }$ is dimensionally consistent.\\
(iv) Find the radius of a capillary tube in which water will rise to a height of 25 cm . (The density of water is 1000 in SI units.)
When liquid is poured onto a horizontal surface, it forms puddles of depth $d$. You are given that $d = k S ^ { \alpha } \rho ^ { \beta } g ^ { \gamma }$ where $k$ is a dimensionless constant.\\
(v) Use dimensional analysis to find $\alpha , \beta$ and $\gamma$.
Water forms puddles of depth 0.44 cm . Mercury has surface tension 0.487 and density 13500 in SI units.\\
(vi) Find the depth of puddles formed by mercury on a horizontal surface.
\hfill \mbox{\textit{OCR MEI M3 2012 Q1 [18]}}