# Question 13:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dr}=4\pi r^2$ | B1 | Do not accept $\frac{dV}{dr}=\frac{4}{3}\pi\times 3r^2$ |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt} \Rightarrow \frac{20}{V(0.05t+1)^3}=4\pi r^2\times\frac{dr}{dt}$ | M1 | Uses $\frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt}$ correctly. Follow through on their $\frac{dV}{dr}$. Not enough to just state this |
| $\Rightarrow \frac{dr}{dt}=\frac{15}{4\pi^2 r^5(0.05t+1)^3}$ | dM1, A1 | dM1: Makes $\frac{dr}{dt}$ subject and attempts to replace $V$ with $\frac{4}{3}\pi r^3$. A1: Allow $\frac{dr}{dt}=\frac{A}{B\pi^2 r^5(0.05t+1)^3}$ where $A,B$ integers and $\frac{A}{B}$ cancels to $\frac{15}{4}$ |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dt}=\frac{20}{V(0.05t+1)^3} \Rightarrow \int V\,dV=\int\frac{20\,dt}{(0.05t+1)^3}$ | B1 | Separates variables to achieve $\int V\,dV=\int\frac{20\,dt}{(0.05t+1)^3}$ (with or without integral sign) |
| $\frac{V^2}{2}=\frac{20(0.05t+1)^{-2}}{-0.1}+c$ | M1M1A1 | M1: Integrates lhs to form $aV^2$. M1: Integrates rhs to form $b(0.05t+1)^{-2}$. A1: Correct including constant |
| Sub $V=1$, $t=0$: $0.5=-200+c \Rightarrow c=200.5$ | M1 | Substitutes $V=1$, $t=0$ into integrated expression of form $pV^2=q(0.05t+1)^n+c$ to find $c$ |
| $\Rightarrow V^2=401-\frac{400}{(0.05t+1)^2}$ | A1 | Or equivalent e.g. $V^2=401-\frac{160000}{(t+20)^2}$ |

## Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sub $t=20$: $V^2=401-\frac{400}{(1+1)^2}(=301)$ | M1 | Substitutes $t=20$ into equation for $V^2$ which includes a numerical constant; finds value for $V$ or $V^2$ (cannot score for impossible values i.e. $V^2<0$) |
| Sub $V=\sqrt{301}$ into $V=\frac{4}{3}\pi r^3 \Rightarrow r\approx 1.61$ m | dM1, A1 | dM1: Substitutes their $V$ into $V=\frac{4}{3}\pi r^3$, or their $V^2$ into $V^2=\frac{16}{9}\pi^2 r^6$. A1: $r=$ awrt $1.61$ (m) |

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