2023 A/L Physics Past Paper · Question 26 (English Medium)

Answer: (3) $2$

What the question asks

What we know:

A liquid flows through a horizontal tube. The tube has an inner radius $a$ and a length $l$.

The pressure difference between the two ends of the tube is $\Delta p$.

The speed of the liquid is $v = \dfrac{Ca^n\Delta p}{\eta l}$, and $C$ has no dimensions.

We want: the number $n$, the power of the radius $a$.

What each letter means:

$v$ ($\text{m s}^{-1}$): the speed of the liquid.

$a$ ($\text{m}$): the inner radius of the tube.

$l$ ($\text{m}$): the length of the tube.

$\Delta p$ ($\text{Pa}$): the pressure difference between the two ends.

$\eta$ ($\text{Pa s}$): the coefficient of viscosity.

$M$, $L$, $T$: the dimensions of mass, length and time.

Relevant theory

Symbolic expressions which show the relation of a physical quantity to basic units is called its dimension.

If the equation is correct, the dimension of its both sides should be equal. If the equation has several terms, the dimensions of all the separate terms should be the same.

The coefficient of the viscosity of a fluid is defined as the tangential force on a unit area

under a unit velocity gradient of the fluid in uniform stream line motion.

The first quote says a dimension shows how a quantity is made from mass, length and time. The second quote says that in a correct equation both sides have the same dimensions. The third quote tells us what $\eta$ is, so we can write its dimensions.

For example, think of a cook following a recipe. If the recipe says "2 cups of rice make 2 cups of cooked rice", it makes sense. If it says "2 cups of rice make 2 minutes", it makes no sense.

An equation is the same. The left side and the right side must be the same kind of thing. So we count the kind of each side, and then find the $n$ that makes them match.

$[v] = LT^{-1}$

$[a] = [l] = L$

$[\Delta p] = \dfrac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$

$[\eta] = \dfrac{MLT^{-2}}{L^2 \times T^{-1}} = ML^{-1}T^{-1}$

The dimensions of $C$ are $1$, because $C$ has no dimensions. So $C$ does not change the count.

Reading the options

$L^{-1}T^{-1}$

$\dfrac{[\Delta p]}{[\eta][l]} = \dfrac{ML^{-1}T^{-2}}{ML^{-1}T^{-1} \times L}$

$L^{n-1}T^{-1}$

$[a^n] \times L^{-1}T^{-1} = L^{n} \times L^{-1}T^{-1}$

$L^{1}T^{-1}$

$[v] = LT^{-1}$

$2$

$n - 1 = 1$

$p + q$

The right side must have the dimensions of a speed, L T^{-1}. Option 3 gives that.

asked

L T^{-1}

$the speed v$

1

L^{-1/2} T^{-1}

$n = 1/2$

2

T^{-1}

$n = 1$

3

$n = 2$

4

L^{2} T^{-1}

$n = 3$

5

L^{3} T^{-1}

$n = 4$

Now match it with the five options.

(1) $n = \dfrac{1}{2}$ gives $L^{-1/2}T^{-1}$. It does not match.

(2) $n = 1$ gives $T^{-1}$. It does not match.

(3) $n = 2$ gives $LT^{-1}$. This matches.

(4) $n = 3$ gives $L^{2}T^{-1}$. It does not match.

(5) $n = 4$ gives $L^{3}T^{-1}$. It does not match.

Checks

Put $n = 2$ back into the equation and count again.

$$\dfrac{[a^2][\Delta p]}{[\eta][l]} = \dfrac{L^2 \times ML^{-1}T^{-2}}{ML^{-1}T^{-1} \times L} = LT^{-1}$$

This is the dimensions of a speed. And a thicker tube lets the liquid flow faster, so a power of $a$ above zero is what we expect.

Conclusion

$n = 2$

Conclusion

We worked out the dimensions of $\eta$ from its definition: a force on an area for each unit of velocity gradient. The pressure difference is also a force on an area.

Teacher's note

Write the dimensions of each quantity in $M$, $L$ and $T$. Both sides of the equation must have the same dimensions.

Then match the power of $L$ on both sides. A constant with no dimensions does not change the count.

Exam tip