2023 A/L Physics Past Paper · Question 37 (English Medium)

Answer: (1) $\dfrac{V(\beta - \rho)gL}{A}\left(\dfrac{1}{Y_1} + \dfrac{1}{Y_2}\right)$

What the question asks

We want: the change $e$ in the length of the composite rod.

What each letter means:

$e$ ($\text{m}$): the total extension of the composite rod.

$e_1$, $e_2$ ($\text{m}$): the extensions of the first and the second rod.

$L$ ($\text{m}$) and $A$ ($\text{m}^2$): the length and the area of each rod.

$Y_1$, $Y_2$ ($\text{N m}^{-2}$): the Young's moduli of the rods.

$V$ ($\text{m}^3$): the volume of the sphere.

$\beta$, $\rho$ ($\text{kg m}^{-3}$): the density of the sphere and of the liquid.

$T$ ($\text{N}$): the pull in the rods.

$g$ ($\text{m s}^{-2}$): the acceleration of gravity.

Relevant theory

The ratio of tensile stress to tensile strain is called the Young's modulus of the material

of the wire.

When a body is completely or partially immersed in a fluid at rest, it experiences an upthrust equal to the weight of the fluid displaced by the body.

The first quote says what Young's modulus is: the tensile stress divided by the tensile strain. The second quote says that a body in a liquid feels an upthrust equal to the weight of the liquid it pushes aside.

For example, hang a heavy bag from two elastic bands joined end to end. Each band holds the full pull of the bag, and each band stretches.

The bag stretches the pair by the stretch of the first band plus the stretch of the second.

Required equations

$W = V\beta g$

$U = V\rho g$

$T = W - U = V(\beta - \rho)g$

$e_1 = \dfrac{T L}{A Y_1}, \quad e_2 = \dfrac{T L}{A Y_2}$

$e = e_1 + e_2 = \dfrac{V(\beta - \rho)gL}{A}\left(\dfrac{1}{Y_1} + \dfrac{1}{Y_2}\right)$

Reasoning and calculation

The pull in the rods:

The sphere hangs at rest. The rod pulls it up with the pull $T$, and the liquid pushes it up with the upthrust $U$. Its weight $W$ pulls it down.

So $T + U = W$, which gives $T = W - U$.

The rods have a negligible mass, so each rod carries the same pull $T$: the lower rod holds the sphere, and the upper rod holds the lower rod and the sphere.

$V(\beta - \rho)g$

$T = V\beta g - V\rho g = V(\beta - \rho)g$

The two extensions add up:

Each rod has the same $L$, the same $A$ and the same pull $T$. The moduli are different.

The rod of $Y_1$ stretches by $\dfrac{TL}{AY_1}$. The rod of $Y_2$ stretches by $\dfrac{TL}{AY_2}$. The change of the whole length is the sum.

$20$

$T = 0.001 \times 2000 \times 10$

$0.2$

$e_1 = \dfrac{20 \times 2}{10^{-6} \times 2 \times 10^{11}}$

$0.4$

$e_2 = \dfrac{20 \times 2}{10^{-6} \times 10^{11}}$

$0.6$

$e = e_1 + e_2$

$v * (b - r) * g * l / a * (1 / y1 + 1 / y2) * 1000$

Reading the options

Now compare the form with the five options.

Option (1) has the pull on top, $V(\beta - \rho)gL$, over $A$, times the sum $\dfrac{1}{Y_1} + \dfrac{1}{Y_2}$. It matches.

Option (2) has a difference. Option (3) and option (4) have the pull in the bottom. Option (5) has the moduli on top. None of these match.

Checks

Check the units. The pull $V(\beta - \rho)g$ is in $\text{N}$. Then $\dfrac{L}{A}$ is in $\text{m}^{-1}$ and $\dfrac{1}{Y}$ is in $\text{m}^2\,\text{N}^{-1}$.

Multiply: $\text{N} \times \text{m}^{-1} \times \text{m}^2\,\text{N}^{-1} = \text{m}$. The result is a length, as it must be.

Does it make sense?

If $\beta = \rho$, the upthrust holds up the whole weight, so the pull is zero and the extension is zero. The formula gives zero.

If the second rod becomes very stiff, $Y_2$ becomes very big and $\dfrac{1}{Y_2}$ goes to zero. Then the first rod is the one that stretches, as the formula says.

Conclusion

$e = \dfrac{V(\beta - \rho)gL}{A}\left(\dfrac{1}{Y_1} + \dfrac{1}{Y_2}\right)$

The change of the length of the composite rod is $\dfrac{V(\beta - \rho)gL}{A}\left(\dfrac{1}{Y_1} + \dfrac{1}{Y_2}\right)$.

The answer is (1).

The quotes give Young's modulus and Archimedes' principle. The steps $T = W - U$, the same pull in the two massless rods, and the extensions adding are our reasoning from them. We take the weight of the rods as negligible, as the question says.

Teacher's note

Find the pull first: weight minus upthrust. The same pull goes through both rods. Each stretches by $\dfrac{TL}{AY}$, and the stretches add.

Check the units: the answer must come out in metres.

Exam tip