A rod of aluminum has a diameter of 1.000002 cm. A ring of cast steel has an inner diameter of 1.000000 cm. If the rod and the ring are placed in a liquid under high pressure, at what value of the pressure will the aluminum rod fit inside the steel ring?
I was trying:
$$\frac{F}{A} = \frac{\Delta V * B }{V}$$
however the volume is:
$$V = \pi r^2 * h$$
and I dont have the height so I got stuck does any one have an idea?, I was trying to take the height as 1 but didnt work
note:bulk modulus for aluminium and steel is:
$$B_a = 7.5*10^{10}$$ $$B_s = 17*10^{10}$$
update: @goncalo told me aluminuim is isotropic so:
$$\frac{\Delta r}{r} = \frac{\Delta V}{3V}$$
I was trying:
$$3\frac{\Delta r}{r}B = 449999$$
but this is wrong the correct answer is: 810000pa
Let $r_a$ be the radius of the rod of aluminum.
Also, let $r_s$ be the inner radius of the steel ring.
Then, we have $$\frac{B_a(r_a−r)}{r_a}=\frac{B_s(r_s−r)}{r_s}$$ Solving this for $r$, we get $$r=\frac{B_s-B_a}{\frac{B_s}{r_s}-\frac{B_a}{r_a}}$$
Therefore, we see that $\frac{3B_a}{r_a}(r_a-r)$ can be written as $$\begin{align}\frac{3B_a}{r_a}(r_a-r)&=\frac{3B_a}{r_a}\bigg(r_a-\frac{B_s-B_a}{\frac{B_s}{r_s}-\frac{B_a}{r_a}}\bigg) \\\\&=\frac{3B_a}{r_a}\bigg(r_a-\frac{(B_s-B_a)r_sr_a}{B_sr_a-B_ar_s}\bigg) \\\\&=\frac{3B_a}{r_a}\bigg(\frac{r_a(B_sr_a-B_ar_s)-(B_s-B_a)r_sr_a}{B_sr_a-B_ar_s}\bigg) \\\\&=\frac{3B_a}{r_a}\bigg(\frac{B_sr_a(r_a-r_s)}{B_sr_a-B_ar_s}\bigg) \\\\&=\frac{3B_aB_s(r_a-r_s)}{B_sr_a-B_ar_s}\end{align}$$ from which you will get the answer.