= Solution
At the final state of the <slump test for yield stress>, an axisymmetric deposit of height $h(r)$ is just able to support itself. <Hydrostatic pressure> gives radial pressure gradient $\rho g h'(r)$, while <lubrication theory> makes the magnitude of the basal <shear stress>
$$
|\tau_b|=\rho g h\,|h'|.
$$
The marginally yielded final profile therefore satisfies
$$
\rho g h(-h')=\tau_y,
\qquad h(R)=0.
$$
Integration gives
$$
\boxed{
h(r)=\left[\frac{2\tau_y}{\rho g}(R-r)\right]^{1/2}}.
$$
Using <mass conservation>, the known volume is
$$
V=2\pi\int_0^Rrh(r)\,dr
=\frac{8\pi}{15}
\left(\frac{2\tau_y}{\rho g}\right)^{1/2}R^{5/2}.
$$
Solving for the yield stress produces the estimate
$$
\boxed{\tau_y=\frac{225\,\rho gV^2}{128\pi^2R^5}}.
$$
Thus one measures the final radius $R$ and inserts it with $V$ and $\rho$. The estimate assumes a thin deposit, negligible <surface tension>, complete initial yielding, and a spatially uniform yield stress.
Dry sand is a <granular material> governed primarily by frictional stability. Its final free surface reaches the <angle of repose> $\theta_r$, so the deposit is approximately a cone,
$$
\boxed{h(r)=(R-r)\tan\theta_r}.
$$
This constant-slope profile differs from the square-root edge of the yield-stress-fluid model.
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