The Chelyabinsk meteor is estimated to have had a mass of 10,000 – 13,000 metric tons. A black hole of mass $13\times10^6\ \mathrm{kg}$ has a radius of
$$ r = \frac{2GM}{c^2} = \frac{(2 \times 6.674\times10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}) \times (13\times10^6\ \mathrm{kg})}{9\times10^{16}\ \mathrm{m^2\ s^{-2}}} \approx 1.93 \times 10^{-20}\ \mathrm m \approx \frac{1}{22000} r_\text{proton}$$
According to this Hawking Radiation Calculator a black hole of this mass has a lifetime of about 185000 seconds, a little over 2 days.
Assume for a moment that somehow such a black hole could exist and collide with the earth, am I correct in assuming that, since its interaction cross-section is so small, it would sail through the earth without much happening?
Looking it at from another point of view, calculating the gravitational attraction to the black hole at short distances, I find that at $1\ \mathrm m$ the force is negligible ($.0009\ \mathrm N$). However the inverse-square law applies, so at $1\ \mathrm{mm}$ the force is $867\ \mathrm N$, meaning maybe the "cross-section" isn't so small after all.
The Hawking Radiation Calculator also gives a luminosity of $\rlap{\raise{0.5ex}{\rule{17ex}{1px}}}\approx 3.56 \times 10^8\ \mathrm W$. At a distance of $\rlap{\raise{0.5ex}{\rule{5ex}{1px}}}1\ \mathrm{km}$ the intensity would be only $\rlap{\raise{0.5ex}{\rule{10ex}{1px}}}28\ \mathrm{W/m^2}$. You probably wouldn't want to get too close to it. (see below)
So what happens:
- Not much, sails right through
- Lots of fireworks but no lasting damage
- Immediate global cataclysm
- (1) or (2) initially but the black hole settles at the center of the earth and eventually consumes the planet
- (4) but Hawking Radiation prevents net matter inflow and the black hole eventually evaporates in a burst of energy… but then, how much energy?
Correction: I did something wrong the first time in the Hawking Radiation Calculator… the actual luminosity would be $\approx 2.1 \times 10^{18}\ \mathrm W$, greater by 10 orders of magnitude. At $1000\ \mathrm{km}$ the flux would be about $167\ \mathrm{kW/m^2}$. So basically a significant fraction of the Earth's surface under the black hole's path would be sterilized, and as it approached the surface it would induce fusion. Not a pretty sight, and we're getting closer to Option 3 for a lot of people.
Best Answer
Update: The values of luminosity of the blackhole and the composition of the Hawking radiation emitted by a hot blackhole claimed in this answer seem to be inaccurate. Check @A.V.S's answer for a more accurate description.
For a blackhole of a mass of a few thousand metric tons, the Hawking blackbody radiation would correspond to an astronomically high temperature of about $10^{16}$ K ($ \because T = \frac{1.227 \times 10^{23}}{M}$ ). The radiation from such a hot blackhole would mostly be in high energy gamma rays with each photon carrying TeVs of energy. Note that the temperature claimed above is orders of magnitude higher than that is required to start nuclear fusion.
As most meteors have velocities well above the escape velocity of earth, the blackhole might just pass through the earth in a hyperbolic orbit around the core of the earth ( I assume that there is no other mechanism, electromagnetical or otherwise that would cause the blackhole to lose energy but I could be wrong). $2 \times 10^{18} W$ of energy is enormous, infact an order of magnitude more than the amount of solar radiation received by the earth and all that energy is in high energy gamma rays. This could be the end of all life forms on earth and would permanently deface earth.
Update: Not just a fraction of the earth and or a 'lot of people' as you say in your question's update, I believe that the power output would be enough to sterilize the entire planet or atleast the macroscopic life forms. Consider the fact that all the nuclear bombs ever tested and deployed on Earth till now together total about $10^{18} J$. So that would mean the power output of the blackhole would be equivalent to blowing up all those bombs every second. Moreover the temperature of the explosion and the energy of photons produced would be orders of magnitude higher than that produced by any normal fission or a fusion nuclear bomb. The runaway reactions that would follow such an event would have direct and indirect consequences on life on Earth. Also consider that fact that the power output and the temperature of the black hole increase more and more as the black hole evaporates and reduces it mass.