If you look at the right part of the tube, then the bottom of the tube has 8cm of mercury above it (plus atmospheric pressure) so the pressure is $\rho g 0.08$ (+ 1atm). Yes?
If you look at the left side of the tube it has 2cm of mercury above it plus the unknown gas pressure in the vessel. So the pressure is $\rho g 0.02 + P$.
And the two pressures have to be equal because we're calculating both of them from the same place.
If you're sitting at the bottom of the U then the pressure on the left and right sides has to be equal otherwise there would be a net force on you and you would move.
My diagram is misleading because the pressure increases as we move downwards so the red arrows should be bigger on the bottom half than on the top half.
So the upwards component in the bottom half is bigger than the downwards component in the top half. There is a net upwards force.
@harambe If the sides are straight then the force they exert is always horizontal so they cannot exert any net upwards or downwards force on the liquid. In that case the force exerted by the base must equal mg.
Volume of the outer sphere = $\tfrac{4}{3}\pi (0.08^3)$ and volume of inner sphere = $\tfrac{4}{3}\pi (0.08^3)$ so the volume of the material in the spherical shell is $\tfrac{4}{3}\pi (0.08^3 - 0.06^3)$. Yes?
Hi, I bought a copy of Peter Gnadig's 200 puzzling physics problems, the issue is - I don't know where to read up the prerequisites to get a running start at the book. Any suggestions?
@SudarshanaSuri the problems are intended to help physics undergraduates broaden their skills, so if you haven't done a physics degree you'll struggle with them. Since they span pretty much all of classical physics there isn't really any specific reading to cover them. You have to study all of classical physics.
@JohnRennie, Two coherent sources A and B of radio waves are 5m apart. Each source emits waves with wavelength 6m. Consider points along the line between two sources, at what distances, if any, from A is interfernece constructive.
@blue_eyed_... you get constructive interference at a point when the path lengths from that point to the two sources differ by an integral number of wavelengths.
i.e. the difference in the path lengths is $n\lambda$ where $n$ is an integer equal to or greater than zero.
@JohnRennie, The path differences is yd/D where d= separation between the slits, D= separation between the slits and the screen and y=distance of fringe.
From the question it looks as if the line is the straight line between the two sources A and B. If so that line is 5m long. There is only one point on this line where the difference in the distances to the two sources is equal to $n\lambda$.
I agree. There is only the one position between the sources where the path difference is equal to $n\lambda$, and that's the midpoint where the distance to both sources is the same i.e. $n=0$.
A square loop of side 10cm and resistance of 0.7 ohm is placed vertically in the east west plane. A uniform magnetic field of 0.1 T is set up across the plane in north east direction. The magnetic field is decreased to 0 in 0.7 sec at a steady rate. Determine the magnitude of induced emf and current during this time interval.
@sammygerbil, An athlete peddles a stationary tricycle whose pedals are attached to a coil having 100 turns each of area 0.1m^2. The coil, lying in the XY plane is rotated, in this plane, at rate of 50 rpm, about Y axis, in a region where a uniform magnetic field, $\vec {B}=0.01 \vec {k}$ Tesla, is present. Find the (i) maximum emf (ii) average emf, generated in the coil over one complete revolution.
@sammygerbil, I've studied about a case where an alternating emf is induced when a rectangular coil rotates in a magnetic field. Not sure about circular coil
@Dante Write down the conditions for equilibrium, and for the ladder being on the point of slipping. The latter means $f_1=\mu N_1, f_2=\mu N_2$ : 2 equations. The former means balancing forces vertically and horizontally : another 2 equations. Finally take moments about a convenient point : another 1 equation. There are 4 unknowns $f_1, f_2, N_1, N_2$ but 5 equations so you can eliminate all of the unknowns and find a relation between the remaining variables.
@sammygerbil whats the way to do this in 2 mins. I had to: Equate magnitudes of y momentums Conserve KE Conserve x momentums And it took 5 minutes which is ... you know ...
@Abcd I think perhaps you are intended to recognise the result as coming from and elastic collision between particles of equal mass. That is the 'quick' method.
The 'semi-quick' method is to transform to the COM frame. Then the particles approach in a head-on collision and separate at 180 degrees. When you transform back to the lab frame the velocities are 90 degrees apart.
@Abcd In the COM frame both particles move towards the COM with the same speed in opposite directions along the same straight line. The COM is the point of collision. When they separate they go in opposite directions with the same spee, to conserve momentum and KE in the COM frame.
@Abcd Yes. If they have the same mass this is the result. You can confirm the answer by doing the calculation in the COM frame.
That's one of the tricks to doing these questions. You have a hunch about which is the correct answer, then you do a quick calculation to check the answer. Instead of deriving a general formula to get an answer then seeing which option matches.
Even if the masses are different the calculation is quicker in the COM frame.
How does one measure the pressure at the end of the manometer tube when the manometer tube itself is having a variable area?
