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Posts tagged Problems in Physics
Physics: Problem of the day-19
Jun 18th
Q. A gas takes part in two thermal processes in which it is heated from the same initial state to the same final temperature. The two processes are represented on a p – V diagram by straight lines 1 →2 and 1 → 3:
Figure-1
In which process is the amount of heat supplied to the gas larger?
The amount of heat exchange, by the first law of thermodynamics, is given by
Figure-2
Since the final temperature of the gas is the same in both the processes, Δ U must be the same in both cases. We then simply have to compare the work done in the two cases:
Figure-3
Therefore, heat supplied to the gas is more in the process 1 → 3.
Physics: Problem of the day-18
Jun 15th
Topic: Electrostatics
An electric field line emerges from a positive point charge + q1 at an angle α to the straight line connecting it to a negative charge – q2.
Figure-1
At what angle β will the field line enter the charge – q2?
What we see in the diagram here is a cross-sectional diagram of a 3-dimensional situation. Field lines will emanate from +q1 in all directions in space. Although the presence of – q2 will prevent these field lines from being uniformly distributed, we can still assume that in the immediate proximity of +q1, the contribution to the field from –q2 will be relatively negligible, so that we can further assume that the field lines emerge from +q1 and enter –q2 in spatially homogeneous bundles. Thus, the “number of lines” emerging from +q1 and confined to the spatially conical region of vertical angle 2α will be the same as those entering –q2 and confined to the spatially conical region of vertical angle 2β.
Figure-2
The ratio number of lines emanating from +q1 in the depicted region, to the total number of lines emanating from +q1 is the ratio of the corresponding solid angles:
Therefore, the number of lines emanating from +q1 in the depicted spatial region is proportional to 1 – cos α. This immediately leads to:
Physics: Problem of the day – 15
May 11th
Topic: Mechanics
The masses of two stars are m1 and m2 their separation is l. If this star-system is to remain stable, the stars must be revolving about the system’s center of mass, with the force of gravity providing the necessary centripetal force. From the information provided, is it possible to calculate the time period of the stars revolution?
Figure-1
Let X be the center of mass of the system. Obviously,
Let the common angular velocity of the stars about X be ω.
The time period T is inversely dependent on ω:
Therefore, the information we were provided about the star-system was sufficient to calculate the time period of revolutions of the stars about the system’s center of mass.
It might seem to you that we did not even use one of the equations, namely m1l1 = m2 l2 , which means we have one extra equation than required. Is that true? If it is, we could have determined the time period using only two of the three constants provided to us.
Physics: Problem of the day – 13
May 8th
Topic: Mechanics
Q. A cylindrical tube of length L which has inside it a light spring of length l and spring constant k, is placed vertically on a table. A small ball of mass m is released at the top of the tube.
Figure-1
At what height h from the surface of the table does the ball have maximum velocity?
For a reader who is alert enough, this problem is a matter of seconds. We just have to visualize what happens to the ball as it falls. It accelerates until it has reached the top of the spring. Now, as it compresses the spring, its acceleration will decrease due to the spring whose compression causes it to exert a progressively increasing force on the ball. But it will nonetheless accelerate, up to the time when it can no longer have a downwards acceleration because the spring has been highly compressed. At this point, the spring force on the ball balances its weight, and the ball has attained its maximum velocity. After this of course, the ball starts slowing down.
Therefore, at h,
Note that L was of no use to us!
Physics: Problem of the day – 12
May 5th
Topic: Electrostatics
Q: Small identical balls with equal charges are fixed at the vertices of a regular polygon with 2001 sides, each of length l. At a certain instant, one of the balls is released, and a sufficiently long time interval later, the ball adjacent to the first ball is released. The kinetic energies of the released balls are found to differ by K at a sufficiently long distance from the polygon. What is the charge q on each ball?
Physically, each ball is under the effect of potentials due to all the other balls. For example, when the first ball is released, it moves away from the polygon, since it is under the effect of repulsive forces due to all the other charges. The kinetic energy that the first ball gains when it reaches infinity, is by energy conservation the same as its electrostatic potential energy when it was in its original position.
Similarly, the kinetic energy that the second ball gains when it reaches infinity is the same as its electrostatic potential energy when it was part of the polygon, excluding the effect of the first ball.
The difference in the original potential energies of the two balls is simply q2/4πε0a, which is due to the interaction between the first and second balls. This difference is given to be equal to K. The value of q therefore becomes known.