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Posts tagged 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 – 16
May 17th
TOPIC : Mechanics
Q. Three loads of mass m1, m2 and M are suspended on a string passed through two pulleys as shown:
Figure-1
The pulleys are at the same distance from the point of suspension. What is the ratio of these masses at which the system is in equilibrium?
The easiest way to reach a solution is to consider the point at which the three strings join each other:
Figure-2
Since this point is in equilibrium,
which means that the three vectors should be able to form at triangle:
Figure-3
Therefore, we only require that:
T1 + T2 > T3, T2 + T3 > T1, T3 + T1 > T2
=> m1 + m2 > M, m2 + M > m1, M + m1 > m2
If these conditions hold, the system can attain equilibrium.
The same conditions can be derived more analytically by considering the angles the two inclined strings form with the horizontal, and writing the force equations in more details – that is an exercise for you.
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 – 14
May 10th
Topic: Current Electricity
Q. A wire frame in the form of a tetrahedron ABCD is connected to a d.c. source.
Figure-1
The resistances of all the edges of the tetrahedron are equal. Which edge of the frame should be eliminated to obtain the maximum change in the current in the circuit?
The first observation to be made is that by symmetry, there is no current in CD. If the resistance of each edge is R, then the net resistance Rnet between A and B is
Now, we have to remove an edge. Note that the edges AD, AC, BD, BC are equivalent in the sense that the effect will be the same for any of them being removed. Suppose that the edge AD is removed. The net resistance R1 between A and B now is:
Now suppose that AB is removed. In this case, the net resistance R2 between A and B is:
Note that no current flows in this case in the edge CD, by symmetry. The new current is
Therefore, the change in current is more in the second case:
