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Physics, Chemistry and Math!
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-17
Jun 7th
Topic: Mechanics
Threads of lengths h1, h2 and h3 are fastened to the vertices of a homogeneous triangular plate of weight W. The other ends of the threads are fastened to a common point, as shown in the figure below.
Figure-1
What is the tension in each thread, expressed in terms of the lengths of the threads and the weight of the plate?
Clearly, the centre of mass S of the triangle has to be below the point of suspension. Denote the vectors pointing from the centre of mass S to the vertices of the triangle by r1, r2 and r3, and that to the suspension point by m (see figure).
Figure-2
The forces F1, F2 and F3 exerted on the plate by the threads can now be expressed in terms of the vectors defining the thread directions:
Fi = λi (m – ri), i = 1, 2, 3
Since the plate is in equilibrium, the vector sum of the forces acting on it is zero, i.e.
F1 + F2 + F3 + W = 0
Making use of the fact that the vector pointing to the center of mass of the triangle (the origin of the vector reference frame) is the arithmetic mean of the vectors pointing to its vertices, we have
r1 + r2 + r3 = 0
We note that W and m are parallel and, therefore, W = –km. Eliminating r3 from the above equations gives
(λ3 – λ1) r1 + (λ3 – λ2) r2 + (λ1 + λ2 + λ3 – k) m = 0
Since r1, r2 and m, are not in the same plane, a linear combination of them can only be zero if the coefficient of each is zero.
Thus, λ1 = λ2 = λ3 which implies that the tensions in the threads are proportional to their lengths. This deduction would become invalid if one of the threads were slack, since the plane of the plate would then become vertical and m would lie in it.
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.