Bonus problem – Page 2

An infinite ladder of resistors

January 16th, 2022jeefirst0 Comments

Problem 3.152 of Irodov

The figure below shows an inifite circuit formed by the repetition of the same link, consisting of resistance $R_1 = 4.0 \ \Omega$ and $R_2 = 3.0 \ \Omega$ . Find the resistance of this circuit between points $A$ and $B$ .

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Solution

Let’s denote the resistance between the points $A$ and $B$ by $R_{AB}$ . Since the circuit is infinite, removing the first $R_1$ and $R_2$ resistors gives the same arrangement back again — the arrangement is self-similar. That means, the resistance between the points $C$ and $D$ is just $R_{AB}$ without the left-most $R_1$ and $R_2$ resistors, and we may redraw the circuit as shown below.

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It is now straightforward to calculate the resistance,

(1) …

Electric field in a hollow region

January 9th, 2022jeefirst0 Comments

IIT-JEE 2007 Paper 2, Question 6

(This problem is was repeated in JEE Advanced 2015 Paper 2, Question 11)

A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in the figure. The electric field inside the emptied space is

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zero everywhere
non-zero and uniform
non-uniform
zero only at its center

Solution

As explained in a related problem we can think of the given charge distribution as a …

Delay in a fiber optic cable

January 9th, 2022jeefirst0 Comments

JEE Advanced 2019 Paper 1, Question 18

A planar structure of length $L$ and width $W$ is made of two different optical media of refractive indices $n_{1}=1.5$ and $n_{2}=1.44$ as shown in figure. If $L \gg W$ , a ray entering from end ${\rm AB}$ will emerge from end ${\rm CD}$ only if the total internal reflection condition is met inside the structure. For $L=9.6$ m, if the incident angle $\theta$ is varied, the maximum time taken by a ray to exit the plane $\mathrm{CD}$ is $t \times 10^{-9}$ s. Determine $t$ .

[ Speed of light $c=3 \times 10^{8}$ m/s. ]

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Solution

To ensure that the ray spends the maximum possible time inside the media, we should arrange it so …

Charge at one corner of a cube

January 8th, 2022jeefirst0 Comments

A charge $q$ sits at one corner of a cube of side length $a$ as shown in the figure below. What is the electric flux through the shaded side?

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Solution

Since the problem is asking us to find the electric flux it is natural to guess that we need to apply Gauss’s Law. However, we need a closed surface with appropriate symmetry to be able to use Gauss’s Law (see article). So we consider the situation shown below.

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We have placed …

Electric field from a sphere with a cavity

January 8th, 2022jeefirst0 Comments

A sphere of radius $a$ is filled with positive charge with uniform density $\rho$ . Then a smaller sphere of radius $a/2$ is carved out, as shown in the figure below, and left empty. What are the direction and magnitude of the electric field at $A$ ? At $B$ ?

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Solution

This problem can be solved by using the principle of superposition. For instance, consider a point charge $+q$ at some point $P$ in space. It creates an electric field everywhere. However, if you place a …