Sliding on a block with a circular cut

JEE Advanced 2017 Paper 1, Question 2

A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x=0, in a co-ordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following options is/are correct?

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  1. The position
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An infinite ladder of resistors

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)

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Pulleys and masses connected to a spring

JEE Advanced 2019 Paper 2, Question 2

A block of mass 2 M is attached to a massless spring with spring-constant k. This block is connected to two other blocks of masses M and 2 M using two massless pulleys and strings. The accelerations of the blocks are a_{1}, a_{2} and a_{3} as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is x_{0}. Which of the following option(s) is/are correct?

[g is the acceleration due to gravity. Neglect friction]

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  1. x_{0}=\frac{4 M g}{k}
  2. When spring achieves an extension of \frac{x_{0}}{2} for the first time, the speed of the
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Electric field in a hollow region

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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  1. zero everywhere
  2. non-zero and uniform
  3. non-uniform
  4. zero only at its center

Related Problems:
Electric field from a sphere with a cavity
Flux from a charged shell
Charge at one corner of a cube

Solution

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

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Charge at one corner of a cube

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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Related problems:
Electric field from a sphere with a cavity
Electric field in a hollow region
Flux from a charged shell

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 …

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