Home » 2010 » December (Page 2)

Monthly Archives: December 2010

Resolution of Vectors

From: Vakas

Subject: Resolution with geometry
Message Body:
Dear sir,
I m not able to understand that when we resolute mg in a inclined plane that plane having angle theta,how does the vertical component goes through inclined surface & cos comp opp to N.
explain with geometry so that i don’t get confused in future in determining the two comps is right direction.

Ans:

In a right angled triangle,

$cos\theta =\frac{adjacent side}{hypotenuse}$

and

$sin\theta =\frac{opposite side}{hypotenuse}$

When mg is resolved into mutually perpendicular directions, one parallel to plane and the other perpendicular to the plane (opposite to the Normal reaction), the component which is adjacent to θ is the cos component and the one away from θ is the sine component.

Students find difficulty ,mainly in identifying the angle .

See the diagram below and try to analyze

Why a cyclotron cannot be used to accelerate an electron?

why a cyclotron cannot be used to accelerate an electron?

The mass of electron is very less. So, when it is accelerated in a cyclotron, its speed increases rapidly and reaches relativistic values. (i.e. speeds comparable to the speed of light) Then the mass of the electron increases according to the equation,

$m=\frac{m_{0}^{}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

This change in mass changes the frequency of revolution of electron. Since

Therefore the electron goes out of phase with the cyclotron frequency and acceleration stops.

This happens  because as the relativistic mass increases due to increase in speed, the frequency decreases which means that the time period of revolution increases. Therefore the polarity of the Dees reverses before the electron completes half revolution and the accelerator will not function properly.

IISc Launched A New UG Programme For Future Researchers

Indian Institute of Science (IISc), Bangalore is all set to launch a new four-year undergraduate (UG) Bachelor of Science (BS) programme in the coming academic session in 2011. The main aim of the course is to prepare young minds for research and doctoral studies in the upcoming areas of science and technology. Besides motivating students to take up higher research, mentoring and tutorials, the programme will introduce rigorous specializations in fourth, fifth, and sixth semesters. The students will be allowed to choose electives from other streams as well during this period. In the seventh and eighth semesters, students will work on a research-oriented project supervised by a faculty member. Admissions to the programme will be based on national entrance examinations like IIT JEE, AIEEE, AIPMT, and Kishore Vaigyanik Protsahan Yojana.

AIEEE Will Finally Go Online For The First Time in 2011

All aspiring candidates of engineering and architecture courses prefer to take online test instead of regular pen-paper based format. AIEEE is finally about to conduct its online entrance exams on 1st May, 2011. The online exams will follow the standardized pattern of paper-based test which is being used for nine years. However, Central Board of Secondary Education (CBSE), the organizer of the exam, will only allow a limited number of students to appear in the online exams in 2011. Students from 20 cities will be able to give their exams online with a maximum of 5,000 students per city. The students will be allotted seats for online exams on first-come-first-served basis.

Now anyone can post comment without registration!

As per feedback received, students want to be able to post comments to the articles, questions and answers without registering and logging in.

Now it is enabled on trial basis. Anyone can post a comment without logging in. But the comment will appear only after approved by the site administrator.

A Numerical from electrostatics

TIM posted:

Treat the sphere as a solid non-conducting sphere of radius R=6000 km with uniform charge Q=4000 C. The sphere has a conducting spherical shell. The shell has inner radius r1= 6005 km, outer radius r2= 6010 km and zero net charge. Assume gravity is negligible for all parts of this problem.
a) How much charge is on the inner and outer surfaces of the shell layer?
b) What is the potential difference deltaV between the base of the shell and the sphere?

Difference between emf and pd

EMF stands for electromotive force.

EMF of a cell is equal to the potential difference between the two terminals of a cell when no current is drawn from it.

Pd (potential difference) between any two points on a circuit is defined as the work done per unit charge in carrying a positive test charge from one point to the other. With reference to a cell, the potential difference between the terminals of a cell when a current is drawn from it is called terminal voltage.

• 2,157,261 hits

Subscribe to Blog via Email

Join 4,006 other subscribers