ICSE Board 10th Exam Date Sheet 2010 | ICSE Board Class X Time table

Council for the Indian School Certifcate Examinations (CISCE)will be announcing the time table for ICSE 2010  soon. We will make the date sheet as soon as it is published.

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Chat brain-to-brain at the speed of thought Future Of The Web? Thoughts, Images Can Be Transmitted To Minds Without Using An Interface

London: Ever wanted to communicate your thoughts without speaking a word? British scientists have created a system for “brain to brain communication”, a development researchers claimed will allow people to send thoughts, words and images directly to the minds of others.
    The system, developed by the researchers at the British University of Southampton, has been hailed as the future of the internet, which would provide a revolutionary way to communicate without the need for keyboards and telephones.
    It was claimed the technology, the first of its kind, would allow people to send thoughts, words and images directly to the minds of others, particularly people with a disability.
    “This could be useful for those people who are locked into their bodies, who can’t speak, can’t even blink,” said the lead scientist Christopher James.
    However, he cautioned that his experiments were “the first baby steps” towards technologies that would allow people instantly to send thoughts, words, and images directly into the minds of others.
    Scientists used “brain-computer interfacing”, a technique that allows computers to analyse brain signals, that enabled them to send messages formed by a person’s brain signals though an internet connection to another person’s brain miles away.
    According to James, during transmission two people were connected to electrodes that measure activity in specific parts of the brain. The first person generated a series of zeros and ones, where they imagined moving their left arm for zero and right arm for one.
    After the first person’s computer recognises the binary thoughts, it sends them to the internet and then to the other person’s computer.
    A lamp is then flashed at two different frequencies for one and zero. “It’s not telepathy,” James said. e added: “There’s no conscious thought forming in one person’s head and another conscious thought appearing in another person’s mind.
    “The next experiments are to get that second person to be aware of the information that is being sent to them. For that, I need to get my thinking cap on, so to speak.” AGENCIES

PHYSICS PROBLEM SOLVING STRATEGY

PHYSICS PROBLEM SOLVING STRATEGY

Dr. Mark Hollabaugh

Normandale Community College

Two factors can help make you a better physics problem solver. First of all, you must know and understand the principles of physics. Secondly, you must have a strategy for applying these principles to new situations in which physics can be helpful. We call these situations problems. Many students say, “I understand the material, I just can’t do the problems.” If this is true of you as a physics student, then maybe you need to develop your problem-solving skills. Having a strategy to organize those skills can help you.

Physics problem solving can be learned just like you learned to drive a car, play a musical instrument, or ride a bike. What can aid you more than anything is to have a general approach to follow with each problem you encounter. You may use different tools or tactics with differing areas of physics, but the overall strategy remains the same. Most likely, you have already acquired some problem-solving skills and habits from previous courses in physics, chemistry, or mathematics. Like other areas of learning and life, some of these habits may be beneficial and some may actually hinder your progress in learning how to solve physics problems.

So, in learning this new approach, be willing to try new ideas and to discard old habits that may in fact be hindering your understanding. As you mature as a physics problem solver, you will find that the approach will become second nature to you. You will begin automatically to do those things that will lead you to construct an effective solution to the problem.

As with so many other learning activities, it is useful to break a problem solving strategy into major and minor steps. The strategy we would like you to learn has five major steps: Focus the Problem, Physics Description, Plan a Solution, Execute the Plan, and Evaluate the Solution. Let’s take a detailed look at each of these steps and then do an sample problem following the strategy. At this stage of our discussion, do not worry if there are physics terms or concepts that you do not understand. You will learn these concepts as they are needed. Then, refer back to this discussion.

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BERNOULLI’s PRINCIPLE

In fluid dynamics, Bernoulli's principle states that for an inviscid flow, an increase in the speed of the fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy. Bernoulli's principle is named after the Dutch-Swiss mathematician Daniel Bernoulli who published his principle in his book Hydrodynamica in 1738.

Bernoulli's principle can be applied to various types of fluid flow, resulting in what is loosely denoted as Bernoulli's equation. In fact, there are different forms of the Bernoulli equation for different types of flow. The simple form of Bernoulli's principle is valid for incompressible flows (e.g. most liquid flows) and also for compressible flows (e.g. gases) moving at low Mach numbers. More advanced forms may in some cases be applied to compressible flows at higher Mach numbers.

Bernoulli's principle can be derived from the principle of conservation of energy. This states that in a steady flow the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline. This requires that the sum of kinetic energy and potential energy remain constant. If the fluid is flowing out of a reservoir the sum of all forms of energy is the same on all streamlines because in a reservoir the energy per unit mass (the sum of pressure and gravitational potential ρ g h) is the same everywhere.

Fluid particles are subject only to pressure and their own weight. If a fluid is flowing horizontally and along a section of a streamline, where the speed increases it can only be because the fluid on that section has moved from a region of higher pressure to a region of lower pressure; and if its speed decreases, it can only be because it has moved from a region of lower pressure to a region of higher pressure. Consequently, within a fluid flowing horizontally, the highest speed occurs where the pressure is lowest, and the lowest speed occurs where the pressure is highest.

APPLICATIONS OF BERNOULLI’s PRINCIPLE

1. The carburetor used in many reciprocating engines contains a venturi to create a region of low pressure to draw fuel into the carburetor and mix it thoroughly with the incoming air. The low pressure in the throat of a venturi can be explained by Bernoulli's principle; in the narrow throat, the air is moving at its fastest speed and therefore it is at its lowest pressure.

A carburetor is a device that blends air and fuel for an internal combustion engine. It was invented by Karl Benz before 1885 and patented in 1886. It is colloquially called a carb. The word carburetor comes from the French carbure meaning "carbide". Carburer means to combine with carbon. In fuel chemistry, the term has the more specific meaning of increasing the carbon (and therefore energy) content of a fuel by mixing it with a volatile hydrocarbon.

2. The Pitot tube and static port on an aircraft are used to determine the airspeed of the aircraft. These two devices are connected to the airspeed indicator which determines the dynamic pressure of the airflow past the aircraft. Dynamic pressure is the difference between stagnation pressure and static pressure. Bernoulli's principle is used to calibrate the airspeed indicator so that it displays the indicated airspeed appropriate to the dynamic pressure.

A pitot tube is a pressure measurement instrument used to measure fluid flow velocity. The pitot tube was invented by the French engineer Henri Pitot in the early 1700s and was modified to its modern form in the mid 1800s by French scientist Henry Darcy. It is widely used to determine the airspeed of an aircraft and to measure air and gas velocities in industrial applications.

The basic pitot tube consists of a tube pointing directly into the fluid flow. As this tube contains fluid, a pressure can be measured; the moving fluid is brought to rest (stagnates) as there is no outlet to allow flow to continue. This pressure is the stagnation pressure of the fluid, also known as the total pressure or (particularly in aviation) the pitot pressure.

The measured stagnation pressure cannot of itself be used to determine the fluid velocity (airspeed in aviation). However, Bernoulli's equation states:

Stagnation Pressure = Static Pressure + Dynamic Pressure

Which can also be written

Solving that for velocity we get:

Where V is fluid velocity

and p_t is stagnation or total pressure

and p_s is static pressure

and ρ is fluid density

The dynamic pressure, then, is the difference between the stagnation pressure and the static pressure. The static pressure is generally measured using the static ports on the side of the fuselage. The dynamic pressure is then determined using a diaphragm inside an enclosed container. If the air on one side of the diaphragm is at the static pressure, and the other at the stagnation pressure, then the deflection of the diaphragm is proportional to the dynamic pressure, which can then be used to determine the indicated airspeed of the aircraft. The diaphragm arrangement is typically contained within the airspeed indicator, which converts the dynamic pressure to an airspeed reading by means of mechanical levers.

Instead of static ports, a pitot-static tube may be employed, which has a second tube coaxial with the pitot tube with holes on the sides, outside the direct airflow, to measure the static pressure.

Pitot tubes on aircraft commonly have heating elements called pitot heat to prevent the tube from becoming clogged with ice. The failure of these systems can have catastrophic consequences, as in the case of Austral Líneas Aéreas Flight 2553, Birgenair Flight 301 (investigators suspected that some kind of insect could have created a nest inside the pitot tube. The prime suspect is a species called the Black and yellow mud dauber wasp), Northwest Orient Airlines Flight 6231, AeroPeru Flight 603 (blocked static port), and of one X-31.

class 11 second terminal examination physics

1. What do you mean by systematic error ?
2. Each side of a cube is measured to be 7.023m.  What is the total surface area of the cube to appropriate significant figures ?
3. State work-energy theorem.
4. Find the vector product of two vectors a=( 3i^-4j^+5k^) and  b =(-2i^ + j^-3k^)
5. Define universal gravitational constant.
6. State Newton's 2nd law of motion and explain force from it
7. Show the range of a projectile is maximum when θ=45
8. Io one of the satellite of Jupiter,has an orbital period of 1.769 days and radius of orbit is 4.22 x 10 ⁸ m . Show that mass of Jupiter is about one-thousandth that of the sun.
9. What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to is through one end. Given moment of inertia about its own axis is Ml2 /12.
10. Find the angle force F=(3i^+4j^-5k^) unit and displacement d=(5i^+4j^-3k^ ) unit.Also find the projection of F on d.
11. Derive an expression for potential energy of a spring
12. How will you find the centre of mass of triangular lamina ?
13. State Kepler’s law of planetary motion
14. Derive an expression for the relation between torque and angular momentum of rotation about a fixed axis.
15. Differentiate translational motion and rotational motion.
16. Derive an expression for  linear velocity and angular velocity of v and w.
17. Find the torque of a force 7i^ +3j^-5k^about the origin. the force acts on the particle whose position vector is i^-j^+k^.
18. A 1kg block is situated on a rough inclined is connected to a spring of spring constant 100 Nm-1 as shown in fig. The block is released   from rest with the spring in the unstretched position.The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the black and the incline .Assume that the spring has negligible  mass and the pulley is frictionless.
19. Derive an expression for orbital velocity and time period of a satellite
20. A rocket is fired vertically from the surface of mars with speed of 2km/s. If 20% of its initial energy is lost due to Martian atmospheric resistance , how far will the rocket go from surface or mars before returning to it ? Mass of mars =6.4x10 23 kg , radius of mars =3395 km ,G = 6.67x10 11 Nm2kg-2.
21. State and explain law of conservation of energy in the case of a freely falling body.
22. State and explain the laws of (a) kinetic energy (b) static friction.
23. Derive an expression for acceleration due to gravity with respect to (a) variation with altitude ‘h’  (b) variation with depth ‘h’
24. State law of conservation of angular momentum  for a rotating body.
25. Briefly describe an elastic collision in one dimension and find the relation between initial and final velocities before and after collision and discuss the following cases.   (a)The masses of colliding bodies are equal.  (b) The target body is initially at rest.
26. A rear side of a truck is open and a box of 40 kg mass is placed 5km away from the open end as shown in fig.the co-efficient of friction between the box and the surface below it is 0.15 . On a straight road the truck starts from rest and accelerates with 2m/s 2. At what distance from the starting point does the box fall off the truck ? Ignore the size of the box.
27. a. State perpendicular axis  theorem and parallel axis theorem.   b .Derive an expression for moment of inertia of a ring about its diameter. c.The oxygen molecule has a mass of 5.30 x10 –26 kg and a moment of inertia of 1.94 x10 –46 kgm 2 about an axis through its centre , perpendicular to the line joining the two atoms. Suppose the mean speed of such a molecule in gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation.Find the average angular velocity of the molecule.
28.