CBSE ADDA

CBSE PHYSICS: LAW OF CONSERVATION OF ENERGY(9th physics) : LAW OF CONSERVATION OF ENERGY Energy can neither be created nor destroyed, but it is transformed from one form to another. Alternatively, whenever energy gets transformed, the total energy remains unchanged. Proof – Freely falling body It may be shown that in the absence of external frictional force the total mechanical energy of a body remains constant. Let a body of mass m falls from a point A, which is at a height h from the ground as shown in fig. At A, Kinetic energy kE = 0 Potential energy Ep = mgh Total energy E = Ep + Ek = mgh + 0= mgh During the fall, the body is at a position B. The body has moved a distance x from A. At B, velocity v 2 = u 2 + 2as applying, v 2 = 0 + 2ax = 2ax Kinetic energy Ek = 1/2 m v 2 = 1/2 m x 2gx = mgx Potential energy Ep = mg (h – x) Total energy E = Ep + Ek = mg (h-x) + mgx = mgh – mgx + mgx= mgh If the body reaches the position C. At C, Potential energy Ep = 0 V

ASSIGNMENT AREAS RELATED TO CIRCLES CLASS X : 1. The radius of the circle is 3 m. What is the circumference of another circle, whose area is 49 times that of the first? 2. Two circles touch externally. The sum of their areas is 130 p sq. cm and the distance between their centres is 14 cm. Find the radii of the circles. 3. A wire when bent in the form of an equilateral triangle encloses an area of 121 √3 cm2 . If the same wire is bent in the form of a circle, find the area of the circle. 4. The area enclosed between the two concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, calculate the radius of the inner circle. 5. A wheel of diameter 42 cm, makes 240 revolutions per minute. Find : (i) the total distance covered by the wheel in one minute. (ii) the speed of the wheel in km/hr. Read more » 10th Maths SA-2 Chapter  Quick links Quadratic Equations Circles Co-ordinate Geometry Arithmetic Progression

Multiple type questions 1).what type of wave is sound wave? (a) radio waves (b) electro magnetic wave (c) mechanical wave (d)transverse wave 2.SONAR works on principle of (a) echo (b)resonance(c)particle nature of wave (d) none 3. Loudness depends on (a) wave length (b) frequency(c) amplitude (d)intensity 4. Pitch increases with increase in (a) energy of sound (b) frequency (c) wave length (d) amplitude 5. speed of sound greatest in (a) milk (b) wood(c) iron (d) air 6. The sound whose frequency is between 20 Hz to 20000 Hz is (a)infrasonic wave (b) audible sound (c)ultrasonic wave (d) radio waves Very Short Answer questions 1. What is echo? 2. what are conditions necessary to hear echo? 3. What is SONAR ? Give its use? 4. How density of mediam affect speed of sound? 5. What is pitch of sound? 6. What is amplitude of sound waves? 7. What is resonance? 8. What is sonometer? 9. Define the terms

Work When a force applied on an object and the object moves in the direction of force, we say that the force has done work on the object. Conditions essential for work to be done are :   1. Some force must act on the object   2. Object move in the direction of force   Hence, the product of the force and the distance moved measures work done.   W = F x S   Where W is the work done, F is the force applied and S is the distance covered by the moving object.  Work is the product of the magnitudes of the force and the displacement, and direction is not taken into account. So work is a  scalar  quantity What is the work done when the force on the object is zero? The work done is zero What would be the work done when the displacement of the object is zero? The work done is zero A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force.  If the force acts on the object all through the displacement, what is the work done in this case? Work do

10th Maths Test Paper Q1 A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60°with the wall, find the height of the wall. (7.5 √3 ) Q 2 A pole 12 m high casts a shadow 4 √3 m long on the ground. Find the angle of elevation (60°) Q 3 The angle of elevation of the top of a tower from a point on the ground is 30° if on walking 30m towards the tower, the angle of elevation becomes 60°.Find the height of the tower.(15√3 ) Q 4 An observer 1.5m tall is 20.5m away from a tower 22m high. Determine the angle of elevation of the top of the tower from the eye of the observer. (45°) Q 5 An aero plane when flying at a height of 5000m from the ground passes vertically above another aero plane at an instant when the angles of the elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertic

The Central Board of Secondary Education (CBSE) intends to keep the Central Teacher’s Eligibility Test (CTET) in online mode across 22 Cities in India. The traditional paper pencil mode of exam will also be in practice; last year nearly 790,000 students have registered for CTET. CTET 2011 was conducted by CBSE on 26 June 2011 (Sunday) at 1178 Centers in 86 Cities across the Country and  2 Centers are in abroad. CTET 2012 is likely to be held during January to June 2012; the test could go online across all major cities in India including Bangalore. CTET will have two papers, Paper I will be for teacher posts to classes I to V, while Paper II will be for teacher posts for classes VI to VIII. You can learn more about CTET. Last year only 14% managed to qualify in the written test; as many as 617,131 candidates couldn’t clear the test.

Class 9 Chapter Work and energy Solved Numerical problems 1. A force of 10N causes a displacement or 2m in a body in its own direction. Calculate the work done by force. 20j Solution: the work done by force = F x S = 10 N x 2m = 20 J 2. How much force is applied on the body when 150 joule of work is done in displacing the body through a distance of 10m in the direction of force?(15 N) Solution: W = F x S Þ F = w/s = 150/10 = 15 N 3. A body of 5kg raised to 2m find the work done(98j) Solution: T he work done to raise a body = PE = mgh = 5kg x 9.8 x 2 = 98 joule 4. A work of 4900j is done on road of mass 50 kg to lift it to a certain height. Calculate the height through which the load is lifted. (10m) Solution: work done on road to lift = mgh Þ 4900 = 50 x 9.8 h Þ h = 10m 5. An engine work 54,000J work by exerting a force of 6000N on it. What is the displacement of the force . (9m) Solution: S = W/F = 54,000J/6000N = 9m 6. A force of 10N acting on a body at an angle of 60 deg. with the