1. If a net charge Q, flows across any crosssection of a conductor in time t, then the current I, through the crosssection is I = Q/t Q = It 2. The electric potential difference between two points in an electric circuit carrying some current is the work done to move a unit charge from one point to the other – Potential difference (V) between two points = Work done/Charge V = W/Q W = VQ 3. Q = n x Charge on 1 electron When a steady current flows through a conductor, the electrons in it move with a certain average ‘drift speed’. 4. If the current I, flowing in a metallic wire and the potential difference across its terminals is V . Then potential difference, V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. This is called Ohm’s law. V a I Þ V = RI or, I = V/R 5. Resistance of the conductor depends (i) on its length, (ii) on its area of crosssection, (iii) on the nature of its material and (iv) temperature R a l/A R = r l/A Or, r = RA/l 6. If resistors joined in series: V = V1 + V2 + V3 but I = I1 = I 2 = I 3 Then R = R 1 + R 2 + R 3 6. If resistors joined in Parallel: V = V1 = V2 = V3 but I = I1 + I 2 + I 3 Then 1/R = 1/R 1 + 1/R 2 + 1/R 3 7. If a current I flowing through a resistor of resistance R. and the potential difference across is V for time t sec Then, the work done in moving the charge Q through a potential difference V is W= VQ. But, Q = It W = V I t (i) Now, Power = work done / Time P = W/t {or, W = Pt [The energy supplied to the circuit by the source in time t is P × t = V I t] P = V I t /t [ Using eq. (i) ] P = VI  (ii) The amount of heat produced in time t = H H = the energy supplied to the circuit by the source in time t = V I t Applying Ohm’s law, V = IR H = I2 R t Note: heat produced in a resistor is (i) Directly proportional to the square of current for a given resistance, (ii) Directly proportional to resistance for a given current, and (iii) Directly proportional to the time for which the current flows through the resistor. 8. Electric Power: The rate at which electric energy is dissipated or consumed in an electric circuit is called electric power. The power P is given by P = VI Using, V = IR Þ P = I2R = V2/R Also using, V/R = I Þ P = V2/R 9. The commercial unit of electric energy is kilowatt hour (kW h) = 1 unit. 10. 1 kW h = 1000 watt × 3600 second = 3.6 × 106 watt second = 3.6 × 106 joule (J) Now start practicing these Numerical: Click here To View
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If the polynomial x4 – 6x3 + 16x2 – 26x + 10 – a is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a. Find k and a.
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1. A force can cause a change in state and direction of an object. 2. An interaction between two objects causes a force. 3. Force applied by direct touching an object is called contact force. 4. Gravitational force is an example of non contact force. 5. Liquid exerts pressure in all direction. 6. Air exerts pressure on the earth due to its weight. 7. Atmospheric pressure is measured by using barometer is made by E. Torricelli. 8. Aneroid barometer does not use any liquid 9. Gases and Liquid are collectively called Fluid. Match the Coolum 1. Earth revolving around the sun>gravitational force 2. A force applied by touching> Contact force 3. Force between two charged object>electrostatic force 4. Bloating of the tube of cycle tyre > air pressure 5. Force applied per unit area> pressure State true or false 1. A player kicking football in an example of a non contact force> F 2. Liquid and gases do not exert pressure> F 3. Pressure in liquid decreases with depth > F 4. Like charges repel and unlike chare attract each other> T 5. Magnetic pole attract magnetic substance like iron> T Think Zone Q. What is the role of air pressure in the filling of a syringe with a liquid medicine, by a doctor? Answer: The air pressure plays an important role in the filling of syringe as the liquid medicine rushes into the syringe when air pressure inside it decreases on pulling piston out. Q. Why do people with high blood pressure sweat a lot? Answer: This is because the pressure of their body fluid becomes more than atmospheric pressure .This forces the water to ooze out easily. Q. There is a famous saying n Hindi a sword cannot replace a needle .Give reason for the saying in the light of physics Answer: In the physics, the pressure applied on needle gets concentrated its tip i.e. on very small area and hence it pierces the cloth easily. But to make the sword cut a surface; we have to apply much higher pressure due to large surface area of the cutting edge of the sword. Hence a sword cannot replace a needle. Q. Every square centimeter of our body experience atmospheric force equal to mass of 100kg that is enough to grind us. Then why do people cannot crushed by atmospheric pressure? Answer: this is because there is pressure exerted by our body due to flow of blood that equalizes atmospheric pressure. Q. A plastic comb when rubbed with hair can attract piece of paper. Name force and its nature? Answer: Electrostatic Force. It is generated when a charged body meets to another charged or by rubbing two uncharged body. Q. Why it is not easy to walk wearing high heels shoe? Ans: High heels shoe distribute a large amount force in a Small area hence making her feel uncomfortable while walking on ground. Q. Give two uses of fluid pressure? Ans: Cooling, Cooking, filling LPG in cylinder Q. Give reason: (a)A rolling ball stops after moving some distance Answer: This is because of friction force of the ground that opposes motion. (b) Every object left above the surface of the earth without a support, fall downwards. Answer: This is due to gravitational pull of the earth. (c) If gravitational force act between you and your friend. Then why should not you pull each other? Answer: Gravity only becomes noticeable when there is a really massive object like a moon, planet or star. Due to small masses no force of gravity is noticed. Class VIII Mathematics [Click on links given below] VIII Algebra VIII Commercial Maths VIII Geometry & Menstruation CBSE Class VIII Science Physics Chemistry Biology 8th Sample paper 8th Social Science School Details By Jsunil[Maths and Science Teacher]
http://www.centralpublicschool.net School Name : CENTRAL PUBLIC SCHOOL Principal: MR. MD. ARIF Day or Residential: DAY Available Classes: 12 Medium: ENGLISH Type: COED Board of Education: CBSE Address : TAJPUR ROAD , SAMASTIPUR848101 BIHAR ; Phone 6274  222970 Q. Prove that no number of the type 4k+2 can be a perfect square.
Ans: If p is a prime factor of a perfect square, p2 must also be a factor of that perfect square. 4k+2 = 2(2k+1) 2 is a factor of 4k+2 but 2k+1 is odd and cannot have factor 2, so 4k+2 is not divisible by 4, and therefore cannot be a perfect square. Q. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point? Ans: We need to calculate the LCM to find the answer. 18 = 2x3x3 ;12= 2x2x3 ; LCM = 36 Q. show that only one out of n, n+2 or n+4 is divisible by 3 where n is positive integer. Solution: When n = 1, exactly one of 1, 1+2, and 1+4 is divisible by 3, namely 1+2, since 3 is divisible by 3, and the other two 1 and 5 are not. Suppose for n < k, only one out of n, n+2, n+4 is divisible by 3 For n = k, we consider k, k+2, k+4. By the induction hypothesis, only one of k1, k+1 and k+3 is divisible by 3. We look at the three possible cases. Case 1: k1 is the one which is divisible by 3. Then k1 = 3m, for some positive integer m. Then add 1 to both sides of k1 = 3m: k1+1=3m+1 k = 3m+1 then k is NOT divisible by 3 Now add 3 to both sides of k1 = 3m: k1+3=3m+3 k+2 = 3m+3 k+2 = 3(m+1) then k+2 IS divisible by 3 Now add 5 to both sides of k1 = 3m: k1+5=3m+5 k+4 = 3m+5 k+4 = 3m+3 + 2 = 3(m+1)+2 then k+4 is NOT divisible by 3. So, we have proved case 1 for n = k Case 2: k+1 is the one which is divisible by 3. Then k+1 = 3m, for some positive integer m. Then add 1 to both sides of k+1 = 3m: k+11=3m1 k = 3m1 then k is NOT divisible by 3 Now add 1 to both sides of k+1 = 3m: k+1+1=3m+1 k+2 = 3m+1 then k+2 is NOT divisible by 3 Now add 3 to both sides of k+1 = 3m: k+1+3=3m+3 k+4 = 3m+3 k+4 = 3(m+1) so k+4 IS divisible by 3. So, we have proved case 2. Case 3: k+3 is the one which is divisible by 3. Then k+3 = 3m, for some positive integer m. Then add 3 to both sides of k+3 = 3m: k+33=3m3 k = 3(m1) then k IS divisible by 3 Now add 1 to both sides of k+3 = 3m: k+31=3m1 k+2 = 3m1 then k+2 is NOT divisible by 3 Now add 1 to both sides of k+3 = 3m: k+3+1=3m+1 k+4 = 3m+1 then k+4 is NOT divisible by 3. So, we have proved case 3. Q.If the H C F of 210 and 55 is expressible in the form 210 × 5 + 55y then find y Answer: Let us first find the H C F of 210 and 55. Applying Euclid division lemna on 210 and 55, we get 210 = 55 × 3 + 45 ....(1) Since the remainder 45 ≠ 0. So, again applying the Euclid division lemna on 55 and 45, we get 55 = 45 × 1 + 10 .... (2) Again, considering the divisor 45 and remainder 10 and applying division lemna, we get 45 = 4 × 10 + 5 .... (3) We now, consider the divisor 10 and remainder 5 and applying division lemna to get 10 = 5 × 2 + 0 .... (4) We observe that the remainder at this stage is zero. So, the last divisor i.e., 5 is the HCF of 210 and 55. ∴ 5 = 210 × 5 + 55y ⇒ 55y = 5  1050 = 1045 ⇒ y = 19 Q. Finds the H.C.F. of 65 and 117 and express it in the form of 65m+117n. Answer: Among 65 and 117; 2117 > 65 Since 117 > 65, we apply the division lemma to 117 and 65 to obtain 117 = 65 × 1 + 52 … Step 1 Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain 65 = 52 × 1 + 13 … Step 2 Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain 52 = 4 × 13 + 0 … Step 3 In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers The H.C.F. of 65 and 117 is 13 From Step 2: 13 = 65 – 52× 1 … Step 4 From Step 1: 52 = 117 – 65 × 1 Thus, from Step 4, it is obtained 13 = 65 – (117 – 65 × 1) ⇒13 = 65 × 2 – 117 ⇒13 = 65 × 2 + 117 × (–1) In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1 Q.Find all positive integral values of n for which n2+96 is perfect square. Answer: Let n2 + 96 = x2 ⇒ x2 – n2 = 96 ⇒ (x – n) (x + n) = 96 ⇒ both x and n must be odd or both even on these condition the cases are x – n = 2, x + n = 48 x – n = 4, x + n = 24 x – n = 6, x + n = 16 x – n = 8, x + n = 12 and the solution of these equations can be given as x = 25, n = 23 x = 14, n = 10 x = 11, n = 5 x = 10, n = 2 So, the required values of n are 23, 10, 5, and 2. Q. Prove that one of every three consecutive integers is divisible by 3. Ans: n,n+1,n+2 be three consecutive positive integers We know that n is of the form 3q, 3q +1, 3q + 2 So we have the following cases Case – I when n = 3q In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by 3 Case  II When n = 3q + 1 put n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not divisible by 3 Case – III When n = 3q +2 put n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. But n and n+1 are not divisible by 3 Hence one of n, n + 1 and n + 2 is divisible by 3 More solved Questions: 
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