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Best of Luck Q1. Hairs of shaving brush cling together when it is removed from water due to- (a) force of attraction between hair (b) surface tension (c) viscosity of water (d) characteristic property of hair Q2. Coatings used on raincoat are waterproof because they - (a) increases angle of contact (b) decreases angle of contact (c) does not alters angle of contact (d) forms a smooth surface Q3. A body measures 5 N in air and 2 N when put in water. The buoyant force is- (a) 7 N (b) 9 N (c) 3 N (d) None of these Q4. A hole is made at the bottom of the tank filled with water (density = 1000 kg/m³). If the total pressure at the bottom of the tank is three atmospheres (1 atmosphere = 10⁵ N/m²), then the velocity of efflux is nearest to- (a) 20 m/s (b) 10√2 m/s (c) 20√2 m/s (d) √500 m/s Q5. When the temperature is increased the angle of contact of a liquid? (a) increases (b) decreases (c) remains the same (d) first increases and then decreases Q6. The surface tension of a soap solution is 2

Electricity - Extra questions Q1. Find the heat generated in a wire of resistance 100Ω connected to a constant voltage supply of 80 volts for 15 seconds. Q2. An electric iron of rating 200V, 4A is connected to a suitable 200V voltage source. If a fan with power consumption 100W is connected to the iron with an independent switch. Calculate - (i) Current for the entire circuit. (ii) Resistance of the entire circuit. Q3. Find the resistivity of a wire with resistance 200Ω, length 2 m and having cross-sectional area 2.5 mm². Q4. Which has more resistivity : a thin copper wire at 30°C or a thick copper wire at 30°C? Q5. Which has more resistivity : a copper wire at 30°C or an aluminium wire at 30°C? Q6. Which has more resistivity : a copper wire at 30°C or a copper wire at 42°C with same length and diameter? Q7. Calculate the resistance of a wire with resistivity 2 × 10-⁶ Ωm, length 5 m and diameter 2 mm. Take, π = 3.14 Q8. How many 10 Ohms resistors are required to make an equivale

Before the beginning, first discuss about aliens. Aliens are those creatures which do not belong to our world. Now a question comes to your mind. Can we contact aliens? The answer is most probably NO. But why? Because first we have to find that type of planets (or natural satellites) which can support life. This is not an easy task. Also, a planet which can support life is not enough. Why? Because life can exist in various forms. One form is unicellular organisms. For example, in our solar system it is believed that Titan (Saturn's moon) can support unicellular life. But the disadvantage is we cannot contact them by our communication methods. But yes, there is a high possibility in future that a rover mission can take some samples of the unicellular or may be small multicellular organisms and send the data to Earth. Now comes the interesting part. If a planet is supporting complex life like animals, plants or even species like us (humans), then also there are very very

Living on Mars... worse than we think From last two decades Mars has been a prominent candidate to settle humans in our solar system other than our home planet Earth. But there lot of things about Mars and complex life that an average person hardly knows. Now, lets dig deep in to this topic and discuss what would be the big problems in future when humans want to live on Mars permanently. 1. Low Gravity : The red planet has about 38% surface gravity compared to Earth's surface. What's a big deal in this? The answer is that humans will face heavy muscle and bone density loss because their bodies will tell their brains that they require very less bone density and muscles to walk, jump or to do any other physical activity. So those will live on Mars have to do very heavy weight training sessions on a regular basis, just only to survive on Mars. 2. Helplessness : From  our current technology, it takes about seven to eight months to reach Mars from Earth.  If we upgrade

 Majority of the students around the world find Physics difficult. Why it is happening on such a large scale? Lets dig deep in to this topic. There are plenty of reasons why students from each and every country are afraid of Physics. But as a Physics teacher, I will give the solutions to love Physics. Solution 1 : Upgrade your mathematical skills to your maximum potential. Topics like Calculus, Vectors, Differential equations, Algebra, Trigonometry, Logarithm are the key concepts of physics. If you have a lose grip on any of these topics, then this is the time to fix it. If you are from a mathematical background(Non-medical in India), then you have an upper hand compared to the students from non-mathematics background(Medical in India). Solution 2 : Improve your imagination to the next level. This is the reason why Albert Einstein was excellent in Physics. Students who are born after year 2,000 have spend their major time in things such as smartphones, computers, TVs, etc. Thi

Multiple Choice Questions Q1. The removal of Hydrogen from a substance is known as ---- (i) oxidation (ii) dehydration (iii) reduction (iv) crystallisation Q2. The development of unpleasant smell and taste in fat and oil containing foods due to aerial oxidation is ---- (i) acidity (ii) rancidity (iii) displacement (iv) sublimation Q3. In order to obtain a yellow precipitate from Lead Nitrate , you should mix a solution of ---- (i) Potassium Chloride (ii) Potassium Iodide (iii) Potassium Bromide (iv) Potassium Cyanide Q4. In exothermic reactions ---- (i) thermostat is required (ii) energy is released (iii) Energy is absorbed (iv) None of the above Q5. Find a & b in - (i) a = 1, b = 1 (ii) a = 1, b = 2 (iii) a = 3, b = 1 (iv) a = 2, b = 1 Q6. Which gas/gases are released when FeSO 4 is heated? (i) Sulphur Dioxide (ii) Sulphur Trioxide (iii) Both (i) & (ii) (iv) None of these Q7. Which compound is used in black and white photography? (i) Silver Chloride (ii) Sil

REFRACTION - Multiple Choice Questions  Q1. The refractive indexes of medium A, B and C are 1.8 , 1.6 and 1.7 respectively. Which of them has smallest critical angle? (i) A (ii) B (iii) C (iv) Insufficient data Q2. A convex lens has focal length of  15 cm & forming an erect image. Position of object may be at ____. (i) 10 cm (ii) 15 cm (iii) 20 cm (iv) 40 cm Q3. The refractive index of a medium is 1.2 . The speed of light in it is ____ m/s. (i) 3 × 10⁸ (ii) 2 × 10⁸ (iii) 2.5 × 10⁸ (iv) None of these Q4. If the focal length of a lens becomes half, its power becomes ____. (i) half (ii) 4 times (iii) 8 times (iv) None of these Q5. For a convex lens, the distance of image is ____. (i) positive (ii) negative (iii) both (i) & (ii) (iv) Insufficient data Q6. A small bulb is placed at the focal point of a converging lens. When the bulb is switched ON , the lens produces ____. (i) a convergent beam of light (ii) a divergent beam of light (iii) a parallel beam of light (iv) a patch

Moving air is called wind. Wind has energy. The energy possessed by wind is due to its high speed. So we can say the wind possesses kinetic energy . It is the kinetic energy of wind which is utilized for doing work. Solar energy or sun's energy is responsible for the blowing of wind, due to the uneven heating of earth by the sun in different regions. Now let's discuss about what is a wind mill. Wind mill The energy of wind is harnessed by using a windmill. A windmill consists of big sized table fan like blades which are fixed over the top of a tall pole in such a way that they are free to rotate when the fast moving wind strikes on the blades of the windmill. It makes the blades rotate continuously. The rotatory motion of the windmill is then used to do Mechanical work through a shaft connected to the rotating blades. The traditional use of wind energy has now been modified by the improvement in technology to generate electricity through wind powered generators. Wind Generator

Capacitors Numericals Q1. Two metal plates having charges Q , -Q face each other at some separation and are dipped into an oil tank, if the oil is pumped out, the electric field between the plates will _____. Q2. Three capacitors of capacitance 6 μF each are available. The minimum and maximum capacitances, which may be obtained are respectively _____. Q3. The plates of a parallel plate capacitor are made of circular discs of radii 5 cm each. If the separation between the plates is 1 mm . What is the capacitance of the capacitor? Ans. 6.95 × 10-⁵ μF Q4. Three capacitors having capacitances 20 μF, 30 μF and 40 μF are connected in series with a 12V battery. Find the charge on each of the capacitors. How much work has been done by the battery in charging the capacitors? Ans. 110 μC on each approximately, 1.32 milli Joule Q5. The distance between the plates of a parallel plate capacitor is 4 cm . A field of 5,000 volt per metre is established between the plates and an uncharged metal pl

Q1. How much energy is needed to relocate an electron from first excited state to fourth excited state in an Hydrogen atom? Also, find the type of light needed in this process. Q2. Find the wavelength of light released when an electron jumped from 4th excited state to ground state in an Hydrogen atom. Also, name the series. Q3. Find the wavenumber of light released when an electron jumped from 5th excited state in bracket series (in H - atom). www.advguruji.blogspot.com Q4. An electron absorbed 2.55 eV energy and jumped from first excited state to nth excited state. Find n . Q5. when an electron jumped from second excited state to some different state, UV ray was released. Name the state and find the frequency of the UV ray. --------------------------------------------------------------------------------------------------- ANSWERS Ans1. 2.86 eV Ans2. 99 nm Ans3. 0.375 × 10⁶ m-¹ Ans4. n = 3 Ans5. 2.90 × 10¹⁵ Hz by    www.advguruji.blogspot.com

  Q1. Melting and boiling are examples of - A. adiabatic process B. isobaric process C. isothermal process D. cyclic process Q2. For an isothermal process the change in internal energy is - A. positive B. negative C. zero D. infinite Q3. Specific heat capacity of any gas for isothermal process is - A. zero B. infinite C. 1 D. log10⁵ Q4. For a cyclic process the change in internal energy is - A. positive B. negative C. Zero D. infinite Q5. Which is not an extensive variable ? A. Pressure B. mass C. volume D. internal energy Q6. First law of thermodynamics is based on - A. Law of conservation of momentum B. Law of conservation of angular momentum C. Law of conservation of mass D. None of these Q7. The slope of pressure volume curve in an adiabatic process is - A. Equal to the slope of isothermal process B. 'γ' times the slope of isothermal process C. "1/γ" times the slope of isothermal process D. Negative to the slope of isothermal process Q8. An ideal gas of volume 0.

QUESTIONS ON SATELLITES Q1. The orbital velocity of an artificial satellite in a circular orbit just above the Earth's surface is 'v'. For a satellite orbiting at an altitude of half the earth's radius the orbital velocity is ____. Q2. Two satellites P and Q ratio of masses 2:1 are in circular orbits of radii 'r' and '2r' respectively. Then ratio of total mechanical energy of satellite P to Q is ____. A. 1:4 B. 1:2 C. 4:1 D. 1:1 Q3. The period of a satellite in a circular orbit around a planet is independent of ____. A. The mass of the satellite B. The radius of the planet C. The mass of the planet D. All of these Q4. Two identical satellites are orbiting at distances 'R' and '7R' respectively from the surface of Earth, R is the radius of Earth. The ratio of their _ A. Kinetic energies is 4 B. Potential energies is 4 C. Total energies is 4 D. All of these Q5. The orbital velocity of a body close to Earth's surface is ____.

Q1. Calculate the percentage decrease in the weight of a body when it is taken to a height of 40 km above the Earth's surface. Radius of Earth = 6,400 km Ans. 1.25% Q2. Calculate the depth below the surface of Earth where the acceleration due to gravity is 25% of its value at the Earth's surface. Take, radius of Earth = 6,400 km. Ans. 4.8 × 10⁶ m Q3. A body weighs 100N on Earth's surface. Calculate the gravitational force on the body at a height equal to one-fourth the radius of Earth. Ans. 64 N Q4. Calculate the percentage decrease in weight of a body when taken to a tunnel 38.4 km below the Earth's surface. Ans. 0.6% Q5. Calculate the height from the Earth's surface at which the acceleration due to gravity becomes 75% compared at Earth's surface. Take, radius of Earth = 6,400 km. Ans. 990 km ---------------------------------------------------------------------------------------------- EXTRA QUESTIONS Q6. Find the percentage increase in acceleration due to grav

Q1. A bus covered 40% distance of its initial journey at a speed of 15 m/s, next 20% of its journey at a speed of 40 m/s and remaining journey at a speed of 25 m/s. Find the average speed of the bus in the entire journey. Ans. 20.9 m/s Q2. A car maintained a speed of 30 m/s in half of its journey time, 20 m/s in quarter time of its journey and 10 m/s in remaining time of its journey. Find the average speed of the car in its full journey. Ans. 22.5 m/s Q3. Aakash while driving to school, computes the speed for his trip to be 25 km/h. On his return trip along the same route, there was less traffic and the speed was 40 km/h. Find the average speed of Aakash's trip. Ans. 30.7 km/h by    www.advguruji.blogspot.com

Q1. An electron beam has an aperture of 2 mm². A total of 7 × 10¹⁵ electrons flow through any perpendicular cross-section per second. Calculate the current density of the electron beam. Q2. The charge in a given wire varies with time, given by -     q = 3t³ - t² Where, q in 'C' and t in 'secs'. Find the current flowing in the wire at 4th second. Q3. The current flowing in a conductor is the function of time, given by -    I = 8t - 3 Where, I in 'Ampere' and t in'secs'. Find the total charge flowed in first 5 seconds in the conductor. Also, find the charge which flowed from 6th to 8th second. ----------------------------------------------------------------------             ANSWERS A1. 560 Am-² A2. 136 A A3. 85 C, 106 C Introduction to CURRENT ELECTRICITY (Basic Concepts) CLICK HERE by  www.advguruji.blogspot.com

Electric field due to an electric dipole on its - Consider, an electric dipole of length 2a having charges +q and  -q. A point P lies on the equatorial line of the dipole at a distance r from the centre of the dipole. The electric fields due to '-q' and '+q' are E1 and E2 respectively. Now, | E1 |  = kq/[(r² + a²)½]² = kq/(r² + a²)    .....(1) | E2 |  = kq/[(r² + a²)½]² = kq/(r² + a²)    .....(2) From equation (1) and (2) | E1 |  =  | E2 |  =   x (say) Clearly, components E1sinθ and E2sinθ cancel each other. Therefore, Net electric field at point P is - E = xCosθ + xCosθ = 2xCosθ From equation (1) E = 2kq.Cosθ/(r² + a²)         = kq2a/[(r² + a²).(r² + a²)½]   = kq2a/[(r² + a²)³]½ Also, dipole moment - Questions for Practice Q1. Find the ratio of electric fields due to a short dipole on the axial line to equatorial line at same distances from the centre. Ans.    Ea/Ee = (2kp/r³)/(kp/r³)                   = 2

Electric Current It is defined as the rate of flow of positive charge. It has the direction towards the motion of positive charge and direction opposite to the motion of negative charge. It is denoted by I. Its SI unit is ampere(A) or C/s. It is of two types :- Direct current [DC] Alternating current [AC] Direct Current ( DC) The current which has a same direction of flow. It is of two types : 1. fluctuating current The current whose magnitude changes again and again is called fluctuating current. 2. Steady current or ideal current The current whose magnitude do not changes with time is called steady current. Alternating current (AC) The current which changes its direction of flow periodically. (In detail, chapter 7) On the basis of calculation Current is of two types : 1. Average current [ I = q/t ] It is defined as the ratio of total charge flow to the total time of flow. 2. Instantaneous current [ I = dq/dt ] It is defined as the ratio of small change in c

Q1. Two isolated spherical conductors having radii 5 cm and 10 cm. They have charges of 12nC and -3nC respectively. Find the charges after they are connected by a conducting wire. Also, find the common potential after redistribution. Ans. 3 nC,  6 nC,  540V Q2. Two parallel plate capacitors A and B having capacitance of 1μF and 5 μF are charged separately to the same potential of 100V. Now, the positive plate of A is connected to the negative plate of B, and the negative plate of A is connected to the positive plate of B. Find the final charges on each capacitor. Ans. (200/3 μC),  (1000/3) μC This is the video for the Lovely students CLICK HERE by    www.advguruji.blogspot.com

Extra information on capacitors Q:1 Why there is low potential difference between the two conductors of a capacitor? Ans. As we know,              C = q/V So, for large capacitance of a capacitor and large amount of charge storing in a capacitor, V must be low. Also, in practicality if 'V' is large it will be result to a strong electric field between the conductors which can ionize the medium between the conductors resulting to acceleration in the charges towards the opposite charges and the charges inside the capacitor can fully or partially leaks away. Q:2 What is a capacitor? Ans. A capacitor is an energy storing device which stores energy in the form of charge hold inside the capacitor. Q:3 What is dielectric strength? Ans. It is the maximum amount of electric field of a dielectric medium so that the medium will not ionize. For example, dielectric strength of air = 3 × 10⁶ N/C. Pyaare bacchon, tumhari help   k liye mene video daali hai Q:4 List some uses of

Linear Equations Q1. Ten years ago, father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be. Find their present ages. Ans. Present age of father = 34 years          Present age of son = 12 years Q2. Three years ago, Priya's mother was six times older than her. Four years from now, Priya's mother will be one year older than thrice her age. Find the present age of Priya and her mother. Ans.  Present age of Priya = 8 years       Present age of mother = 33 years Q3. Five years ago, Rakesh was five times younger than his mother. Ten years from now, his age will be half of his mother's age. Find the present age of Rakesh and his mother. Ans. Present age of Rakesh = 10 years      Present age of mother = 30 years Q4.           (equation in one variable) A  is elder to  B  by two years. A's father  F  is twice as old as  A .  B  is twice as old as his sister  S . If the ages of the father and sister differ by 40 years,  find th

Applications of Electrostatic shielding - 1. In a lightning thunderstorm, it is safe to sit inside a car rather than near a tree or in open ground. The metallic body of the car act as electrostatic shielding from the lightning to inside the car.  (Important point for boards) 2. In coaxial cable, the outer conductor connected to the ground provides an electrical shield to the signals carried by the central conductor. For electric potential due to an electric dipole CLICK HERE by    www.advguruji.blogspot.com

Cross multiplication method This can be generally written as - For previous class questions CLICK HERE by      www.advguruji.blogspot.com

Q1. A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get 18/11, but if the numerator is increased by 8 and the denominator is doubled, we get 2/5. Find the fraction. Ans.   12/25 Q2. The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction. Ans.  3/7 Q3. A fraction becomes 1/3 if 1 is subtracted from both its numerator and denominator. If 1 is added to both the numerator and denominator, it becomes 1/2. Find the fraction. Ans.  3/7 BEST OF LUCK.. to my dear students.. by           www.advguruji.blogspot.com

Consider, an electric dipole of length 2a having charges +q and -q . A point P is at a distance r from the centre of the dipole such that OP is making an angle θ with dipole moment p . CONSTRUCTION :- Extend OP to BP such that BC is perpendicular to BP, Draw AD perpendicular to BP. PROOF :- Now, β = r + aCosθ      .........(1)          α = r + aCosθ      .........(2) Let, 'V1' and 'V2' be the potential due to +q and -q respectively. Net electric potential at 'P' is -               V = V1 + V2             V = kq/α + k(-q)/β             From equation (1) and (2)            Also,       p = q.2a Therefore,               [  V = kpCosθ/(r² - a²Cos²θ)   ] Note :- For axial line, θ = 0           For equatorial line, θ = π/2 Pyaare bacchon, tumhari help k liye video daali hai by  www.advguruji.blogspot.com

Q1. The sum of two digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number. Answer.    47 or 74 Q2. The sum of the digits of a two digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. Answer.    18 Q3. When a two digit number and its reverse order digit number are added, the result is 99. The tens digit of the number is greater to the ones digit by 5. Find the number. Answer.  72 LOVELY STUDENTS... PLEASE COMMENT IF YOU HAVE ANY DOUBT               by    www.advguruji.blogspot.com

Consider, an uniformly charged infinite sheet having surface charge density σ . Let, A be the small area of the sheet containing charge q . As, charge lie on both the sides of the sheet, so electric field E is also on both the sides. using Gauss theorem, net electric flux through the cylinder enclosing the charge 'q' is -                       Φ    =  q/εο          ......(1) But,                Φ   =  Φ1 + Φ2 + Φ3                             = EACos0° + EACos0° + 0                        Φ  =  2EA      .......(2)           From equation (1) & (2)                 2EA = q/εο                  E    =   q/2Aεο  Also,          σ  = q/A  Therefore,                      [  E = σ/2εο  ] Still not clear click here to watch     by    www.advguruji.blogspot.com

Consider, an electric dipole having dipole moment p in an uniform electric field E such that the dipole moment is making an angle θ with the field. Now, small amount of work done dW in rotating the dipole by small angle dθ is - Total work done when dipole is rotated from angle θ1 to angle θ2 with the field is -                             =  -pE[Cosθ2 - Cosθ1]                        W =  pE[Cosθ1 - Cosθ2] This work done will be stored as potential energy of the dipole. Therefore,  U =  pE[Cosθ1 - Cosθ2] NOTE :- As we know, more the potential energy of a system, less it will be stable and vice versa. 1. For most stable condition    [U = minimum]        θ1 = 180° ,    θ2 = 0° Therefore, U = -2pE 2. For most unstable condition    [U = maximum]       θ1 = 0°  ,     θ2 = 180° Therefore, U = pE[1 - (-1)]                    U = 2pE Still not clear click here to watch      by  www.advguruji.blogspot.com