CBSE Class 12 Physics Question Papers 2024

<p>diffusion of both electrons and holes</p> Signup and view all the answers

<p>6 m</p> Signup and view all the answers

<p>more stable nucleus than its neighbours</p> Signup and view all the answers

<p>1.6 x 10^16</p> Signup and view all the answers

<p>C</p> Signup and view all the answers

<p>7.2 W</p> Signup and view all the answers

<p>2.4 x 10^10 m^-3</p> Signup and view all the answers

<p>(E-V)/R</p> Signup and view all the answers

<p>4.5 x 10^-20 J</p> Signup and view all the answers

<p>repel each other</p> Signup and view all the answers

<p>1.5</p> Signup and view all the answers

<p>1/2</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>True</p> Signup and view all the answers

<p>False</p> Signup and view all the answers

<p>d = λ/2</p> Signup and view all the answers

<p>A diagram with the following features: (i) <strong>Conduction band</strong> is the outermost band, which is either partially filled or completely vacant (for n-type/p-type) (ii) <strong>Valence band</strong> is the innermost band, which is either partially filled or completely filled (for n-type/p-type) (iii) <strong>Forbidden gap</strong> is the energy gap between the conduction band and valence band. (iv) n-type semiconductor has donor energy level just below the conduction band while p-type semiconductor has acceptor energy level just above the valence band. The donor level has one electron and acceptor level has one hole.</p> Signup and view all the answers

<p>(i) A strong current can cause excessive heating in the p-n junction due to high resistance of the depletion layer. This excessive heat can damage the junction and create a short circuit leading to failure of the diode. (ii) Impurities are added to intrinsic semiconductors to control their conductivity. The addition of impurities creates either excess free electrons (n-type) or holes (p-type), resulting in a significant increase in the number of charge carriers and hence conductivity. This makes them suitable for use in electronic devices.</p> Signup and view all the answers

<p>(a) Infrared waves are produced when a body is heated. When a body is heated, the atoms and molecules inside it vibrate at higher frequencies. This movement of charged particles generates changing electric and magnetic fields, leading to the emission of infrared radiation. Infrared waves penetrate the skin and tissues, leading to the generation of internal energy and increase in temperature, hence these waves are also called heat waves. Applications of infrared waves: (i) Infrared imaging is used in night vision devices that detect infrared radiation from warm objects. (ii) In healthcare, infrared radiation is used in thermal imaging to detect the heat pattern of the body, which helps in diagnosing various medical conditions.</p> <p>(b) X-rays are produced by bombarding a metal target (usually tungsten) with high-speed electrons. The electrons are accelerated using a high potential difference. When the high-energy electrons hit the target, they interact with the atoms of the target material, causing the emission of X-rays. Applications of X-rays: (i) X-rays are used in medical imaging for diagnosing various diseases by providing images of the bones and organs.<br /> (ii) X-rays are used in industrial applications such as detecting internal defects in materials, analyzing the structure of crystals, and examining welds.</p> Signup and view all the answers

<p>A galvanometer is a device that measures small electric current. However, its range is limited. A galvanometer can be converted into an ammeter by connecting a low resistance shunt in parallel with the galvanometer coil. The shunt allows a large portion of the total current to pass through it, reducing the current flowing through the galvanometer coil to a measurable value. The shunt should be made of a low resistance wire and its value should be adjusted so that the galvanometer coil carries only a small fraction of the total current.</p> Signup and view all the answers

<p>(a) Ionisation energy is the minimum amount of energy required to remove an electron from the outermost shell of an isolated gaseous atom in its ground state. The ionization energy of a hydrogen atom is 13.6 (b) Mass defect is the difference between the actual mass of the nucleus and the sum of the masses of the individual protons and neutrons that make up the nucleus. The mass defect is related to the stability of the nucleus because it's a measure of the binding energy that holds the nucleus together. A larger mass defect implies a higher binding energy, making the nucleus more stable.</p> Signup and view all the answers

<p>The image formed is at a distance of 40 cm from the spherical surface.</p> Signup and view all the answers

<p>(a) The impedance of the circuit is Z = √(R^2 + (ωL)^2) = √(10^2 + (100π * (100/π * 10^-3))^2) = √(100 + 100) = √200 = 10√2 Ω. (b) The phase angle is given by tan φ = (ωL)/R = (100π * (100/π * 10^-3))/10 = 1. Hence φ = 45° (c) The voltage drop across the inductor is given by V_L = IωL = I * 100π * (100/π * 10^-3) = 10I. To find I, we use the relation V/Z = I, hence I = 141/10√2 = 10√2. Therefore V_L = 10 * 10√2 = 100√2 V</p> Signup and view all the answers

<p>Applying Snell's law to the first surface, we get sin i = µ sin r₁, and applying Snell's law to the second surface, we get µ sin r₂ = sin e. For total internal reflection at the second surface, angle of incidence r₂ must be greater than the critical angle C. Therefore, µ sin C = sin e. Simplifying these three equations, we get sin i = µ sin r₁, µ sin (A - r₁) = sin C, and µ sin C = sin e. Simplifying further, we get sin i = µ sin r₁ , µ sin (A - r₁) = sin C, and sin i = µ sin (A - r₁) Hence, sin i = µ sin (A - r₁)</p> Signup and view all the answers

<p>(a) <strong>(i) Nuclear Fission</strong> is a nuclear reaction in which a heavy nucleus, such as Uranium or Plutonium, splits into two or more lighter nuclei, releasing a huge amount of energy in the process. For example, the spontaneous fission of Uranium-235 results in the formation of Barium-141 and Krypton-92. <strong>Nuclear Fusion</strong> is a nuclear reaction in which two light nuclei, such as Deuterium and Tritium, combine to form a heavier nucleus, also releasing a huge amount of energy. An example is the fusion of Deuterium and Tritium to produce Helium-4, which is the process that powers the sun. <strong>(ii) Energy Release in Fission and Fusion</strong> The binding energy per nucleon curve is a plot that shows the binding energy per nucleon (the energy required to remove a nucleon from the nucleus) as a function of the mass number. The curve has a peak around iron (Fe), indicating that iron has the highest binding energy per nucleon and is hence the most stable nucleus. When a heavy nucleus splits into two or more lighter nuclei during fission (moving towards the peak of the binding energy per nucleon curve), the binding energy per nucleon increases, resulting in the release of energy. Similarly, when two light nuclei fuse to form a heavier nucleus during fusion (moving towards the peak), there is an increase in the binding energy per nucleon, leading to the release of energy.</p> <p>(b) <strong>(i) Size of Nucleus</strong> The size of a nucleus is typically measured using Rutherford scattering experiments. In these experiments, alpha particles are scattered by a thin foil. The observed scattering angles depend on the radius of the nucleus. By analyzing the scattering data, scientists obtain information about the nucleus's size. The radius of a nucleus is related to its mass number (A) by the following relationship: R = RaA⅓ , where R is the radius of the nucleus, Ra is the constant, and A is the mass number. <strong>(ii) Density of Nucleus</strong> The density of a nucleus is very high. The density of a nucleus is independent of its mass number. The formula for density is ρ = M/V. Since the mass of a nucleus is proportional to its mass number (A) and the volume is proportional to the cube of its radius, the density of the nucleus becomes ρ = (A/R³) = (kA/R³) = k/(RaA⅓)³, which simplifies to k/Ra³, where k is a constant. This indicates that the density of the nucleus is independent of its mass number.</p> Signup and view all the answers

<p>The equivalent emf of the combination is given by: E = (E₁ /r₁ + E₂ /r₂) / (1/r₁ + 1/r₂). This formula represents the combined effect of the two cells when connected in parallel, taking into account their individual emf and internal resistances.</p> Signup and view all the answers

<p>(a) The electric field (E) at the surface of a charged sphere is given by E = kQ/R², where Q is the charge and R is the radius of the sphere. Thus, the ratio of the electric fields at the surfaces of two spheres having charges Q₁ and Q₂ and radii a and b is E₁/E₂ = (kQ₁/a²) / (kQ₂/b²) = (Q₁/a²) / (Q₂/b²). Since the spheres are connected by a wire, they will reach a common potential. The potential of a sphere is given by V = kQ/R. Therefore, we also know that V₁ = kQ₁/a = V₂ = kQ₂/b. Solving for the ratio of charges Q₁/Q₂ = a/b. Substituting this ratio into the electric field ratio, we get E₁/E₂ = (a/b)/(a²) / (b²) = (a/b)/(a²/b²) = b/a.</p> <p>(b) (i) When capacitors A and B are connected, the total charge (Q) in the system remains constant. Q = C * V. Therefore, the final charges on capacitors A and B are Q₁ and Q₂ respectively, and Q₁ + Q₂ = Q. After connecting capacitor A, the potential across the combination is given by V' = Q/C₁ = (Q/3C). Using this value, we can find the final charge on capacitor A, Q₁ = C * V' = C * (Q/3C) = Q/3. Likewise, the final charge on capacitor B is Q₂ = (2C) * V' = (2C) * (Q/3C) = 2Q/3. Therefore, the ratio of final charges on capacitors A and B is Q₁/Q₂ = (Q/3)/(2Q/3) = 1/2. (ii) The energy stored in a capacitor is given by E = 1/2<em>C</em>V². Initial stored energy in capacitor A, E₁ = ½ * C * V² = ½ * CV². The final energy stored in capacitors A and B is E₂ = ½ * C* V₁² + ½ * C₂ * V₂² = ½ * (Q/9 * C) + ½ * (2Q/9 * C) = ½ * (Q²/3C) + ½ * (2Q²/9 * C) = (1/6) * (Q²/C) + (1/9) * (Q²/C) = (5Q²/18C). Therefore, the ratio of final energy to initial energy is E₂/E₁= (5Q²/18C) / (½ * CV²) = (5Q²/18C) / (½ * C * (Q/C)²) = 5/9</p> Signup and view all the answers

<p>(a) <strong>(i) Huygen's Principle:</strong> Huygen's principle states that every point on a wavefront can be considered as a new source of secondary wavelets that spread out in all directions with the same speed as the original wave. The new wavefront is the envelope of these secondary wavelets. When the wavefront in a rarer medium strikes a denser medium, it slows down because the speed of light is slower in a denser medium. The light rays diverge to form a diverging wavefront. The secondary wavelets from each point of the wavefront will then proceed to form a reflected wavefront at the boundary between two mediums.<br /> <strong>(ii) Concave Mirror:</strong> Given that the focal length (f) of the mirror is 12 cm and magnification (M) is 3. Since the image is virtual, M will be positive. Using the formula for magnification M = -v/u, we get v = -3u. Using the mirror formula, 1/f = 1/u + 1/v, we have 1/12 = 1/u - 1/(3u) = 2/(3u). Simplifying this gives 3u = 24, hence u = 8 cm. Therefore, the object is 8 cm away from the concave mirror.</p> <p>(b) <strong>(i) Refracting Telescope:</strong> A refracting telescope uses two convex lenses, the objective lens and the eyepiece lens. The objective lens forms a real and inverted image of a distant object at its focal point, which lies close to the focal point of the eyepiece lens. The eyepiece lens further magnifies this image, resulting in a magnified and virtual image of the distant object. The magnifying power of a refracting telescope is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the eye.<br /> <strong>(ii) Limitations:</strong> (1) <strong>Chromatic aberration:</strong> Refracting telescopes suffer from chromatic aberration, this means the different colors of light get focused at different points causing a blurred image.<br /> (2) <strong>Large size:</strong> Because the objective lens needs to be large to collect enough light, refracting telescopes can become bulky and difficult to support. Reflecting telescopes are more efficient than refracting telescopes in terms of resolving the image.</p> Signup and view all the answers

<p>(a) (i) <strong>Electric field due to a thin straight wire:</strong> Consider a cylindrical Gaussian surface of radius r and length l, coaxial with an infinitely long thin straight wire. Using Gauss's law, the flux through the Gaussian surface is given by Φ = ∫E.dA = E * 2πrl. The total charge enclosed by the Gaussian surface is Q = λl. Hence, applying Gauss's law, we get Φ = Q/ε₀. E * 2πrl = λl/ε₀. Therefore, E = λ/(2πε₀r). This expression indicates that the electric field due to an infinitely long thin straight wire is inversely proportional to the distance from the wire. (ii) <strong>Kinetic energy of an electron:</strong> Since the wire is infinitely long and positively charged, the electric field experienced by the electron will be directed radially inward. In this case, the centripetal force needed to keep the electron revolving in a circle is provided by the electric force. Therefore, we can write: mv²/r = eE. Substituting E = λ/(2πε₀r) into the equation, we get mv²/r = eλ/(2πε₀r). Solving for v², we get v² = eλ/(2πε₀m). The kinetic energy of the electron is given by K.E = ½ * mv² = ½ * m * eλ/(2πε₀m) = eλ/(4πε₀). Therefore, the kinetic energy of the electron is proportional to the linear charge density and inversely proportional to the dielectric constant of the medium surrounding the wire. (iii) <strong>Graph of kinetic energy vs linear charge density:</strong> The graph of kinetic energy vs linear charge density will be a straight line passing through the origin. The slope of this line is positive and equal to e/(4πε₀)</p> <p>(b) (i) <strong>Electric field and potential zero:</strong> (1) Consider two identical point charges of magnitude q placed at distances 0 and a from a point P. Let the distance of point P from the charge at x = a be x. The electric field due to charge 1 at P is given by E₁ = Kq/(x²), and due to charge 2 at P is given by E₂ = Kq/((a-x)²). For the electric field to be zero at point P, E₁ + E₂ = 0. Kq/(x²) + Kq/((a-x)²) = 0. The electric field is zero at a point located at x = a/2. (2) The electric potential due to a point charge is given by V = Kq/d, where d is the distance from the charge. For the potential to be zero, the potential due to both charges must be equal and opposite in sign. Considering a point P on the line joining the two charges. The electric potential due to charge 1, V₁ = Kq/x and potential due to charge 2, V₂ = Kq/(a-x). For the potential to be zero, V₁+V₂ =0. Thus, Kq/x + Kq/(a-x) = 0. This equation has no solution as the two terms on the left side are never equal and opposite in sign, hence no point on the line joining two charges will have zero potential.<br /> (ii) <strong>Significance of negative potential energy</strong> The negative value of electrostatic potential energy indicates that the system of charges is in a stable configuration. This is because work is done by the system to move the charges apart, causing the potential energy to increase towards zero.<br /> <strong>(ii) Potential energy of three charges:</strong> Consider an equilateral triangle ABC, with side length 2 m. The potential energy of the system is the sum of the potential energies due to each pair of charges.</p> Signup and view all the answers