Updated on, Dec 25, 2024
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KCET 2024 Physics Question Paper is available for download here.. KCET Physics question paper was conducted on April 19, 2024 from 10.30 AM to 11.50 PM by Karnataka Examination Authority (KEA). KCET 2024 Physics question paper consists of 60 questions to be attempted in 80 minutes for a total of 60 marks.
KCET 2024 Physics Question Paper with Solution PDF
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KCET 2024 Physics Question Paper with Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 1. The ratio of molar specific heats of oxygen is: (A) 1.4 (B) 1.67 (C) 1.33 (D) 1.28 |
(A) 1.4 | For a diatomic gas like oxygen, the ratio of molar specific heats (γ) is approximately 1.4, reflecting its degrees of freedom and molecular structure. |
| 2. For a particle executing simple harmonic motion (SHM), at its mean position: (A) Velocity is zero and acceleration is maximum (B) Velocity is maximum and acceleration is zero (C) Both velocity and acceleration are maximum (D) Both velocity and acceleration are zero |
(B) Velocity is maximum and acceleration is zero | At the mean position in SHM, the velocity is maximum due to maximal kinetic energy, while acceleration is zero as the restoring force is minimal. |
| 3. A motor-cyclist moving towards a huge cliff with a speed of 18 km/h, blows a horn of source frequency 325 Hz. If the speed of the sound in air is 330 m/s, the number of beats heard by him is: (A) 5 (B) 4 (C) 7 (D) 6 |
(A) 5 | Using the Doppler effect formula, the observed frequency due to motion towards the source is calculated, leading to 5 beats when compared with the original frequency. |
| 4. A body has a charge of -3.2 μC. The number of excess electrons it has is: (A) 1.2 × 1015 (B) 5 × 1010 (C) 3.2 × 1016 (D) 5.12 × 1013 |
(C) 3.2 × 1016 | The number of excess electrons is calculated by dividing the total charge by the charge of an electron, giving 3.2 × 1016 electrons. |
| 5. A point charge A of +10μC and another point charge B of +20μC are kept 1m apart in free space. The electrostatic force on A due to B is vector F1, and the electrostatic force on B due to A is F2 ). Then: (A) vector F2 = 2 vector F2 (B) vector F1 = -vector F2 (C) 2vector F1 = -vector F2 (D) vector F1= vector F2 |
(B) vector F1 = -vector F2 | According to Newton's Third Law, the forces between two charges are equal in magnitude and opposite in direction, confirming vector F1 = -vector F2. |
| 6. A uniform electric field E = 3×10 3 is acting along the positive Y-axis. The electric flux through a rectangle of area 10m×10m , where the plane of the rectangle is parallel to the Z-X plane is: (A) ( 12× 103,Vm) (B) ( 9× 103 ,Vm) (C) ( 15× 103 ,Vm) (D) ( 18 ×103 ,Vm) |
(B) ( 9 ×103 ,Vm) | The electric flux through a surface parallel to the field direction is zero since the field lines are perpendicular to the normal of the surface. |
| 7. The total electric flux through a closed spherical surface of radius r enclosing an electric dipole of dipole moment 2aq is: (A) Zero (B) q/ϵ0 (C)2q/ϵ0 (D) 8πr2q/ϵ0 |
(A) Zero | According to Gauss's Law, the net flux through a surface enclosing a dipole is zero, as the enclosed net charge is zero. |
| 8. Under electrostatic condition of a charged conductor, which among the following statements is true? (A) The electric field on the surface of a charged conductor is σ/2ϵ0, where σ is the surface charge density (B) The electric potential inside a charged conductor is always zero (C) Any excess charge resides on the surface of the conductor (D) The net electric field is tangential to the surface of the conductor |
(C) Any excess charge resides on the surface of the conductor | Excess charge on a charged conductor distributes itself to reside on the surface, minimizing the potential energy of the system. |
| 9. A cube of side 1 cm contains 100 molecules each having an induced dipole moment of 0.2 × 10−16C in an external electric field of 4 × 104 N/C. The electric susceptibility of the materials is: (A) 50 (B) 5 (C) 0.5 (D) 0.05 |
(B) 5 | Using the polarization and the external electric field, the electric susceptibility of the material is calculated as 5. |
| 10. A capacitor of capacitance 5μF is charged by a battery of emf 10V. At an instant of time, the potential difference across the capacitors is 4V and the time rate of change of potential difference across the capacitor is 0.6 V/s. Then the time rate at which energy is stored in the capacitor at the instant is: (A) 12μW (B) 3μW (C) Zero (D) 30μW |
(A) 12μW | The rate of energy storage in the capacitor is determined by the product of the capacitance, the potential difference, and the rate of change of the potential, yielding 12μW. |
| 11. vector E is the electric field inside a conductor whose material has conductivity σ and resistivity rho. The current density inside the conductor is vector j. The correct form of Ohm's law is: (A) vector E = σ vector j (B) vector E = rho vector j (C) vector j = σ E (D) vector j = rho vector E |
(C) vector j = σ E | Ohm's law states that the current density vector j is proportional to the electric field vector E, with the conductivity σ as the constant of proportionality. |
| 12. In the circuit shown, the end A is at potential V0, and end B is grounded. The electric current indicated in the circuit is: (A) V0/R (B) 2V0/R (C) V0/3R (D) V0/2R |
(C) V0/3R | Using Ohm’s law, the current is calculated considering the total resistance in the series circuit, yielding V0/3R. |
| 13. The electric current flowing through a given conductor varies with time as shown in the graph below. The number of free electrons which flow through a given cross-section of the conductor in the time interval ( 0≤ t ≤ 20 , s) is: (A) 3.125 × 1019 (B) 1.6 × 1019 (C) 6.25 × 1018 (D) 1.625 × 1019 |
(C) 6.25 × 1018 | The total charge and consequently the number of electrons is determined by the integral of the current over time, resulting in 6.25 × 1018 electrons. |
| 14. The I-V graph for a conductor at two different temperatures 100°C and 400°C is as shown in the figure. The temperature coefficient of resistance of the conductor is about (in per degree Celsius): (A) 3 x 10-7 (B) 6 x 10-7 (C) 9 x 10-7 (D) 12 x 10-7 |
(A) 3 x 10-7 | Calculating the change in resistance with temperature, the temperature coefficient of resistance is found to be 3 x 10-7 °C-1. |
| 15. An electric bulb of 60 W, 120 V is to be connected to a 220 V source. What resistance should be connected in series with the bulb, so that the bulb glows properly? (A) 50 Ω (B) 100 Ω (C) 200 Ω (D) 288 Ω |
(C) 200 Ω | To ensure the correct voltage drop across the bulb for safe operation, a series resistor is calculated to balance the voltage from the source to match the bulb's design voltage. |
| 16. In an experiment to determine the temperature coefficient of resistance of a conductor, a coil of wire X is immersed in a liquid. It is heated by an external agent. A meter bridge set up is used to determine resistance of the coil X at different temperatures. The balancing points measured at temperatures t1 = 0°C and t2 = 100°C are 50 cm and 60 cm respectively. If the standard resistance taken out is S = 4 in both trials, the temperature coefficient of the coil is: (A) 0.05°C-1 (B) 0.02°C-1 (C) 0.005°C-1 (D) 2.0°C-1 |
(C) 0.005°C-1 | From the change in resistance with temperature, the temperature coefficient of resistance is calculated using the known lengths and standard resistance to measure the relative change in resistivity. |
| 17. A moving electron produces: (A) Only electric field (B) Both electric and magnetic field (C) Only magnetic field (D) Neither electric nor magnetic field |
(B) Both electric and magnetic field | A moving electron, due to its charge, creates an electric field and, due to its motion, a magnetic field, following the principles of electromagnetism. |
| 18. A coil having 9 turns carrying a current produces a magnetic field B1 at the centre. Now the coil is rewound into 3 turns carrying the same current. Then the magnetic field at the centre B2 is: (A) B1/9 (B) 3B1 (C) 2B1 (D) B1/3 |
(D) B1/3 | The magnetic field produced by a coil is directly proportional to the number of turns; therefore, reducing the number of turns to a third reduces the magnetic field to a third. |
| 19. A particle of specific charge q/m is projected from the origin towards the positive x-axis with the velocity 10 m/s in a uniform magnetic field B = -2 hat{k} T. The velocity v of the particle after time t = 5 ms will be (in m/s): (A) 5hat{i} + 5hat{j} (B) 5hat{i} - 5hat{j} (C) 5hat{i} + 5hat{k} (D) 5hat{i} - 5hat{k} |
(B) 5hat{i} - 5hat{j} | Applying the right-hand rule to the velocity and magnetic field vectors, the resulting motion of the particle is circular in the xy-plane, with the velocity vector given as 5hat{i} - 5hat{j}. |
| 20. The magnetic field at the centre of a circular coil of radius R carrying current I is 64 times the magnetic field at a distance x on its axis from the centre of the coil. Then the value of x is: (A) R/sqrt(5) (B) R/3 (C) R/4 (D) R/sqrt(3) |
(A) R/sqrt(5) | Using the ratio of magnetic fields and the properties of the magnetic field along the axis of a circular coil, x is calculated as R/sqrt(5) to satisfy the given condition. |
| 21. Magnetic hysteresis is exhibited by magnetic materials. (A) only para (B) only dia (C) only ferro (D) both para and ferro |
(C) only ferro | Magnetic hysteresis is a phenomenon observed in ferromagnetic materials, where the magnetic induction lags behind the magnetizing force. |
| 22. Magnetic susceptibility of Mg at 300 K is 1.2 x 10-5. What is its susceptibility at 200 K? (A) 18 x 10-5 (B) 180 x 10-5 (C) 1.8 x 10-5 (D) 0.18 x 10-5 |
(C) 1.8 x 10-5 | Assuming that magnetic susceptibility inversely varies with temperature, the susceptibility at 200 K is higher than at 300 K. |
| 23. A uniform magnetic field of strength B = 2 mT exists vertically downwards. These magnetic field lines pass through a closed surface as shown in the figure. The closed surface consists of a hemisphere S1, a right circular cone S2, and a circular surface S3. The magnetic flux through S1 and S2 are respectively: (A) φ1 = -20 Wb, φ2 = +20 Wb (B) φ1 = +20 Wb, φ2 = -20 Wb (C) φ1 = -40 Wb, φ2 = +40 Wb (D) φ1 = +40 Wb, φ2 = -40 Wb |
(A) φ1 = -20 Wb, φ2 = +20 Wb | The magnetic flux through the hemisphere (S1) is negative because it enters the surface, and the flux through the cone (S2) is positive because it exits the surface. |
| 24. In the figure, a conducting ring of certain resistance is falling towards a current-carrying straight long conductor. The ring and conductor are in the same plane. Then the: (A) induced electric current is zero (B) induced electric current is anticlockwise (C) induced electric current is clockwise (D) ring will come to rest |
(C) induced electric current is clockwise | According to Lenz's law, the induced current in the ring will be clockwise to oppose the change in magnetic flux as the ring approaches the conductor. |
| 25. An induced current of 2 A flows through a coil. The resistance of the coil is 100 Ω. What is the change in magnetic flux associated with the coil in 1 ms? (A) 2 x 10-2 Wb (B) 2 x 10-3 Wb (C) 22 x 10-2 Wb (D) 2 x 10-4 Wb |
(B) 2 x 10-3 Wb | The change in magnetic flux can be calculated using Faraday's law of electromagnetic induction, factoring in the induced voltage and the time interval. |
| 26. A square loop of side length a is moving away from an infinitely long current-carrying conductor at a constant speed v as shown. Let x be the instantaneous distance between the long conductor and side AB. The mutual inductance M of the square loop - long conductor pair changes with time t according to which of the following graphs? (A) Graph A (B) Graph B (C) Graph C (D) Graph D |
(C) Graph C | The mutual inductance decreases as the square loop moves away from the conductor, hence a graph showing a decline over time accurately reflects this relationship. |
| 27. Which of the following combinations should be selected for better tuning of an LCR circuit used for communication? (A) R = 20 Ω, L = 1.5 H, C = 35 μF (B) R = 25 Ω, L = 2.5 H, C = 45 μF (C) R = 25 Ω, L = 1.5 H, C = 45 μF (D) R = 15 Ω, L = 3.5 H, C = 30 μF |
(D) R = 15 Ω, L = 3.5 H, C = 30 μF | For effective tuning, it is preferable to choose components that provide a high Q-factor, which enhances the sharpness and efficiency of the circuit's resonance, particularly for communication purposes. |
| 28. In an LCR series circuit, the value of only capacitance C is varied. The resulting variation of resonance frequency f0 as a function of C can be represented as: (A) Graph A (B) Graph B (C) Graph C (D) Graph D |
(C) Graph C | The resonance frequency of an LCR circuit inversely varies with the square root of the capacitance; thus, as capacitance increases, the resonance frequency decreases, as shown in Graph C. |
| 29. The figure shows variation of R, XL, and XC with frequency f in a series LCR circuit. Then for what frequency point is the circuit capacitive? (A) B (B) C (C) D (D) A |
(C) D | At point D, the capacitive reactance (XC) is higher than the inductive reactance (XL), which makes the circuit behavior predominantly capacitive. |
| 30. Electromagnetic waves are incident normally on a perfectly reflecting surface having surface area A. If I is the intensity of the incident electromagnetic radiation and c is the speed of light in vacuum, the force exerted by the electromagnetic wave on the reflecting surface is: (A) 2IA/c (B) IA/c (C) IA/2c (D) I/2Ac |
(A) 2IA/c | When electromagnetic waves strike a perfectly reflecting surface, they impart momentum to the surface. The force exerted is calculated as 2IA/c, where I is the intensity and c is the speed of light, reflecting the principle of conservation of momentum which accounts for the wave's reflection. |
| 31. The final image formed by an astronomical telescope is (A) real, erect and diminished (B) virtual, inverted and diminished (C) real, inverted and magnified (D) virtual, inverted and magnified |
(D) virtual, inverted and magnified | An astronomical telescope forms a virtual, inverted, and magnified image of distant objects, allowing detailed observation of celestial bodies. |
| 32. If the angle of minimum deviation is equal to the angle of a prism for an equilateral prism, then the speed of light inside the prism is (A) 3 x 10^8 m/s (B) 2 x 10^8 m/s (C) sqrt(3) x 10^8 m/s (D) 1.5 x 10^8 m/s |
(C) sqrt(3) x 10^8 m/s | For an equilateral prism, the minimum deviation equals the angle of the prism when the refractive index is such that the speed of light inside the prism is sqrt(3) x 10^8 m/s, based on the refractive properties of the prism material. |
| 33. A luminous point object O is placed at a distance 2R from the spherical boundary separating two transparent media of refractive indices n1 and n2, where R is the radius of curvature of the spherical surface. If n1 = 1.5, n2 = 3, and the image is obtained at a distance from P equal to (A) 30 cm in the rarer medium (B) 30 cm in the denser medium (C) 18 cm in the rarer medium (D) 18 cm in the denser medium |
(D) 18 cm in the denser medium | Using the lens maker's formula and the specific refractive indices provided, the calculations indicate that the image forms at 18 cm in the denser medium, illustrating the effects of refraction at a curved interface. |
| 34. An equiconvex lens of radius of curvature 14 cm is made up of two different materials. Left half and right half of vertical portion is made up of material of refractive index 1.5 and 1.2 respectively as shown in the figure. If a point object is placed at a distance of 40 cm, calculate the image distance. (A) 25 cm (B) 50 cm (C) 35 cm (D) 40 cm |
(C) 35 cm | The unique construction of the lens requires applying the lens maker’s formula to each material segment, leading to a complex focusing effect where the final image distance is calculated to be 35 cm. |
| 35. A galaxy is moving away from the Earth so that a spectral line at 6000 Å is observed at 6300 Å. Then the speed of the galaxy with respect to the Earth is (A) 500 km/s (B) 150 km/s (C) 200 km/s (D) 1500 km/s |
(A) 500 km/s | By applying the Doppler shift formula to the observed change in wavelength, we calculate the galaxy's recession speed relative to Earth, which provides evidence for the expanding universe theory. |
| 36. Three polaroid sheets are co-axially placed as indicated in the diagram. Pass axes of the polaroids 2 and 3 make 30° and 90° with pass axes of polaroid sheet 1. If I1 is the intensity of the incident unpolarised light entering sheet 1, the intensity of the emergent light through sheet 3 is (A) zero (B) I1 cos^2 30° (C) I1 / 2 (D) I1 / 4 |
(C) I1 / 2 | The arrangement of the polaroids causes the light's intensity to diminish as it passes through each layer, with the final emerging intensity being half the initial, based on the orientations and properties of polarized light interaction. |
| 37. In Young's double slit experiment, an electron beam is used to produce interference fringes of width β1. Now the electron beam is replaced by a beam of protons with the same experimental setup and same speed. The fringe width obtained is β2. The correct relation between β1 and β2 is (A) β1 = β2 (B) No fringes are formed (C) β1 < β2 (D) β1 > β2 |
(D) β1 > β2 | The difference in mass between electrons and protons affects their respective de Broglie wavelengths, resulting in different fringe widths where electrons, being lighter, produce wider fringes compared to heavier protons. |
| 38. Light of energy E falls normally on a metal of work function W0. The kinetic energies K of the photoelectrons are (A) K = 2E/3 (B) K = E/3 (C) K < 2E/3 (D) 0 < K < E/3 |
(B) K = E/3 | The energy of the incident light minus the work function of the metal determines the kinetic energy of the ejected photoelectrons, illustrating the fundamental principle of the photoelectric effect where excess energy is converted into kinetic energy. |
| 39. The photoelectric work function for photo metal is 2 eV. Among the four wavelengths, the wavelength of light for which photoemission does not take place is (A) 200 nm (B) 300 nm (C) 700 nm (D) 400 nm |
(C) 700 nm | Given the work function, wavelengths longer than a certain threshold will not have sufficient energy to overcome the work function barrier, hence no photoemission occurs at 700 nm, illustrating the quantum nature of light. |
| 40. In alpha particle scattering experiment, if v is the initial velocity of the particle, then the distance of closest approach is (A) 4 d (B) 2 d (C) d (D) d/2 |
(D) d/2 | The distance of closest approach in alpha particle scattering provides insights into the nuclear structure and the repulsive force between the nucleus and the alpha particles, being inversely proportional to the initial kinetic energy. |
| 41. The ratio of the area of the first excited state to the ground state of the orbit of a hydrogen atom is (A) 1:16 (B) 1:8 (C) 4:1 (D) 16:1 |
(D) 16:1 | The area of the orbit is proportional to the square of the radius. The radius of the first excited state is four times that of the ground state, so the ratio of areas is 16:1. |
| 42. The ratio of the volume of an Al27 nucleus to its surface area is (Given R0 = 1.2 x 10-15 m) (A) 2.1 x 10-15 (B) 1.3 x 10-15 (C) 0.22 x 10-15 (D) 1.2 x 10-15 |
(D) 1.2 x 10-15 | The volume V of the nucleus is proportional to R3 and the surface area A is proportional to R2. Therefore, the ratio of volume to surface area is proportional to R, which results in R0 being approximately 1.2 x 10-15 m. |
| 43. Consider the nuclear fission reaction n1 + U235 -> Ba144 + Kr89 + 3 n. (A) 200 MeV (B) 180 MeV (C) 67 MeV (D) 60 MeV |
(D) 60 MeV | The kinetic energy per fast neutron is calculated using the binding energy of the products and reactants: KE = (B.E. of products - B.E. of reactants)/3. Substituting the given binding energies: KE = (1980 - 1800)/3 = 180/3 = 60 MeV. |
| 44. The natural logarithm of the activity R of a radioactive sample varies with time t as shown. At t = 0, there are N0 undecayed nuclei. Then N0 is equal to [Take e7.5] (A) 7,500 (B) 3,500 (C) 75,000 (D) 150,000 |
(C) 75,000 | From the graph, at t = 0, the activity R0 is e2. Thus, R0 = e2 = 7.5. Now, we calculate the decay constant lambda: lambda = 1/(10 x 103) = 10-4 sec-1. Using the formula for activity: N0 = R0/lambda = 7.5/10-4 = 75,000. |
| 45. Depletion region in an unbiased semiconductor diode is a region consisting of (A) both free electrons and holes (B) neither free electrons nor holes (C) only free electrons (D) only holes |
(B) neither free electrons nor holes | The depletion region in an unbiased semiconductor diode is formed by the recombination of holes and electrons at the junction. As a result, this region does not contain free charge carriers. |
| 46. The upper level of the valence band and the lower level of the conduction band overlap in the case of (A) silicon (B) copper (C) carbon (D) germanium |
(B) copper | In copper, the valence band and conduction band overlap, allowing free electrons to flow and making copper a good conductor. This is not the case in silicon, germanium, or carbon, where there is a gap between the two bands. |
| 47. In the diagram shown, the Zener diode has a reverse breakdown voltage Vz. The current through the load resistance RL is IL. The current through the Zener diode is (A) Vo = Vz (B) Vo = Vz - IL RL (C) Vo = Vz (D) Vo = Vz - IL RL |
(D) Vo = Vz - IL RL | The voltage across the Zener diode is the difference between the Zener voltage and the voltage drop across the load resistance: Vo = Vz - IL RL. |
| 48. A p-n junction diode is connected to a battery of emf 5.7 V in series with a resistor 5k ohm such that it is forward biased. If the barrier potential of the diode is 0.7 V, neglecting the diode resistance, the current in the circuit is (A) 1.14 mA (B) 1 mA (C) 1 A (D) 1.14 A |
(B) 1 mA | Using Ohm’s law, the current in the circuit is: I = (V - Vb)/R where V = 5.7 V is the battery voltage, Vb = 0.7 V is the barrier potential, and R = 5 x 10^3 ohm. Substituting the values, we get: I = (5.7 - 0.7)/5 x 10^3 = 1 mA. |
| 49. An athlete runs along a circular track of diameter 80m. The distance travelled and the magnitude of displacement of the athlete when he covers 3/4 of the circle is (in m) (A) 60, 40 (B) 40, 60 (C) 120, 80 (D) 80, 120 |
(A) 60, 40 | Distance travelled = 3/4 x 2 pi R = 3/4 x 2 pi x 40 = 60 m, Displacement = sqrt(R^2 + R^2) = 40 m. |
| 50. Among the given pair of vectors, the resultant of two vectors can never be 3 units. The vectors are (A) 1 unit and 2 units (B) 2 units and 5 units (C) 3 units and 6 units (D) 4 units and 8 units |
(D) 4 units and 8 units | The resultant of two vectors lies between the maximum P + Q and minimum P - Q resultant. Therefore, for 4 and 8 units, the resultant can never be 3 units. |
| 51. A block of certain mass is placed on a rough inclined plane. The angle between the plane and the horizontal is 30°. The coefficients of static and kinetic frictions between the block and the inclined plane are 0.6 and 0.5 respectively. Then the magnitude of the acceleration of the block is [Take g = 10 m/s2] (A) 2 m/s2 (B) Zero (C) 0.196 m/s2 (D) 0.67 m/s2 |
(B) Zero | Since the force of friction f = μN and μs, the static friction coefficient, is high enough to counteract any motion, the block does not move. Therefore, the acceleration is zero. |
| 52. A particle of mass 500 g is at rest. It is free to move along a straight line. The power delivered to the particle varies with time according to the following graph: (A) 2 J/s (B) 5 J/s (C) 5 N (D) 5.5 N |
(D) 5.5 N | From the graph, we can observe that the work done by the force is given by the area under the curve. The momentum p of the particle at t = 5 s can be calculated as: p2 = 2 x 500 x 103 = 25 m/s. Thus, the momentum of the particle is 5 kg·m/s. |
| 53. Dimensional formula for activity of a radioactive substance is (A) M1L0T-1 (B) M0L1T0 (C) M0L0T-1 (D) M1L0T-2 |
(C) M0L0T-1 | The activity is the inverse of time, which has the dimensional formula T-1. Thus, the dimensional formula for activity is M0L0T-1. |
| 54. A ceiling fan is rotating around a fixed axle as shown. The direction of angular velocity is along (A) +j (B) -j (C) +k (D) -k |
(D) -k | The direction of the angular velocity vector follows the right-hand rule. If the fan is rotating counterclockwise, the angular velocity will point in the -k-direction. |
| 55. A body of mass 1 kg is suspended by a weightless string which passes over a frictionless pulley of mass 2 kg as shown in the figure. The mass is released from a height of 1.6m from the ground. With what velocity does it strike the ground? (A) 16 m/s (B) 8 m/s (C) 4√5 m/s (D) 4 m/s |
(B) 8 m/s | Using the conservation of energy, the potential energy at the start is converted into kinetic energy: mgh = 1/2 mv2. Substitute values: 1 x 9.8 x 1.6 = 1/2 x 1 x v2. Solving for v, we get v = 8 m/s. |
| 56. What is the value of acceleration due to gravity at a height equal to half the radius of the Earth, from its surface? (A) 4.4 m/s2 (B) 6.5 m/s2 (C) Zero (D) 9.8 m/s2 |
(A) 4.4 m/s2 | At height h, the acceleration due to gravity is given by: gh = g0 / (1 + h/R)2. Substituting h = R/2: gh = g0 / (1 + 1/2)2 = 9.8 / 2.25 = 4.4 m/s2. |
| 57. A thick metal wire of density ρ and length L is hung from a rigid support. The increase in length of the wire due to its own weight is (A) ρL2g / 2Y (B) gρL2 / 2Y (C) ρgL2 / 2Y (D) ρgL2 / 4Y |
(B) gρL2 / 2Y | The increase in length due to the weight of the wire is given by the formula for the elongation of a wire under a force: ΔL = FL / AY where F = ρgL2 is the force due to the weight of the wire, and Y is the Young's modulus. Substituting the values, we get: ΔL = ρgL2 / 2Y. |
| 58. Water flows through a horizontal pipe of varying cross-section at a rate of 0.314 m³/s. The velocity of water at a point where the radius of the pipe is 10 cm is (A) 5 m/s (B) 10 m/s (C) 100 m/s (D) 103 m/s |
(B) 10 m/s | Using the principle of conservation of mass and the continuity equation, A1v1 = A2v2, where A is the cross-sectional area and v is the velocity, the velocity at the given point can be found by: v = Q / A. Substituting Q = 0.314 m³/s and A = πr2, we get v = 10 m/s. |
| 59. A solid cube of mass m at a temperature θ0 is heated at a constant rate. It becomes liquid at temperature θ1, and vapor at temperature θ2. Let s1 and s2 be specific heats in its solid and liquid states respectively. If L and Lv are latent heats of fusion and vaporization respectively, then the minimum heat energy supplied to the cube until it vaporises is (A) mc1(θ1 - θ0) + mc2(θ2 - θ1) + mLv (B) mLf + mc2(θ2 - θ1) + mLv (C) m(θ2 - θ0) + mLv (D) mc1(θ1 - θ0) + mLf + mLv |
(A) mc1(θ1 - θ0) + mc2(θ2 - θ1) + mLv | The total heat energy required is the sum of the energy needed to raise the temperature of the solid, the energy to melt the solid into liquid, and the energy to vaporize the liquid: Q = mc1(θ1 - θ0) + mc2(θ2 - θ1) + mLv. |
| 60. One mole of an ideal monatomic gas is taken round the cyclic process MNOM. The work done by the gas is (A) 4.5 P0V0 (B) 9 P0V0 (C) 10 P0V0 (D) 2 P0V0 |
(C) 10 P0V0 | The work done by a gas in a cyclic process is the area enclosed by the path on a PV diagram. For the given process, the work done can be calculated by: W = P0V0. |
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