HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 1 September 12, 2021 CREATIVE LEARNING CLASSES KARKALA PHYSICS NEET EXAM 2021 DETAILED SOLUTIONS 1. A cup of coffee cools from 90 0 C to 80 0 C in ‘t’ minutes, when the room temperature is 20 0 C. the time taken by a similar cup of coffee to cool from 80 0 C to 60 0 C at a room temperature same at 20 0 C is (1) 13/5 t (2) 10/13t (3) 5/13t (4) 13/10t Answer: (1) By Newton’s law of cooling 𝑑𝑇/𝑑𝑡 = 𝑘 [ 𝑇 + 𝑇 2 − 𝑇 ] ଽି଼ ௧ = 𝐾 [ ଽା ଶ − 20 ] ....... (1) ଼ ି ௧ = 𝐾[ ଼ ା ଶ − 20] ............. (2) 𝑏𝑦 𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (1) 𝑏𝑦 (2) 𝑤𝑒 𝑔𝑒𝑡 𝑡′ = 13 10 𝑡 2. Polar molecules are the molecules: (1) acquire a dipole moment only in the presence of electric field due to displacement of charges. (2) acquire a dipole moment only when magnetic filed is absents (3) having a permanent electric dipole moment (4) having zero dipole moment Answer: (3) 3. A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6MeV while that of fragments is 8.5MeV. The total gain in the binding energy in the process is (1) 9.4 MeV (2) 804 MeV (3)216 MeV (4) 0.9 MeV Answer: (3) 240 X → 2 [ 120 Y] Binding energy of X(Ex)=240x 7.6MeV=1824MeV Binding energy of X(Ey)=120x 8.5MeV=1020MeV Energy gain = 2 (E Y )-(E x ) = 2x1020 – 1824= 216MeV 4. An electromagnetic wave of wavelength λ is incident on a photo sensitive surface of negligible work function. If ‘m’ mass is of photo electron emitted from the surface as de Broglie wavelength λ d . then (1) λ d =(2mc/h) λ 2 (2) λ =(2mc/h) λ d 2 (3) λ =(2h/mc) λ d 2 (4) λ =(2m/hc) λ d 2 Answer: (2) ℎ𝜗 = ℎ𝜗 + 𝐾. 𝐸 work function is zero so ℎ𝜗 = 𝐾. 𝐸 ℎ𝑐/𝜆 = 𝐾. 𝐸 HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 2 September 12, 2021 ℎ𝑐/𝜆 = 1 2 𝑚𝑣 ଶ ℎ𝑐/𝜆 = 𝑃 ଶ 2𝑚 ℎ𝑐/𝜆 = 𝑃 ଶ 2𝑚 ℎ𝑐/𝜆 = (ℎ/𝜆 ௗ ) ଶ 2𝑚 𝜆 = (2𝑚𝑐/ℎ)𝜆 ௗ ଶ 5. A radio-active nucleus z X A , undergoes spontaneous decay in the sequence z X A → z-1 B → Z-3 C → Z-2 D Where z is the atomic number of element X. the possible decay particles in the sequence are (1) α, β + , β - (2) β + , α, β - (3) β - , α, β + (4) α, β - , β + Answer: (2) In β + decay atomic number decreases by one In β- decay atomic number increases by one In alpha decay atomic number decreases by two 6. Column -I gives certain physical terms associated with flow of current through a metallic conductor. Column-II gives some mathematical relation involving electrical quantities Column -I Column - II (A) Drift velocity (P) మ ఘ (B) Electrical resistivity (Q) nev d (C) relaxation period (R) eEτ/m (D) Current density ( S ) E/J (1) (A)-(R), (B)-(S),(C)-(Q), (D)-(P) (2) (A)-(R), (B)-(P),(C)-(S), (D)-(Q) (3) (A)-(R), (B)-(Q),(C)-(S), (D)-(P) (4) (A)-(R), (B)-(S),(C)-(P), (D)-(Q) Answer: (4) 7. In potentiometer circuit a cell of EMF 1.5V gives balance point at 36cm length of wire. If another cell of EMF 2.5V replaces the first cell, then at what length of the wire balance point occurs? (1) 21.6cm (2) 64cm (3) 62cm (4) 60cm Answer: (4) E 𝛼 l 𝐸 ଵ 𝐸 ଶ = 𝑙 ଵ 𝑙 ଶ ଵ ହ ଶ ହ = ଷ మ So l 2 = 60cm 8. The velocity of a small ball of mass m and density d, when dropped in a container filled with glycerin becomes constant after some time. If the density of the glycerin is d/2, then the viscous force acting on the ball will be (1) Mg (2) 3Mg/2 (3) 2Mg (4) Mg/2 Answer: (4) HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 3 September 12, 2021 Where F v = viscous force F b = buoyant force Mg= weight F v + F b =mg F v = mg- F b = (mg) B – (mg) l = (ρ B V B g) – (ρ l V B g) = (ρ B- ρ l ) V B g = (d- ௗ ଶ ) V B g = ௗ ଶ V B g= mg/2 9. A lens of large focal length and large aperture is best suit at as an objective of an astronomical telescope since: (1) a large aperture contributes to the quality and visibility of the images (2) a large area of the objective ensures better light gathering power (3)a large aperture provides a better resolution (4) all of the above Answer: (4) 10. Two charged spherical conductors of radius R 1 and R 2 are connected by a wire. then the ratio of surface charge density is of the spheres( ఙ భ ఙ మ )is: (1) R 2 / R 1 (2) ට ோ భ ோ మ (3) R 1 2 / R 2 2 (4) ோ భ ோ మ Answer: (1) Suppose two charged spherical conductors of radius R 1 and R 2 are connected by a wire then potential become same V 1 =V 2 𝜎 ଵ 𝑅 ଵ 𝜀 = 𝜎 ଶ 𝑅 ଶ 𝜀 𝜎 ଵ 𝜎 ଶ = 𝑅 ଶ 𝑅 ଵ 11. Water falls from a height of 60m at the rate of 15kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine?(g= 10m/s 2 ) (1) 8.1kW (2) 12.3𝑘𝑊 (3) 7.0kW (4) 10.2𝑘𝑊 Answer: (1) Mass of water falling/second = 15 kg/s h = 60 m, g = 10 m/s 2 , loss = 10 % i.e., 90% is used. Power generated =mgh/t = (15 x 10 x 60 x 0.9)/1 = 8100 W = 8.1 kW. HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 4 September 12, 2021 12. The effective resistance of a parallel connection that consist of 4 wires of equal length, equal area of cross section and same material is 0.25Ω. What will be the effective resistance if they are connected in series? (1) 0.5Ω (2)1 Ω (3) 4 Ω (4) 0.25 Ω Answer: (3) Suppose we taken 4 wires of equal length, equal area of cross section and same material then the resistance offered by this wire is same. So R 1 = R 2 = R 3 = R 4 =R So, for parallel combination ଵ ோ ು = ଵ ோ + ଵ ோ + ଵ ோ + ଵ ோ = ସ ோ 𝑅 = 𝑅 4 0.25 = 𝑅 4 R= 1Ω If they are connected in series, R S = 4R= 4x1Ω=4Ω 13. Match column I and column II and choose the correct match from the given choices Column -I Column - II (A) Root mean square speed of gas molecules (P) ଵ ଷ 𝑛𝑚 𝑣 ̅ ଶ (B) Pressure exerted by ideal gas (Q) ඥ 3 𝑅𝑇 / 𝑀 (C) Average kinetic energy of molecule (R) 5 RT /2 (D) Total internal energy of 1 mole of a diatomic gas (S) 3k B T/2 (1) (A)-(Q), (B)-(R),(C)-(S), (D)-(P) (2) (A)-(Q), (B)-(P),(C)-(S), (D)-(R) (3) (A)-(R), (B)-(Q),(C)-(P), (D)-(S) (4) (A)-(R), (B)-(P),(C)-(S), (D)-(Q) Answer: (2) 14. A particle is released from height S from the surface of the earth. At a certain height its kinetic energy is 3 times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively (1) S/4, ඥ ଷௌ ଶ (2) S/2, ඥ ଷௌ ଶ (3) S/4, ට ଷௌ ଶ (4) S/4, ଷௌ ଶ Answer: (3) HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 5 September 12, 2021 K= 3mgh ଵ ଶ 𝑚𝑣 ଶ = 3𝑚𝑔ℎ 𝑣 ଶ = 6𝑔ℎ ...........(1) 𝑣 ଶ = 𝑢 ଶ + 2𝑔𝑦 𝑣 ଶ = 0 + 2𝑔𝑦 ............(2) 6𝑔ℎ = 2𝑔𝑦 3ℎ = 𝑦 3ℎ = 𝑆 − ℎ 4ℎ = 𝑆 h= S/4 𝑣 = ඥ 6𝑔ℎ 𝑣 = ඨ 6𝑔 𝑆 4 𝒗 = ඨ 𝟑𝒈𝑺 𝟐 15. A convex lens A of focal length 20cm and concave lens B of focal length 5 cm are kept along the same axis with a distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then the distance d in cm will be (1) 15 (2)50 (3) 30 (4) 25 Answer: (1) The second focus of the convex lens should coincide with the first focus of concave lens for the given image formation. d = 20 - 5 or d = 15 cm 16. Find the value of the angle of emergence from the prism. Refractive index of the glass is √ 3 (1) 30 0 (2)45 0 (3) 90 0 (4) 60 0 Answer: (4) HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 6 September 12, 2021 17. A capacitor of capacitance C, is connected across an AC source of voltage V, given by V= V 0 sin ωt The displacement current between the plates of the capacitor, would then be given by (1) 𝐼 ௗ = బ ௦ఠ௧ ఠ (2) 𝐼 ௗ = బ ௦ఠ௧ ఠ (3) 𝐼 ௗ = 𝑉 𝜔𝐶 𝑠𝑖𝑛𝜔𝑡 (4) 𝐼 ௗ = 𝑉 𝜔𝐶𝑐𝑜𝑠𝜔𝑡 Answer: (4) 𝐼 ௗ = 𝜀 𝑑𝛷 𝑑𝑡 𝐼 ௗ = 𝜀 𝑑(𝐸𝐴) 𝑑𝑡 𝐼 ௗ = 𝜀 𝐴 𝑑(𝑉 𝑑 ⁄ ) 𝑑𝑡 𝐼 ௗ = 𝜀 𝐴 𝑑 𝑑(𝑉) 𝑑𝑡 𝐼 ௗ = 𝜀 𝐴 𝑑 𝑑(𝑉 𝑠𝑖𝑛𝜔𝑡) 𝑑𝑡 𝐼 ௗ = 𝜀 𝐴 𝑑 𝑉 𝜔𝑐𝑜𝑠𝜔𝑡 𝐼 ௗ = 𝐶𝑉 𝜔𝑐𝑜𝑠𝜔𝑡 18. A dipole is placed in an electric field as shown. In which direction will it move? HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 7 September 12, 2021 (1) towards the right as its potential energy will decrease (2) towards the left as its potential energy will decrease (3) towards the right as its potential energy will increase (4) towards the left as its potential energy will increase Answer: (1) 19. If force [F], acceleration[A], and time[T] are chosen as the fundamental physical quantity. Find the dimensions of energy (1) [F][A] [T 2 ] (2) [F][A][T -1 ] (3) [F][A -1 ][T] (4) [F][A][T] Answer: (1) Energy= F α A β T γ M 1 L 2 T −2 =(MLT −2 ) α (LT −2 ) β (T) γ M 1 L 2 T −2 =M α L α+β T −2α−2β+γ α=1 α+β=2 ⇒ β=1 −2α−2β+γ=−2 ⇒ γ=2 ⇒ Energy= F 1 A 1 T 2 20. An infinitely long straight conductor carries a current of 5A as shown. An electron is moving with a speed of 10 5 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20cm at an instant. Calculate the magnitude of the force experience by the electron at that instant (1) 8 π × 10 –20 N (2) 4 π × 10 –20 N (3) 8 × 10 –20 N (4) 4 × 10 –20 N Answer: (3) Magnetic field produced by wire is perpendicular to the motion of electron and it given by B= 2i μ 0 /4πa=(4π x10 −7 ×5)/(2π×20×10 -2 ) =0.5x10 −5 Wb/m 2 HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 8 September 12, 2021 Hence force on electron F=qvB=(1.6×10 −19 )×10 5 ×0.5×10 −5 =8×10 −20 N 21. A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is (1) 2n (2) 3n (3) 4n (4) n Answer: (1) Let, x=Asinωt v=dx/dt=Aωcosωt Kinetic energy, K=mv 2 /2 ⟹ K=(mω 2 A 2 cos 2 ωt)/2 ⟹ K= ( 𝟏 𝟐 𝐦𝛚 𝟐 𝐀 𝟐 ( 𝟏 ା 𝐜𝐨𝐬𝟐𝛚𝐭 𝟐 ) ⟹ K= 𝟏 𝟒 𝐦𝛚 𝟐 𝐀 𝟐 (𝟏 + 𝐜𝐨𝐬 𝟐 𝛚𝐭 ) ∴ ω K =2ω ∴ Frequency of oscillation of K.E =2n 22. An inductor of inductance L, a capacitor of capacitance C and resistor of resistance R are connected in series to an AC source of potential difference V volts as shown in the figure. The potential difference across L, C and R is 40V, 10V and 40V respectively. The amplitude of current flowing through LCR series circuit is 10 √ 2 A. The impedance of the circuit is (1) ହ √ ଶ Ω (2) 4Ω (3) 5Ω (4) 4 √ 2 Ω Answer: (3) 𝑉 = ට 𝑉 ோ ଶ + (𝑉 − 𝑉 ) ଶ 𝑉 = ඥ 40 ଶ + (40 − 10) ଶ V=50Volt 𝑍 = 𝑉 ௦ 𝐼 ௦ = 50 𝐼 √ 2 ൘ = 50 10 √ 2 √ 2 ൘ = 5Ω 23. The electron concentration is an n- type semiconductor is the same as hole concentration in a p- type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them (1) current in p- type is > current in n-type (2) current in n- type is > current in p-type (3) no current will flow in p-type, current will only flow in n-type (4) current in n- type is = current in p-type Answer: (2) HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 9 September 12, 2021 24. The half- life of radioactive nuclei is 100hours. The fraction of original activity that will remain after 150 hours would be (1) ଵ ଶ √ ଶ (2) ଶ ଷ (3) ଶ ଷ √ ଶ (4) ଵ ଶ Answer: (1) 𝑡 𝑇 ଵ ଶ ൗ = 150 100 = 3 2 𝐴 = 𝐴 2 ଷ ଶ ൗ 𝐴 𝐴 = 1 2 ଷ ଶ ൗ = 1 √ 8 = 1 2 √ 2 25. The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600nm, when it delivers the power of 3.3x10 -3 W will be (H=6.6x10 -34 Js) (1) 10 17 (2) 10 16 (3) 10 15 (4) 10 18 Answer: (2) 𝑃 = 𝐸 𝑡 𝑃 = 𝑛ℎ𝑐 𝜆 (1/𝑡) 𝑛 𝑡 = 𝑃( 𝜆 ℎ𝑐 ) 𝑛 𝑡 = 3.3 × 10 ି ଷ × ( 600 × 10 ି ଽ 6.6 × 10 ି ଷସ × 3 × 10 ଼ ) = 10 ଵ 26. A thick current carrying cable of radius R carries current I uniformly distributed across its cross- section. The variation of magnetic field B(r) due to the cable with the distance r from the axis of the cable is represented by; Answer: (4) 27. For a plane electromagnetic wave propagating x-direction, which one of the following combinations gives the correct possible directions for electric field (E) and magnetic field (B) respectively? (1) −𝚥 ̂ + 𝑘 , −𝚥 ̂ − 𝑘 (2) 𝚥 ̂ + 𝑘 , −𝚥 ̂ − 𝑘 (3) −𝚥 ̂ + 𝑘 , −𝚥 ̂ + 𝑘 (4) 𝚥 ̂ + 𝑘 , 𝚥 ̂ + 𝑘 Answer: (1) Propogation direction = 𝐸 ሬ ⃗ × 𝐵 ሬ ⃗ = ( −𝚥 ̂ + 𝑘 ) x( −𝚥 ̂ − 𝑘 ) = 2𝚤 ̂ Here 𝚤 ̂ denotes plane electromagnetic wave propagating along x-direction HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 10 September 12, 2021 28. A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and the area of each plate is A, the energy stored in the capacitor is; ( 𝜀 = permittivity of free space) (1) 𝜀 𝐸𝐴𝑑 (2) ଵ ଶ 𝜀 𝐸 ଶ 𝐴𝑑 (3) ா మ ௗ ఌ బ (4) ଵ ଶ 𝜀 𝐸 ଶ Answer: (2) Energy U= ଵ ଶ 𝐶𝑉 ଶ = ଵ ଶ ( ఌ బ ௗ )(𝐸𝑑) ଶ = ଵ ଶ 𝜀 𝐸 ଶ 𝐴𝑑 29. Consider the following statements (A ) and (B) and identify the correct answer (A) A Zener diode is connected in reverse bias, when used as a voltage regulator. (B) The potential barrier of p-n junction lies between 0.1V to 0.3V (1) (A) and (B) both are incorrect (2) (A) is correct and (B) is incorrect (3) (A) is incorrect but (B) is correct (4) (A) and (B) both are correct Answer: (2) Where statement-B is wrong, because potential barrier for Si is 0.7V and for Ge is 0.3V. 30. A screw gauge gives the following readings when used to measure the diameter a wire. Main scale reading 0mm, circular scale reading 52divisions. Given that 1mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is (1) 0.026cm (2) 0.26cm (3) 0.052cm (4) 0.52cm Answer: (3) Given M.S.R =0mm CSR= 52 LC= 1/100mm Diameter= MSR+(CSR x LC) = 0+(52x0.01) =0.052cm 31. The equivalent capacitance of the combination shown in the figure is (1) 2C (2) C/2 (3) 3C/2 (4) 3C Answer: (1) Since the potential at point A is equal to the potential at point B, no current will flow along arm AB. Hence, the capacitor on arm AB will not contribute to the circuit. Also, because remaining two capacitor are connected in parallel, the net capacitance of the circuit is given by HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 11 September 12, 2021 C eq = C+C=2C 32. A spring is stretched by 5cm by a force 10N. The time period of the oscillations when a mass of 2kg is suspended by it is (1) 6.28s (2) 3.14s (3) 0.628s (4) 0.0628s Answer: (3) F= 10N ∆𝑥 = 5𝑐𝑚 𝐾 = ி భ ∆ ௫ = ଵ ହ × ଵ షమ = 200𝑁/𝑚 𝑇𝑖𝑚𝑒 𝑝𝑒𝑟𝑖𝑜𝑑 𝑇 = 2𝜋 ට 𝑚 𝐾 𝑇 = 2𝜋 ඨ 2 200 = 2𝜋/10 = 0.628 𝑠 33. A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let S n be the distance travelled by the block in the interval t=n-1 to t=n. Then, the ratio S n /S n+1 is (1) ଶିଵ ଶାଵ (2) ଶାଵ ଶିଵ (3) ଶ ଶିଵ (4) ଶିଵ ଶ Answer: (1) u=0 a= gsinθ S n =u+ ଶ (2𝑛 − 1) S n = ௦ఏ ଶ (2𝑛 − 1) And S n+1 = ௦ఏ ଶ (2(𝑛 + 1) − 1) Solving ௌ ௌ శభ = ଶିଵ ଶାଵ 34. The escape velocity from the earth’s surface is ‘v’. The escape velocity from the surface of another planet having a radius, four times that of earth and same mass density is, (1) 2v (2) 3v (3) 4v (4)v Answer: (1) 𝑉 = ට ଶீெ ோ and ρ’=ρ 𝑉′ = ඨ 2𝐺𝑀′ 𝑅′ 𝑀′ = 𝜌 4 3 𝜋𝑅′ ଷ M’= ெ ర య గோ య ସ ଷ 𝜋𝑅′ ଷ 𝑀′ = 𝑀( 𝑅′ 𝑅 ) ଷ 𝑅 ᇱ = 4𝑅 M’=M4 3 M’=64M 𝑉′ = ඨ 2𝐺64𝑀 4𝑅 HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 12 September 12, 2021 𝑉′ = ඨ 2𝐺16𝑀 𝑅 V’=4V 35. If E and G respectively denote energy and gravitational constant, then E/G has the dimensions of: (1) [M][L -1 ][T -1 ] (2) [M][L 0 ][T 0 ] (3) [M 2 ][L -2 ][T -1 ] (4) [M 2 ][L -1 ][T 0 ] Answer: (4) Dimension of energy[ML 2 T -2 ] Dimension of G= [MLT -2 ][L 2 ] /[M 2 ] ா ீ = [𝑀 ଶ 𝐿 ି ଵ ] 36. A uniform rod of length 200cm and mass 500g is balanced on a wedge placed at 40cm mark. A mass of 2kg is suspended from the rod at 20cm and another unknown mass ‘m’ is suspended from the rod at 160cm mark as shown in the figure. Find the value of ‘m’ such that the rod is in equilibrium. (g=10m/s 2 ) (1) 1/3 kg (2) 1/6kg (3) 1/12kg (4) 1/2kg Answer: (3) For equilibrium clockwise torque= anti clock wise torque M (100-40) +m (160-40) =2(40-20) But M= 500g =0.5kg 0.5(60) +m (120) =2(20) 30+120m=40 120m=10 m= 1/12kg 37. A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle 𝜃 to the horizontal, the maximum height attained by it equals 4R. The angle of projection 𝜃 is then given by (1) 𝜃 = cos ି ଵ ( గ మ ோ ் మ ) ଵ ଶ ൗ (2) 𝜃 = sin ି ଵ ( గ మ ோ ் మ ) ଵ ଶ ൗ (3) 𝜃 = sin ି ଵ ( ଶ் మ గ మ ோ ) ଵ ଶ ൗ (4) 𝜃 = cos ି ଵ ( ் మ గ మ ோ ) ଵ ଶ ൗ Answer: (3) HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 13 September 12, 2021 𝑇 = 2𝜋𝑅 𝑈 𝐻 = 𝑢 ଶ sin ଶ 𝜃 2𝑔 ; 4𝑅 = 𝐻 4𝑅 = ( 2𝜋𝑅 𝑇 ) ଶ sin ଶ 𝜃 2𝑔 sin ଶ 𝜃 = 2𝑔 𝑇 ଶ 𝜋 ଶ 𝑅 𝜃 = sin ି ଵ ( 2𝑔 𝑇 ଶ 𝜋 ଶ 𝑅 ) ଵ ଶ ൗ Answer: (2) 38. A series LCR circuit containing 5H inductor, 80μF capacitor and 40Ω resistor is connected to 230V variable frequency AC source. The angular frequencies of the source at which power transfer to the circuit is half the power at the resonant angular frequency is likely to be (1) 50rad/s and 25rad/s (2) 46rad/s and 54rad/s (3) 42rad/s and 58rad/s (3) 25rad/s 75 rad/s Answer: (2) 𝑉 ௦ ଶ 𝑍 = 𝑅 𝑍 = 1 2 𝑥 𝑉 ௦ ଶ 𝑅 2 𝑅 ଶ = 𝑍 ଶ 2 𝑅 ଶ = 𝑅 ଶ + ( ଵ ఠ − 𝜔𝐿) ଶ 𝑅 ଶ = ( 1 𝜔𝐶 − 𝜔𝐿) ଶ 𝑅 = 1 𝜔𝐶 − 𝜔𝐿 HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 14 September 12, 2021 LCω 2 +RωC-1=0 .....(1) or LCω 2 -RωC-1=0..........(2) Solving we get ω=46rad/s and ω=54rad/s 39. For the given circuit, the input digital signals are applied at the terminals A, B and C. What would be the output at the terminal Y? (1) 0.4A (2) 2A (3) 4A (4) 0.2A Answer: (3) A B C A.B B.C B C ത ത ത ത ത A.B+ B C ത ത ത ത ത 1 1 1 1 1 0 1 1 1 0 1 0 1 1 1 0 1 0 0 1 1 1 0 0 0 0 1 1 0 1 1 0 1 0 0 0 1 0 0 0 1 1 0 0 1 0 0 1 1 0 0 0 0 0 1 1 compare truth table with the diagram gives answer 40. A step-down transformer connected to an AC mains supply of 220V is made to operate at 11V, 44W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit? (1) 0.4A (2) 2A (3)4A (4)0.2A Answer: (4) P=VI Ignoring power losses V P I P =V S I S 220(I P )=44 I P =44/220=0.2A 41. Two conducting circular loops of radii R 1 and R 2 are placed in the same plane with their centres coinciding. If R 1 >>R 2 , the mutual inductance M between them will be directly proportional to (1) ோ మ ோ భ (2) ோ భ మ ோ మ (3) ோ మ మ ோ భ (4) ோ భ ோ మ Answer: (3) HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 15 September 12, 2021 Field due to current in loop of radius R 1 if current ‘i’ flow is B= ఓ బ ଶோ భ Flux through small loop of radius R 2 is 𝜙 =B(A)= ఓ బ ଶோ భ ( 𝞹𝑅 ଶ ଶ )...... (1) Also 𝜙 = Mi.............................................. (2) Comparing 1 and 2 M ∝ ோ మ మ ோ భ 42. Three resisters having resistances r 1 , r 2 and r 3 are connected as shown in the given circuit. The ratio య భ of currents in terms of resistances used in the circuit is (1) మ మ ା య (2) భ భ ା మ (3) మ భ ା య (4) భ మ ା య Answer: (1) Three resistors having resistance r 1, r 2 and r 3 are connected as shown in the given circuit. The ratio 3 1 i i of currents in terms of resistance used in the circuit is From junction law at A For parallel resistors r 2 and r 3 i 1 = i 2 + i 3 ............. (1) i 2 r 2 = i 1 r 3 i 3 = 3 3 2 2 2 3 2 i r i r i r r ......... (2) Substitute (2) in (1) i 1 = 3 3 3 2 3 1 3 2 2 i r r r i i i r r 3 2 1 2 3 i r i r r 43. In the product 𝐹 ⃗ = 𝑞 ൫ 𝑣 ⃗ 𝑥𝐵 ሬ ⃗ ൯ = 𝑞𝑣 ⃗ 𝑥(𝐵𝚤 ̂ + 𝐵𝚥 ̂ + 𝐵 𝑘 ) . For q=1 and 𝑣 ⃗ = 2𝚤 ̂ + 4𝚥 ̂ + 6𝑘 and 𝐹 ⃗ = 4𝚤 ̂ − 20𝚥 ̂ + 12𝑘 . What will be the complete expression for 𝐵 ሬ ⃗ ? (1) −6𝚤 ̂ − 6𝚥 ̂ − 6𝑘 (2) 8𝚤 ̂ + 8𝚥 ̂ − 6𝑘 (3) 6𝚤 ̂ + 6𝚥 ̂ − 8𝑘 (4) −8𝚤 ̂ − 8𝚥 ̂ − 6𝑘 Answer: (1) 𝐹 ⃗ = 𝑞(𝑣 ⃗ × 𝐵 ሬ ⃗ ) ⟹ 4𝚤 ̂ − 20𝚥 ̂ + 12𝑘 = 1{ ൫ 2𝚤 ̂ + 4𝚥 ̂ + 6𝑘 ൯ × ൫ 𝐵𝚤 ̂ + 𝐵𝚥 ̂ + 𝐵 𝑘 ൯ } HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 16 September 12, 2021 ⟹ 4𝚤 ̂ − 20𝚥 ̂ + 12𝑘 = 𝚤 ̂ 𝚥 ̂ 𝑘 2 4 6 𝐵 𝐵 𝐵 ⟹ 4𝚤 ̂ − 20𝚥 ̂ + 12𝑘 = ൣ ( 4𝐵 − 6𝐵 ) 𝚤 ̂ − ( 2𝐵 − 6𝐵 ) 𝚥 ̂ + ( 2𝐵 − 4𝐵 ) 𝑘 ൧ ⟹ 4𝚤 ̂ − 20𝚥 ̂ + 12𝑘 = ( 4𝐵 − 6𝐵 ) 𝚤 ̂ − ( 2𝐵 − 6𝐵 ) 𝚥 ̂ − (2𝐵)𝑘 Comparing equation both sides, For 𝑘 ⇒ −2𝐵 = 12 ∴ 𝐵 = −6 For 𝚤 ̂ , ⟹ ( 4𝐵 − 6𝐵 ) = 4 ⟹ 4𝐵 + 36 = 4 ⟹ 4𝐵 = −32 ∴ 𝐵 = −8 44. 27 drops of same size are charged at 220V each. They combined to form a bigger drop. Calculate the potential of the bigger drop. (1) 1320V (2) 1520V (3) 1980V (4) 660V Answer: (3) r – radius of small drop R – radius of bigger drop Let Q be the charge on each small drop ⟹ 220 = ொ ସగఌ బ ................(1) If 27 drops combine the volume of the big drop = volume of 27 small drop Charge on big drop = charge on 27 small drops ⟹ 27 ൬ 4 3 𝜋𝑟 ଷ ൰ = 4 3 𝜋𝑅 ଷ ⟹ 𝑅 = 3𝑟 ...................(2) ⟹ 𝑄 = 27. 𝑄 ⟹ 𝑉 = 𝑄 4𝜋𝜀 (𝑅) = 27. 𝑄 4𝜋𝜀 (3𝑟) = 9 ൬ 𝑄 4𝜋𝜀 𝑟 ൰ = 9 ( 220 ) = 1980 𝑉 45. A ball of mass 0.15kg is dropped from a height 10m, strikes the ground and rebounds to the same height. The magnitude if impulse imparted to the ball is (g=10m/s 2 ) nearly (1) 4.2kgm/s (2) 2.1kgm/s (3) 1.4kgm/s (4) 0kgm/s Answer: (1) A ball of mass 0.15 kg is dropped from a height 10m, strike the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g=10 m/s 2 ) nearly Given m=0.15 kg HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 17 September 12, 2021 Collision must be elastic if ball goes to the same height after impact Velocity before impact=velocity after impact V= 2 2 10 10 200 10 2 / gh m s Momentum before collision momentum after collision Impulse=change in the momentum = 1.5 2 ( 1.5 2) 3 2 kg. / 4.2 m/s m s kg 46. A car starts from rest and accelerates at 5m/s 2 . At t=4s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t=6s? (take g=10m/s 2 ) (1) 2m/s, 0 (2) 20 √ 2 m/s, 0 (3) 20 √ 2 m/s, 10m/s 2 (4) 20m/s, 5m/s 2 Answer: (3) Speed of the car at t=4s 𝑣 ௫ = 𝑢 + 𝑎𝑡 = 0 + ( 5 × 4 ) = 20m/s Since ball is dropped from car, it will have no acceleration in x axis. Y axis, (acceleration is due to gravity) 𝑣 ௬ = 𝑢 + 𝑎𝑡 = 0 + (10𝑥2) = 20𝑚/𝑠 Therefore, speed= 20 √ 2 m/s 47. A point object is placed at a distance of 60cm from a convex lens of focal length 30cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40cm from it, the final image would be formed at a distance of (1) 30cm from the lens, it would be a real image HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 18 September 12, 2021 (2) 30cm from the plane mirror, it would be a virtual image (3) 20cm from the plane mirror, it would be a virtual image (4) 20cm from the lens, it would be a real image Answer: (3) Refraction at lens, ଵ − ଵ = ଵ ⟹ ଵ − ଵ ି = ଵ ଷ ⟹ ଵ = ଵ ∴ 𝑉 = 60 𝑐𝑚 (Image form at 60cm from lens which is 20cm behind mirror) 𝑟𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 𝑝𝑙𝑎𝑛𝑒 𝑚𝑖𝑟𝑟𝑜𝑟 𝑖𝑚𝑎𝑔𝑒 𝑖𝑠 𝑓𝑜𝑟𝑚𝑒𝑑 20𝑐𝑚 𝑖𝑛 𝑓𝑟𝑜𝑛𝑡 of mirror which is real. refraction through lens ଵ − ଵ ି ଶ = ଵ ଷ ⟹ ଵ = ଵ ଷ − ଵ ଶ ∴ 𝑉 = −60𝑐𝑚. {final image is from at 60cm to right of lens 20cm behind which is virtual} 48. From a circular ring of mass M and radius R an arc corresponding to a 90 0 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the center of the ring and perpendicular to the plane of the ring is K times MR 2 . Then the value of K is (1) ଼ (2) ଵ ସ (3) ଵ ଼ (4) ଷ ସ Answer: (4) After a quarter of ring is removed, ∴ Mass of remaining part = యಾ ర ∴ Moment of inertia of remaining part = యಾ ర ( 𝑅 ଶ ) = యಾೃ మ ర ⟹ ଷெோ మ ସ = 𝐾𝑀𝑅 ଶ ∴ 𝐾 = ଷ ସ 49. A uniform conducting wire of length 12a and resistance R is wound up as a current carrying coil in the shape of, (A) an equilateral triangle of side ‘a’ (B) a square of side ‘a’ The magnetic dipole moments of the coil in each case respectively are (1) 3Ia 2 and Ia 2 ( 2) 3Ia 2 and 4Ia 2 (3) 4Ia 2 and 3Ia 2 (4) √ 3 𝐼𝑎 ଶ and 3Ia 2 Answer: (4) Perimeter = 3a a a Total number of loops = ୲୭୲ୟ୪ ୪ୣ୬୲୦ ୭ ୵୧୰ୣ ୪ୣ୬୲୦ ୮ୣ୰ ୪୭୭୮ = ଵଶ ଷ = 4 magnetic dipole moment = 𝜂𝐼𝐴 = 4 ( 𝐼 ) ቀ √ ଷ ସ 𝑎 ଶ ቁ = √ 3 (𝐼)𝑎 ଶ a a total number of loops = ଵଶ ସ = 3 a a magnetic dipole moment = 𝜂𝐼𝐴 = 3 ( 𝐼 )( 𝑎 ଶ ) = 3(𝐼)𝑎 ଶ HKS PU COLLEGE, HASSAN CREATIVE PU COLLEGE, KARKALA 19 September 12, 2021 50. A particle of mass m is projected with a velocity v=kV e (k<1) from the surface of the earth (V e = escape velocity). The maximum height above the surface reached by the particle is (1) 𝑅( ଵା ) ଶ (2) ோ మ ଵା (3) మ ோ ଵି మ (4) 𝑅( ଵି ) ଶ Answer: (3) Using conservation of mechanical energy 𝑈 + 𝐾 𝑖 = 𝑈 𝑓 + 𝐾 𝑓 ି ீ ெ ோ + ଵ ଶ 𝑚𝑘 2 v e 2 = ି ீ ெ ோା ∴ 𝑣 = ଶୋ ோ ⟹ ଵ ோ - మ ோ = ଵ ோା ∴ ℎ = ோ మ ଵି మ