Practice Questions

Q24.An object is located in a fixed position in front of a screen. Sharp image is obtained on the screen for two positions of a thin lens separated by 10 cm . The size of the images in two situations are in the ratio 3 : 3. What is the distance between the screen and the object? (1) 124.5 cm (2) 144.5 cm (3) 65.0 cm (4) 99.0 cm

Q24.The refractive index of the material of a concave lens is μ. It is immersed in a medium of refractive index μ1 . A parallel beam of light is incident on the lens. The path of the emergent rays when μ1 > μ is: JEE Main 2014 (12 Apr Online) JEE Main Previous Year Paper (1) (2) (3) (4)

Q24.A thin convex lens made from crown glass (μ = 32 ) has focal length f . When it is measured in two different liquids having refractive indices 4 and 5 , it has the focal lengths f1 and f2 respectively. The correct relation 3 3 between the focal lengths is : (1) f1 = f2 < f (2) f1 > f and f2 becomes negative (3) f2 > f and f1 becomes negative (4) f1 and f2 both become negative

Q25.In a compound microscope the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm. When object is kept at 1.25 cm in front of objective, final image is formed at infinity. Magnifying power of the compound microscope should be: (1) 200 (2) 100 (3) 400 (4) 150

Q25.Interference pattern is observed at ' P ' due to superimposition of two rays coming out from a source ' S ' as shown in the figure. The value of ' l ' for which maxima is obtained at ' P ' is: ( R is perfect reflecting surface) (1) 1 = 2nλ (2) 1 = (2n−1)λ √3−1 2(√3−1 ) (2n−1)λ 1 = (3) 1 = (2n−1λ)√3 √3−1 4(2−√3 ) (4)

Q25.In Young’s double-slit experiment, the distance between the two identical slits is 6. 1 times larger than the slit width. Then the number of intensity maxima observed within the central maximum of the single-slit diffraction pattern is : (1) 3 (2) 6 (3) 24 (4) 12

Q26.Two monochromatic light beams of intensity 16 and 9 units are interfering. The ratio of intensities of bright and dark parts of the resultant pattern is: (1) 16 (2) 4 9 3 (3) 7 (4) 49 1 1

Q26.Two beams, A and B of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30o makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB respectively, then IAIB equals : (1) 3 (2) 3 2 (3) 1 (4) 1 3

Q26.In an experiment of single slit diffraction pattern, first minimum for red light coincides with first maximum of some other wavelength. If wavelength of red light is 6600Å , then wavelength of first maximum will be: (1) 3300Å (2) 4400Å (3) 5500Å (4) 6600Å JEE Main 2014 (12 Apr Online) JEE Main Previous Year Paper

Q26.Match List ‐I (Experiment performed) with List ‐ II (Phenomena discovered/associated)and select the correct option from the options given below the lists List-I List-II (a) Davisson and Germer experiment (i) Wave nature of electrons (b) Millikan's oil drop experiment (ii) Charge of an electron (c) Rutherford experiment (iii) Quantisation of energy levels (d) Franck-Hertz experiment (iv) Existence of the nucleus (1) (a)‐(iii), (b)‐(iv), (c)‐(i), (d)‐(ii) (2) (a)‐(i), (b)‐(ii), (c)‐(iv), (d)‐(iii) (3) (a)‐(iv), (b)‐(iii), (c)‐(ii), (d)‐(i) (4) (a)‐(i), (b)‐(ii), (c)‐(iii), (d)‐(iv)

Q27.A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of the same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If the half-life of C14 is 5730 years, then the age of the wooden piece placed in the museum is approximately [This question was awarded a bonus and proper correction was made to avoid that] (1) 10439 years (2) 39049 years (3) 19042 years (4) 13094 years

Q27.A photon of wavelength λ is scattered from an electron, which was at rest. The wavelength shift Δλ is three times of λ and the angle of scattering θ is 60∘ . The angle at which the electron recoiled is ϕ. The value of tan ϕ JEE Main 2014 (11 Apr Online) JEE Main Previous Year Paper is : (electron speed is much smaller than the speed of light) (1) 0.16 (2) 0.22 (3) 0.25 (4) 0.28

Q27.A beam of light has two wavelengths of 4972Å and 6216Å with a total intensity of 3.6 × 10−3 Wm−2 equally distributed among the two wavelengths. The beam falls normally on an area of 1 cm2 of a clean metallic surface of work function 2.3eV. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in 2 s is approximately: (1) 6 × 1011 (2) 9 × 1011 (3) 11 × 1011 (4) 15 × 1011

Q28.Match the List-I (Phenomenon associated with electromagnetic radiation) with List-II (Part of electromagnetic spectrum) and select the correct code from the choices given below this lists: (1) (I)-(A), (II)-(B), (III)-(B),,(IV)-(C) (2) (I)-(A), (II)-(B), (III)-(C), (IV)-(C) (3) (I)-(D), (II)-(C), (III)-(A), (IV)-(B) (4) (I)-(B), (II)-(A), (III)-(D), (IV)-(A)

Q28.For LED's to emit light in visible region of electromagnetic light, it should have energy band gap in the range of: (1) 0.1eV to 0.4eV (2) 0.5eV to 0.8eV (3) 0.9eV to 1.6eV (4) 1.7eV to 3.0eV

Q28.Hydrogen (1H1), Deuterium (1H2), singly ionised Helium (2He4)+ and doubly ionised lithium (3Li6)++ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wave lengths of emitted radiation are λ1, λ2, λ3 and λ4 respectively then approximately which one of the following is correct ? (1) 4λ1 = 2λ2 = 2λ3 = λ4 (2) λ1 = 2λ2 = 2λ3 = λ4 (3) λ1 = λ2 = 4λ3 = 9λ4 (4) λ1 = 2λ2 = 3λ3 = 4λ4 JEE Main 2014 (06 Apr) JEE Main Previous Year Paper

Q28. Given, A and B are input terminals Logic 1 is > 5 V Logic 0 is < 1 V JEE Main 2014 (19 Apr Online) JEE Main Previous Year Paper Which logic gate operation, the following circuit does? Note: This question was awarded a bonus. C option changed. (1) OR gate. (2) NOR gate. (3) Output will always be one. (4) XOR gate.

Q29.A radioactive nuclei with decay constant 0.5/s is being produced at a constant rate of 100 nuclei/s. If at t = 0 there were no nuclei, the time when there are 50 nuclei is: (1) 1 s (2) 2 ln ( 43 )s (3) ln 2 s (4) ln ( 34 )s

Q29.Identify the gate and match A, B, Y in the bracket to check. JEE Main 2014 (09 Apr Online) JEE Main Previous Year Paper (1) OR (A = 1, B = 1, Y = 0) (2) XOR (A = 0, B = 0, Y = 0) (3) NOT (A = 1, B = 1, Y = 1) (4) AND (A = 1, B = 1, Y = 1)

Q29.A piece of bone of an animal from a ruin is found to have 14C activity of 12 disintegrations per minute per gm of its carbon content. The 14C activity of a living animal is 16 disintegrations per minute per gm. How long ago nearly did the animal die? (Given half life of 14C is t1/2 = 5760 years) (1) 1672 years (2) 2391 years (3) 3291 years (4) 4453 years

Q30. In an experiment to determine the gravitational acceleration g of a place with the help of a simple pendulum, the measured time period squared is plotted against the string length of the pendulum in the figure. What is the value of g at the place? (1) 10. 0 m s−2 (2) 9. 87 m s−2 (3) 9. 91 m s−2 (4) 9. 81 m s−2

Q30.A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it ? (1) A meter scale (2) A vernier calliper where the 10 divisions in vernier scale matches with 9 division in main scale and main scale has 10 divisions in 1 cm (3) A screw gauge having 100 divisions in the (4) A screw gauge having 50 divisions in the circular circular scale and pitch as 1 mm scale and pitch as 1 mm

Q30.A Zener diode is connected to a battery and a load as show below: The currents, I, IZ and IL are respectively. (1) 15 mA, 5 mA, 10 mA (2) 15 mA, 7.5 mA, 7.5 mA (3) 12.5 mA, 5 mA, 7.5 mA (4) 12.5 mA, 7.5 mA, 5 mA

Q30.For sky wave propagation, the radio waves must have a frequency range in between: (1) 1MHz to 2MHz (2) 5MHz to 25MHz (3) 35MHz to 40MHz (4) 45MHz to 50MHz

Q1. Let [∈0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then : (1) [∈0] = [M−1 L2 T−1 A−2] (2) [∈0] = [M−1 L2 T−1 A] (3) [∈0] = [M−1 L−3 T2 A] (4) [∈0] = [M−1 L−3 T4 A2] + m s−1 , where ˆi is along the ground and ˆj is along the vertical