Practice Questions

Q14.A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? [Given reference intensity of sound as 10−12 W/m2 ] (1) 40 cm (2) 20 cm (3) 10 cm (4) 30 cm

Q14.Shown in the figure is a shell made of a conductor. It has inner radius 𝑎 and outer radius 𝑏, and carries charge 𝑄 . At its centre a dipole →𝑝 is placed as shown then: (1) Surface charge density on the inner surface of the (2) Electric field outside the shell is the same as that shell is zero everywhere. of a point charge at the centre of the shell. (3) Surface charge density on the inner surface is (4) Surface charge density on the outer surface 𝑄 ( ) depends on →𝑝. uniform and equal to 2 . 4π𝑎2 JEE Main 2019 (12 Apr Shift 1) JEE Main Previous Year Paper is kept at the origin. The potential and electric field due to this dipole on the 𝑦- axis

Q15.Four point charges −q, + q, + q and −q are placed on y-axis at y = −2d, y = −d, and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D ≫d, will behave JEE Main 2019 (09 Apr Shift 2) JEE Main Previous Year Paper as: (1) E ∝ 1 (2) E ∝ D1 D4 (3) E ∝ 1 (4) E ∝ 1 D3 D2

Q16.A system of three charges are placed as shown in the figure: If D >> d, the potential energy of the system is best given by: (1) + q D2 4πϵ0 1 [−q2d −q D2Q d ] (2) 4πϵ01 [+ q2d Q d ] 2q (3) + D2 4πϵ0 1 [−q2d −q2D2Q d ] (4) 4πϵ01 [−q2d Q d ]

Q16.A parallel plate capacitor is made of two square plates of side 𝑎, separated by a distance 𝑑 𝑑≪𝑎. The lower triangular portion filled with a dielectric of dielectric constant 𝐾, as shown in the figure. Capacitance of this JEE Main 2019 (09 Jan Shift 1) JEE Main Previous Year Paper capacitor is: (1) 𝐾𝜀0𝑎2 (2) 𝐾𝜀0𝑎2 2𝑑𝐾+ 1 𝑑𝐾- 1ln⁡𝐾 (3) 𝐾𝜀0𝑎2 ln⁡𝐾 (4) 1 𝐾𝜀0𝑎2 𝑑 2 𝑑

Q18.In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may conclude that 0 (2) R(T) = R0 (1) R(T) = R0eT2/T2 T2 (3) R(T) = R0e−T2/T2 0 (4) R(T) = R0e−T20/T2

Q19.An electron, moving along the x− axis with an initial energy of 100 eV, enters a region of magnetic field → B = (1.5 × 10−3 T)ˆk at S (see figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is: (electron’s charge 1.6 × 10−19 C , mass of electron = 9.1 × 10−31 kg ) (1) 1.22 cm (2) 12.87 cm (3) 11.65 cm (4) 2.25 cm

Q20.The region between y = 0 and y = d contains a magnetic field B = B^z. A particle of mass m and charge q mv , the acceleration of the charged particle at the point of its enters the region with a velocity →v = v^i. if d = 2qB emergence at the other side is : (1) qv m B ( 12^i −√32 ^j) (2) qvmB ( √32 ^i + 12 ^j) (3) qv B −^j+^i (4) None of the above m ( √2 )

Q20.A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5 N m-1 (see figure). The assembly is kept in a uniform magnetic field of 0.1 T . If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N . If the mass of the strip is 50 grams, its resistance 10Ω and air drag negligible, JEE Main 2019 (08 Apr Shift 1) JEE Main Previous Year Paper N will be close to: (1) 1000 (2) 5000 (3) 10000 (4) 50000

Q21.An insulating thin rod of length l has a linear charge density ρ(x) = ρ0 xl on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is: (1) π n ρ0l3 (2) n ρ0l3 4 (3) π n ρ0 l3 (4) π3 n ρ0l3

Q23.An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. The radius of the loop is 𝑎 and distance of its centre from the wire is 𝑑 𝑑≫𝑎. If the loop applies a force 𝐹 on the wire then: 2 (1) 𝐹∝𝑎 (2) 𝐹= 0 𝑑 𝑎 (3) 𝐹∝𝑎2 (4) 𝐹∝ 𝑑 𝑑3

Q24.What is the position and nature of image formed by lens combination shown in figure? ( f1, f2 are focal lengths) (1) 40 cm from point B at right; real (2) 203 cm from point B at right, real (3) 70 cm from point B at right; real (4) 70 cm from point B at left; virtual JEE Main 2019 (12 Jan Shift 1) JEE Main Previous Year Paper

Q24.The magnetic field of a plane electromagnetic wave is given by B = B0ˆi[cos(kz −ωt)] + B1ˆjcos(kz + ωt) , where B0 = 3 × 10–5 T and B1 = 2 × 10–6 T . The RMS value of the force experienced by a stationary charge Q = 10–4 C at z = 0 is closest to (1) 0.1 N (2) 0.9 N (3) 3 × 10–2 N (4) 0.6 N

Q24.In figure, the optical fiber is 𝑙= 2 m long and has a diameter of d = 20 μm . If a ray of light is incident on one end of the fiber at angle θ1 = 40° , the number of reflections it makes before emerging from the other end is close to: (refractive index of fiber is 1.31 , sin 40° = 0.64 and sin-1⁡0.49 = 30° .) (1) 55000 (2) 57000 (3) 45000 (4) 66000

Q25.A thin convex lens L (refractive index = 1.5 ) is placed on a plane mirror M. When a pin is placed at A , such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When liquid of refractive index μl is put between the lens and the mirror, the pin has to be moved to A', such that OA' = 27 cm, to get its inverted real image at A' itself. The value of μl will be (1) 4 (2) √3 3 (3) 3 (4) √2 2

Q25.An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20 cm . A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be: JEE Main 2019 (08 Apr Shift 1) JEE Main Previous Year Paper (1) 40 cm from the (2) 20 cm from the (3) 40 cm from the (4) 20 cm from the convergent lens, twice convergent mirror, convergent lens, same convergent mirror, the size of the object twice the size of the size of the object same size of the object object

Q25.In a double-slit experiment, green light (5303 A) falls on a double slit having a separation of 19.44μm and a width of 4.05μm. The number of bright fringes between the first and the second diffraction minima is (1) 10 (2) 5 (3) 4 (4) 9

Q26.Consider a Young's double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength λ such that the first minima occurs directly in front of the slit (S1) ? (1) λ (2) λ (5−√2) 2(√5−2) (3) λ (4) λ 2(5−√2) √5−2

Q26.In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 10−12 m, the minimum electron energy required is close to: (1) 100 keV (2) 25 keV (3) 1 keV (4) 500 keV

Q26.A particle of mass m moves in a circular orbit in a central potential field U(r) = 12 kr2. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as: (1) rn ∝n2, En ∝ n21 (2) rn ∝√n, En ∝n (3) rn ∝n, En ∝n (4) rn ∝√n, En ∝1n

Q28.In Li+ + ,electron in first Bohr orbit is excited to a level by a radiation of wavelength λ.When the ion gets de excited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of λ ? (Given: ℎ= 6.63 × 10-34 J s ; 𝑐= 3 × 108 m s-1 ) (1) 10.8 nm (2) 9.4 nm (3) 11.4 nm (4) 12.3 nm

Q28.Consider a tank made of glass (refractive index 1.5 ) with a thick bottom. It is filled with a liquid of refractive index 𝜇. A student finds that, irrespective of what the incident angle 𝑖 (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of 𝜇 is: 3 4 (1) (2) √5 3 5 5 (3) (4) √3 √ 3

Q28.An NPN transistor operates as a common emitter amplifier, with a power gain of 60 dB . The input circuit resistance is 100 Ω and the output load resistance is 10 kΩ . The common emitter current gain β is: (1) 6 × 102 (2) 102 (3) 104 (4) 60

Q30.The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is, RL = 4 kΩ . The series resistance of the circuit is Ri = 1 kΩ . If the battery voltage VB varies from 8 V to 16 V , what are the minimum and maximum values of the current through Zener diode? (1) 0.5 mA; 6 mA (2) 1 mA; 8.5 mA (3) 0.5 mA; 8.5 mA (4) 1.5 mA; 8.5 mA

Q30.In a meter bridge, the wire of length 1 m has a non-uniform cross-section such that, the variation dRdl of its resistance R with length l is dRdl ∝ √l1 . Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP? JEE Main 2019 (12 Jan Shift 1) JEE Main Previous Year Paper (1) 0.2 m (2) 0.35 m (3) 0.3 m (4) 0.25 m o