Practice Questions

Q12.A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is (Take, g = 10 m s−2 ) (1) 2√2 s (2) √2 s (3) 2π√2 s (4) 2 s

Q12.Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T , respectively. At time t = 0 one particle has displacement A while the other one has displacement −A2 and they are moving towards each other. If they cross each other at time t, then t is: (1) 5T (2) T 6 3 (3) T (4) T 4 6

Q13.The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge density ρ = Ar , where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, JEE Main 2016 (03 Apr) JEE Main Previous Year Paper is: (1) 2Q (2) 2Q π(a2−b2) πa2 (3) Q (4) Q 2πa2 2π(b2−a2)

Q13.Two engines pass each other moving in opposite directions with uniform speed of 30 m/s. one of them is blowing a whistle of frequency 540 Hz. Calculate the frequency heard by driver of second engine before they pass each other. speed of sound is 330 m/sec: (1) 450 Hz (2) 540 Hz (3) 270 Hz (4) 648 Hz

Q14.Within a spherical charge distribution of charge density ρ(r), N equipotential surfaces of potential V0, V0 + ΔV , V0 + 2 Δ V , … V0 + N Δ V (ΔV > 0), are drawn and have increasing radii r0, r1, r2, … rN, respectively. If the difference in the radii of the surfaces is constant for all values of V0 and Δ V then : (1) ρ(r) = constant (2) ρ (r) ∝ 1 r2 (3) ρ (r) ∝ 1r (4) ρ (r) ∝ r

Q14.A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 μF and 9 μF capacitors), at a point distant 30 m from it, would equal: (1) 420 N/C (2) 480 N/C (3) 240 N/C (4) 360 N/C

Q14.The potential (in volts) of a charge distribution is given by V(z) = 30 −5z2 for |z| ≤1 m V(z) = 35 −10 |z| for |z| ≥1 m . V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume ρ0 (in units of ϵ0 ) which is spread over a certain region, then choose the correct statement. (1) ρ0 = 20 ϵ0 in the entire region (2) ρ0 = 10 ϵ0 for |z| ≤1 m and ρ0 = 0 else where (3) ρ0 = 20 ϵ0 for |z| ≤1 m and ρ0 = 0 else where (4) ρ0 = 40 ϵ0 in the entire region JEE Main 2016 (09 Apr Online) JEE Main Previous Year Paper

Q15.Three capacitors each of 4 μF are to be connected in such a way that the effective capacitance is 6 μF . This can be done by connecting them (1) all in series (2) all in parallel (3) two in parallel and one in series (4) two in series and one in parallel

Q15.A galvanometer having a coil resistance of 100 Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is: (1) 0.1 Ω (2) 3 Ω (3) 0.01 Ω (4) 2 Ω JEE Main 2016 (03 Apr) JEE Main Previous Year Paper

Q16. In the circuit shown, the resistance r is a variable resistance. If for r = fR , the heat generation in r is maximum then the value of f is (1) 1 (2) 1 2 (3) 1 (4) 3 4 4

Q16.Hysteresis loops for two magnetic materials A and B are given below: These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use: (1) A for transformers and B (2) B for electromagnets and transformers. for electric generators. (3) A for electric generators and transformers. (4) A for electromagnets and B for electric generators.

Q16.A galvanometer has a 50 division scale. Battery has no internal resistance. It is found that there is deflection of 40 divisions when R.B. = 2400 Ω . Deflection becomes 20 divisions when resistance taken from resistance box is 4900 Ω. Then we can conclude : Note: This question is awarded as the bonus. Now the question is corrected. (1) Current sensitivity of galvanometer is (2) Resistance of galvanometer is 200 Ω. 20 μA/division (3) Resistance required on R.B. for a deflection of 10 (4) Full scale deflection current is 2 mA. divisions is 9800 Ω.

Q17.Two identical wires A and B, each of length l, carry the same current I . Wire A is bent into a circle of radius R and wire B is bent to form a square of side a . If BA and BB are the values of magnetic field at the centres of the circle and square respectively, then the ratio BA is BB (1) π2 (2) π2 16 8√2 (3) π2 (4) π2 8 16√2

Q17.To know the resistance G of a galvanometer by half deflection method, a battery of emf VE and resistance R is used to deflect the galvanometer by angle θ . If a shunt of resistance S is needed to get half deflection the G, R and S are related by the equation: (1) S(R + G) = RG (2) 2S(R + G) = RG (3) 2G = S (4) 2S = G

Q17.The resistance of an electrical toaster has a temperature dependence given by R(T) = R0[1 + α(T −T0)] in its range of operation. At T0 = 300 K, R = 100 Ω and at T = 500 K, R = 120 Ω . The toaster is connected JEE Main 2016 (10 Apr Online) JEE Main Previous Year Paper to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is : Note: This question was awarded as the bonus since all options were incorrect in the exam. (1) 60000 ln( 65 ) J (2) 200 ln 32 J (3) 300 J (4) 400 ln( 1.31.5 ) J

Q18.A fighter plane of length 20 m, wing span (distance from tip of one wing to the tip of the other wing) of 15 m and height 5 m is flying towards east over Delhi. Its speed is 240 ms−1. The earth's magnetic field over Delhi is 5 × 10−5T with the declination angle ~0o and dip of θ such that sin θ = 32 . If the voltage developed is VB between the lower and upper side of the plane and VW between the tips of the wings then VB and VW are close to : (1) VB = 40 mV ; VW = 135 mV with left side of (2) VB = 45 mV ; VW = 120 mV with right side of pilot at higher voltage pilot at higher voltage (3) VB = 40 mV ; VW = 135 mV with right side of (4) VB = 45 mV ; VW = 120 mV with left side of pilot at high voltage pilot at higher voltage

Q18.An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to: (1) 0.044 H (2) 0.065 H (3) 80 H (4) 0.08 H

Q18.A 50 Ω resistance is connected to a battery of 5 V . A galvanometer of resistance 100 Ω is to be used as an ammeter to measure current through the resistance, for this a resistance rS is connected to the galvanometer. Which of the following connections should be employed if the measured current is with in 1% of the current without the ammeter in the circuit? (1) rS = 0.5Ω in series with galvanometer (2) rS = 1 Ω in series with galvanometer (3) rS = 1 Ω in parallel with galvanometer (4) rS = 0.5 Ω in parallel with the galvanometer

Q19.Arrange the following electromagnetic radiations per quantum in the order of increasing energy: A : Blue light B : Yellow light C : X-ray D : Radiowave (1) C, A, B, D (2) B, A, D, C (3) D, B, A, C (4) A, B, D, C

Q19.A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of 75o . One of the fields has a magnitude of 15 mT. The dipole attains stable equilibrium at an angle of 30o with this field. The magnitude of the other field (in mT) is close to (1) 1 (2) 11 (3) 36 (4) 1060

Q19.A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with τ , where B0 and τ are constants at time t = 0 . If the resistance of the loop is R , then the time as B = B0e−t heat generated in the loop after a long time (t →∞) is 0 (1) π2r4B4 0 (2) π2r4B2 2τR 2τR (3) π2r4B20R (4) π2r4B20 τ τR

Q20.An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer, the tree appears as (1) 20 times taller. (2) 20 times nearer. (3) 10 times taller. (4) 10 times nearer. JEE Main 2016 (03 Apr) JEE Main Previous Year Paper

Q20.Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed v in a uniform magnetic field B going into the plane of the paper (see figure). If charge densities σ1 and σ2 are induced on the left and right surfaces respectively of the sheet, then (ignore fringe effects) (1) σ1 = −ϵ02 vB , σ2 = ϵ0 2vB (2) σ1 = ϵ0 vB, σ2 = −ϵ0 vB (3) σ1 = ϵ0 2vB , σ2 = −ϵ02 vB (4) σ1 = σ2 = ϵ0 vB

Q20.A series LR circuit is connected to a voltage source with V (t) = V0 sin(ωt) . After a very large time, current I(t) behaves as (t0 ≫LR ): JEE Main 2016 (09 Apr Online) JEE Main Previous Year Paper (1) (2) (3) (4)

Q21.Consider an electromagnetic wave propagating in vacuum. Choose the correct statement: JEE Main 2016 (10 Apr Online) JEE Main Previous Year Paper (1) For an electromagnetic wave propagating in +y (2) For an electromagnetic wave propagating in + y → → direction the electric field is E = 1 Eyz(x, t)ˆz direction the electric field is E = 1 Eyz(x, t)ˆy √2 √2 → 1 → 1 and the magnetic field is B = Bz(x, t)ˆy and the magnetic field is B = Byz(x, t)ˆz √2 √2 (3) For an electromagnetic wave propagating in + x (4) For an electromagnetic wave propagating in + x direction the electric field is direction the electric field is → → E = 1 Eyz(y, z, t) (ˆy + ˆz) and the magnetic E = 1 Eyz(x, t) (ˆy −ˆz) and the magnetic field √2 √2 → → field is B = 1 Eyz(y, z, t) (ˆy + ˆz) is B = 1 Byz(x, t)(ˆy + ˆz) √2 √2