Practice Questions

Q12.A pendulum with the time period of 1 s is losing energy due to damping. At a certain time, its energy is 45 J . If after completing 15 oscillations its energy has become 15 J, then its damping constant (in s−1 ) will be JEE Main 2015 (11 Apr Online) JEE Main Previous Year Paper (1) 1 (2) 1 ln 3 2 15 (3) 1 ln 3 (4) 2 30

Q12.For a simple pendulum, a graph is plotted between its kinetic energy (K.E.) and potential energy (P.E.) against its displacement d. which one of the following represents these correctly? (graphs are schematic and not drawn to scale) (1) (2) (3) (4)

Q12. x and y displacements of a particle are given as x(t) = a sin ωt and y(t) = a sin 2ωt. Its trajectory will look like: (1) (2) (3) (4)

Q13.A cylindrical block of wood (density = 650 kg m−3 ), of base area 30 cm2 and height 54 cm, floats in a liquid of density 900 kg m−3 . The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly) : (1) 52 cm (2) 26 cm (3) 39 cm (4) 65 cm

Q13.A train is moving on a straight track with speed 20 m s−1 . It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 m s−1 ) close to: (1) 24% (2) 6% (3) 12% (4) 18% JEE Main 2015 (04 Apr) JEE Main Previous Year Paper

Q14. Two long currents carrying thin wires, both with current I , are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ' θ ' with the vertical. If wires have a mass λ per unit length then the value of I is: ( g = gravitational acceleration) (1) (2) tanθ √πλgLμ0 sinθ√πλgLμ0 cosθ (3) (4) tanθ 2sinθ√πλgLμ0cosθ 2√πgLμ0

Q14.A source of sound emits sound waves at frequency f0 . It is moving towards an observer with fixed speed vs (vs < v) , where v is the speed of sound in air). If the observer were to move towards the source with speed v0 , one of the following two graphs (A and B) will give the correct variation of the frequency f heard by the observer as v0 is changed. The variation of f with v0 is given correctly by: (1) Graph A with slope = f0 (2) Graph A with slope = f0 (v−vs) (v+vs) (3) Graph B with slope = f0 (4) Graph B with slope = f0 (v−vs) (v+vs)

Q14.A bat moving at 10 m s−1 towards a wall sends a sound signal of 8000 Hz towards it. On reflection, it hears a sound of frequency f . The value of f in Hz is close to (speed of sound = 320 m s−1) JEE Main 2015 (10 Apr Online) JEE Main Previous Year Paper (1) 8258 (2) 8424 (3) 8000 (4) 8516

Q15.A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞ ) on its surface. For this sphere the equipotential surfaces with potential 3V0 2 , 5V04 , 3V04 and V04 have radius R1 , R2, R3 and R4 respectively. Then Note : This question had two option correct at the time of examination. Proper corrections are made in the question to avoid it. (1) 2R > R4 (2) R1 = 0 and R2 > (R4 −R3) (3) R1 ≠0 and (R2 −R1) > (R4 −R3) (4) R1 = 0 and R2 < (R4 −R3)

Q15.Shown in the figure are two point charges +Q and −Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If σ1 is the surface charge on the inner surface and Q1 net charge on it and σ2 the surface charge on the outer surface and Q2 net charge on it then: (1) σ1 = 0, Q1 = 0 , σ2 = 0, Q2 = 0 (2) σ1 ≠0, Q1 = 0 , σ2 ≠0, Q2 = 0 (3) σ1 ≠0, Q1 ≠0 , σ2 ≠0, Q2 ≠0 (4) σ1 ≠0, Q1 = 0 , σ2 = 0, Q2 = 0

Q16.An electric field E = (25 ˆi ˆj) be zero then the potential at x = 2 m, y = 2 m is: (1) −110 J C−1 (2) −140 J C−1 (3) −130 J C−1 (4) −120 J C−1

Q16.A thin disc of radius b = 2a has a concentric hole of radius a in it (see figure). It carries uniform surface charge σ on it. If the electric field on its axis at a height h(h << a) from its centre is given as Ch then the value of C is (1) 4 aϵ0σ (2) aϵ0σ (3) σ (4) σ 5aϵ0 2aϵ0

Q16.A long cylindrical shell carries positive surface charge σ in the upper half and negative surface charge –σ in the lower half. The electric field lines around the cylinder will look like figure given in: (figures are schematic and not drawn to scale) JEE Main 2015 (04 Apr) JEE Main Previous Year Paper (1) (2) (3) (4)

Q17.In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF. Q2 as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale) JEE Main 2015 (04 Apr) JEE Main Previous Year Paper (1) (2) (3) (4)

Q17.In the figure is shown a system of four capacitors connected across a 10 V battery. The charge that will flow from switch S when it is closed is: JEE Main 2015 (11 Apr Online) JEE Main Previous Year Paper (1) Zero (2) 20 μC from a to b (3) 5 μC from b to a (4) 5 μC from a to b

Q18. In the circuit shown, the current in the 1 Ω resistor is: (1) 0. 13 A , from P to Q (2) 1. 3 A , from P to Q (3) 1. 3 A , from Q to P (4) 0. 13 A , from Q to P

Q18.A 10 V battery with internal resistance 1 Ω and a 15 V battery with internal resistance 0.6 Ω are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to: (1) 11.9 V (2) 13.1 V (3) 12.5 V (4) 24.5 V

Q18.In the electric network shown, when no current flows through the 4 Ω resistor in the arm EB, the potential difference between the points A and D will be: (1) 5 V (2) 3 V (3) 4 V (4) 6 V

Q19.When 5 V potential difference is applied across a wire of length 0. 1 m, the drift speed of electrons is 2.5 × 10−4 m s−1 . If the electron density in the wire is 8 × 1028 m−3 , the resistivity of the material is close to: (1) 1 .56 ×10−5 Ω m (2) 1 .6 ×10−8 Ω m (3) 1 .6 ×10−7 Ω m (4) 1 .6 ×10−6 Ω m

Q19.Suppose the drift velocity vd in a material varied with the applied electric field E as vd ∝√E . Then V −I graph for a wire made of such a material is best given by: (1) (2) (3) (4)

Q19.A short bar magnet is placed in the magnetic meridian of the earth with North Pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East-West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in Am2 is close to: (Given μ0 = 10−7 in SI units and BH = Horizontal component of earth's magnetic field = 3.6 × 10−5 Tesla.) 4π (1) 4.9 (2) 14.6 (3) 19.4 (4) 9.7

Q20.A 25 cm long solenoid has the radius 2 cm and 500 turns. It carries a current of 15 A. If it is equivalent to a −−→ Magnetic moment → M is: magnet of the same size and magnetization M ( volume ),then (1) 3π A m−1 (2) 30000 A m−1 (3) 30000π A m−1 (4) 300 A m−1 JEE Main 2015 (10 Apr Online) JEE Main Previous Year Paper

Q20.A rectangular loop of sides 10 cm and 5 cm, carrying a current I of 12 A , is placed in different orientations as shown in the figure below. JEE Main 2015 (04 Apr) JEE Main Previous Year Paper (a) (b) (c) (d) If there is a uniform magnetic field of 0. 3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium? (1) (b) and (c), respectively. (2) (a) and (b), respectively. (3) (a) and (c), respectively. (4) (b) and (d), respectively. −

Q21.A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B . The field occupies a region of space by width d . If α be the angle of deviation of proton from the initial direction of motion (see figure), the value of sin α will be: (1) B (2) q 2 √qdmV Bd√ 2mV (3) B q (4) qV d √ 2mV √Bd2m

Q21.Two coaxial solenoids of different radii carry current I in the same direction. Let F1→ be the magnetic force on −→ the inner solenoid due to the outer one and F2 be the magnetic force on the outer solenoid due to the inner one. Then: −−−−(1) → → (2) → → F1 is radially outwards and F2 = 0 F1 = F2 = 0 −−−−(3) → → (4) → → F1 is radially inwards and F2 is radially F1 is radially inwards and F2 = 0 outwards