Practice Questions
14,828 questions across 23 years of JEE Main — find and practise any topic!
Q13.A simple harmonic oscillator of angular frequency 2 rad s−1 is acted upon by an external force F = sin t N . If the oscillator is at rest in its equilibrium position at t = 0, its position at later times is proportional to: (1) sin t + 21 cos 2t (2) cost −12 sin 2t (3) sin t −12 sin 2t (4) sin t + 12 sin 2t
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 wire of length L = 20 cm is bent into a semi-circular arc and the two equal halves of the arc are uniformly charged with charges +Q and −Q as shown in the figure. The magnitude of the charge on each half is |Q| = 103ε0 , where ε0 is the permittivity of free the space. The net electric field at the centre O is (1) (25 × 103)ˆi N C−1 (2) (50 × 103)ˆi N C−1 (3) (25 × 103)ˆj N C−1 (4) (50 × 103)ˆj N C−1 → + 30 N C−1 exists in a region of space. If the potential at the origin is taken to
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
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)
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 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)
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
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
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 given circuits (a) and (b), switches S1 and S2 are closed at t = 0 and kept close for a long time. The variation of currents in the two circuits for t ≥0 are shown in the options. (Figures are schematic and not drawn to scale.) JEE Main 2015 (10 Apr Online) JEE Main Previous Year Paper (1) (2) (3) (4)
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 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.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.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
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.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
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. −
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 wire carrying current I is tied between points P and Q and is in the shape of a circular arc of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, as shown in the figure) in the vicinity of the wire. If the wire subtends an angle 2θo at the center of the circle (of which it forms an arch) then the tension in the wire is: JEE Main 2015 (11 Apr Online) JEE Main Previous Year Paper (1) IBR (2) IBR sinθ0 (3) IBR (4) IBRθ0 2sinθ0 sinθ0
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
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