Practice Questions

Q15.The total length of a sonometer wire fixed between two bridges is 110 cm. Now, two more bridges are placed to divide the length of the wire in the ratio 6 : 3 : 2 . If the tension in the wire is 400 N and the mass per unit length of the wire is 0. 01 kg m−1 , then the minimum common frequency with which all the three parts can vibrate, is (1) 1000 Hz (2) 1100 Hz (3) 100 Hz (4) 110 Hz →

Q15.Two factories are sounding their sirens at 800 Hz. A man goes from one factory to other at a speed of 2 m/s. The velocity of sound is 320 m/s. The number of beats heard by the person in one second will be: (1) 2 (2) 4 (3) 8 (4) 10 →

Q15.A transverse wave is represented by : y = 10π sin ( 2πT t −2πλ x) For what value of the wavelength the wave velocity is twice the maximum particle velocity? (1) 40 cm (2) 60 cm (3) 10 cm (4) 20 cm

Q15.A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s. (1) 12 (2) 8 (3) 6 (4) 4 →

Q16.The electric field in a region of space is given by, E = E0ˆi + 2E0ˆj where E0 = 100 N C−1 . The flux of this field through a circular surface of radius 0. 02 m parallel to the Y ‐Z plane is nearly (1) 0. 02 N m2 C−1 (2) 0. 005 N m2 C−1 (3) 0. 125 N m2 C−1 (4) 3. 14 N m2 C−1

Q16.A spherically symmetric charge distribution is characterised by a charge density having the following variations: ρ(r) = ρo (1 −rR ) for r < R ρ(r) = 0 for r ≥R Where r is the distance from the centre of the charge distribution ρo is a constant. The electric field at an internal point (r < R) is: ρo (1) (2) ρ0 3 3 4εo ( r − 4Rr2 ) ε0 ( r −r24R ) (3) (4) 3 3 3εo 12εo ρo ( r − 4Rr2 ) ρo ( r − 4Rr2 )

Q16.Assume that an electric field E = 30x2ˆi exists in space. Then the potential difference VA −VO , where VO is the potential at the origin and VA the potential at x = 2 m is : (1) 120 J C−1 (2) −120 J C−1 (3) −80 J C−1 (4) 80 J C−1

Q16.The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about 150 N/C , directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be : [Given : ∈O= 8.85 × 10−12 C2/N-m2, RE = 6.37 × 106m] (1) −680 kC (2) + 670 kC (3) + 680 kC (4) − 670 kC

Q16.A cone of base radius R and height h is located in a uniform electric field E parallel to its base. The electric flux entering the cone is: (1) 1 EhR (2) EhR 2 (3) 2EhR (4) 4EhR

Q17.The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation: K(x) = Ko + λx(λ = a constant ) The capacitance C, of the capacitor, would be related to its vacuum capacitance Co for the relation : (1) C = λd (2) C = λ ln(1+Koλd Co) d⋅ln(1+Koλd Co) (3) C = λd (4) C = λ ln(1+λd/Ko d.ln(1+Ko/λd Co) Co)

Q17.A parallel plate capacitor is made of two plates of length 1 , width w and separated by distance d. A dielectric slab (dielectric constant K ) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force F = −∂U∂x where U is the energy of the capacitor when dielectric is inside the capacitor up to distance x (See figure). If the charge on the capacitor is Q then the force on the dielectric when it is near the edge is: (1) Q2 d K (2) Q2 W ( K −1) 2wl2εo 2dl2ε0 (3) Q2 d (K −1) (4) Q2w K 2wl2εo 2dl2εo JEE Main 2014 (11 Apr Online) JEE Main Previous Year Paper

Q18.The circuit shown here has two batteries of 8.0 V and 16.0 V and three resistors 3Ω, 9Ω and 9Ω and a capacitor of 5.0μF. How much is the current I in the circuit in steady state? JEE Main 2014 (12 Apr Online) JEE Main Previous Year Paper (1) 1.6 A (2) 0.67 A (3) 2.5 A (4) 0.25 A

Q18.A d.c. main supply of e.m.f. 220 V is connected across a storage battery of e.m.f. 200 V through a resistance of 1 Ω. The battery terminals are connected to external resistance R. The minimum value of R, so that a current passes through the battery to charge it is: (1) 9 Ω (2) 7 Ω (3) 11 Ω (4) 0

Q18.In the circuit shown, current (in A) through 50 V and 30 V batteries are, respectively: (1) 2.5 and 3 (2) 3.5 and 2 (3) 4.5 and 1 (4) 3 and 2.5

Q18. Four bulbs B1 , B2 , B3 and B4 of 100 W each are connected to 220 V main as shown in the figure. The reading in an ideal ammeter will be (1) 0. 90 A (2) 1. 35 A (3) 0. 45 A (4) 1. 80 A

Q18.In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be: JEE Main 2014 (06 Apr) JEE Main Previous Year Paper (1) 8 A (2) 10 A (3) 12 A (4) 14 A

Q19.The coercivity of a small magnet, where the ferromagnet gets demagnetised is 3 × 103 A/m . The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetised when inside the solenoid is (1) 30 mA (2) 60 mA (3) 3 A (4) 6 A

Q19.In the experiment of calibration of voltmeter, a standard cell of e.m.f. 1.1 volt is balanced against 440 cm of potential wire. The potential difference across the ends of resistance is found to balance against 220 cm of the wire. The corresponding reading of voltmeter is 0.5 volt. The error in the reading of volmeter will be: (1) −0.15 volt (2) 0.15 volt (3) 0.5 volt (4) −0.05 volt

Q19.The mid points of two small magnetic dipoles of length d in end-on positions, are separated by a distance x (x ≫d). The magnitude of force between them is proportional to x−n where n is : JEE Main 2014 (09 Apr Online) JEE Main Previous Year Paper (1) 3 (2) 4 (3) 2 (4) 1

Q19.In the circuit diagrams (A, B, C and D) shown below, R is a high resistance and S is a resistance of the order of galvanometer resistance G. The correct circuit, corresponding to the half deflection method for finding the resistance and figure of merit of the galvanometer, is the circuit labelled as: (a) (b) JEE Main 2014 (11 Apr Online) JEE Main Previous Year Paper (C) (d) (1) Circuit A with G = (R−S)RS (2) Circuit B with G = S (3) Circuit C with G = S (4) Circuit D with G = (R−SRS )

Q20.A square frame of side 10 cm and a long straight wire carrying current 1 A are in the plane of the paper. Starting from close to the wire, the frame moves towards the right with a constant speed of 10 m s−1 (see figure). The e.m.f induced at the time the left arm of the frame is at x = 10 cm from the wire is (1) 0.75 μV (2) 1 μV (3) 2 μV (4) 0.5 μV JEE Main 2014 (19 Apr Online) JEE Main Previous Year Paper Q21. → The figure shows a circular area of tthe radius R where a uniform magnetic field B is going into the plane of the paper and increasing in magnitude at a constant rate. In that case, which of the following graphs, drawn schematically, correctly shows the variation of the induced electric field E(r)? (1) (2) (3) (4)

Q20.A positive charge ' q ' of mass ' m ' is moving along the +x axis. We wish to apply a uniform magnetic field B for time Δt so that the charge reverses its direction crossing the y axis at a distance d. Then: (1) B = mvqd and Δt = πdv (2) B = 2qdmv and Δt = πd2v (3) B = 2mvqd and Δt = πd2v (4) B = 2mvqd and Δt = πdv

Q20.The magnetic field of earth at the equator is approximately 4 × 10−5T. The radius of earth is 6.4 × 106m. Then the dipole moment of the earth will be nearly of the order of : (1) 1020 A m2 (2) 1016 A m2 (3) 1010 A m2 (4) 1023 A m2

Q21.In the circuit shown here, the point ' C ' is kept connected to point A '' till the current flowing through the circuit becomes constant. Afterward, suddenly, point ' C ' is disconnected from point A '' and connected to point ' B' at time t = 0. Ratio of the voltage across resistance and the inductor at t = RL will be equal to : (1) e (2) 1 1−e (3) - 1 (4) 1−e e

Q21.When the rms voltages VL, VC and VR are measured respectively across the inductor L, the capacitor C and the resistor R in a series LCR circuit connected to an AC source, it is found that the ratio VL : VC : VR = 1 : 2 : 3. If the rms voltage of the AC source is 100 V, then VR is close to : (1) 50 V (2) 70 V (3) 100 V (4) 90 V