IIT JAM Physics MCQ Test 5

30 Questions MCQ Test Mock Test Series for IIT JAM Physics | IIT JAM Physics MCQ Test 5

This mock test of IIT JAM Physics MCQ Test 5 for Physics helps you for every Physics entrance exam. This contains 30 Multiple Choice Questions for Physics IIT JAM Physics MCQ Test 5 (mcq) to study with solutions a complete question bank. The solved questions answers in this IIT JAM Physics MCQ Test 5 quiz give you a good mix of easy questions and tough questions. Physics students definitely take this IIT JAM Physics MCQ Test 5 exercise for a better result in the exam. You can find other IIT JAM Physics MCQ Test 5 extra questions, long questions & short questions for Physics on EduRev as well by searching above.

What is the electric field at the center of a uniformly charged semicircular arc i.e., at point P in Fig.


Let λ is the charge per unit length of arc of radius ‘a' splitting the arc in to small segment Δs1  and point charge on each is λΔs1,. From coulomb’s law, the electric field due to the Δs1  is
From fig. it is clear that at point PEx will be zero because i E, will be canceled by the contribution from a symmetrically placed Δt on the left half of the arc. So in order to find total E we need only compute Ey at point P. so.
-ve sign arises Because ΔEv is in - y direction sum over the whole are
Along the circle of radius a. ds = adθ


Two electric dipoles p1 and p2 are placed at (0, 0, 0) and (2, 0, 0) respectively with both of them pointing in the +z direction. Without changing the orientations of the dipoles. P2 is moved to (0. 4. 0). The ratio of the electrostatic potential energy of the dipoles after moving to that before moving is.


In first case


In new case P2 is shifted to (0, 4, 0). then the electric field at the site of P2 is

So from equation (2) & (4)


A spherical piece of radius much less than the radius of a charged spherical shell (charge density σ) is removed from the shell itself then electric field intensity at the mid point of aperture is


For a charged spherical shell.
Let  be the electric field due to small removed portion and  the electric field due to remaining portion.
Comparing (i) and rii) we get
and  (iv)
⇒  (from (iii) and (iv)


A cubical box sits in a uniform electric field as shown in Fig. What is the net flux coming out of the box.


Because the field lines simply skim the four sides of the box. the flux through them is zero. The flux through the top is since cosθ = cos 00 = 1 in this case .
For the bottom.  The negative sign arises because the flux is into the bottom area, not out of it.
Because A1 = A0. the net flux from the box is 


A charge q is placed at the centre of the line joining two equal charges O. The system of three charges will be in equilibrium if q equal to


Let two equal charges Q each, be held at A and B. where AB = 2x. C is the centre of AB. where charge q is held.

For the three charges to be in equilibrium, net force on each charge must be zero.
Now. total force on Q at B is


Two point charges q1 = - 2 C and q2 = - 1 C are separa ted by a distance d. The position on the line joining the two charges where a third charge q = + 1C will be in equilibrium is at a distance


When two charges are of same nature (both positive or both negative) then neutral point is always between the two charges on the line joining them. But when two charges are of opposite nature (one positive and other negative) then neutral point is always outside the two charges on the extended line joining the two charges.
So. we conclude that options (A) and (C) are absolutely ruled out.



A long cylinder carries a volume charge density that is given by p = kr where k is constant and r is the distance from the axis. Then the electric field inside the cylinder is




Does the potential function φ = q(x2 + y2 + z2)-1/2  satisfies the laplace’s equations.


We have 

and  ...(iii)
Adding (1), (2) and (3)
i.e.  which is Laplace's equation. Thus the function
 satisfies Laplace's equation.


If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by


Equivalent circuit is as shown in Figure

Since  we get



Two slabs of the same dimensions, having dielectric constants Kand K2. completely fill die space between the plates of a parallel plate capacitor as shown in the figure. If C is the original capacitance of the capacitor, the new capacitance is


The arrangement is equivalent to two capacitors, each of plate area A and separation d/2, connected in series having capacitances


Original capacitance 


An electron moving with a speed 'u‘ along the position x - axis at y = 0 enters a region of uniform magnetic field which exists to the right of y - axis . The electron exists from the region after sometimes with the speed V at coordinate y then


Magnetic force does not change the speed of charged particle. Hence, v = u. Further magnetic field on the electron in the given condition is along negative y-axis in the starting. Or it describes a circular path in clockwise direction. Hence, when it exits from the field, y < 0.


A long straight conductor, carrying a current L is bent into the shape shown in the figure. The radius of the circular loop is r. The magnetic field at the centre of the loop is


Field due to straight part.  out of the page
Field due to circular part.   to the page 
Net field B =  into the page.


A battery of emf V volt is connected across a coil of uniform wire as shown. The radius of the coil is 'a' metres and the total resistance of the coil is R ohm. The magnetic field at the centre O of the coil (in tesla) is



In terms of basic units of mass (M). Length (L). time (Tr and charge 0. the dimensions of magnetized field (H» are


We know

and  M1T-1Q-1


A magnetic needle is kept in a non-uniform magnetic field. It experiences:-


Option(A) is correct as both the magnetic force & torque will be applied on it. (torque): 

*Multiple options can be correct

A particle of charge +q and mass m moving under the influence of a uniform electric field E and a uniform magnetic field B k follows a trajectory from P to Q as shown in the figure. The velocities at P and Q are  Which of the following statement is correct?


Work done by tire electric force = Change of kinetic energy 

Now. rate of work done = F.v
∴ Rate of work done by the electric field at P

Rate of work done by the electric field at Q 
= qE 2v cos 90" = 0
Work done by the magnetic field is obviously zero at it points. 
So C is not possible

*Multiple options can be correct

A series LCR circuit containing a resistance of 120 Ω has angular lesonance frequency 4 * 105 rad s-1. At resonance the voltages across resistance and inductance are 60 V and 40 V respectively.


Now in case of series LCR circuit,

So current will lag the applied voltage by 45if

*Multiple options can be correct

A positively charged thin metal ring of radius R is fixed in the xy plane with its centre at the origin 0. A negatively charged particle P is released from rest at the point (0. 0. Z0) where Z0 > 0. Then the motion of P is :


Let o be the charge on the ling, the negative charge -q is released from point P (0, 0, Z0). The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be 

E = 0 at centre of the ring because Z0= 0

Therefore, force on charge P will be towards centre as shown, and its magnitude is

Similarly, when it crosses the origin . the force is again towards centre O.
Thus tire motion of the particle is period ic for all values of Z0 lying between 0 and 
 (From equation 1)
i.e. there storing force 0. Hence the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position.)

*Multiple options can be correct

In the series circuit of Fig. suppose   Then


The inductive and capacitive reactances are 

The impedance Z of the circuit is

Z = 500 Ω
Witfi source voitage amplitude V = 50 V the current amplitude is

The phase angle is

from fig. the voltage amplitudes V R, V L. and Vc across the resistor, inductor, and capacitor. respectively, are

*Multiple options can be correct

In a nonmagnetic medium E = 4sin  Then,


we know electric field vector
Since  the medium is not free space but a lossless medium.

On plane 2x + y = 5 
Hence the total power is  

*Answer can only contain numeric values

Three point charges are placed at the following points on the x axis: + 2μC at x = 0, - 3μC at x = 40 cm. - 5μC at x = 120 cm. What is the force (in Newton) on the - 3μC chaige ?


Figure is a diagram of the step with . The force on q2 is the vector sum of two contributions, the attractive force due to q1(toward q1) and the repulsive force due to q3 (also toward q1). The sum of these two forces, taken algebraically, since they are along the same line, is 
F = F1 + F2

= - 0 . 55N or 0.55 N to left

*Answer can only contain numeric values

Four capacitors, each of 25μF.are connected as shown in the following circuit. A d.c. voltmeter reads 200 volts. The charge (in coulomb) on each capacitor is _____


All the capacitors are in parallel and P.D. across each is 200 V. 
∴ O = CV = 25 * 10-6 * 200 = 5 * 10-3 C

*Answer can only contain numeric values

The figure shows two identical parallel plate capacitors connected to a battery with switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of constant 3. The ratio of the total energy stored in both capacitors before and after tire introduction of the dielectric is ____


Initial energy 
Final energy of A 
Final p.d. across B. V . is given by (3C) V = C V or V = V/3
∴ Final energy of B

∴ Total final energy

*Answer can only contain numeric values

The electric field intensity at a point on tine surface of a conductor is given by   What is the surface charge density (in Pc/m2) at the point.


Supposing the conductor to be surrounded by free space.

The ambiguity in sign arises from that in the direction of the outer normal to the surface at the given point.

*Answer can only contain numeric values

A copper rod of length 0.19 m is moving with uniform velocity 10 ms-1 parallel to a long straight wire carrying a current of 5A. The rod is perpendicular to the wire with its ends at distance 0.01 m and 0.2 m from it. Calculate the e.m.f. induced 


Here  Due to motion of element dx in the field. B induced e.m. f. dE = Bvdx Total e.m. f.

= 30 x 10-6 V = 30 μV

*Answer can only contain numeric values

In an LCR circuit, the capacitance is made one - fourth when in resonance, then how many times the inductance should be changed, so that the circuit remains in resonance?


At resonance ; XL = Xc


Thus change in inductance should be 4 times.

*Answer can only contain numeric values

A 50 Hz A.C. current of peak value 1A flows through the primary coil of a transformer. If the mutual inductance between primary and secondary is 1.5 H. the mean value of the induced voltage (in V) is ________ .


Frequency of A.C. n = 50 Hz
time period T 
Time in which current changes from peak
value (ii) to zero (ii
then Δi- change in current 
= ii — ii = 0 — 1 = - 1 A .
∴ mean induced e.m.f.

*Answer can only contain numeric values

A long solenoid of diameter 0.1 m has 2 * 104 turns per meter. At the centre of, the solenoid a 100 turn coil of radius 0.01m is placed with its axis coinciding with that of the solenoid. The current in the solenoid is decreasing at a constant rate from +2A to -2A in 0.05 sec. Find the total charge (in μC ) flowing through the coil during this time if the resistance of the coil is 10 π2 ohm.


As field inside, a solenoid is given by

so here.

And hence flux linked with the coil

Further as the coil has a resistance of 10π2 ohm. the current induced in it. 

and as, 

*Answer can only contain numeric values

Find the ratio of average and rms value for the saw-tooth voltage of peak value Vas shown in Fig.


As the equation of the saw-tooth wave shown in Fig. will be.

so (a) 
and (b) 

*Answer can only contain numeric values

A 750 hertz, 20 V source is connected to a resistance of 100 ohm. an inductance of 0.1803 henry and a capacitance of 10 microfarad all in series. Calculate the time (in sec .) in which the resistance (thermal capacity 2J/oC ) will get heated by 10oC.


As in this problem,

and hence. 

but as in case of ac.