IIT JAM Physics MCQ Test 5 - Physics MCQ

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
...(i)
From fig. it is clear that at point PE_x will be zero because i E, will be canceled by the contribution from a symmetrically placed Δ_t on the left half of the arc. So in order to find total E we need only compute E_y at point P. so.
...(ii)
-ve sign arises Because ΔE_v is in - y direction sum over the whole are
...(iii)
Along the circle of radius a. ds = adθ
So. 

In first case


...(i)

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

...(iii)
So, 
 ...(iv)
So from equation (2) & (4)

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

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 

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

So, 
Now, 

We have 


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

Equivalent circuit is as shown in Figure


or 
Since  we get

∴ 
i.e. 

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

∴ 

Original capacitance 
∴  

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.

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

We know

and  M¹T^-1Q^-1
So, 
 

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

Work done by tire electric force = Change of kinetic energy 

or 
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

Now in case of series LCR circuit,

So current will lag the applied voltage by 45⁰ if

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 
Secondly   
 (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.)

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 V_c across the resistor, inductor, and capacitor. respectively, are

we know electric field vector
Since  the medium is not free space but a lossless medium.
 (nonmagnetic). 
Hence 
or 



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

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

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

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

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.

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

At resonance ; X_L = X_c

Now 

Thus change in inductance should be 4 times.

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

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, 
So,  

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

so (a) 
i.e. 
and (b) 
i.e. 
i.e. 

As in this problem,


and hence. 

but as in case of ac.