AP EAMCET Mock Test - 7 - JEE MCQ

The direction of propagation of the electromagnetic wave is perpendicular to the variation of the electric field  and as well as to the magnetic field  as shown below.

By Applying the Right-hand rule for vector cross product, the direction of propagation of the wave is 

Here, Electric field oscillations is along Z-axis, Magnetic field oscillations is along X-axis and thus the propagation of electromagnetic wave will be along Y-axis.

Given,

Number of turns in the coil, N = 400

The inductance of the coil, L = 8 mH

Current in the coil, i = 5 mA

Let the magnetic flux through each coil be ϕ, then

Due to current in the wire, the flux through the ring will be zero. Hence the mutual inductance is zero.

Situation is shown in figure. When mass m falls vertically on spring, then spring is compressed by distance d

Hence, net work done in the process is W = Potential energy stored in the spring+ Loss of potential energy of mass

Given, σ & −σ be the surface charge denisty, electric field between the plates,

Equilibrium position will be at the point where net force is 00.

Consider, X_o be the amplitude.

We know at equilibrium net force is 00.

So,

The ball is thrown at an angle θ with the horizontal. Thus,

u_x = 15cosθ

u_y = 15sinθ

In order to return at the same point, net displacement will be zero. Thus, in the horizontal direction,

and in the vertical direction,

⇒ t = 3sinθ ...(3)

from (1) we get, 

t = 10 cosθ  ...(4)
from Eq. (3) and (4), we get,

The force experienced by the current element is 
dl is the small current element of the loop and the force is acting perpendicular to the magnetic field as well as current element. So, the loop will expand.
Fleming's left hand rule: At any point, the force exerted by the magnetic field on the current element is directed outwards, and hence loop will expand.
Thus, the correct choice is (c).

I_g = Current for full-scale deflection = 4 x 10^-4 x 25 = 10^-2 ohm
R =  ohm in series

Let

L₁: x + 2y = 8

L₂: 2x + y = 2

L₃: x − y = 1

Since, the shaded area is below the line L₁, therefore we have x + 2y ≤ 8

Since, the shaded area is above the line L₂, therefore we have 2x + y ≥ 2

Since, the common region is to the left of the line L₃, therefore we have x − y ≤ 1.

Also, the solution region lies in first quadrant, therefore x ≥ 0 & y ≥ 0.

 passes through (3, 0) and (3√2, 2)
 
and 
 
 

Given,

p and q Are two statements.

Let us construct a truth table for p∨q

From the above truth table, we see that,

p∨q is false when both p and q are false.

As x³ + y³ + z³ − 3xyz = (x + y + z) + (x² + y² + z² − xy − yz − zx)
so  U = (x + y + z)(x² + y² + z² − xy − yz − zx)
logU = log(x + y + z) + log(x² + y² + z² − xy − yz − zx)

The given equation can be written as

By given condition −2 < m − 1 < m + 1 < 4
⇒ − 1 < m < 3

It is clear from the given options that option (4) is false.
Let a = 2, b = 4 and c = 6
Now, consider option (4).
a² = 4, b² = 16 and c² = 36, which is not an AP.

As   is not a linear function of x i.e., of the type ax+b, the given function is not periodic.

Apply C₁  C₁ + C₂ + C₃

Apply R₂ - R₁, R₃ - R₁
 = 0
=> Expanding along C₁
=> (3a - x) = 0
=> x²(3a - x) = 0
=> x = 0, 3a