JEE Main Practice Test- 4 - JEE MCQ

At some distance from centre inside core

Now time for A to B = 

Magnetic flux passing through inner loop due to current in outer loop-

So, emf developed, 

As there is no loss of energy
Initial mechanical energy = Final mechanical energy
For initial Mechanical energy

Where  and v_relative =5 m/s

= 25 + 25 = 50
For final Mechanical energy
When there is maximum extension relative velocity between block = 0

As permittivity is greater than the permittivity of free space so the refractive index of medium is

Wavelength of the electromagnetic wave in medium will be 
Velocity of wave will be. 

Let u be the speed of projection and v be the speed at B

Applying law of conservation of energy,

Using equation (i) in equation (ii),
we have 

Resistance of B₁, B₂ and B₃ are same which is = 2Ω
As P ∝i² (when ‘R’ constant)
So, maximum power consumed by bulb B₃
Which is = 32 W
⇒i²R = 32
⇒i² . (2) = 32
⇒i = 4 Ampere
So, current passing through B₁ + B₂ = i/2 = 2
Total power consumed in circuit
P = (i/2)² (2) + (i/2)² (2) + i²(2)
= (2)² (2) + (2)² (2) + (4)² (2)
= 48W

As you know that, sideband power is,  Where P_C is power of carrier wave, P_S s sideband power and ‘m’ is modulation rate,

When sphere connected by wire, charge will flow from one sphere to another until potential of both sphere become same.
Or,
As both sphere is identical in dimension and material, charge will divide equally

Now, Initial potential energy of system
(Self energy of spherical conductor )
Final potential energy of system

Heat generation in conduction = U_i - u_f

When S₁ and S₂ both open and jockey has no deflection then let current in upper circuit is ” i”.

So, R_A = 1Ω
R_B = 3Ω
R_C = 6Ω
R_D = 1Ω

When jockey is touched, and there is no deflection then potential difference between ‘A’ and the point should be same in both path-

So, in lower circuit as S₂ is open
No current will flow in this circuit
⇒Potential difference between A and P
Will be = 2V

⇒i₁ . 3= i₂ . 6 and i₁ + i₂ = 1

Potential difference between A and P

For this point ‘P’ should be half way in wire C.
Similarly when jockey touched on wire ‘B’, Null point will be obtained at middle point of wire B.

Distance of object from the spherical surface

For the refraction at the spherical Surface

Let E_i (0 < i < 2) denote the event that urn contains 'i' white and '(2 – i)' black balls.
Let A denote the event that a white ball is drawn from the urn.
We have P(E_i) = 1/3 for i = 0, 1, 2 
P (A|E₁) = 1/3, P(A|E₂) = 2/3, P(A|E₃) = 1.
By the total probability rule,
P(A)= P(E₁)P(A|E₁) + P(E₂)P(A|E₂) + P(E₃)P(A|E₃)

a, b, c are in GP b² = ac
 are in AP.
2(log 2b – log 3c) = log a – log 2b + log 3c – log a
log 2b = log 3c 2b = 3c
∴ b² = ac & 2b = 3c
 b = 2a/3 and c = 4a/9
Since (a+b) = 5a/3 > c, (b+c) = 10a/9 > a and
(c + a) = 13a/9 > b, therefore, a, b, c are the sides
of a triangle. Also, as ‘a’ is the greatest side, let us
find angle A of ΔABC

Hence ΔABC is an obtuse angled triangle.

Put 2x + y = X ⇒Therefore, the given equation is reduced to

 = x + constant
2(2x + y) + log (2x + y – 1) = 3x + constant
x + 2y + log (2x + y – 1) = C

1. Area bounded by curve y = f(x), x =1 and x = b is

Now differentiating both sides with respect to b we get
fB. = sin (3b + 4) + 3(b – 1) cos (3b + 4)

1. sin 5x + sin x + sin 3x = 0
2 sin 3x . cos2x + sin3x = 0
sin 3x (2 cos 2x + 1) = 0, 0 ≤ x ≤π/2

From both x = π / 3 (other than 0).

Let  Hence the given series S is
S = 

(K–2)x² + 8x + K + 4 = 0

Let 
If the equation f(x) = 0, has real distinct and negative roots then:
D > 0 & f(0) > 0 & 
(i) D > 0

64 – 4 (K² + 2K – 8) > 0
– K² – 2K + 24 < 0 K² + 2K – 24 < 0
(K + 6)(K – 4) < 0
–6 < K < 4
(ii)

⇒ K < –4 or K > 2
(iii)

K – 2 > 0
K > 2
∴D > 0 & f(0) > 0 &

–6 < K < 4 & (K < –4 or K > 2) & K > 2
–6 < K < 4 & _{K < –4 & K > 2) or (K > 2 & K > 2}
– 6 < K < 4 & 
2 < K < 4
K∈(2, 4)
Among the given options
3 lies in the given range,
Hence K = 3

Equation of any tangent to parabola  

If the tangent to parabola is also tangent to the circle (x-3)² + y² = 9 then:

9m⁴ + 6m² + 1 = 9m⁴ + 9m²
3m² – 1 = 0

From the figure it is evident that for the tangent above x-axis   Hence equation of required common tangent is:

Let 

Adding (i) and (ii):

Equation of line joining A (1,0) and B (3,0) is y = 0. Equation of family of circles passing through A and B is:
(x – 1)(x – 3) + (y – 0) (y – 0) + λy = 0
x² + y² – 4x + λy + 3 = 0
If above circle touches y-axis then x = 0 is a tangent. Substituting x = 0:
y² + λy + 3 = 0
Discriminant of above quadratic must be zero.
λ² – 12 = 0

Hence the circles are:

Thus, the coordinates of C₁ and C₂ arerespectively. Also the radii of each of the circles is r₁ = r₂ = 2
We know that the angle of intersection of two circles of radii r₁ and r₂ is given by
 where d is the distance between their centers.

Since between two lines whenever there is angle θ there is always the angle π–θ, therefore,

We have


Also for the greatest value of abc² the numbers have to be equal, i.e a = b = c/2
Also given that maximum value = 1/64
so, a + b + c = 1
i.e. a = b = 1/4, c = 1/2

B = –A^–1 BA
AB = A (–A^–1BA)

AB = –BA
AB + BA = 0
Now, (A + B)² = A² + B² + AB + BA
(A + B)² = A² + B², (∵AB + BA = 0)

 x² + 5x + 6 < 0 –3 < x < –2 
y = 1 + a² x – x³

(i) Let a > 0:
Hence  Hence y has minima at  for a > 0.

(ii) Let a < 0:
Hence Hence y has minima at  for a < 0.

From (i) & (ii) 

Substitute (x – 1) = t. Hence

Hence given limit is:

f (x) is an even function.
NOTE: Integration of every even function is odd function and integration of every odd function is even function, provided the integrands are integrable and the limit is 0 to x. Similarly differentiation of every odd function is even function and that of every even function is odd function, provided the they are differentiable. Here  which is an odd function, therefore  is even function.

a^–2 + b^–2 = c^–2 
 bx + ay – ab = 0
The foot of perpendicular to the above line from O(0,0) is given by:

Using a^–2 + b^–2 = c^–2

x² + y² = c²

Length of latus rectum of hyperbola = Length of the transverse axis