UGEE Mock Test - 1 - JEE MCQ

v = 5x² + 10x - 4

At x = 1 m, E = -20V/m

Given:

• v = 3 × 10⁵ m/s
• B = 0.3 T
• θ = 30°
• e/m = 10⁸ C/kg

Short Solution:
1. Find the perpendicular velocity component:

2. Calculate the radius of curvature:

Conclusion:
The radius of curvature is:
0.5 cm

Letm = KF^aL^bT^c
Substituting the dimensions of
[F] = [MLT^-2], [L] = [L] and [T] = [T] and
comparing both side, we get m = FL^-1T²

Induced emf produced between the centre
and a point on the disc is given by

ω = 60 rad/s, B = 0.05Wb/m²
and R = 100 cm = 1m

For stone to be dropped from rising balloon of velocity 29 m/s, u = 29 m /s, t = 10 sec.

The phase angle between voltage V and current I is π/2. Therefore, power factor cos φ = cos (π/2) = 0. Hence the power consumed is zero.

The bullets are fired at the same initial speed

The equivalent system is shown in figure.

The equivalent force constant of which

The equivalent circuit is shown in figure.

If I is length of pendulum and θ be angular amplitude then from principle of conservation of energy

From eq. (3) and (4)
 

The figure is showing I - V characteristics of non ohmic or non linear conductors.

Given that, f(x) = [x] and g (x) = IX

Understanding the Problem

We are given the equation of a circle:
x² + y² = 8
This represents a circle centered at the origin (0, 0) with a radius of √8, which is equal to 2√2.

There is a chord of this circle that makes equal intercepts of length a on the coordinate axes. That means the chord cuts the x-axis and y-axis at the same distance from the origin, like at points (a, 0) and (0, a), or even (-a, 0) and (0, -a). Only the magnitude matters here, so we’re concerned with |a|.

Our goal is to find which inequality about |a| is correct.

Visualizing the Chord

A chord that makes equal intercepts on the axes has a line equation in intercept form:

(x / a) + (y / a) = 1
Multiplying through by a:
x + y = a

This is one possible form. Other forms like x - y = a or -x - y = a can also represent chords with equal intercepts, but the logic for all of them will be the same in terms of calculating the magnitude |a|.

Using the Distance Formula

For a line to be a chord of a circle, it must intersect the circle at two distinct points. This happens only if the distance from the center of the circle to the line is less than the radius of the circle.

The distance from the origin (0, 0) to the line x + y - a = 0 is given by:
d = |a| / √2

We already know the radius of the circle is 2√2.

So we set:
|a| / √2 < 2√2

Multiply both sides by √2:
|a| < 2√2 × √2
|a| < 4

Considering Other Forms

Try a different line with the same intercepts:
x - y = a

Again, the distance from the origin to this line is:
|a| / √2, which gives the same inequality:
|a| < 4

So, regardless of whether the line is x + y = a, x - y = a, -x + y = a, or -x - y = a, the required condition is always:
|a| < 4

Checking the Options

Now let’s look at the options:

a) |a| < 8
b) |a| < 4√2
c) |a| < 4
d) |a| > 4

From all our reasoning, the correct answer is:
c) |a| < 4

Final Answer: c) |a| < 4

We have x² + y² + 4y - 5 = 0. Its centre is C1 (0, -2)

We have, 

Required number of ways
= coeff of x¹⁶ in (x³ + x⁴ + x⁵ + x⁶ + x⁷)⁴

AM ≥ HM

For minimum value = 0.
Let and   are anti-parallel so

The equation represents a pair of lines if
1·1·4 + 2·f·g·1-1·f²-1·g² - 4·1² = 0
⇒ (f - g)2 = 0 ⇒ f = g
The equation becomes (x + y)² + 2g(x + y) + 4 = 0
Which represents pair of parallel lines, which
are real provided (2g)² - 4·4·1 ≥ 0 ⇒ |g| ≥ 2.

Let R be the set of families having a radio and T the set families having a T.V., then n (R ∪ T) The number of families having at least on of the radio and T.V. 1003 - 63 = 940
n(R) = 794 and n(T) = 187
Let x families have both a radio and a T.V.
Then number of families who have only radio = 794 - x
And the number of families who have only T.V. = 187 - x
From Venn diagram, 794 - x + x - 187 - x = 940
⇒ 981 - x = 940 or x = 981 - 940 = 41
Hence, the required number of families having both a radio and a T.V = 41

For (x) to be defined, we must have

The centre of the given circle is C (4, 2)
Let A ≡ (-3,2)

If (α, β) are the coordinates of the other end of the diameter, then, as the middle ploint of the diameter is the centre,

Thus, the coordinates of the other end of diameter are (11, 2)

Total number of ways = 6 × 6 × ..... to n times = 6ⁿ.
Total number of ways to show only to even number = 3 × 3 × ........ to n times = 3ⁿ.
∴ required number of ways = 6ⁿ - 3ⁿ.