VITEEE: Subject Wise and Full Length VITEEE PCME Mock Test - 7 Free Online

Option C is correct.

Let the heights be h₁ = 20 m and h₂ = 80 m, and place their bases at x = 0 and x = d on a horizontal axis.

The line from the top of the first tower to the foot of the second has equation y = h₁(1 - x/d).

The line from the top of the second tower to the foot of the first has equation y = h₂(x/d).

At intersection set them equal: h₁(1 - x/d) = h₂(x/d). Solving gives x = d·h₁/(h₁ + h₂).

Substitute into y = h₂(x/d) to obtain y = h₁·h₂/(h₁ + h₂).

Numerically, y = (20 × 80)/(20 + 80) = 1600/100 = 16 m. Hence the height of intersection is 16 m.

P(X = k) =
Now, P(X = k + 1) = r(k) P(X = k)
⇒
⇒ r(k) = λ/(k+1)

2log(y + 3) = logx + logc 
(y + 3)² = xc, which is parabola.

Given, Mean = 4, Variance = 3
np = 4, npq = 3

Mode is an integer x such that

⇒ 3.25 < x < 4.25
∴ x = 4

Let α, β be the roots of x² + bx + c = 0, and γ, δ be the roots of x² + qx + r = 0.

For the first equation: α + β = -b, αβ = c

For the second equation: γ + δ = -q, γδ = r

Given that αβ = γδ (i.e., c = r), we can proceed with the following relationship:

(α - β)²(α + β)² = (γ - δ)²(γ + δ)²
(Squaring both sides and applying components and division)

Then, αβ(α + β)² = γδ(γ + δ)²
(After subtracting -1 from both sides)

This simplifies to:
c * b² = r * q²

Hence, the correct answer is D) b²r = q²c.

Let O be the center of the circle.

Since the hexagon is regular, we have:
∠A₀OA₁ = 360° / 6 = 60°

Thus, △A₀OA₁ is an equilateral triangle. From the unit circle, A₀O = 1.

Hence, A₀A₁ = 1.

Also, A₀A₂ = A₀A₄.

We know that a perpendicular drawn from the center bisects the chord, so:

A₀A₂ = A₀A₄ = 2 × A₀D
= 2 (OA₀ sin 60°)
= 2 × (1) × (√3/2) = √3

Thus,
(A₀A₁) × (A₀A₂) × (A₀A₄) = (1) × (√3) × (√3) = 3

THEOREM : Number of permutations (arrangements) of 'n' dissimilar things taken 'r' at a time, when repetition of things is not allowed is given by

Here we have 5 seats and two people 

= 20

Let, g(x) = ax + b

Now at x = 0, equating both L.H.L and R.H.L we get

now for

Taking log both side we get,

arg = arg(z + i) - arg(z - i) = tan^-1- tan^-1 = tan^-1 = , i.e. x² + y² - 2x = 1 is the equation of a circle.

The given lines are coplanar if the normal to the plane containing these lines is perpendicular to both of them.
Since the given lines are parallel to the vectors and ,
so, the normal to the plane is parallel to , which is perpendicular to the line joining the points on the plane with position vectors and
⇒ , which is the required condition for the given lines to be coplanar.

Correct answer: Option B

• Mobile uses radio waves for cellular communication; mobile networks operate in radio-frequency bands.
• Sonar uses ultrasound (mechanical sound waves above 20 kHz) for underwater detection; this is not an electromagnetic wave.
• Radar uses microwaves (high-frequency radio waves in the GHz range) to detect the position and speed of objects.
• Optical fibre transmits signals using light, commonly in the infrared region for telecommunication (near-infrared wavelengths are typically used).

The resulting matching is (A → R), (B → S), (C → P), (D → Q), which corresponds to Option B.

The phase angle between voltage V and current I is π/2.
Power factor,

Hence, the power consumed is zero.

Slope of the graph will give us the reciprocal of resistance.

Here, resistance at temperature T₁ is greater than that at T₂.
Since the resistance of metallic wire is more at higher temperatures than at lower temperatures, therefore T₁ > T₂.

In beta decay, atomic number increases by 1 whereas the mass number remains the same. Therefore, following equation can be possible.

D is the converging lens.

1/f = (μ - 1) (1/R1 - 1/R2)

Here μ is the refractive index of the lens material relative to the surrounding medium, i.e. μ = n_lens / n_medium. Use the sign convention: radii of curvature are taken positive if the centre of curvature lies to the right of the surface (light assumed to travel left → right).

For a thin lens the sign of 1/f (and hence whether the lens is converging or diverging) depends on the signs of both (μ - 1) and (1/R1 - 1/R2).

Option D: the diagram shows a lens that is thicker at the centre (biconvex shape). From the figure the lens material has n_lens = 1.5 and the surrounding medium has n_medium = 1.3, so μ = 1.5/1.3 ≈ 1.154 > 1. For a biconvex lens with light left → right, we take R1 > 0 and R2 < 0, so (1/R1 - 1/R2) > 0. Therefore 1/f > 0, so the focal length is positive and the lens is converging.

Options A, B and C: in each of these the combination of shape and refractive-index arrangement makes the product (μ - 1)(1/R1 - 1/R2) negative, so 1/f < 0 and the lens behaves as diverging. Briefly:

• Option A: although the lens material is optically denser than the outside medium (μ > 1), the geometry is such that (1/R1 - 1/R2) < 0 (the lens is effectively thinner at the centre for the chosen light direction), giving f < 0.

• Option B: the surrounding medium is air (n = 1.0) and the drawn shape is double-concave (thinner at centre); with μ > 1 but (1/R1 - 1/R2) < 0, the net focal length is negative.

• Option C: although numerical values of refractive indices differ on the two sides, the diagram shows a shape that produces (1/R1 - 1/R2) < 0 for the indicated light direction and so the net power is negative.

Hence only option D gives a positive focal length and is the converging lens.

The slope of the (I-V) graph is given by 1/R. In the CD region, the slope of the I-V curve is negative, meaning R is negative. Therefore, the portion corresponding to negative resistance is the CD region.

The separation between the images formed by extreme wavelengths of the visible range is called the longitudinal chromatic aberration.

It is given by the formula:

Longitudinal chromatic aberration = Dispersive power (ω) × Mean focal length (fy)

Given:

• Dispersive power (ω) = 0.02
• Mean focal length (fy) = 20 cm

Substituting the values:

Longitudinal chromatic aberration = 0.02 × 20 = 0.40 cm

When object is placed between F and pole of a convex lens then a virtual, erect and magnified image will be formed on the same side behind the object.

Given,

m = 25 kg,
v_initial = 2 m/s.

So, the initial kinetic energy of the body:

KE_initial = (1/2) m v²
KE_initial = (1/2) × 25 × 4 = 50 J.

Here, work done = area of F−x graph.

⇒ W = area of triangle

⇒ W = (1/2) × base × height
⇒ W = (1/2) × 4 × 20 = 40 J.
Since the force is resistive, W = −40 J.

By the work-energy theorem,
Work done = change in kinetic energy.

W = KE_final − KE_initial
−40 = KE_final − 50
KE_final = 10 J.

For polyfunctional acids, pI is also the pH midway between the pK_a values on either side of the isoionic species. For example, the pI of aspartic acid is 2.95.
pI = (pK₁ + pK₃)/2 = (2.0 + 3.9)/2 = 2.95

Since only 50% of the expected product is obtained:
5 L of N₂ reacts with 15 L of H₂ to form 10 L of NH₃.
used = 5 L, left = 30 - 5 = 25 L
used = 15 L, left = 30 - 15 = 15 L
NH₃ formed: 10 L

Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones. >C=O and −OH are functional groups of typical ketose, while −CHO and −OH are functional groups of typical aldose. Both aldose and ketose contain a carbonyl group. Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones, which give aldehydic as well as ketonic groups upon hydrolysis.

'Dig into' means to find out her past to know the reality.

We cannot use 'Smoke' as a subject to the verb. The correct way to use it as a noun/subject is by converting it to a gerund.
Gerund is a verb form which functions as a noun. Thus, 'smoke' will get converted to 'smoking.'

In the given sentence, we have to find what will be best suited in place of the underlined part in the sentence.
My opinion for the film is that it will bag the national award.
Here, ''opinion for'' should be replaced with ''opinion about''. ''Opinion about'' means ''any judgement or view about something or place or person''. In this sentence, it is about the National award.
Hence, 'opinion about' is the correct answer. 

As per the given above table, we see

In school A, income from grants = ₹60 thousands.

Income from tuition fee = ₹120 thousands.

School A = 60/120 = 0.5

In school B, income from grants = ₹54 thousands.

Income from tuition fee = ₹60 thousands.

School B = 54/60 = 0.9
In school C, income from grants = ₹120 thousands.

Income from tuition fee = ₹210 thousands.

School C = 120/210 = 0.57

In school D, income from grants =₹42 thousands.

Income from tuition fee =₹90  thousands.

School D = 42/90 = 0.47

In school E, income from grants = ₹55 thousands.

Income from tuition fee = ₹120 thousands.

School E = 55/120 = 0.46 lowest ratio income)

Therefore, school E has the lowest ratio of income by way of grants and tuition fees.

Given that ∠A = 90° in △ABC and P is a point on side AC, we use the Pythagoras' theorem to find AB:

AB² = BC² - AC²
AB² = 10² - 8²
AB² = 100 - 64
AB² = 36
AB = 6 cm

Now, applying Pythagoras' theorem in △ABP:

AP² = BP² - AB²
AP² = 9² - 6²
AP² = 81 - 36
AP² = 45
AP = √45 = 3√5 cm

Final Answer:
AP = 3√5 cm (Option D).