UGEE REAP Mock Test- 2 - JEE MCQ

Initial speeds: Train = 45 km/h, Motorbike = 70 km/h.
When the train passes point A completely, its rear is at 0 km, front at 0.3 km, midpoint at 0.15 km. Motorbike starts at 0 km.
Time to reach midpoint (0.15 km relative to train’s rear):
Relative speed = 70 - 45 = 25 km/h
Time = 0.15 / 25 = 0.006 hours
Distance by motorbike = 70 × 0.006 = 0.42 km
At t = 0.006 h, speeds change to 60 km/h (train) and 65 km/h (motorbike). Train front at 0.57 km, motorbike at 0.42 km.
Time to cover remaining 0.15 km:
Relative speed = 65 - 60 = 5 km/h
Time = 0.15 / 5 = 0.03 hours
Distance = 65 × 0.03 = 1.95 km
Total distance = 0.42 + 1.95 = 2.37 km.

To determine the maximum amount Manoj is willing to pay each time to throw the dice, given that he wants to make an average profit of Rs 5 per throw, we first need to calculate his expected earnings per throw.
Calculation of Expected Earnings
1. Dice Outcomes:

• There are 6 faces on a fair die, each equally likely with probability

2. Earnings Based on Dice Outcomes:

• If the number is odd (1, 3, 5), Manoj earns thrice the number.
• If the number is even (2, 4, 6), Manoj earns twice the number.

Expected Earnings Calculation:

Expected earnings = ∑(probability of outcome × earnings for that outcome)
For odd numbers (1, 3, 5):
Earnings 

For even numbers (2, 4, 6):

simplify

Profit and Cost Calculation
Manoj wants an average profit of Rs 5 per throw.
Profit = Expected earnings - Cost per throw
Given the profit of Rs 5:
5 = 8.5 - Cost per throw
Cost per throw = 8.5 -  5
Cost per throw = 3.5
Conclusion
The maximum amount Manoj is willing to pay each time to throw the dice, in order to make an
average profit of Rs 5 per throw, is Rs 3.50.
Therefore. the correct answer is: Rs 3.50.

To determine the probability of at least one tail showing up when two biased coins are tossed, we first need to calculate the probabilities for each coin individually and then use these to find the combined probability.
Coin 1:

• Given that heads occur four times as frequently as tails.
• Let P(T₁) be the probability of getting tails and P(H₁) be the probability of getting heads for the first coin.

Since heads occur four times as frequently as tails:
P(H₁) 4P(T₁)
We know that the sum of the probabilities for heads and tails must be I:
P(H₁) + P(T₁) = 1
Substitute P(H₁) + 4P(T₁) into the equation:
4P(T₁) + P(T₁) = 1
5P(T₁) = 1
P(T₁) = 1/5 = 0.2
Therefore,
P(H₁) = (1 - P(T₁) = 1 - 0.2 = 0.8
Coin 2:

• Given that heads occur 65% of the time.
• Thus, P(H₂) = 0.65 and P(T₂) = 1 - P(H₂) = 1 - 0.65 = 0.35

Combined Probability:
To find the probability of at least one tail showing up when both coins are tossed, we use the complement of the probability that no tails show up (i.e., both coins show heads).
1. Probability that both coins show heads:

2. Probability of at least one tail:

Final Answer:
The probability of at least one tail turning up is: 48%
So, the correct option is (c) 48%.

To find the number of positive integer solutions to abc = 30,
where 30 = 2¹ × 3¹ × 5¹, assign the exponent of each prime to one of a, b, c
For prime 2 (exponent 1), there are 3 choices (give to a, b, or c); similarly for 3 and 5.
Total solutions = 3 × 3 × 3 = 27.

Sum of the 10th, 20th, 30th terms of an AP is equal to the 58th term. Hence, 3a + 57d = a + 57d
So, a = 0 and d can be anything not equal to zero. Now, we have to find out the ratio 3a + 57d and 3a + 27d, or, we have to find out the ratio 57d : 27d= 19 : 9.

Shishu’s rate: 1/30​ per hour (since he takes 30 hours).
Mayank’s rate: 1.5 × 1/30 = 1/20  per hour.
Shishu’s work in 12 hours: 12 × 1/30 ​ = 2/5.
Remaining work: 1 − 2/5 = 3/5.
Combined rate: 1/30 + 1/20 = 2/60 + 3/60 = 5/60 = 1/12 per hour.
Time together: 

Using Ampère's Law, we can analyse the magnetic field in a thin-walled pipe with a surface current. Here are the key points to consider:

• The magnetic field inside the hollow region (where _r < _a) is
• Zero because no current is enclosed within that area.

Thus, we conclude that B = 0 everywhere inside the pipe.

The magnetic field inside the inner conductor increases linearly with radius, decreases as 1/r between the conductors, and becomes zero outside. The equations for the magnetic field B in different regions are as follows:

• For r < a: B = ^{μ₀ I r} / (2π a²)
• For a ≤ r < b: B = (μ₀ I) / (2π r)
• For r ≥ b: B = 0

In transition II,
E₂ = -3.4 ev, E₄ = -0.85 eV,

The Balmer series for the hydrogen atom refers to the set of spectral lines that are emitted when an electron transitions from a higher energy level (n ≥ 3) to the second energy level (n = 2). These transitions result in the emission of light in the visible spectrum.
The options given are:
(a) if we measure the frequencies of light emitted when an excited atom falls to the ground state
(b) if we measure the frequencies of light emitted due to transitions between excited states and the first excited state
(c) in any transition in a H-atom
(d) none of these
The correct observation for the Balmer series is:
(b) if we measure the frequencies of light emitted due to transitions between excited states and
the first excited state (which is n = 2)
Thus, the correct answer is: (b)

First, calculate the differences:

• From 4 to 6: 6 - 4 = 2
• From 6 to 9: 9 - 6 = 3
• From 9 to 14: 14 - 9 = 5

The differences are 2, 3, and 5. These numbers suggest a pattern, as they resemble terms from a known sequence. One possibility is that the differences follow the Fibonacci sequence, where each number is the sum of the two preceding ones: 1, 1, 2, 3, 5, 8, 13, and so on. Notice that 2, 3, and 5 appear in this sequence (the 3rd, 4th, and 5th Fibonacci numbers if we index starting with F₁ = 1, F₂ = 1, F₃ = 2, F₄ = 3, F₅ = 5). If the differences correspond to consecutive Fibonacci numbers starting from F₃ = 2, the sequence of differences would be:

• F₃ = 2
• F₄ = 3
• F₅ = 5
• F₆ = 8 (since 5 + 3 = 8)

Applying this pattern:

• 4 + 2 = 6
• 6 + 3 = 9
• 9 + 5 = 14
• 14 + 8 = 22

Each number is calculated by taking half of the previous number and then adding 10. Starting with 36:

• 36 → (36/2) + 10 = 28
• 28 → (28/2) + 10 = 24
• 24 → (24/2) + 10 = 22
• 22 → (22/2) + 10 = 21

∴ VX is the code for TV.

∴ CALLER is coded DBMMFS.

As we observe that the letters and their corresponding codes are given in order, i.e., the code for C is $, R is Ω and so on.
Hence, the code for ACCIDENT is '#$$86θ74'.

1. Total area of the circle:
Since the radius of the circle is 1 unit, the total area A of the circle is:
A = π·1² = π
2. Total area of the seven sectors:
The total area of the seven sectors S₁,S₂, .....,S₇ is given to be 1/8 of the area of the circle:
Total area of the sectors 
3. Relationship between the areas of the sectors:
It is given that the area of the j-th sector is twice that of the (j - 1)-th sector for j = 2, 3,....,7.
Let the area of S₁ be A₁. Then:
A₂ = 2A₁, A₃ = 2A₂ = 4A₁, A₄ = 2A₃ = 8A₁, A₅, = 2A₄ = 16A₁, A₆ = 2A 
4. Sum of the areas of the sectors:
The sum of the areas of all seven sectors is: 
A₁ + A₂ + A₃ + A₄ + A₅ + A₆ + A₇ = A₁ + 2A₁ + 4A₁ + 8A₁ + 16A₁ + 32A₁ + 6A
Since this total area is π/8:
 
5. Finding the angle subtended by the arc of S₁:
The area of a sector is given by  where r is the radius and θ is the angle in radians.
Since r = 1

To determine the shortest among A, B, C, D, E, F, G, H:
Each has a unique height.
E is second tallest (does not play cricket).
F > A, D but < H, B.
E > B.
G < D but > A.
H is fourth tallest.
Shortest does not play cricket.
Height order (tallest to shortest):

• C (tallest, no football restriction).
• E (second tallest).
• B (E > B, B > F).
• H (given fourth).
• F (F > A, D, < H, B).
• D (F > D, D > G).
• G (D > G > A).
• A (G > A, shortest).

A does not play cricket (consistent with shortest).
Thus, A is the shortest.

The section of the Venn diagram representing prime and not even is shown below.

There are 3 numbers in the relevant section out of a possible 10 numbers altogether. The probability, as a fraction, is 3/10.

All mothers are women and some mothers and some women may be engineers.

Given: Total population of state A = 3000, 12% below poverty line, male-to-female ratio above poverty line = 4:3.

Population above poverty line = 3000 * (100 - 12)/100 = 3000 * 0.88 = 2640.

Male-to-female ratio = 4:3, total parts = 4 + 3 = 7, female fraction = 3/7.

Number of females = 2640 * (3/7) ≈ 1131.

Nearest option: 1200 (A).

1. Statement Analysis:

• Statement (i): 'No manager is a leader' implies that the set of managers and leaders are disjoint.
• Statement (ii): 'All leaders are executives' means every leader is within the set of executives.

2. Visual Representation:

Imagine three circles:

• Managers (M)
• Leaders (L)
• Executives (E)

From statement (i), M and L do not overlap. From statement (ii), L is entirely within E.

3. Conclusion Evaluation:

• Conclusion (i): 'No manager is an executive' does not necessarily follow because managers could potentially be executives without being leaders.
• Conclusion (ii): 'No executive is a manager' also doesn't follow since an executive could be a manager who isn't a leader.

4. Logical Reasoning:

The premises do not provide sufficient information to establish any direct relationship between managers and executives beyond their relation to leaders.

Therefore, neither conclusion can be definitively drawn from the given statements. Thus, the correct answer is that neither conclusion follows.

• Paraphrase is to rewrite something in a more concise manner, in order to make the meaning clear. Hence this is the right answer.
• Paradox is a contradiction
• Paradigm is a new view point on something
• Paraffin is a flammable oily liquid used in medicine / or wax

The sentence under quotes is denoting figures and hence this sentence is figurative.

Morphology:
The present tense of each verb starts with m-. To make it past tense, change the m- to n-.
Word order:
The basic word order of the sentence is Verb-Object-Subject.
Adjectives go after nouns.
"The" goes before the noun.
Demonstratives ("this" and "that") appear twice, once before and once after the noun. If there is an adjective, it appears inside of the demonstrative pair with the noun (e.g., "this frog green this").

Morphology:
The present tense of each verb starts with m-. To make it past tense, change the m- to n-.
Word order:
The basic word order of the sentence is Verb-Object-Subject.
Adjectives go after nouns.
"The" goes before the noun.
Demonstratives ("this" and "that") appear twice, once before and once after the noun. If there is an I adjective, it appears inside of the demonstrative pair with the noun (e.g., "this frog green this").

Robert Millikan conducted the oil drop experiment in 1909, accurately measuring the charge of an electron.

J.J. Thomson discovered the electron itself in 1897 but did not measure its charge.

The photoelectric effect occurs when light of a high enough frequency hits a photosensitive material, causing it to emit electrons. Here’s how it works:

• A negative electrode (C) emits electrons when exposed to light.
• These electrons are attracted to a positive electrode (A), creating a flow of electric current.
• The current, known as photoelectric current, depends on the number of electrons emitted.

Historically, Heinrich Hertz discovered that when ultraviolet light illuminated the cathode, it increased the ease with which high voltage sparks passed between metal electrodes. This was due to the emission of electrons caused by the light.

In electric discharge through gases at low pressure, a coloured glow appears because:

• Electrons from the cathode collide with gas atoms.
• This collision excites electrons in the gas atoms, causing the glow.

This demonstrates the principle of electron excitation in gases.

As,
R + 3 = U
S + 3 = V
Q + 3 = T
Similarly,
X + 3 = A
Y + 3 = B
W + 3 = Z
Hence, ABZ is the correct answer.