GATE PI 2018

This question tests your understanding of two frequently confused words: "complemented" and "complimented." It's crucial to distinguish their meanings based on context.

Let's define each term precisely:

• Complemented: This is the past tense of "complement." To complement means to complete, enhance, or improve something by adding to it. Think of it as making something better or perfect.
  • Example: The new software complemented the existing system, making it more efficient.
• Complimented: This is the past tense of "compliment." To compliment means to express praise, admiration, or a positive remark to someone. It's about giving positive feedback.
  • Example: She complimented him on his hard work.

Now, let's apply these definitions to the sentence: "The dress _________ her so well that they all immediately _________ her on her appearance."

1. First Blank: "The dress _________ her so well."
   • If the dress fit "so well," it means it enhanced her appearance or made her look better. The word that signifies enhancement or improvement is "complemented."
2. Second Blank: "they all immediately _________ her on her appearance."
   • When people react positively to someone's appearance and express admiration, they are giving praise. The word for expressing praise or admiration is "complimented."

Therefore, the first blank requires "complemented" and the second blank requires "complimented."

This question tests your understanding of vocabulary and sentence structure, specifically the use of conjunctions to establish contrast.

The sentence states, "The judge's standing in the legal community, though shaken by false allegations of wrongdoing, remained _________."

1. "Though" indicates contrast: The word "though" signals that what follows will be contrary to what one might expect given the initial phrase ("shaken by false allegations"). If something is "shaken," you might expect it to be damaged or lessened.
2. "Remained" implies persistence: This word means the state continued or persisted. So, despite being shaken, the standing stayed in a particular condition.

Let's evaluate the options:

• A. undiminished: This means "not reduced or lessened." This fits perfectly. Even though the standing was shaken, it remained not reduced. This highlights the resilience and persistence of the standing despite the challenge.
• B. damaged: If the standing "remained damaged," it would contradict the contrast implied by "though." The sentence structure suggests that the damage did not persist, or was overcome.
• C. illegal: This is irrelevant to a judge's "standing" or reputation.
• D. uncertain: While false allegations might initially cause uncertainty, "remained" with "though shaken" points to a more definitive outcome regarding the standing. "Undiminished" provides a stronger contrast and resolution.

Therefore, "undiminished" best conveys that despite being challenged, the judge's reputation did not decrease.

To find the missing group, we first analyze the pattern in the number of letters and then the pattern in the starting letters of each group.

1. Pattern in the Number of Letters:

• Group 1: BC has $2$ letters.
• Group 2: FGH has $3$ letters.
• Group 3: LMNO has $4$ letters.
  The number of letters increases by $1$ in each subsequent group. Thus, the missing Group 4 should have $4+1=5$ letters.

2. Pattern in the Starting Letters:
Let's identify the position of the starting letter of each group in the alphabet:

• Group 1 starts with B (2nd letter).
• Group 2 starts with F (6th letter).
• Group 3 starts with L (12th letter).

Now, observe the difference in their positions:

• Difference between F and B: $6 - 2 = 4$.
• Difference between L and F: $12 - 6 = 6$.
  The difference itself is increasing by $2$ ($4, 6$). Therefore, the next difference should be $6 + 2 = 8$.
  To find the position of the starting letter of the next group, we add this difference to the position of the last known starting letter (L): $12 + 8 = 20$.
  The $20^{th}$ letter of the English alphabet is T.

Combining both patterns, the missing group must start with T and contain $5$ consecutive letters. These letters are T, U, V, W, X. Therefore, the missing group is TUVWX.

This problem asks us to compare the areas of a circle, a square, and an equilateral triangle when they all have the same perimeter. The fundamental principle here is that for a fixed perimeter, the circle encloses the maximum area among all two-dimensional shapes.

Let's assume the common perimeter for all three shapes is $P$.

1. Circle Area Calculation:
For a circle with radius $r$, the perimeter (circumference) is $C = 2\pi r$.
Since $P = 2\pi r$, we get $r = \frac{P}{2\pi}$.
The area of the circle is $A_{circle} = \pi r^2$.
Substituting $r$:
$A_{circle} = \pi \left( \frac{P}{2\pi} \right)^2 = \pi \left( \frac{P^2}{4\pi^2} \right) = \frac{P^2}{4\pi}$.
Using $\pi \approx 3.14159$, $A_{circle} \approx \frac{P^2}{4 \times 3.14159} \approx \frac{P^2}{12.56636}$.

2. Square Area Calculation:
For a square with side $s$, the perimeter is $P = 4s$.
So, $s = \frac{P}{4}$.
The area of the square is $A_{square} = s^2$.
Substituting $s$:
$A_{square} = \left( \frac{P}{4} \right)^2 = \frac{P^2}{16}$.

3. Equilateral Triangle Area Calculation:
For an equilateral triangle with side $a$, the perimeter is $P = 3a$.
So, $a = \frac{P}{3}$.
The area of an equilateral triangle is $A_{triangle} = \frac{\sqrt{3}}{4} a^2$.
Substituting $a$:
$A_{triangle} = \frac{\sqrt{3}}{4} \left( \frac{P}{3} \right)^2 = \frac{\sqrt{3}}{4} \left( \frac{P^2}{9} \right) = \frac{\sqrt{3}P^2}{36}$.
Using $\sqrt{3} \approx 1.732$, $A_{triangle} \approx \frac{1.732 P^2}{36} \approx \frac{P^2}{20.784}$.

4. Area Comparison:
Let's list the areas and their approximate denominators:

• $A_{circle} = \frac{P^2}{4\pi} \approx \frac{P^2}{12.566}$
• $A_{square} = \frac{P^2}{16}$
• $A_{triangle} = \frac{\sqrt{3}P^2}{36} \approx \frac{P^2}{20.784}$

To compare these fractions, since the numerator ($P^2$) is the same and positive, the fraction with the smallest denominator will have the largest value.
Comparing the denominators: $12.566 < 16 < 20.784$.
Therefore, $A_{circle} > A_{square} > A_{triangle}$.
The circle has the largest area.

To solve this problem, we'll simplify each term of the expression by applying fundamental logarithm properties. The expression is:

$\frac{1}{{1 + {{\log }_u}vw}} + \frac{1}{{1 + {{\log }_v}wu}} + \frac{1}{{1 + {{\log }_w}uv}}$

We'll use three key logarithm properties:

1. Logarithm Identity: $\log_b b = 1$
2. Logarithm Product Rule: $\log_b M + \log_b N = \log_b (MN)$
3. Change of Base Formula (Reciprocal Form): $\frac{1}{{\log_b a}} = \log_a b$

Let's simplify each term step-by-step:

Term 1: Simplifying $\frac{1}{{1 + {{\log }_u}vw}}$
The denominator is $1 + {{\log }_u}vw$.
Using the Logarithm Identity, we can write $1$ as $\log_u u$.
So, $1 + {{\log }_u}vw = \log_u u + {{\log }_u}vw$.
Applying the Logarithm Product Rule, $\log_u u + {{\log }_u}vw = \log_u (u \cdot vw) = \log_u (uvw)$.
Thus, the first term becomes $\frac{1}{{\log_u (uvw)}}$.
Now, using the Change of Base Formula, $\frac{1}{{\log_u (uvw)}} = \log_{uvw} u$.

Term 2: Simplifying $\frac{1}{{1 + {{\log }_v}wu}}$
Similarly, for the second term's denominator: $1 + {{\log }_v}wu$.
Rewrite $1$ as $\log_v v$: $\log_v v + {{\log }_v}wu$.
Apply the Logarithm Product Rule: $\log_v (v \cdot wu) = \log_v (uvw)$.
So, the second term is $\frac{1}{{\log_v (uvw)}}$.
Using the Change of Base Formula, $\frac{1}{{\log_v (uvw)}} = \log_{uvw} v$.

Term 3: Simplifying $\frac{1}{{1 + {{\log }_w}uv}}$
For the third term's denominator: $1 + {{\log }_w}uv$.
Rewrite $1$ as $\log_w w$: $\log_w w + {{\log }_w}uv$.
Apply the Logarithm Product Rule: $\log_w (w \cdot uv) = \log_w (uvw)$.
So, the third term is $\frac{1}{{\log_w (uvw)}}$.
Using the Change of Base Formula, $\frac{1}{{\log_w (uvw)}} = \log_{uvw} w$.

Now, substitute these simplified terms back into the original expression:
$\log_{uvw} u + \log_{uvw} v + \log_{uvw} w$
Since all terms have the same base ($uvw$), we can apply the Logarithm Product Rule:
$\log_{uvw} (u \cdot v \cdot w)$
This simplifies to:
$\log_{uvw} (uvw)$
Finally, using the Logarithm Identity ($\log_b b = 1$), where the base and the argument are both $uvw$:
$\log_{uvw} (uvw) = 1$
The value of the expression is $1$.

This problem requires us to find the number of students who watched all three films, given a specific condition about their viewing patterns.

The crucial condition is: "Each student watched either only one film or all three." This means there are no students who watched exactly two films.

Let's define our variables:

• $N_{Total} = 40$ (Total students)
• $N_A = 13$ (Students who watched film A)
• $N_B = 16$ (Students who watched film B)
• $N_C = 19$ (Students who watched film C)
• $N_{OnlyA}$, $N_{OnlyB}$, $N_{OnlyC}$ (Students who watched only film A, B, or C, respectively)
• $N_{AllThree}$ (Students who watched all three films)

Based on the special condition, the total number of students watching a specific film ($N_A$, $N_B$, $N_C$) is the sum of those who watched only that film and those who watched all three films.
So, we can write:
$N_A = N_{OnlyA} + N_{AllThree}$
$N_B = N_{OnlyB} + N_{AllThree}$
$N_C = N_{OnlyC} + N_{AllThree}$

From these, we can express the "only one film" categories:
$N_{OnlyA} = N_A - N_{AllThree} = 13 - N_{AllThree}$
$N_{OnlyB} = N_B - N_{AllThree} = 16 - N_{AllThree}$
$N_{OnlyC} = N_C - N_{AllThree} = 19 - N_{AllThree}$

The total number of students is the sum of those who watched only one film and those who watched all three films:
$N_{Total} = N_{OnlyA} + N_{OnlyB} + N_{OnlyC} + N_{AllThree}$

Substitute the expressions for $N_{OnlyA}$, $N_{OnlyB}$, and $N_{OnlyC}$ into the total students equation:
$40 = (13 - N_{AllThree}) + (16 - N_{AllThree}) + (19 - N_{AllThree}) + N_{AllThree}$

Combine the constant terms and the $N_{AllThree}$ terms:
$40 = (13 + 16 + 19) + (-N_{AllThree} - N_{AllThree} - N_{AllThree} + N_{AllThree})$
$40 = 48 + (-3 N_{AllThree} + N_{AllThree})$
$40 = 48 - 2 N_{AllThree}$

Now, solve for $N_{AllThree}$:
$2 N_{AllThree} = 48 - 40$
$2 N_{AllThree} = 8$
$N_{AllThree} = \frac{8}{2}$
$N_{AllThree} = 4$

To verify, let's calculate the number of students who watched only one film:
$N_{OnlyA} = 13 - 4 = 9$
$N_{OnlyB} = 16 - 4 = 12$
$N_{OnlyC} = 19 - 4 = 15$

Total students = $N_{OnlyA} + N_{OnlyB} + N_{OnlyC} + N_{AllThree} = 9 + 12 + 15 + 4 = 40$. This matches the given total.

This problem requires us to calculate the total length of a wire, divide it based on given ratios, and then find the sum of the areas enclosed by a square and a circle formed from these pieces.

First, we determine the total wire length. If the wire is bent into a square with an area of $1936 \text{ m}^2$, its side length $s$ is found using $s^2 = 1936$, so $s = \sqrt{1936} = 44$ meters. The total length of the wire is the perimeter of this square, given by $4 \times s = 4 \times 44 = 176$ meters.

Next, the $176 \text{ m}$ wire is cut into two pieces, where the longer piece is thrice the shorter piece. Let the shorter piece be $x$ and the longer piece be $3x$. Their sum is $x + 3x = 176$, which simplifies to $4x = 176$. Solving for $x$, we get $x = \frac{176}{4} = 44$ meters. So, the shorter piece is $44 \text{ m}$, and the longer piece is $3 \times 44 = 132$ meters.

The longer piece ($132 \text{ m}$) is bent into a square. Its perimeter is $132 \text{ m}$. Let its side be $s_1$. Then $4 \times s_1 = 132$, so $s_1 = \frac{132}{4} = 33$ meters. The area of this square is $s_1^2 = 33^2 = 1089 \text{ m}^2$.

The shorter piece ($44 \text{ m}$) is bent into a circle. Its circumference is $44 \text{ m}$. Using the formula $C = 2\pi r$, where $r$ is the radius, we have $2\pi r = 44$. Using $\pi = \frac{22}{7}$, we get $2 \times \frac{22}{7} \times r = 44$, which simplifies to $\frac{44}{7} \times r = 44$. Solving for $r$, we find $r = \frac{44 \times 7}{44} = 7$ meters. The area of this circle is $\pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154 \text{ m}^2$.

Finally, the sum of the areas enclosed by the two pieces is the area of the square plus the area of the circle: $1089 \text{ m}^2 + 154 \text{ m}^2 = 1243 \text{ m}^2$.

This problem tests your understanding of work-rate concepts, specifically how the number of workers (robots), days, and hours per day relate to the total work completed. We'll use the idea of "robot-hours" as a standard unit of work.

First, let's calculate the work done in the initial phase.
Given: $M_1 = 125$ robots, $D_1 = 39$ days, $H_1 = 7$ hours/day.
The work completed ($W_1$) is given by:
$W_1 = M_1 \times D_1 \times H_1$
$W_1 = 125 \times 39 \times 7 = 34125 \text{ robot-hours}$
This $34125$ robot-hours represents $5/7$ of the total contract work.

Next, we find the total work for the entire contract and the remaining work.
Since $34125$ robot-hours is $5/7$ of the total work:
$\text{Total Work} = 34125 \times \frac{7}{5} = (34125 \div 5) \times 7 = 6825 \times 7 = 47775 \text{ robot-hours}$
The remaining work is the total work minus the work done:
$\text{Remaining Work} = 47775 - 34125 = 13650 \text{ robot-hours}$
Alternatively, the remaining work is $1 - \frac{5}{7} = \frac{2}{7}$ of the total work:
$\text{Remaining Work} = \frac{2}{7} \times 47775 = 2 \times 6825 = 13650 \text{ robot-hours}$

Now, let's determine the conditions for completing the remaining work.
Total contract duration is $52$ days. $39$ days have passed, so remaining days ($D_2$) are:
$D_2 = 52 - 39 = 13 \text{ days}$
New hours per day ($H_2$) for each robot are $8$ hours/day.

Let $M_2$ be the total number of robots required to complete the remaining work.
$M_2 \times D_2 \times H_2 = \text{Remaining Work}$
$M_2 \times 13 \times 8 = 13650$
$M_2 \times 104 = 13650$
Solving for $M_2$:
$M_2 = \frac{13650}{104} = 131.25 \text{ robots}$
Since we cannot have a fraction of a robot and the work must be completed on time, we round up to $132$ robots.

Finally, we calculate the additional robots required.
Total robots needed for the second phase ($M_2$) $= 132$.
Initial robots employed $= 125$.
$\text{Additional Robots} = 132 - 125 = 7 \text{ robots}$

This problem involves conditional statements, often written as "If P, then Q." For such statements, remember:

• If P is true and Q is true, the statement holds.
• If P is true and Q is false, the statement is contradicted (or violated).
• If P is false, the statement holds (it's not violated because the "if" condition wasn't met).

Let's test each option against the three given statements:

Statement 1: If the house number is a multiple of 3, then it is a number from 50 to 59.
Statement 2: If the house number is NOT a multiple of 4, then it is a number from 60 to 69.
Statement 3: If the house number is NOT a multiple of 6, then it is a number from 70 to 79.

Evaluating Option A: House Number 54

• Statement 1:
  • Is 54 a multiple of 3? Yes, $54 \div 3 = 18$. (P is True)
  • Is 54 a number from 50 to 59? Yes. (Q is True)
  • Conclusion: Statement 1 holds (True $\rightarrow$ True).
• Statement 2:
  • Is 54 NOT a multiple of 4? Yes, $54 \div 4 = 13$ with a remainder of 2. (P is True)
  • Is 54 a number from 60 to 69? No. (Q is False)
  • Conclusion: Statement 2 is contradicted (True $\rightarrow$ False).
    Since Statement 2 is contradicted, 54 cannot be the house number.

Evaluating Option B: House Number 65

• Statement 1:
  • Is 65 a multiple of 3? No, $65 \div 3 = 21$ with a remainder of 2. (P is False)
  • Conclusion: Statement 1 holds (False P-part).
• Statement 2:
  • Is 65 NOT a multiple of 4? Yes, $65 \div 4 = 16$ with a remainder of 1. (P is True)
  • Is 65 a number from 60 to 69? Yes. (Q is True)
  • Conclusion: Statement 2 holds (True $\rightarrow$ True).
• Statement 3:
  • Is 65 NOT a multiple of 6? Yes, $65 \div 6 = 10$ with a remainder of 5. (P is True)
  • Is 65 a number from 70 to 79? No. (Q is False)
  • Conclusion: Statement 3 is contradicted (True $\rightarrow$ False).
    Since Statement 3 is contradicted, 65 cannot be the house number.

Evaluating Option C: House Number 66

• Statement 1:
  • Is 66 a multiple of 3? Yes, $66 \div 3 = 22$. (P is True)
  • Is 66 a number from 50 to 59? No. (Q is False)
  • Conclusion: Statement 1 is contradicted (True $\rightarrow$ False).
    Since Statement 1 is contradicted, 66 cannot be the house number.

Evaluating Option D: House Number 76

• Statement 1:
  • Is 76 a multiple of 3? No, $76 \div 3 = 25$ with a remainder of 1. (P is False)
  • Conclusion: Statement 1 holds (False P-part).
• Statement 2:
  • Is 76 NOT a multiple of 4? No, $76 \div 4 = 19$. So, 76 IS a multiple of 4. (P is False)
  • Conclusion: Statement 2 holds (False P-part).
• Statement 3:
  • Is 76 NOT a multiple of 6? Yes, $76 \div 6 = 12$ with a remainder of 4. (P is True)
  • Is 76 a number from 70 to 79? Yes. (Q is True)
  • Conclusion: Statement 3 holds (True $\rightarrow$ True).
    All three statements hold for 76. Therefore, 76 is the house number.

Summary:

• 54: Statement 2 is contradicted.
• 65: Statement 3 is contradicted.
• 66: Statement 1 is contradicted.
• 76: All statements hold.

The house number is 76.

For an unbiased coin, each toss is an independent event, meaning prior outcomes don't affect future ones. The probability of getting a Head (H) or a Tail (T) is equal: $P(H) = \frac{1}{2}$ and $P(T) = \frac{1}{2}$.

We need to find the most probable outcome for the last two tosses of the fourth trial (HHHT__ __). Let's list all possible outcomes for two tosses and their probabilities:

• $P(HH) = P(H) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• $P(HT) = P(H) \times P(T) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• $P(TH) = P(T) \times P(H) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• $P(TT) = P(T) \times P(T) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Now, let's evaluate the probability of each given option:

• Option A: "Two T will occur" corresponds to TT, so its probability is $P(TT) = \frac{1}{4}$.
• Option B: "One H and one T will occur" corresponds to HT or TH. Its probability is $P(HT) + P(TH) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.
• Option C: "Two H will occur" corresponds to HH, so its probability is $P(HH) = \frac{1}{4}$.
• Option D: "One H will be followed by one T" corresponds to HT, so its probability is $P(HT) = \frac{1}{4}$.

Comparing these probabilities, $\frac{1}{2}$ (from Option B) is the highest.

The median is the middle value in a dataset when it's arranged in ascending or descending order.

• If the number of terms ($n$) is odd, the median is the $\left[\frac{(n+1)}{2}\right]^{th}$ term.
• If the number of terms ($n$) is even, the median is the average of the $\left[\frac{n}{2}\right]^{th}$ term and the $\left[\frac{n}{2} + 1\right]^{th}$ term.

Calculation:
Given weights: $3, 7, 6, 2, 3, 4$.
First, arrange them in ascending order: $2, 3, 3, 4, 6, 7$.
Here, $n = 6$ (an even number of terms).
The $\left[\frac{n}{2}\right]^{th}$ term is the $\left[\frac{6}{2}\right]^{th} = 3^{rd}$ term, which is $3$.
The $\left[\frac{n}{2} + 1\right]^{th}$ term is the $\left[\frac{6}{2} + 1\right]^{th} = 4^{th}$ term, which is $4$.
Median $= \frac{3 + 4}{2} = 3.5$.
Therefore, the median weight is $3.5$ kg.

The vector triple product $a \times (b \times c)$ is an important operation in vector algebra, often seen in physics and engineering for quantities like torque or the Lorentz force. The result of this operation is always a vector. To simplify $a \times (b \times c)$, we use a standard identity known as the "BAC-CAB" rule, which states:

$a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$

Here, $a \cdot c$ is the dot product of vector $a$ and vector $c$, resulting in a scalar. Similarly, $a \cdot b$ is the dot product of vector $a$ and vector $b$, also a scalar. This identity shows that $a \times (b \times c)$ equals vector $b$ scaled by $(a \cdot c)$, minus vector $c$ scaled by $(a \cdot b)$.

Comparing this identity with the given options:
Option A: $(a \cdot b)b - (a \cdot b)c$ is incorrect as the scalar coefficients do not match.
Option B: $(b \cdot c)a - (a \cdot c)b$ represents $(a \times b) \times c$, which is generally different from $a \times (b \times c)$. The order of operations in cross products is critical.
Option C: $(a \cdot c)b - (a \cdot b)c$ perfectly matches the BAC-CAB rule.
Option D: $(b \cdot c)a - (a \cdot b)c$ is incorrect as its terms and scalar coefficients do not align with the established formula for $a \times (b \times c)$.

Therefore, the correct expression for the vector triple product $a \times (b \times c)$ is $(a \cdot c)b - (a \cdot b)c$.

Let's analyze the function $y = 5 |x|$ at $x = 0$ for continuity and differentiability.

Continuity Check at $x = 0$:
A function $f(x)$ is continuous at $x = a$ if $f(a)$ exists, $\lim_{x \to a} f(x)$ exists, and $\lim_{x \to a} f(x) = f(a)$.

1. Function Definition: Evaluate $y$ at $x = 0$.
   $y(0) = 5 |0| = 5 \times 0 = 0$
   The function is defined at $x = 0$.

2. Limit Existence: Evaluate the limit as $x$ approaches 0.

   • Right-hand limit ($x \to 0^+$): When $x > 0$, $|x| = x$.
     $\lim_{x \to 0^+} 5 |x| = \lim_{x \to 0^+} 5x = 5 \times 0 = 0$
   • Left-hand limit ($x \to 0^-$): When $x < 0$, $|x| = -x$.
     $\lim_{x \to 0^-} 5 |x| = \lim_{x \to 0^-} 5(-x) = 5 \times 0 = 0$
     Since the left-hand limit ($0$) and the right-hand limit ($0$) are equal, the limit exists:
     $\lim_{x \to 0} 5 |x| = 0$

3. Limit equals Function Value:
   We found $\lim_{x \to 0} 5 |x| = 0$ and $y(0) = 0$. Since $\lim_{x \to 0} 5 |x| = y(0)$, the function is continuous at $x = 0$.

Differentiability Check at $x = 0$:
A function $f(x)$ is differentiable at $x = a$ if the derivative $f'(a)$ exists, defined as $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$.
For $y = 5 |x|$ at $x = 0$, with $f(0) = 0$, we evaluate:
$y'(0) = \lim_{h \to 0} \frac{5 |0+h| - 0}{h} = \lim_{h \to 0} \frac{5 |h|}{h}$
We examine the one-sided derivatives:

• Right-hand derivative ($h \to 0^+$): Here, $|h| = h$.
  $\lim_{h \to 0^+} \frac{5 |h|}{h} = \lim_{h \to 0^+} \frac{5h}{h} = \lim_{h \to 0^+} 5 = 5$
• Left-hand derivative ($h \to 0^-$): Here, $|h| = -h$.
  $\lim_{h \to 0^-} \frac{5 |h|}{h} = \lim_{h \to 0^-} \frac{5(-h)}{h} = \lim_{h \to 0^-} -5 = -5$
  Since the right-hand derivative ($5$) is not equal to the left-hand derivative ($-5$), the limit $\lim_{h \to 0} \frac{5 |h|}{h}$ does not exist. Therefore, the function $y = 5 |x|$ is not differentiable at $x = 0$.

Conclusion:
The function $y = 5 |x|$ is continuous but not differentiable at $x = 0$.

To find the y-component of the force, we use the formula for the magnitude of a force in a 2D coordinate system: $F = \sqrt{F_x^2 + F_y^2}$.
Given $F = 10 \text{ kN}$ and $F_x = -6 \text{ kN}$, we substitute these values:
$10 = \sqrt{(-6)^2 + F_y^2}$
Squaring both sides gives $100 = 36 + F_y^2$, which leads to $F_y^2 = 100 - 36 = 64$.
Therefore, $F_y = \pm 8 \text{ kN}$.

When austenite, a single solid phase (FCC structure), transforms into two different solid phases, ferrite (BCC structure) and cementite ($Fe_3C$), upon cooling, this specific type of reaction is known as a eutectoid transformation. A eutectoid reaction is defined as a solid phase transforming directly into two other solid phases upon cooling, represented as $\text{Solid 1} \rightarrow \text{Solid 2} + \text{Solid 3}$. This reaction is crucial in the iron-carbon phase diagram, occurring at $727^\circ C$ ($1341^\circ F$) and $0.76\%$ carbon, where austenite converts into pearlite (a lamellar mixture of ferrite and cementite). Therefore, the decomposition of austenite into ferrite and cementite is a eutectoid reaction.

Let's match the geometric tolerances with their standard symbols.

• P: Concentricity is represented by the symbol of two concentric circles, which is 3.
• Q: Parallelism is indicated by the symbol with two parallel lines, which is 1.
• R: Circularity is denoted by a simple circle symbol, which is 4.
• S: Perpendicularity is shown by the symbol for perpendicular lines, which is 2.

Therefore, the correct matching is P-3, Q-1, R-4, S-2.

Light interference happens when two or more light waves combine, creating a new pattern with bright (constructive interference) and dark (destructive interference) areas, often seen as fringes. Let's analyze the given instruments:

• Optical flat: This precision instrument uses a highly flat surface to check the flatness of other surfaces. When monochromatic light illuminates an optical flat placed on a test surface, the light reflects from the bottom of the flat and the top of the test surface. These reflected waves interfere due to the thin air gap between them, producing interference fringes that visualize flatness deviations (similar to Newton's rings). This directly uses light interference.

• Autocollimator: This device measures angular alignment using reflection and image formation principles. It projects a pattern and measures the reflected image's deviation, but it does not rely on light interference.

• Optical projector: Also known as a profile projector, it magnifies and displays the profile of small parts on a screen using lenses and light sources for projection. Light interference is not its operating principle.

• Coordinate measuring machine (CMM): A CMM measures physical dimensions using a probe and precise positioning systems (like linear encoders). Its operation is based on mechanics, geometry, and accurate sensing, not light interference.

Therefore, the optical flat is the instrument that fundamentally utilizes the principle of interference of light for its operation.

In competitive exams like GATE, understanding process charts and their symbols is crucial for topics like Industrial Engineering. Process charts help visualize workflows for analysis. The American Society of Mechanical Engineers (ASME) established a common set of five symbols for these charts.

Let's look at the standard ASME symbols:

• Operation: Represents an action that changes the product's location or condition.
• Storage: Indicates a waiting stage for materials or finished goods.
• Delay: Occurs when the process stops, and an item waits for the next event.
• Transport: Shows the movement of an item from one place to another.
• Inspection: Denotes checking the quantity or quality of an item.

The symbol given in the question (which is a square) specifically represents Inspection. This is an act of checking for the correctness of the quantity or the quality of the items, for example, gauging a piston pin. Therefore, from the common ASME symbols, this symbol clearly corresponds to Inspection.

Control charts help us monitor processes to spot unusual variations. For data that involves counting or classifying items, we use "Attribute Control Charts." When we want to track the fraction of defective items in a sample, we use a P-chart. The P-chart specifically monitors the proportion of items that don't meet specifications, calculated as: $p = \frac{\text{Number of Defectives}}{\text{Total Number Inspected}}$.

Let's quickly see why the other options are incorrect for this specific scenario:

• Range chart (R-chart) and Mean chart (X-bar chart) are "Variable Control Charts" used for continuous measurements (like length or weight), not for counts of defective items.
• C-Chart monitors the number of defects per unit, not the fraction of defective items. For instance, a C-chart tracks how many flaws are on one product, while a P-chart tracks the proportion of products that have any flaws.

Therefore, for monitoring the fraction of defective items, the P-chart is the correct choice.

For products produced in very large quantities where machines and auxiliary facilities are arranged strictly according to the processing sequence, a Product Layout is used. This layout, also known as a line layout or assembly line layout, dedicates machines and equipment to a specific product's operations, ensuring a predictable, linear material flow. This design optimizes efficiency and speed for high-volume, low-variety production by minimizing material handling, work-in-process inventory, and production time.

In contrast, a Process Layout groups machines by function and is suitable for low-volume, high-variety production, not a single high-volume product's fixed sequence. A Fixed Position Layout is for immobile products, bringing resources to the product, which is not applicable for items produced in very large quantities that move through a process. A Cellular Layout groups dissimilar machines for families of parts with similar processing requirements, differing from the single product, extremely high-volume scenario described.

The casting process involves pouring molten metal into a mold and allowing it to solidify. Defects like Hot tear, Porosity, and Blister are commonly associated with this process. A hot tear is a crack occurring during cooling due to thermal stresses when the casting is still weak. Porosity refers to voids from trapped gases or shrinkage during solidification. A blister is a surface defect (raised area) caused by trapped gas from vaporized moisture on the mold surface.

A Central burst, however, is a defect typically found in products like pipes or bars, appearing as a crack along the central axis, and is characteristic of mechanical working processes such as extrusion or drawing, not the solidification phase of casting.

Oxy-acetylene welding flame type is determined by the volume ratio of oxygen ($\text{O}_2$) to acetylene ($\text{C}_2\text{H}_2$).
\begin{itemize}
\item Neutral flame: Ratio close to $1:1$ ($\text{O}_2 : \text{C}_2\text{H}_2$).
\item Carburizing flame: Acetylene in excess ($\text{O}_2 : \text{C}_2\text{H}_2 < 1:1$).
\item Oxidizing flame: Oxygen in excess ($\text{O}_2 : \text{C}_2\text{H}_2 > 1:1$).
\end{itemize}
Given the oxygen to acetylene ratio is $1.5:1$, which means $\text{O}_2 > \text{C}_2\text{H}_2$. This indicates an excess of oxygen, thus resulting in an oxidizing flame.

The Six Sigma standard aims for a defect rate of no more than 3.4 defects per million opportunities (DPMO). This means that for every million units produced, there should be fewer than 3.4 defective units.

Each firm manufactures 100 lakhs components, which is $10,000,000$ components. We need to calculate the DPMO for each firm and compare it with the $3.4$ DPMO threshold.

The formula for DPMO is:
$\text{DPMO} = \frac{\text{Number of defective components}}{\text{Total components manufactured}} \times 1,000,000$

For Firm P:
$\text{DPMO}_P = \frac{30}{10,000,000} \times 1,000,000 = 3$
Since $3 < 3.4$, Firm P conforms to the Six Sigma standard.

For Firm Q:
$\text{DPMO}_Q = \frac{42}{10,000,000} \times 1,000,000 = 4.2$
Since $4.2 > 3.4$, Firm Q does not conform to the Six Sigma standard.

For Firm R:
$\text{DPMO}_R = \frac{47}{10,000,000} \times 1,000,000 = 4.7$
Since $4.7 > 3.4$, Firm R does not conform to the Six Sigma standard.

Therefore, only Firm P meets the Six Sigma standard.

We need to select the best process for drilling a very small hole ($0.5$ mm diameter) that is also very deep ($30$ mm) in mild steel. This means we're looking for a process that can achieve a high aspect ratio, which is the ratio of depth to diameter.

Let's calculate the aspect ratio for this hole:
Aspect Ratio ($AR$) = $\frac{\text{Hole Depth (L)}}{\text{Hole Diameter (d)}} = \frac{30 \text{ mm}}{0.5 \text{ mm}} = 60$.

Now, let's evaluate the given machining processes for creating such a high aspect ratio hole:

1. Electrochemical Machining (ECM):

   • Principle: ECM removes material through an electrochemical dissolution process (like reverse electroplating) where the workpiece is the anode, and a shaped tool is the cathode, with an electrolyte flowing between them. There's no physical contact between the tool and the workpiece.
   • Suitability: ECM is excellent for high aspect ratio holes because there's no tool wear, and the material removal rate isn't significantly affected by depth. It works well with electrically conductive materials like mild steel, produces burr-free surfaces, and avoids thermal distortion. Its precision is high enough for a $0.5$ mm diameter.

2. Chemical Machining (CM):

   • Principle: CM uses chemical etchants to selectively dissolve material from the workpiece.
   • Suitability: While CM can create fine features, achieving an aspect ratio of 60 for a $0.5$ mm diameter hole is extremely challenging. Control over etching depth and prevention of over-etching or undercutting at such small scales and depths is difficult.

3. Abrasive Jet Machining (AJM):

   • Principle: AJM uses a high-velocity stream of abrasive particles mixed with a gas to erode material.
   • Suitability: For a $0.5$ mm diameter and $30$ mm deep hole, AJM struggles with precision. The abrasive stream tends to widen the hole at greater depths (trailing edge erosion), making it difficult to maintain the desired diameter and accuracy throughout.

4. Plasma Arc Machining (PAM):

   • Principle: PAM uses a high-temperature plasma jet to melt and eject material.
   • Suitability: PAM is generally used for cutting thicker materials and is much less precise than ECM for small-diameter holes. It creates a significant Heat Affected Zone (HAZ) and cannot achieve the required precision or smooth finish for a $0.5$ mm hole; its effective cutting width (kerf) is typically larger.

Considering the high aspect ratio of 60, the requirement for high precision ($0.5$ mm diameter), and working with mild steel, Electrochemical Machining (ECM) is the most suitable process due to its ability to handle deep, narrow features without tool wear or thermal effects, while maintaining high accuracy and surface finish.

Let's break down these processes to understand which one doesn't produce powders.

Atomization is a method where molten metal is sprayed using high-velocity gas or liquid jets, or centrifugal force, to create fine metal droplets that solidify into powder.
Ball milling involves grinding materials in a rotating drum with grinding media (balls), which break down larger particles into finer powder through impact and attrition.
Electrolysis can produce metal powders by depositing metal ions from an electrolyte onto a cathode, which are then collected and processed into powder.

Sintering, on the other hand, is a heat treatment process that takes existing powder particles and bonds them together into a solid, denser object. It involves heating a compacted powder form to a high temperature, below its melting point, allowing atomic diffusion to fuse the particles. Therefore, sintering is a consolidation process for powders, not a production process.

The question asks to identify the process where molten thermoplastic is forced between rolls to produce thin sheets. Calendering is a mechanical process that uses heat and pressure from a series of rotating rollers. In calendering, molten thermoplastic is fed into the nip (the gap between rollers) and progressively squeezed to form a continuous thin sheet of controlled thickness. This precisely matches the description.

Blow moulding creates hollow plastic objects by inflating a molten plastic tube inside a mold. Compression moulding involves placing plastic material into a heated mold cavity and then compressing it to fill the cavity. Extrusion forces molten material through a shaped die to create continuous profiles; while film extrusion exists, the specific action of passing material between rolls to form thin sheets is characteristic of calendering, which explicitly uses the 'nip' action of rollers to control thickness. Therefore, calendering is the correct answer.

A fundamental property of matrices states that the sum of the eigenvalues of a matrix is equal to its trace (the sum of its diagonal elements). For a $3 \times 3$ matrix

, the trace is $tr(A) = a_{11} + a_{22} + a_{33}$. If $\lambda_1, \lambda_2, \lambda_3$ are the eigenvalues, then $\lambda_1 + \lambda_2 + \lambda_3 = tr(A)$. Given the diagonal elements $a_{11} = -10$, $a_{22} = 5$, $a_{33} = 0$, and two eigenvalues $\lambda_1 = -15$, $\lambda_2 = -15$, we can find the third eigenvalue $\lambda_3$.
$(-15) + (-15) + \lambda_3 = (-10) + 5 + 0$
$-30 + \lambda_3 = -5$
$\lambda_3 = 25$

Okay class, let's break down this probability question.

The core concept here is the Addition Theorem of Probability. This theorem helps us find the probability of at least one of two events ($A$ or $B$) happening. It's given by the formula:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
where $P(A \cup B)$ means the probability of event $A$ or event $B$ (or both) occurring, and $P(A \cap B)$ means the probability of both event $A$ and event $B$ occurring simultaneously.

We are given the following probabilities:
$P(F) = 0.3$ (Probability of event $F$ occurring)
$P(G) = 0.4$ (Probability of event $G$ occurring)
$P(F \cap G) = 0.2$ (Probability of both $F$ and $G$ occurring simultaneously)

We need to find the probability of occurrence of at least one event, which is $P(F \cup G)$. Using the Addition Theorem of Probability:
$P(F \cup G) = P(F) + P(G) - P(F \cap G)$
Plugging in the given values:
$P(F \cup G) = 0.3 + 0.4 - 0.2$
$P(F \cup G) = 0.5$
So, the probability of at least one event occurring is $0.5$.

The kinetic energy of a flywheel, which helps regulate engine speed variations, is given by the formula $KE = \frac{1}{2}I{\omega ^2}$. For a solid circular disc, the moment of inertia $I$ is calculated as $0.5 \times mR^2$.

Given:
Mass ($m$) = $10 \, \text{kg}$
Radius ($R$) = $100 \, \text{mm} = 0.1 \, \text{m}$
Angular velocity ($\omega$) = $20 \, \text{rad/s}$

First, calculate the moment of inertia ($I$):
$I = 0.5 \times mR^2$
$I = 0.5 \times 10 \times (0.1)^2$
$I = 0.05 \, \text{kg.m}^2$

Now, substitute this into the kinetic energy formula:
$KE = \frac{1}{2}I{\omega ^2}$
$KE = \frac{1}{2} \times 0.05 \times (20)^2$
$KE = 10 \, \text{Joule}$

Concept: For any object to float in a fluid, its weight must be perfectly balanced by the buoyant force acting on it. This is known as the Principle of Flotation. Mathematically, this means $W = F_B$, where the buoyant force $F_B = \rho_f \times g \times V$, with $\rho_f$ being the density of the fluid and $V$ the volume of the fluid displaced (which is the immersed volume of the object).

Calculation: We are given a cylindrical wooden block with length $l = 50 \text{ cm}$, density $\rho_c = 800 \text{ kg/m}^3$, floating in water with density $\rho_w = 1000 \text{ kg/m}^3$. Let $y$ be the depth of immersion (in cm) and $A$ be the cross-sectional area of the cylinder (in cm$^2$).
Applying the principle of flotation:
Weight of cylinder = Weight of displaced water volume
$\rho_c \times A \times l \times g = \rho_w \times A \times y \times g$
We can cancel $A$ and $g$ from both sides, simplifying the equation to:
$y = \frac{\rho_c l}{\rho_w}$
Substituting the given values:
$y = \frac{800 \times 50}{1000} = 40 \text{ cm}$
Therefore, the submerged depth of the cylinder is $40 \text{ cm}$.

Let's break down this reliability problem step-by-step.

Key Concepts:

• Series Connection: If components are connected in series, the system works only if all components work. The overall reliability is the product of individual reliabilities: $R_s = R_1 \times R_2 \times \ldots \times R_n$
• Parallel Connection (with identical standby units): If components are connected in parallel, the system works if at least one component works. If you have $n$ identical units in parallel, the overall reliability is: $R_p = 1 - \left[ (1 - R_1)(1 - R_2) \ldots (1 - R_n) \right]$

Calculation:

Given:

• Reliability of P, $R_P = 0.97$
• Reliability of Q, $R_Q = 0.86$
• Reliability of R, $R_R = 0.93$

Initially, P, Q, and R are in series. To improve reliability, two additional stand-by units of Q are added. This means we now have three Q units (the original plus two stand-bys) connected in parallel.

1. Reliability of the Q subsystem ($R_{PQ}$): Since the three Q units are in parallel, the reliability of this Q subsystem is:
   $R_{PQ} = 1 - \left[ (1 - R_Q)(1 - R_Q)(1 - R_Q) \right]$
   $R_{PQ} = 1 - \left[ (1 - 0.86)(1 - 0.86)(1 - 0.86) \right]$
   $R_{PQ} = 1 - \left[ (0.14)(0.14)(0.14) \right] = 1 - [0.002744] = 0.997256$
   Rounding to four decimal places as seen in the original: $R_{PQ} = 0.9973$.

2. Overall Reliability of the Machine ($R_s$): Now, the machine consists of P, the improved Q subsystem ($R_{PQ}$), and R connected in series.
   $R_s = R_P \times R_{PQ} \times R_R$
   $R_s = 0.97 \times 0.9973 \times 0.93$
   $R_s = 0.89930985 \approx 0.899$