TS EAMCET 2026 May 9 Shift 2 Engineering Question Paper with Solution PDF

Step 1: Understanding the Question:

We need to find the percentage error in the area of a circle given the percentage error in its radius.


Step 2: Key Formula or Approach:

The area \( A \) of a circle is related to its radius \( r \) by the formula \( A = \pi r^2 \).

Taking the natural logarithm on both sides gives \( \ln(A) = \ln(\pi) + 2 \ln(r) \).

Differentiating both sides gives the relative error: \( \frac{dA}{A} = 2 \frac{dr}{r} \).

Multiplying by 100 gives the percentage error: \( \left(\frac{dA}{A} \times 100\right) = 2 \times \left(\frac{dr}{r} \times 100\right) \).


Step 3: Detailed Explanation:

Given that the percentage error in the radius is \( \left(\frac{dr}{r} \times 100\right) = 3 \).

Substitute this value into the percentage error formula for the area:
\[ Percentage error in area = 2 \times 3 = 6 \]

Step 4: Final Answer:

The percentage error in the area is 6.
Quick Tip: For any physical quantity dependent on a power \( n \), i.e., \( y = kx^n \), the percentage error is approximately \( n \times (percentage error in x) \) for small errors.

Step 1: Understanding the Question:

We are asked to find the total area of the region bounded by the curve \( y = x^3 \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 4 \).


Step 2: Key Formula or Approach:

The area \( A \) bounded by a curve \( y = f(x) \) and the x-axis from \( x = a \) to \( x = b \) is given by the integral of the absolute value of the function:
\[ A = \int_{a}^{b} |f(x)| dx \]

Step 3: Detailed Explanation:

The curve \( y = x^3 \) is negative for \( x \in [-2, 0] \) and positive for \( x \in [0, 4] \).

Therefore, the total area is the sum of two definite integrals:
\[ A = \int_{-2}^{0} (-x^3) dx + \int_{0}^{4} x^3 dx \]
Now, compute each integral separately.

For the first interval \( [-2, 0] \):
\[ \int_{-2}^{0} (-x^3) dx = \left[ -\frac{x^4}{4} \right]_{-2}^{0} = 0 - \left( -\frac{(-2)^4}{4} \right) = 0 - \left( -\frac{16}{4} \right) = 4 \]
For the second interval \( [0, 4] \):
\[ \int_{0}^{4} x^3 dx = \left[ \frac{x^4}{4} \right]_{0}^{4} = \frac{4^4}{4} - 0 = \frac{256}{4} = 64 \]
Total Area = \( 4 + 64 = 68 \).


Step 4: Final Answer:

The area of the bounded region is 68.
Quick Tip: When finding the area bounded by a curve and the x-axis, always check for the roots of the function within the given interval to ensure you split the integral and take the absolute value for negative regions.

Step 1: Understanding the Question:

The problem asks for the possible number of black balls in the first box based on conditional probability, where the probability of selecting a specific box given the color of the drawn ball is known.


Step 2: Key Formula or Approach:

We use Bayes' Theorem to find the probability of the ball being drawn from the second box given it is black:
\[ P(B_2 | Black) = \frac{P(Black | B_2) \cdot P(B_2)}{P(Black | B_1) \cdot P(B_1) + P(Black | B_2) \cdot P(B_2)} \]

Step 3: Detailed Explanation:

Let \( B_1 \) and \( B_2 \) be the events of choosing the first and second box, respectively.

Since the box is chosen at random, \( P(B_1) = P(B_2) = \frac{1}{2} \).

Let \( b_1 \) and \( b_2 \) be the number of black balls in the first and second box, respectively.

The probability of drawing a black ball from the first box is \( P(Black | B_1) = \frac{b_1}{10} \).

The probability of drawing a black ball from the second box is \( P(Black | B_2) = \frac{b_2}{10} \).

Using Bayes' Theorem, the probability that the black ball came from the second box is:
\[ P(B_2 | Black) = \frac{\frac{b_2}{10} \cdot \frac{1}{2}}{\frac{b_1}{10} \cdot \frac{1}{2} + \frac{b_2}{10} \cdot \frac{1}{2}} = \frac{b_2}{b_1 + b_2} \]
We are given that this probability is \( \frac{1}{5} \).
\[ \frac{b_2}{b_1 + b_2} = \frac{1}{5} \]
Cross-multiplying yields:
\[ 5b_2 = b_1 + b_2 \] \[ b_1 = 4b_2 \]
Since \( b_1 \) and \( b_2 \) represent the number of black balls, they must be integers from 1 to 10.

The problem states "few of them are black balls", which implies the number of black balls is strictly greater than 0.

If \( b_2 = 1 \), then \( b_1 = 4(1) = 4 \).

If \( b_2 = 2 \), then \( b_1 = 4(2) = 8 \).

If \( b_2 = 3 \), then \( b_1 = 4(3) = 12 \), which is impossible because a box only contains 10 balls.

Thus, the possible number of black balls in the first box is 4 or 8.


Step 4: Final Answer:

The number of black balls in the first box is 4 or 8.
Quick Tip: Bayes' Theorem simplifies significantly when the prior probabilities (like choosing Box 1 or Box 2) and the denominators (total balls per box) are identical. It reduces to a ratio of the favorable items.

Step 1: Understanding the Question:

We need to determine how many of the 4th roots of the complex number \( (-1 - \sqrt{3}i)^3 \) are purely real numbers.


Step 2: Key Formula or Approach:

Represent the complex number \( z = -1 - \sqrt{3}i \) in polar form \( z = r e^{i\theta} \), where \( r = \sqrt{x^2 + y^2} \) and \( \theta = \tan^{-1}(y/x) \).

Then, evaluate \( z^3 \) and find its four 4th roots using De Moivre's Theorem.


Step 3: Detailed Explanation:

Let \( z = -1 - \sqrt{3}i \).

The magnitude is \( r = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2 \).

The argument is \( \theta = \pi + \tan^{-1}\left(\frac{-\sqrt{3}}{-1}\right) = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \).

Alternatively, it can be written as \( -\frac{2\pi}{3} \).

So, the polar form is \( z = 2 e^{-i 2\pi/3} \).

Now, calculate \( z^3 \):
\[ z^3 = \left(2 e^{-i 2\pi/3}\right)^3 = 2^3 e^{-i 2\pi} = 8(1) = 8 \]
The expression reduces to evaluating \( 8^{1/4} \).

The equation to solve is \( w^4 = 8 \).

The 4th roots of 8 are given by \( w_k = 8^{1/4} e^{i (2k\pi)/4} \), for \( k = 0, 1, 2, 3 \).

Evaluating for each \( k \):

For \( k = 0 \): \( w_0 = 8^{1/4} e^{0} = 8^{1/4} \) (Real).

For \( k = 1 \): \( w_1 = 8^{1/4} e^{i\pi/2} = 8^{1/4} i \) (Imaginary).

For \( k = 2 \): \( w_2 = 8^{1/4} e^{i\pi} = -8^{1/4} \) (Real).

For \( k = 3 \): \( w_3 = 8^{1/4} e^{i3\pi/2} = -8^{1/4} i \) (Imaginary).

There are exactly 2 real values among the four roots.


Step 4: Final Answer:

The number of real values is 2.
Quick Tip: Whenever a complex expression simplifies to finding the \( n \)-th roots of a real number, there will be exactly 1 real root if \( n \) is odd, 2 real roots if \( n \) is even and the number is positive, and 0 real roots if \( n \) is even and the number is negative.

Step 1: Understanding the Question:

We need to find the number of distinct normal lines to the parabola \( y^2 = 7x \) that pass through a specific point \( (2, 0) \).


Step 2: Key Formula or Approach:

The equation of the normal to the parabola \( y^2 = 4ax \) in terms of its slope \( m \) is given by:
\[ y = mx - 2am - am^3 \]

Step 3: Detailed Explanation:

Comparing the given parabola \( y^2 = 7x \) with the standard form \( y^2 = 4ax \), we find \( 4a = 7 \), which means \( a = \frac{7}{4} \).

Substitute \( a \) into the normal equation:
\[ y = mx - 2\left(\frac{7}{4}\right)m - \left(\frac{7}{4}\right)m^3 \] \[ y = mx - \frac{7}{2}m - \frac{7}{4}m^3 \]
Since this normal line must pass through the point \( (2, 0) \), substitute \( x = 2 \) and \( y = 0 \) into the equation:
\[ 0 = 2m - \frac{7}{2}m - \frac{7}{4}m^3 \]
Factor out \( m \):
\[ 0 = m \left( 2 - \frac{7}{2} - \frac{7}{4}m^2 \right) \] \[ 0 = m \left( -\frac{3}{2} - \frac{7}{4}m^2 \right) \]
This yields two possibilities:

1) \( m = 0 \)

2) \( -\frac{3}{2} - \frac{7}{4}m^2 = 0 \implies \frac{7}{4}m^2 = -\frac{3}{2} \implies m^2 = -\frac{6}{7} \)

Since \( m \) represents the slope of a real line, \( m^2 \) must be non-negative.

Therefore, \( m^2 = -\frac{6}{7} \) provides no real solutions.

The only real solution is \( m = 0 \).

Since there is only one valid real slope, only 1 normal can be drawn.


Step 4: Final Answer:

The number of normals that can be drawn is 1.
Quick Tip: For a point \( (h, 0) \) on the axis of the parabola \( y^2 = 4ax \), if \( h > 2a \), then 3 real normals can be drawn. If \( h \leq 2a \), only 1 real normal (the axis itself) can be drawn. Here \( 2a = 3.5 \) and \( h = 2 \), so \( h < 2a \), confirming only 1 normal exists.

Step 1: Understanding the Question:

We are throwing 3 standard six-sided dice simultaneously.

We need to find the probability that the sum of the three numbers appearing on the dice is a prime number.


Step 2: Key Formula or Approach:

The total number of possible outcomes when throwing 3 dice is \( 6^3 = 216 \).

The minimum sum is \( 1+1+1 = 3 \) and the maximum sum is \( 6+6+6 = 18 \).

The prime numbers in the range [3, 18] are 3, 5, 7, 11, 13, and 17.

Probability is the number of favorable outcomes divided by the total number of outcomes.


Step 3: Detailed Explanation:

Let \( N(S) \) be the number of ways to obtain a sum \( S \).

We count the outcomes for each prime sum:

- For sum = 3: The combination is (1, 1, 1), which gives 1 way.

- For sum = 5: The combinations are (1, 1, 3) and (1, 2, 2).

Number of arrangements: \( \frac{3!}{2!} + \frac{3!}{2!} = 3 + 3 = 6 \) ways.

- For sum = 7: The combinations are (1, 1, 5), (1, 2, 4), (1, 3, 3), and (2, 2, 3).

Number of arrangements: \( \frac{3!}{2!} + 3! + \frac{3!}{2!} + \frac{3!}{2!} = 3 + 6 + 3 + 3 = 15 \) ways.

- For sum = 11: The combinations are (1, 4, 6), (1, 5, 5), (2, 3, 6), (2, 4, 5), (3, 3, 5), and (3, 4, 4).

Number of arrangements: \( 6 + 3 + 6 + 6 + 3 + 3 = 27 \) ways.

- For sum = 13: Using symmetry of dice sums, \( N(S) = N(21 - S) \).

So, \( N(13) = N(21 - 13) = N(8) \).

Let's find combinations for 8: (1, 1, 6), (1, 2, 5), (1, 3, 4), (2, 2, 4), (2, 3, 3).

Number of arrangements: \( 3 + 6 + 6 + 3 + 3 = 21 \) ways.

- For sum = 17: By symmetry, \( N(17) = N(21 - 17) = N(4) \).

The only combination for 4 is (1, 1, 2).

Number of arrangements: \( \frac{3!}{2!} = 3 \) ways.

Total favorable outcomes = \( 1 + 6 + 15 + 27 + 21 + 3 = 73 \).

Total possible outcomes = 216.

Probability = \( \frac{73}{216} \).


Step 4: Final Answer:

The probability is \( \frac{73}{216} \).
Quick Tip: Using the symmetry property \( N(S) = N(21 - S) \) for the sums of three standard 6-sided dice drastically cuts down calculation time for larger sums.

Step 1: Understanding the Question:

The question asks for the order of magnitude of the range of the weak nuclear force, one of the four fundamental forces in physics.


Step 3: Detailed Explanation:

Fundamental forces have different operational ranges:

- Gravity and Electromagnetism have an infinite range.

- The strong nuclear force is a short-range force, operating at distances on the order of \( 10^{-15} \) m (roughly the size of an atomic nucleus).

- The weak nuclear force, which is responsible for processes like beta decay, has an even shorter range.

Its effective range is extremely small, typically on the order of \( 10^{-16} \) m to \( 10^{-17} \) m.

Therefore, among the given options, \( 10^{-16} \) m is the correct order of magnitude.


Step 4: Final Answer:

The range of the weak nuclear force is of the order of \( 10^{-16} \) m.
Quick Tip: Memorize the operational ranges of fundamental forces: Weak Nuclear (\( \sim 10^{-16} \) m), Strong Nuclear (\( \sim 10^{-15} \) m), Electromagnetic (Infinite), Gravity (Infinite).

Step 1: Understanding the Question:

We need to find a specific temperature where the numerical value on the Fahrenheit scale is exactly 90% greater than the numerical value on the Celsius scale.


Step 2: Key Formula or Approach:

The relationship between Fahrenheit (\( F \)) and Celsius (\( C \)) is given by:
\[ F = \frac{9}{5}C + 32 \]
The condition given in the problem can be expressed as:
\[ F = C + 0.90C = 1.9C \]

Step 3: Detailed Explanation:

Substitute \( F = 1.9C \) into the temperature conversion formula:
\[ 1.9C = \frac{9}{5}C + 32 \] \[ 1.9C = 1.8C + 32 \]
Subtract \( 1.8C \) from both sides:
\[ 1.9C - 1.8C = 32 \] \[ 0.1C = 32 \]
Multiply by 10 to solve for \( C \):
\[ C = 320 \]
So, the temperature is 320 degree.

Now, convert this back to the Fahrenheit scale to match the given options:
\[ F = 1.9 \times 320 \] \[ F = 19 \times 32 = 608 \]
Thus, the reading on the Fahrenheit scale is 608 degree.


Step 4: Final Answer:

The temperature on the Fahrenheit scale is 608 \(^\circF\).
Quick Tip: Read carefully whether the question asks for the temperature on the Celsius scale or the Fahrenheit scale at the very end to avoid selecting the intermediate value (e.g., 320) by mistake.

Step 1: Understanding the Question:

The question asks us to identify a specific air pollutant known to act as a lung irritant and cause acute respiratory diseases, particularly in children.


Step 3: Detailed Explanation:

Let's evaluate the given options:

- Carbon monoxide (CO) is a toxic gas that interferes with oxygen transport in the blood by binding to hemoglobin, but it is not primarily classified as a lung irritant.

- Carbon dioxide (CO\(_2\)) is a greenhouse gas and is not a lung irritant at typical atmospheric concentrations.

- Sulfur dioxide (SO\(_2\)) is a lung irritant, but its primary notable effect is exacerbating asthma and contributing to acid rain.

- Nitrogen dioxide (NO\(_2\)) is a highly reactive gas and a strong lung irritant.

Prolonged or acute exposure to NO\(_2\), commonly from vehicle emissions or indoor gas stoves, is strongly linked to acute respiratory infections and decreased lung function, especially in children.


Step 4: Final Answer:

The correct lung irritant is NO\(_2\).
Quick Tip: When differentiating between SO\(_2\) and NO\(_2\), remember that NO\(_2\) is classically associated with respiratory infections and lung development issues in children, often emphasized in environmental science contexts.