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TS Polycet (Polytechnic) 2024 Previous Question Paper with Answers And Model Papers With Complete Analysis

21) From a helicopter a person observes an object on the ground at an angle of depression 30°. If the helicopter is flying at a height of 500 m from the ground, then the distance between the person and the object is ఒక హెలికాప్టర్ నుండి ఒక వ్యక్తి భూమి పైనున్న ఒక వస్తువును 30° నిమ్నకోణంలో పరిశీలించాడు. భూమి పై నుండి హెలికాప్టర్ 500 మీ. ఎత్తులో ఎగురుతూ ఉంటే, వ్యక్తికి మరియు వస్తువుకు మధ్య దూరము
A)500 m
B)1000 m
C) $500\sqrt2\;m$
D) $\frac{500}{\sqrt3}m$

A) A

B) B

C) C

D) D

View Answer

B) B
Explanation:Shortcut Method Using Trigonometry (Hypotenuse Calculation)
We are given:
– The angle of depression from the helicopter to the object on the ground is $30^\circ$ .
– The height of the helicopter from the ground is 500 meters.
We need to find the distance between the person and the object (which is the hypotenuse of the right triangle formed by the height, the distance on the ground, and the line of sight).
Step 1: Using the Angle of Depression and Trigonometric Ratio
In this case, the angle of depression forms a right triangle with:
– The height of the helicopter as the opposite side (500 m),
– The distance between the person and the object as the hypotenuse,
– The horizontal distance between the object and the point on the ground directly beneath the helicopter as the adjacent side.
We can use the tangent function to relate the height and the horizontal distance first, then apply the cosine or sine to find the hypotenuse.
$\tan(30^\circ) = \frac{\text{height}}{\text{horizontal distance}}$
Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$ , we have:
$\frac{1}{\sqrt{3}} = \frac{500}{\text{horizontal distance}}$
Solving for the horizontal distance:
$\text{horizontal distance} = 500 \times \sqrt{3}$
Step 2: Apply the Pythagorean Theorem
Now, we can calculate the distance between the person and the object (the hypotenuse). We know:
– The height (opposite side) is 500 m,
– The horizontal distance (adjacent side) is $500 \times \sqrt{3}$ .
Using the Pythagorean theorem:

$\text{distance} = \sqrt{\left(500\right)^2 + \left(500\sqrt{3}\right)^2} <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-5695b77b2d61fcbb2696246b4d7c0f53_l3.png" height="22" width="226" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[\text{distance} = \sqrt{500^2 + 500^2 \times 3}\]" title="Rendered by QuickLaTeX.com"/>\text{distance} = \sqrt{500^2(1 + 3)} <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-63373d10592cf15642b47a67d85a91de_l3.png" height="22" width="170" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[\text{distance} = \sqrt{500^2 \times 4}\]" title="Rendered by QuickLaTeX.com"/>\text{distance} = 500 \times \sqrt{4} = 500 \times 2 = 1000 \, \text{m}$

Final Answer: $\boxed{1000 \, \text{m}}$

22) The product of prime factors of 2024 is 2024 యొక్క ప్రధాన కారణాంకాల లబ్ధము

A) 11 × 23 × 3²

B) 23 × 11 × 2³

C) 7 × 23 × 2³

D) 23 × 112 × 2²

View Answer

B) 23 × 11 × 2³
Explanation:Shortcut Method to Find the Prime Factorization of 2024
We need to find the prime factorization of 2024.
Step 1: Start by dividing 2024 by the smallest prime (2):

$2024 \div 2 = 1012 <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-7b286ff6c1701baf6ab7e0974c539f52_l3.png" height="14" width="115" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[1012 \div 2 = 506\]" title="Rendered by QuickLaTeX.com"/>506 \div 2 = 253$

Now, 253 is not divisible by 2, so we move to the next smallest prime, which is 3. However, 253 is not divisible by 3 either, so we check for divisibility by 5, and it’s not divisible by 5 either.
Step 2: Check divisibility by 11:
$253 \div 11 = 23$
23 is a prime number, so the prime factorization of 2024 is:
$2024 = 2^3 \times 11 \times 23$
Final Answer: $\boxed{23 \times 11 \times 2^3}$

23) {0} is a set which has ______ elements. {0} అనేది ______ మూలకాలను కలిగి ఉన్న సమితి.

A) 0

B) 1

C) 4

D) 3

View Answer

B) 1
Explanation:Explanation:
The set $\{0\}$ is a set containing a single element, which is $0$ .
So, the number of elements in the set $\{0\}$ is 1.
Final Answer: $\boxed{1}$

24) If cosec θ + cot θ = k, then the value of cosec θ is cos ec θ + cot θ = k అయిన, cosec θ యొక్క విలువ
A) $\frac{k^2+1}{2k}$
B)0
C) $\frac{k^2-1}{k^2+1}$
D) $\frac1k$

A) A

B) B

C) C

D) D

View Answer

A) A
Explanation:Shortcut Method to Find $\csc \theta$ in Terms of $k$
We are given that:
$\csc \theta + \cot \theta = k$
We need to express $\csc \theta$ in terms of $k$ .
Step 1: Use the identity for $\csc^2 \theta - \cot^2 \theta$
We know that:
$\csc^2 \theta = 1 + \cot^2 \theta$
Rearranging the given equation:
$\csc \theta = k - \cot \theta$
Now, we square both sides of this equation:
$\csc^2 \theta = (k - \cot \theta)^2$
Expanding the square:
$\csc^2 \theta = k^2 - 2k \cot \theta + \cot^2 \theta$
Using the identity $\csc^2 \theta = 1 + \cot^2 \theta$ , we substitute:
$1 + \cot^2 \theta = k^2 - 2k \cot \theta + \cot^2 \theta$
Step 2: Simplify the equation
Canceling out $\cot^2 \theta$ from both sides:
$1 = k^2 - 2k \cot \theta$
Rearranging:

$2k \cot \theta = k^2 - 1 <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-90bd87d86550f4aaea5e07dbdc5e871f_l3.png" height="119" width="584" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[\cot \theta = \frac{k^2 - 1}{2k}$ Step 3: Substitute $ \cot \theta $ back into the equation for $ \csc \theta $ Now, substitute $ \cot \theta = \frac{k^2 - 1}{2k} $ back into $ \csc \theta = k - \cot \theta $: $\csc \theta = k - \frac{k^2 - 1}{2k}$ Simplify the right-hand side: $\csc \theta = \frac{2k^2 - (k^2 - 1)}{2k}\]" title="Rendered by QuickLaTeX.com"/>\csc \theta = \frac{k^2 + 1}{2k}$

Final Answer: $\boxed{\frac{k^2 + 1}{2k}}$

25) The value of $\log_6^2+\log_6^3$ is $\log_6^2+\log_6^3$ యొక్క విలువ

A) 0

B) 1

C) 2

D) 3

View Answer

B) 1
Explanation:Shortcut Method to Simplify the Expression $\log_6 2 + \log_6 3$
We are asked to simplify the expression:
$\log_6 2 + \log_6 3$
Step 1: Apply the logarithmic property of addition
We use the property of logarithms that states:
$\log_b x + \log_b y = \log_b (xy)$
So, applying this property:
$\log_6 2 + \log_6 3 = \log_6 (2 \times 3) = \log_6 6$
Step 2: Simplify $\log_6 6$
We know that:
$\log_b b = 1$
So:
$\log_6 6 = 1$
Final Answer: $\boxed{1}$

26) Which of the following two sets are equal sets? క్రింది రెండు సమితులలో ఏవి సమాన సమితులు?

A) A = {5, 6}, B = {5}

B) A = {5, 6}, B = {5, 6, 7}

C) A = {5, 6, 7}, B = {7, 5, 6}

D) A = {5, 6, 8}, B = {5, 6, 7}

View Answer

C) A = {5, 6, 7}, B = {7, 5, 6}
Explanation:Understanding Equal Sets
Two sets are equal sets if and only if they have exactly the same elements. The order of the elements in a set does not matter.
Step-by-step analysis of each option:
– 1. A = {5, 6}, B = {5}
– Set A has the elements 5 and 6.
– Set B has only the element 5.
– These are not equal sets, because Set A has an additional element (6) that Set B does not have.
– 2. A = {5, 6}, B = {5, 6, 7}
– Set A has the elements 5 and 6.
– Set B has the elements 5, 6, and 7.
– These are not equal sets, because Set B has an extra element (7) that Set A does not have.
– 3. A = {5, 6, 7}, B = {7, 5, 6}
– Set A has the elements 5, 6, and 7.
– Set B has the same elements, 5, 6, and 7, but in a different order.
– These are equal sets, because they contain exactly the same elements, just in a different order. The order does not matter in sets.
4. A = {5, 6, 8}, B = {5, 6, 7}
– Set A has the elements 5, 6, and 8.
– Set B has the elements 5, 6, and 7.
– These are not equal sets, because Set A contains 8, while Set B contains 7. They have different elements.
Final Answer: $\boxed{3}$

27) Exponential form of $\log_b\;a\;=\;c$ $\log_b\;a\;=\;c$ యొక్క ఘాత రూపము

A) b^a=c

B) a^c=b

C) a^b=c

D) b^c=a

View Answer

D) b^c=a
Explanation:Understanding the Exponential Form of Logarithms
The logarithmic equation $\log_b a = c$ means that the exponent $c$ to which the base $b$ must be raised to produce $a$ is $c$ .
This can be rewritten in exponential form as:
$b^c = a$
Explanation:
– $\log_b a = c$ implies that $b$ raised to the power $c$ gives $a$ .
– So, in exponential form, we have:
$b^c = a$
Final Answer: $\boxed{b^c = a}$

28) If a right circular cylinder has base radius 14 cm and height 21 cm, then its curved surface area is ______. $\left(use\;\pi=\frac{22}7\right)$ ఒక క్రమ వృత్తాకార స్థూపము యొక్క భూవ్యాసార్ధము 14 సెం. మీ. మరియు ఎత్తు 21 సెం.మీ. అయిన, ఆ స్థూపము వక్రతల వైశాల్యము _____. $\left(\pi=\frac{22}7\;అని\;తీసుకొనుము\right)$

A) 924 cm²

B) 2772 cm²

C) 3080 cm²

D) 1848 cm²

View Answer

D) 1848 cm²
Explanation:Shortcut Method to Find the Curved Surface Area of a Cylinder
The formula for the curved surface area (CSA) of a right circular cylinder is:
$\text{CSA} = 2 \pi r h$
Where:
– $r$ = radius of the base
– $h$ = height of the cylinder
– $\pi = \frac{22}{7}$
Given:
– Radius $r = 14 \, \text{cm}$
– Height $h = 21 \, \text{cm}$
– $\pi = \frac{22}{7}$
Step 1: Substitute the values into the formula
$\text{CSA} = 2 \times \frac{22}{7} \times 14 \times 21$
Step 2: Simplify the calculation

$\text{CSA} = 2 \times \frac{22}{7} \times 14 \times 21 = 2 \times \frac{22 \times 14 \times 21}{7} <img src="https://www.mcqbits.com/wp-content/ql-cache/quicklatex.com-5c477987b5f0f759aa500edd595639f8_l3.png" height="36" width="118" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[= 2 \times \frac{22 \times 294}{7}\]" title="Rendered by QuickLaTeX.com"/>= 2 \times 924 = 1848 \, \text{cm}^2$

Final Answer: $\boxed{1848 \, \text{cm}^2}$

29) The volume of a right circular cone with radius 6 cm and height 7 cm is ________. $\left(use\;\pi=\frac{22}7\right)$ 6 సెం.మీ. భూవ్యాసార్ధము మరియు 7 సెం.మీ. ఎత్తు కలిగిన ఒక క్రమ వృత్తాకార శంఖువు యొక్క
ఘనపరిమాణము _______ $\left(\pi=\frac{22}7\;అని\;తీసుకొనుము\right)$

A) 264 cm³

B) 792 cm³

C) 301 cm³

D) 616 cm³

View Answer

A) 264 cm³
Explanation:Shortcut Method to Find the Volume of a Right Circular Cone
The formula for the volume of a right circular cone is:
$\text{Volume} = \frac{1}{3} \pi r^2 h$
Where:
– $r$ = radius of the base
– $h$ = height of the cone
– $\pi = \frac{22}{7}$
Given:
– Radius $r = 6 \, \text{cm}$
– Height $h = 7 \, \text{cm}$
– $\pi = \frac{22}{7}$
Step 1: Substitute the values into the formula
$\text{Volume} = \frac{1}{3} \times \frac{22}{7} \times 6^2 \times 7$
Step 2: Simplify the calculation
$\text{Volume} = \frac{1}{3} \times \frac{22}{7} \times 36 \times 7$
Cancel the 7’s:
$= \frac{1}{3} \times 22 \times 36$
Now, multiply:
$= \frac{1}{3} \times 792 = 264 \, \text{cm}^3$
Final Answer: $\boxed{264 \, \text{cm}^3}$

30) The value of $\sin^215^\circ\;+\;\cos^215^\circ$ is $\sin^215^\circ\;+\;\cos^215^\circ$ యొక్క విలువ
A)0
B)1
C) $\frac12$
D) $\frac1{\sqrt2}$

A) A

B) B

C) C

D) D

View Answer

B) B
Explanation:Simplifying the Expression $\sin^2 15^\circ + \cos^2 15^\circ$
We are asked to find the value of:
$\sin^2 15^\circ + \cos^2 15^\circ$
Step 1: Use the Pythagorean identity
The Pythagorean identity states that:
$\sin^2 \theta + \cos^2 \theta = 1$
This identity holds for any angle $\theta$ , including $15^\circ$ .
Step 2: Apply the identity to the given expression
$\sin^2 15^\circ + \cos^2 15^\circ = 1$
Final Answer: $\boxed{1}$
This is the simplest and quickest method, using the basic trigonometric identity.

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