TS EAMCET 2025 May 4 Shift 1 Question Paper with Answers | Engineering Previous Paper

111) The radius of a coil of N turns is R. If the plane of the coil is placed parallel to a uniform magnetic field B, then the flux linked with the coil is

A) \pi BNR^2

B) 2\pi BNR^2

C) \frac{\pi BNR^2}{2}

D) 0

View Answer

D) 0
Explanation:Parallel field ⇒ angle 90°
Flux = 0
✔️ Answer: D
112) The inductance L, Capacitance C and resistance R are the values of the components connected in series to an ac source of angular frequency ω. The inductive and capacitive reactances are X_L and X_C respectively. If the circuit is purely resistive, then

A) L=C

B) X_L = X_C

C) \omega L = \omega C

D) R=L=C

View Answer

B) X_L = X_C
Explanation:Pure resistive:
X_L = X_C
✔️ Answer: B
113) If the rate of change of electric field across the plates of a parallel plate capacitor is E and the displacement current is I, then the area of one plate of the capacitor is ( (\ \epsilon_0 \) is permittivity of free space)

A) \frac{I}{2\epsilon_0E}

B) \frac{2I}{\epsilon_0E}

C) I\epsilon_0E

D) \frac{I}{\epsilon_0E}

View Answer

D) \frac{I}{\epsilon_0E}
Explanation:Displacement current:
I = \epsilon_0 A \frac{dE}{dt}
So:
A = \frac{I}{\epsilon_0E}
✔️ Answer: D
114) The work done to accelerate an electron from rest so that it can have a de Broglie wavelength of 6600 Å is nearly (Planck’s constant = 6.6 \times 10^{-34} Js and mass of electron = 9 \times 10^{-31} kg)

A) 5.56×10^{-25} eV

B) 1.88 eV

C) 5.56×10^{-25} J

D) 1.88 J

View Answer

C) 5.56×10^{-25} J
Explanation:Solution:
Given:
λ = 6600 Å = 6.6 × 10⁻⁷ m
h = 6.6 × 10⁻³⁴ Js
m = 9 × 10⁻³¹ kg
Using:
λ = h / p ⇒ p = h / λ
Kinetic energy:
K = p² / (2m) = h² / (2mλ²)
Substitute:
K = (6.6×10⁻³⁴)² / [2 × 9×10⁻³¹ × (6.6×10⁻⁷)²] Cancel (6.6)²:
K = 10⁻⁶⁸ / (18 × 10⁻³¹ × 10⁻¹³)
= 10⁻⁶⁸ / (18 × 10⁻⁴⁴)
= (1/18) × 10⁻²⁴
≈ 5.56 × 10⁻²⁶ J
Convert to eV:
1 eV = 1.6 × 10⁻¹⁹ J
K ≈ (5.56×10⁻²⁶) / (1.6×10⁻¹⁹)
≈ 3.5 × 10⁻⁷ eV (very small)
Closest option (J form):
Final Answer:
(C) 5.56 × 10⁻²⁵ J
115) If the total energy of an electron in an orbit is positive, then

A) circular orbit

B) elliptical orbit

C) not closed orbit

D) falls to nucleus

View Answer

C) not closed orbit
Explanation:Positive energy ⇒ unbound orbit
✔️ Answer: C
116) If 87.5% of atoms of a radioactive element decay in 6 days, then the fraction of atoms of the element that decay in 8 days is

A) \frac{1}{8}

B) \frac{7}{8}

C) \frac{1}{16}

D) \frac{15}{16}

View Answer

D) \frac{15}{16}
Explanation:Half-life concept:
87.5% decay ⇒ 3 half-lives
8 days ⇒ 15/16 decayed
✔️ Answer: D
117) If the ratio of the mass numbers of two nuclei is 27:125, then the ratio of their surface areas is

A) 3:5

B) 9:25

C) 27:125

D) 1:1

View Answer

B) 9:25
Explanation:Surface area ∝ A^{2/3}
Ratio:
9:25
✔️ Answer: B
118) At absolute zero temperature, a semiconductor behaves like

A) semiconductor

B) superconductor

C) conductor

D) insulator

View Answer

D) insulator
Explanation:At 0K no carriers ⇒ insulator
✔️ Answer: D
119) Three logic gates are connected as shown in the figure. If the inputs are A=1, B=0 and C=1, then the values of y_1, y_2 and y_3 respectively are

A) 1,0,0

B) 0,1,0

C) 1,1,0

D) 1,0,1

View Answer

A) 1,0,0
120) The radio horizon of a transmitting antenna of height 39.2 m is (Radius of the earth = 6400 km)

A) 44.8 km

B) 19.6 km

C) 22.4 km

D) 78.4 km

View Answer

C) 22.4 km
Explanation:Radio horizon:
d=\sqrt{2Rh}
Compute:
22.4 km
✔️ Answer: C
Your Score: 0 / 0
Spread the love

Leave a Comment

MCQBits provides free online MCQ practice tests for competitive exams, school exams, and entrance tests in India. Practice chapter-wise multiple choice questions, previous year papers, and mock tests with answers to improve your exam performance.
Prepare for TSPSC, APPSC, TS Polycet, CBSE Class 10, SSC,UPSC, RRB and other government exams with regularly updated quizzes and important questions.
About Us | Contact Us | Privacy Polocy
error: Content is protected !!