Mechanics
Units & Measurement and Dimensions
MCQ (Single Correct Answer)
Motion in a Straight Line
MCQ (Single Correct Answer)
Work, Energy and Power
MCQ (Single Correct Answer)
Center of Mass and Collision
MCQ (Single Correct Answer)
Simple Harmonic Motion
MCQ (Single Correct Answer)
Heat and Thermodynamics
MCQ (Single Correct Answer)
Electromagnetism
Current Electricity
MCQ (Single Correct Answer)
Moving Charges and Magnetism
MCQ (Single Correct Answer)
Magnetism and Matter
MCQ (Single Correct Answer)
Electromagnetic Induction
MCQ (Single Correct Answer)
Alternating Current
MCQ (Single Correct Answer)
Electromagnetic Waves
MCQ (Single Correct Answer)
Modern Physics
Dual Nature of Radiation
MCQ (Single Correct Answer)
Semiconductor Devices and Logic Gates
MCQ (Single Correct Answer)
Communication Systems
MCQ (Single Correct Answer)
1
TG EAPCET 2024 (Online) 11th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
In a simple pendulum experiment for the determination of acceleration due to gravity, the error in the measurement of the length of the pendulum is $1 \%$ and the error in the measurement of the time period is $2 \%$. The error in the estimation of acceleration due to gravity is
A
$1 \%$
B
$3 \%$
C
$4 \%$
D
$5 \%$
2
TG EAPCET 2024 (Online) 11th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
A massless spring of length $l$ and spring constant $k$ oscillates with a time period $T$ when loaded with a mass $m$. The spring is now cut into three equal parts and are connected in parallel. The frequency of oscillation of the combination when it is loaded with ${ }_{3}$ mass 4 m is
A
$\frac{2}{T}$
B
$\frac{2}{3 \pi}$
C
$\frac{3}{T}$
D
$\frac{3}{2 T}$
3
TG EAPCET 2024 (Online) 10th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
If a body dropped freely from a height of 20 m reaches the surface of a planet with a velocity of $31.4 \mathrm{~ms}^{-1}$. then the length of a simple pendulum that ticks seconds on the planet is
A
1 m
B
0.625 m
C
2.5 m
D
2 m
4
TG EAPCET 2024 (Online) 9th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
A particle of mass 4 mg is executing simple harmonic motion along $X$-axis with an angular frequency of $40 \mathrm{rad} \mathrm{s}^{-1}$. If the potential energy of the particle is $V(x)=a+b x^2$, where $V(x)$ is in joule and $x$ is in metre, then the value of $b$ is
A
$800 \times 10^{-6} \mathrm{Jm}^{-2}$
B
$1600 \times 10^{-6} \mathrm{Jm}^{-2}$
C
$3200 \times 10^{-6} \mathrm{Jm}^{-2}$
D
$6400 \times 1^{-6} \mathrm{Jm}^{-2}$
TS EAMCET Subjects