List of top Questions asked in AP EAPCET

The angular velocity of a body changes from \(6 \, \text{rad/s}\) to \(21 \, \text{rad/s}\) in a time of \(1.5 \, \text{s}\). If the moment of inertia of the body is \(100 \, \text{g m}^2\), then the rate of change of angular momentum of the body is
If the displacement of a particle executing simple harmonic motion is given by \( x = 0.5 \cos(125.6\,t) \), then the time period of oscillation of the particle is nearly (Here \(x\) is displacement in metre and \(t\) is time in second)
The amplitude of a damped harmonic oscillator becomes 50% of its initial value in a time of 12 s. If the amplitude of the oscillator at a time of 36 s is \(x%\) of its initial amplitude, then the value of \(x\) is
The escape velocity of a body from a planet of mass \(M\) and radius \(R\) is 14 km/s. The escape velocity of the body from another planet having same mass and diameter \(8R\) (in km/s) is
The stress-strain graph of two wires A and B is shown in the figure. If \(Y_A\) and \(Y_B\) are Young’s moduli of materials of wires A and B respectively, then
stress strain
If two soap bubbles each of radius \(2 \, \text{cm}\) combine in vacuum under isothermal conditions, then the radius of the new bubble formed is
A rectangular slab consists of two cubes of copper and brass of equal sides having thermal conductivities in the ratio \(4 : 1\). If the free face of brass is at \(0^\circ \text{C}\) and that of copper is at \(100^\circ \text{C}\), then the temperature of their interface is
If a force \( \vec{F} = (3\hat{i} + 2\hat{j} + 5\hat{k})~\text{N} \) acting on a body displaces it through \( \vec{d} = (2\hat{i} + 2\hat{j} + 1\hat{k})~\text{m} \), then the work done by the force on the body is
If two bodies A and B are projected with same velocity but with different angles \( \theta_1 \) and \( \theta_2 \) respectively with the horizontal such that both will have same range, then the ratio of times of flight of the bodies A and B is
The apparent weight of a girl of mass 30 kg when she is in a lift moving vertically upwards with an acceleration of \( 2~\text{ms}^{-2} \) is
(Acceleration due to gravity = \( 10~\text{ms}^{-2} \))
If a stone of mass \( 0.5~\text{kg} \) tied to one end of a wire is whirled in a circular path of radius \( 2~\text{m} \) with a speed \( 40~\text{rev/min} \) in a horizontal plane, then the tension in the wire is nearly
A body is projected vertically upwards with a velocity of \( 20~\text{ms}^{-1} \). If the potential energy of the body at a height of \( 5~\text{m} \) from the ground is \( 100~\text{J} \), then the kinetic energy of the body at a height of \( 10~\text{m} \) from the ground is
\textit{(Acceleration due to gravity \( g = 10~\text{ms}^{-2} \))}
A body falls freely on to a hard horizontal surface. If the coefficient of restitution between the surface and the body is \( 0.8 \), then the ratio of the maximum height to which the body rises after second impact and the initial height of the body is
Two bodies of masses \( M \) and \( 4M \) initially at rest, start moving towards each other due to their mutual attraction. The velocity of their centre of mass when the first body attains a velocity \( v_0 \) is
Evaluate the integral: \[ \int_0^1 x^{5/2} (1 - x)^{3/2} \, dx = \]
Evaluate the limit: \[ \lim_{n \to \infty} \left[ \frac{1}{n^2} \sec^2 \frac{1}{n^2} + \frac{2}{n^2} \sec^2 \frac{4}{n^2} + \frac{3}{n^2} \sec^2 \frac{9}{n^2} + .s + \frac{1}{n^2} \sec^2 1 \right] = \]
The general solution of the differential equation \( \left(x \sin \frac{y}{x} \right) dy = \left( y \sin \frac{y}{x} - x \right) dx \) is
The general solution of the differential equation \( \cos(x + y) \, dy = dx \) is
If \( Ax^3 + Bxy = 4 \) (A and B are arbitrary constants) is the general solution of the differential equation \[ F(x)\frac{d^2y}{dx^2} + G(x)\frac{dy}{dx} - 2y = 0, \] then \( F(1) + G(1) = \)
The physical quantity having the dimensions of the square root of the ratio of the kinetic energy and surface tension is
If the displacement \( s \) (in metre) of a moving particle in terms of time \( t \) (in second) is \( s = t^3 - 6t^2 + 18t + 9 \), then the minimum velocity attained by the particle is
\( \int x \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \, dx \, (x>0) = \)
\( \int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} = \)
\( \int \sin^{-1}\left(\frac{x}{\sqrt{a+x}}\right) \, dx = \)
If \( \int \frac{x}{x \tan x + 1} \, dx = \log f(x) + k \), then \( f\left(\frac{\pi}{4}\right) = \)