List of top Questions asked in AP EAPCET

A solid sphere of mass 4 kg and radius 28 cm is on an inclined plane. If the acceleration of the sphere when it rolls down without sliding is \( 3.5 \, \text{m s}^{-2} \), then the acceleration of the sphere when it slides down without rolling is
A body of mass 1 kg is suspended from a spring of force constant \( 600 \, \text{N m}^{-1} \). Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of \( 3 \, \text{m s}^{-1} \) and embedded in it. The amplitude of motion is
Three blocks A, B and C are arranged as shown in the figure such that the distance between two successive blocks is 10 m. Block A is displaced towards block B by 2 m and block C is displaced towards block B by 3 m. The distance through which the block B should be moved so that the centre of mass of the system does not change is
Three blocks A, B and C are arranged as shown in the figure
Two satellites A and B are revolving around the earth in orbits of heights \(1.25R_E\) and \(19.25R_E\) from the surface of earth respectively, where \(R_E\) is the radius of the earth. The ratio of the orbital speeds of the satellites A and B is
When a wire made of material with Young's modulus Y is subjected to a stress S, the elastic potential energy per unit volume stored in the wire is
Two bodies A and B of masses 20 kg and 5 kg respectively are at rest. Due to the action of a force of 40 N separately, if the two bodies acquire equal kinetic energies in times \( t_A \) and \( t_B \) respectively, then \( t_A : t_B = \)
If the magnitude of a vector \( \vec{p} \) is 25 units and its y-component is 7 units, then its x-component is
A conveyor belt is moving horizontally with a velocity of \( 2 \, \text{m s}^{-1} \). If a body of mass 10 kg is kept on it, then the distance travelled by the body before coming to rest is (The coefficient of kinetic friction between the belt and the body is 0.2 and acceleration due to gravity is \( 10 \, \text{m s}^{-2} \))
If the displacement 'x' of a body in motion in terms of time 't' is given by \(x = A\sin(\omega t + \theta)\), then the minimum time at which the displacement becomes maximum is
A balloon with mass 'm' is descending vertically with an acceleration 'a' (where a \(<\) g). The mass to be removed from the balloon, so that it starts moving vertically up with an acceleration 'a' is
The height of ceiling in an auditorium is 30 m. A ball is thrown with a speed of \( 30 \, \text{m s}^{-1} \) from the entrance such that it just moves very near to the ceiling without touching it and then it reaches the ground at the end of the auditorium. Then the length of auditorium is [Acceleration due to gravity \( = 10 \, \text{m s}^{-2} \)]
The general solution of the differential equation \( \frac{dy}{dx} = \frac{x+y}{x-y} \) is
The differential equation of the family of circles passing through the origin and having centre on X-axis is
The general solution of the differential equation \( \frac{dy}{dx} + \frac{\sec x}{\cos x + \sin x}y = \frac{\cos x}{1+\tan x} \) is
The number of significant figures in the simplification of \( \frac{0.501}{0.05}(0.312-0.03) \) is
\( \int_{5\pi}^{25\pi} |\sin 2x + \cos 2x| \ dx = \)
\( \int_{-1}^{4} \sqrt{\frac{4-x}{x+1}} \ dx = \)
\( \int_{0}^{\pi/4} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} dx = \)
\( \int \frac{13\cos 2x - 9\sin 2x}{3\cos 2x - 4\sin 2x} dx = \)
If \( k \in N \) then \( \lim_{n\to\infty} \left[ \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \dots + \frac{1}{kn} \right] = \) (Note: The last term should be \( \frac{1}{n+ (k-1)n} = \frac{1}{kn} \) or sum up to \(n+(k-1)n\). The given form \(1/kn\) as the endpoint of the sum means sum from \(r=1\) to \((k-1)n\). The sum is usually \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). If the last term is \( \frac{1}{kn} \), it means \( n+r = kn \implies r = (k-1)n \). So it's \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \).) Let's assume the sum goes up to \( \frac{1}{n+(k-1)n} = \frac{1}{kn} \). So the sum is \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). No, this seems to be \( \frac{1}{n+1} + \dots + \frac{1}{n+(kn-n)} \). The sum should be written as \( \sum_{i=1}^{(k-1)n} \frac{1}{n+i} \). The dots imply the denominator goes up. The last term is \( \frac{1}{kn} \). This means the sum is actually \( \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{n+(k-1)n} \). The number of terms is \( (k-1)n \).
If \( \int \frac{\cos^3 x}{\sin^2 x + \sin^4 x} dx = c - \operatorname{cosec} x - f(x) \), then \( f\left(\frac{\pi}{2}\right) = \)
\( \int \sqrt{x^2+x+1} \ dx \)
\( \int \left( \sum_{r=0}^{\infty} \frac{x^r 2^r}{r!} \right) dx = \)
If the tangent drawn at the point \( (x_1,y_1) \), \(x_1,y_1 \in N \) on the curve \( y = x^4 - 2x^3 + x^2 + 5x \) passes through origin, then \( x_1+y_1 = \)
If \( \beta \) is an angle between the normals drawn to the curve \( x^2+3y^2=9 \) at the points \( (3\cos\theta, \sqrt{3}\sin\theta) \) and \( (-3\sin\theta, \sqrt{3}\cos\theta) \), \( \theta \in \left(0, \frac{\pi}{2}\right) \), then