List of top Questions asked in AP EAPCET

In Young's double slit experiment, the wavelength of monochromatic light is increased by 20% and the distance between the two slits is decreased by 25%. If the initial fringe width is 0.3 mm, then the final fringe width is
Two charged conducting spheres of radii 5 cm and 10 cm have equal surface charge densities. If the electric field on the surface of the smaller sphere is E, then the electric field on the surface of the larger sphere is
As shown in the figure, if the values of the electric potential at three points A, B and C in a uniform electric field (\( \vec{E} \)) are \(V_A\), \(V_B\), and \(V_C\) respectively, then
electric potential at three points A, B and C
As shown in the figure, the work done to move the charge 'Q' from point C to point D along the semi-circle CRD is
work done to move the charge ’Q’ from point C to point D
The length and area of cross-section of a copper wire are respectively 30 m and \( 6 \times 10^{-7} \, \text{m}^2 \). If the resistivity of copper is \( 1.7 \times 10^{-8} \, \Omega \, \text{m} \), then the resistance of the wire 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
An aeroplane of mass \( 4.5 \times 10^4 \) kg and total wing area of \( 600 \, \text{m}^2 \) is travelling at a constant height. The pressure difference between the upper and lower surfaces of its wings is (Acceleration due to gravity \( = 10 \, \text{m s}^{-2} \))
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 \( t_A : t_B = \)
A crane of efficiency 80% is used to lift 8000 kg of coal from a mine of depth 108 m. If the time taken by the crane to lift the coal is one hour, then the power of the crane (in kW) is (Acceleration due to gravity \( = 10 \, \text{m s}^{-2} \))
If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively \( 5 \, \text{m s}^{-1} \) and \( 10 \, \text{m s}^{-2} \), then the time period of oscillation of the particle 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
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
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
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
If the magnitude of a vector \( \vec{p} \) is 25 units and its y-component is 7 units, then its x-component 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 \( = 10 \, \text{m s}^{-2} \)]
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
\( \int \sqrt{x^2+x+1} \ dx \)
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 \( (k-1)n \).
\( \int_{-1}^{4} \sqrt{\frac{4-x}{x+1}} \ dx = \)
\( \int_{0}^{\pi/4} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} dx = \)
\( \int_{5\pi}^{25\pi} |\sin 2x + \cos 2x| \ dx = \)
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 \( \frac{dy}{dx} = \frac{x+y}{x-y} \) is