List of top Questions asked in AP EAPCET

Activation energy for the hydrolysis of sucrose by acid is X kJ mol$^{-1}$ and by sucrase is Y kJ mol$^{-1}$. X and Y respectively are
In triangle \(ABC\), if \(a = 6\), \(b = 8\), and \(c = 10\), then \(\dfrac{2r_1}{r_2} =\)
\( \int \frac{dx}{12\cos x + 5\sin x} = \)
When a current of \( 4 \, \text{mA} \) passes through an inductor, if the flux linked with it is \( 32 \times 10^{-6} \, \text{Tm}^2 \), then the energy stored in the inductor is:
Let \( \vec{a} = 2\vec{i} + \vec{j} - 2\vec{k} \) and \( \vec{b} = \vec{i} + \vec{j} \) be two vectors. If \( \vec{c} \) is a vector such that \( \vec{a} \cdot \vec{c} = |\vec{c}| \), \( |\vec{c} - \vec{a}| = 2\sqrt{2} \) and the angle between \( \vec{a} \times \vec{b} \) and \( \vec{c} \) is \( 30^\circ \), then \( |(\vec{a} \times \vec{b}) \times \vec{c| = \)}
If a resistor of resistance \( 4 \Omega \), a capacitor of capacitive reactance \( 6 \Omega \) and an inductor of inductive reactance \( 9 \Omega \) are connected in series with an AC source, then the impedance of the circuit is:
If $a$ is the determinant of the adjoint of the matrix $\begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 3 \end{bmatrix}$ and $b$ is the determinant of the inverse of the matrix $\begin{bmatrix} 2 & 1 & 3 \\ 1 & -4 & -1 \\ 2 & 1 & 4 \end{bmatrix}$, then $\frac{b+1}{18b} = $
At temperature \(T(K)\), the solubility product of AgBr is \(4 \times 10^{-13}\). What is its solubility in 0.1 M KBr solution?
If the pressure on a body is increased from \( 200 \) kPa to \( 250 \) kPa, the volume of the body decreases by \( 0.25% \). The compressibility of the material of the body (in \( m^2 N^{-1} \)) is:
If the tangent of the curve $$ 4y^3 = 3ax^2 + x^3 $$ drawn at the point $ (a, a) $ forms a triangle of area $\frac{25}{24}$ sq. units with the coordinate axes, then find $ a $.
If \( \int \frac{dx}{2\cos x + 3\sin x + 4} = \frac{2}{\sqrt{3}}f(x) + c \), then \( f\left(\frac{2\pi}{3}\right) = \)
One of the two identical capacitors having the same capacitance C, is charged to a potential \(V_1\) and the other is charged to a potential \(V_2\). If they are connected with their like plates together, then the decrease in the electrostatic potential energy of the combined system is
Let \( P(x) = x^4 + ax^3 + bx^2 + cx + d \) be such that \( x = 0 \) is the only real root of \( P'(x) = 0 \). If \( P(-1)<P(1) \), then in the interval \( [-1,1] \):
The locus of a point at which the line joining the points (-3, 1, 2), (1, -2, 4) subtends a right angle, is
If \( \sinh^{-1}(2) + \sinh^{-1}(3) = \alpha \), then \( \sinh\alpha = \) ?
If $x^2 + y^2 + \sin y = 4$, then the value of $\frac{d^2y}{dx^2}$ at $x=-2$ is
Consider an arithmetic progression where the sum of the first 5 terms is 85, and the sum of the next 5 terms is 235. Find the common difference of the AP.
The difference between the absolute maximum and absolute minimum values of the function \( f(x) = 2x^3 - 15x^2 + 36x - 30 \) on \( [-1, 4] \) is:
For which reaction \( \Delta H \neq \Delta U \)?
A charged particle of mass \( m \) and charge \( q \) moves with a velocity \( \vec{v} \) perpendicular to a magnetic field \( \vec{B} \). The radius of the circular path it follows is:
If the lines \(x+y-2=0\), \(3x-4y+1=0\) and \(5x+ky-7=0\) are concurrent at \((\alpha, \beta)\), then equation of the line concurrent with the given lines and perpendicular to \(kx+y-k=0\) is
If \( y = \tan^{-1} \sqrt{x^2 - 1} + \sinh^{-1} \sqrt{x^2 - 1} \), \( x > 1 \), then \( \frac{dy}{dx} = \)
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 \( \theta \) is the angle between the tangents drawn from the point \( (-1, -1) \) to the circle \( x^2+y^2-4x-6y+c=0 \) and \( \cos\theta = -\frac{7}{25} \), then the radius of the circle is:
Bithionol is added to soaps to impart antiseptic properties. The number of –OH and –Cl groups in its structure are respectively