List of practice Questions

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 \( = 10 \, \text{m s}^{-2} \)]
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 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
If the magnitude of a vector \( \vec{p} \) is 25 units and its y-component is 7 units, then its x-component is
The general solution of the differential equation \( \frac{dy}{dx} = \frac{x+y}{x-y} \) 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
\( \int_{0}^{\pi/4} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} dx = \)
\( \int_{-1}^{4} \sqrt{\frac{4-x}{x+1}} \ dx = \)
The differential equation of the family of circles passing through the origin and having centre on X-axis is
The number of significant figures in the simplification of \( \frac{0.501}{0.05}(0.312-0.03) \) is
\( \int_{5\pi}^{25\pi} |\sin 2x + \cos 2x| \ dx = \)
If \( \int \frac{\cos^3 x}{\sin^2 x + \sin^4 x} dx = c - \operatorname{cosec} x - f(x) \), then \( f\left(\frac{\pi}{2}\right) = \)
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 \( (k-1)n \).
\( \int \frac{13\cos 2x - 9\sin 2x}{3\cos 2x - 4\sin 2x} dx = \)
\( \int \sqrt{x^2+x+1} \ dx \)
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
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 \( x_1+y_1 = \)
Which one of the following functions is monotonically increasing in its domain?
If the area of a right angled triangle with hypotenuse 5 is maximum, then its perimeter is
\( \int \left( \sum_{r=0}^{\infty} \frac{x^r 2^r}{r!} \right) dx = \)
If \( y = \tan^{-1}\left(\frac{x}{1+2x^2}\right) + \tan^{-1}\left(\frac{x}{1+6x^2}\right) \), then \( \frac{dy}{dx} = \)
\( \int \frac{dx}{12\cos x + 5\sin x} = \)
\( \lim_{n\to\infty} \frac{1}{n^3} \sum_{k=1}^{n} k^2 x = \)