\(\frac{\sqrt8}{\Pi}\)
\(\sqrt{\frac{8}{3}}\)
The correct answer is (B): \(\sqrt{\frac{8}{3}}\)
The motion of a particle in the XY plane is given by \( x(t) = 25 + 6t^2 \, \text{m} \); \( y(t) = -50 - 20t + 8t^2 \, \text{m} \). The magnitude of the initial velocity of the particle, \( v_0 \), is given by:
Consider a rope fixed at both ends under tension so that it is horizontal (i.e. assume the rope is along x-axis, with gravity acting along z-axis). Now the right end is continually oscillated at high frequency n (say n=100 Hz) horizontally and in a direction along the rope; amplitude of oscillation is negligible. The oscillation travells along the rope and is reflected at the left end.
Let the total length of rope be l, total mass be m and the acceleration due to gravity be g.
After initial phase (say a mintue or so), the rope has __(BLANK-1)__ wave, which is __(BLANK-2)__ in nature. It results from superposition of left travelling and right travelling __(BLANK-3)__ waves. This resulting wave has a frequency __ (BLANK-4)_ that of oscillation frequency nu. Simple dimensional analysis indicates that the frequency of can be of the form: ___(BLANK-5)__ .
In the following circuit, the reading of the ammeter will be: (Take Zener breakdown voltage = 4 V)
If $10 \sin^4 \theta + 15 \cos^4 \theta = 6$, then the value of $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$ is:
If the area of the region $\{ (x, y) : |x - 5| \leq y \leq 4\sqrt{x} \}$ is $A$, then $3A$ is equal to
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to