Moment of inertia of a rod of mass ' M ' and length ' L ' about an axis passing through its center and normal to its length is ' $\alpha$ '. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :
Let $B_1$ be the magnitude of magnetic field at center of a circular coil of radius $R$ carrying current I. Let $\mathrm{B}_2$ be the magnitude of magnetic field at an axial distance ' $x$ ' from the center. For $x: \mathrm{R}=3: 4, \frac{\mathrm{~B}_2}{\mathrm{~B}_1}$ is :
A light wave is propagating with plane wave fronts of the type $x+y+z=$ constant. Th angle made by the direction of wave propagation with the $x$-axis is :
Match List I with List II.
List - I | List - II | ||
---|---|---|---|
(A) | Coefficient of viscosity | (I) | $\left[\mathrm{ML}^0 \mathrm{~T}^{-3}\right]$ |
(B) | Intensity of wave | (II) | $\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$ |
(C) | Pressure gradient | (III) | $\left[\mathrm{M}^{-1} \mathrm{LT}^2\right]$ |
(D) | Compressibility | (IV) | $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$ |
Choose the correct answer from the options given below: