The problem asks for the moment of inertia of a pair of solid spheres about a tangent passing through their point of contact. We are given the properties of two solid spheres:
The moment of inertia of a solid sphere about an axis passing through its center and perpendicular to a diameter (i.e., about a diameter) is given by the formula: $I_{cm} = \frac{2}{5}MR^2$
To find the total moment of inertia about the specified tangent axis, we calculate the moment of inertia for each sphere using this formula and sum them up.
Using the formula $I = \frac{2}{5}MR^2$ for Sphere 1:
$ I_1 = \frac{2}{5} m_1 r_1^2 $ Substitute the values: $ I_1 = \frac{2}{5} \times (5 \text{ kg}) \times (0.1 \text{ m})^2 $ $ I_1 = \frac{2}{5} \times 5 \times 0.01 $ $ I_1 = 2 \times 0.01 $ $ I_1 = 0.02 \text{ kg.m}^2 $Using the formula $I = \frac{2}{5}MR^2$ for Sphere 2:
$ I_2 = \frac{2}{5} m_2 r_2^2 $ Substitute the values: $ I_2 = \frac{2}{5} \times (10 \text{ kg}) \times (0.2 \text{ m})^2 $ $ I_2 = \frac{2}{5} \times 10 \times 0.04 $ $ I_2 = 4 \times 0.04 $ $ I_2 = 0.16 \text{ kg.m}^2 $The total moment of inertia of the pair of spheres about the tangent passing through the point of contact is the sum of the individual moments of inertia:
$ I_{total} = I_1 + I_2 $ $ I_{total} = 0.02 \text{ kg.m}^2 + 0.16 \text{ kg.m}^2 $ $ I_{total} = 0.18 \text{ kg.m}^2 $A cylindrical tube AB of length $l$, closed at both ends contains an ideal gas of 1 mol having molecular weight $M$. The tube is rotated in a horizontal plane with constant angular velocity $\omega$ about an axis perpendicular to AB and passing through the edge at end A, as shown in the figure. If $P_A$ and $P_B$ are the pressures at $A$ and $B$ respectively, then
(Consider the temperature is same at all points in the tube)

A uniform bar of length 12 cm and mass $20m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in opposite directions with same speed of $v$ and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$. The ratio of $v$ and $\omega$ is :
Match the LIST-I with LIST-II
| List-I: | List-II: |
| A. Magnetic induction | I. $[M L T^{-2} A^{-2}]$ |
| B. Magnetic flux | II. $[M L^2 T^{-2} A^{-2}]$ |
| C. Magnetic permeability | III. $[M L^0 T^{-2} A^{-1}]$ |
| D. Self inductance | IV. $[M L^2 T^{-2} A^{-1}]$ |
Choose the correct answer from the options given below:
A cylindrical tube AB of length $l$, closed at both ends contains an ideal gas of 1 mol having molecular weight $M$. The tube is rotated in a horizontal plane with constant angular velocity $\omega$ about an axis perpendicular to AB and passing through the edge at end A, as shown in the figure. If $P_A$ and $P_B$ are the pressures at $A$ and $B$ respectively, then
(Consider the temperature is same at all points in the tube)
