The moment of inertia \( I \) of each sphere about the central axis (using the parallel axis theorem) is:
\[ I_{\text{total}} = 2 \left( I_{\text{sphere}} + md^2 \right). \]For a solid sphere:
\[ I_{\text{sphere}} = \frac{2}{5}mR^2 = \frac{2}{5} \times 2 \times (0.5)^2 = 0.2 \, \text{kg m}^2. \]Distance \( d \) from the center of each sphere to the midpoint of the rod is \( 0.75 \, \text{m} \).
So,
\[ I_{\text{total}} = 2 \left( 0.2 + 2 \times (0.75)^2 \right) = 2 \left( 0.2 + 1.125 \right) = \frac{53}{20} \, \text{kg m}^2. \]Thus, \( x = 53 \).
List-I | List-II | ||
P | The capacitance between S1 and S4, with S2 and S3 not connected, is | I | \(3C_0\) |
Q | The capacitance between S1 and S4, with S2 shorted to S3, is | II | \(\frac{C_0}{2}\) |
R | The capacitance between S1 and S3, with S2 shorted to S4, is | III | \(\frac{C_0}{3}\) |
S | The capacitance between S1 and S2, with S3 shorted to S1, and S2 shorted to S4, is | IV | \(2\frac{C_0}{3}\) |
\[2C_0\] |
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is: