A body of mass \( m \) and radius \( r \) rolling horizontally with velocity \( V \), rolls up an inclined plane to a vertical height \( \frac{V^2}{g} \). The body is:
a solid cylinder
Step 1: Apply Energy Conservation Using the conservation of mechanical energy: \[ \text{Initial Energy} = \text{Final Energy} \] \[ \frac{1}{2} m V^2 + \frac{1}{2} I \omega^2 = mg h \] Since the body is rolling without slipping: \[ \omega = \frac{V}{r} \] The moment of inertia of a rolling object determines how much kinetic energy is converted into potential energy.
Step 2: Finding the Height \[ \frac{1}{2} m V^2 + \frac{1}{2} \left( I \frac{V^2}{r^2} \right) = mg h \] \[ \frac{1}{2} m V^2 \left( 1 + \frac{I}{m r^2} \right) = mg h \] Solving for \( h \): \[ h = \frac{V^2}{g} \times \frac{1}{1 + \frac{I}{m r^2}} \] For a circular ring, \( I = m r^2 \), so: \[ h = \frac{V^2}{g} \times \frac{1}{1 + 1} = \frac{V^2}{2g} \] The problem states that \( h = \frac{V^2}{g} \), which matches the condition for a circular ring. Thus, the body is a circular ring.
A constant force of \[ \mathbf{F} = (8\hat{i} - 2\hat{j} + 6\hat{k}) \text{ N} \] acts on a body of mass 2 kg, displacing it from \[ \mathbf{r_1} = (2\hat{i} + 3\hat{j} - 4\hat{k}) \text{ m to } \mathbf{r_2} = (4\hat{i} - 3\hat{j} + 6\hat{k}) \text{ m}. \] The work done in the process is:
A ball 'A' of mass 1.2 kg moving with a velocity of 8.4 m/s makes a one-dimensional elastic collision with a ball 'B' of mass 3.6 kg at rest. The percentage of kinetic energy transferred by ball 'A' to ball 'B' is:
A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 9 g, are kept one above the other at the 10 cm mark, the scale is found to be balanced at 35 cm. The mass of the metre scale is:
A liquid cools from a temperature of 368 K to 358 K in 22 minutes. In the same room, the same liquid takes 12.5 minutes to cool from 358 K to 353 K. The room temperature is:
The general solution of the differential equation: \[ (6x^2 - 2xy - 18x + 3y) dx - (x^2 - 3x) dy = 0 \]
The range of gravitational forces is:
In a simple pendulum experiment for the determination of acceleration due to gravity, the error in the measurement of the length of the pendulum is 1% and the error in the measurement of the time period is 2%. The error in the estimation of acceleration due to gravity is:
When the object and the screen are 90 cm apart, it is observed that a clear image is formed on the screen when a convex lens is placed at two positions separated by 30 cm between the object and the screen. The focal length of the lens is: