The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.

The disk has a smaller moment of inertia compared to the ring, hence it will reach the bottom first.

The disk has a smaller moment of inertia than the ring, resulting in greater acceleration down the incline.

The disk has a lower moment of inertia than the ring, allowing it to convert more potential energy into translational kinetic energy.

The disk will reach the bottom first because it has a smaller moment of inertia compared to the ring.

Using the formula ω = ω₀ + αt, we have ω = 10 + 2*5 = 20 rad/s.

The moment of inertia I of a disk about its axis is (1/2)MR^2. Therefore, the kinetic energy K.E. = (1/2)Iω^2 = (1/2)(1/2)MR^2ω^2 = (1/4)MR^2ω^2.

Angular momentum L = Iω = (1/2)MR^2ω, where I = (1/2)MR^2 for a disk.

Angular momentum L = Iω, where I = (1/2)MR^2 for a disk.

At the bottom, the total energy is converted into translational and rotational energy. The translational energy is 2/3 of the total energy.

At the bottom, the total kinetic energy is split equally between translational and rotational for a disk, hence the ratio is 1:1.

Using conservation of energy, potential energy at height h converts to kinetic energy at the bottom. The speed is √(2gh).

The relationship between linear speed and angular speed for rolling without slipping is given by ω = v/R.

To conserve angular momentum, if the radius is doubled, the angular velocity must be halved.

The moment of inertia I of a disk is proportional to r^2, so if the radius is doubled, I becomes 4I. Thus, angular momentum L = Iω becomes 4Iω.

The moment of inertia of a disk is I = (1/2)MR^2. If the radius is doubled, the new moment of inertia becomes I' = (1/2)M(2R)^2 = 4I.

Angular momentum L = Iω. If the radius is doubled, the moment of inertia increases by a factor of 4, thus L = 4Iω.

Angular momentum L = Iω, and if the radius is doubled, the moment of inertia I becomes 4I, thus L = 4Iω.

The linear velocity v = rω. If the radius is doubled, to maintain the same v, the angular velocity must remain ω.

The absolute error is given as 2 m.

Using the lens formula 1/f = 1/v - 1/u, we have 1/(-10) = 1/v - 1/5. Solving gives v = -3.33 cm.

A diverging lens always forms a virtual and upright image when the object is placed in front of it.

Torque (τ) = Force (F) × Distance (r) = 20 N × 0.8 m = 16 Nm.

Torque (τ) = F × d = 20 N × 1 m = 20 Nm.

Torque (τ) = Force (F) × Distance (r) = 50 N × 1 m = 50 Nm.

Torque = Force × Distance = 50 N × 1.2 m = 60 Nm.

Using the lens maker's formula, R = 2f(n-1) = 2*10*(1.5-1) = 20 cm.

Using the lens maker's formula, f = R/(n-1), we find f = 20/(1.5-1) = 40 cm.

For a drop to remain in equilibrium, the upward force due to surface tension must balance the downward gravitational force.

The spherical shape of a liquid drop is due to surface tension, which minimizes the surface area for a given volume.