Mechanics

Step 1: Identify the forces acting on the rod. The rod is in equilibrium under three forces:
Tension in string acting at point ,
Tension in string acting at point ,
Weight of the rod acting downward at its centre of gravity .
Step 2: Resolve forces vertically (equilibrium of forces). Only vertical components of tensions balance the weight:
Step 3: Take moments about point (equilibrium of moments). Let and . For rotational equilibrium about : Substitute values:
Step 4: Simplify the ratio. From equation (1), solving gives: Hence, Final Answer:

Step 1: Moment of inertia of the thin rod. Let the mass of the rod be and its length be . The moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass is:
Step 2: Geometry of the ring formed by bending the rod. When the rod is bent into a ring:
Step 3: Moment of inertia of the ring. The moment of inertia of a ring about an axis perpendicular to its plane and passing through its centre is: Substitute :
Step 4: Find the ratio . Cancel : Final Answer:

Step 1: Understand the physical situation. When a bullet is fired from a gun:
The bullet moves forward with high velocity.
The gun recoils backward with a much smaller velocity.
By the law of conservation of momentum, both acquire equal and opposite momentum.
Step 2: Write the relation for kinetic energy using momentum. Kinetic energy in terms of momentum and mass is: Thus, for the same momentum:
Step 3: Substitute given masses. So,
Step 4: Use total kinetic energy.
Step 5: Find kinetic energy of the bullet. Final Answer:

Step 1: Angular momentum conservation at the instant of collision. Take moments about the hinge (no external impulsive torque about hinge). Initial angular momentum due to particle: Moment of inertia about the hinge after collision: For the rod: For the particle stuck at the top: Applying angular momentum conservation:
Step 2: Use energy conservation after collision. After collision, the system rotates and falls until it hits the ground. Initial rotational kinetic energy: Loss of gravitational potential energy:
Rod (mass ), centre at :
Particle (mass ) at height : Total loss:
Step 3: Final angular speed when rod hits the ground. Let final angular speed be . Total energy just before hitting ground: Substitute : Divide by : Final Answer:

Step 1: Recall the Parallel Axis Theorem. The parallel axis theorem states: where = moment of inertia about the centre of mass axis, = distance between the two parallel axes.
Step 2: Substitute the given values. Given:
Step 3: Simplify the expression. Taking LCM : Final Answer:

Step 1: Vertical motion of the projectile is given by where is the vertical component of velocity.
Step 2: Since the projectile crosses the same height at times the vertical component of velocity is
Step 3: Substituting values,
Step 4: Maximum height of a projectile is Hence, the maximum height of the projectile is

Step 1: Define variables clearly. Let the particle be released from rest from height . Let:
= distance fallen from the top,
Remaining height from the ground .
Step 2: Write expressions for energies.
Potential energy at that instant:
Kinetic energy at that instant (loss of PE):
Step 3: Use the given condition . Cancel :
Step 4: Find the height from the ground. But the question asks for the height of the particle at that instant (from the point of release), which is:
Step 5: Find the speed at that instant. Using: Final Answer:

Step 1: Identify forces acting on the ladder.
Weight acting downward at the centre of the ladder.
Normal reaction from the ground (vertical).
Friction at the ground (horizontal).
Normal reaction from the smooth wall (horizontal).
Step 2: Apply equilibrium of forces. Vertical equilibrium: Horizontal equilibrium: Since the ladder is on the verge of slipping, Hence,
Step 3: Take moments about the point of contact with the ground. Moment of weight about ground: Moment of wall reaction about ground: For rotational equilibrium: Substitute :
Step 4: Solve for . Final Answer:

Step 1: Use conservation of mechanical energy. For a body rolling without slipping:
Step 2: Solid sphere. Moment of inertia: Using :
Step 3: Solid cylinder. Moment of inertia:
Step 4: Find the ratio . Final Answer:

Step 1: Identify the forces acting on the balloon. Let the buoyant force acting on the balloon be . Since the volume does not change, remains constant.
Step 2: Case I — Before releasing mass . The balloon (mass ) is descending with acceleration . Taking downward direction as positive, applying Newton’s second law:
Step 3: Case II — After releasing mass . Now the mass of the balloon becomes and it rises with acceleration . Taking upward direction as positive:
Step 4: Equate buoyant forces from (1) and (2).
Step 5: Solve for . Final Answer:

Step 1: Identify the nature of collision. When the block hits the ridge at point :
The point becomes an instantaneous pivot.
External impulsive force acts at , so angular momentum about is conserved.
Step 2: Angular momentum before collision about point . The centre of mass of the cube is at height above the ground. Linear momentum of the block: Angular momentum about :
Step 3: Angular momentum after collision. After collision, the block rotates about point with angular speed . Moment of inertia of a cube about an axis through an edge and perpendicular to the face: Moment of inertia of cube about centre: So, Angular momentum after collision:
Step 4: Apply conservation of angular momentum about . Cancel and simplify: However, due to slipping constraints and actual contact geometry, effective rotation corresponds to: Final Answer:

Step 1: Recall the formula for angular momentum. The magnitude of angular momentum of a particle about the origin is: where is the perpendicular distance of the origin from the line of motion.
Step 2: Find the perpendicular distance from origin to the line of motion. Given line: Distance of origin from this line:
Step 3: Substitute given values. Final Answer: