Equilibrium Questions

---

Very Short Answers Questions [1 Mark Questions]

Ques. Which type of equilibrium is seen in a ball kept at the bottom of a hemisphere?

1. Neutral equilibrium
2. Unstable equilibrium
3. Stable equilibrium
4. None of the above

Ans. The correct answer is c. Stable equilibrium

Explanation: When a body is in static equilibrium, it will move to its original position when it is displaced.

Ques. A body is in _____ if the object is slightly displaced and it stays in its new position.

1. Stable equilibrium
2. Unstable equilibrium
3. Neutral equilibrium
4. None of the above

Ans. The correct answer is c. Neutral equilibrium

Explanation: When an object is in neutral equilibrium, after being displaced from its original position to a new position, it will remain in that new position.

Ques. A body that is in equilibrium while it persists in moving with a constant velocity is known as

1. Dynamic equilibrium
2. Static equilibrium

Ans. The correct answer is a. Dynamic equilibrium

Explanation: A body is said to be in dynamic equilibrium when it is moving at a constant velocity.

Ques. _______ equilibrium is a state where bodies are at rest.

1. Dynamic
2. Static

Ans. The correct answer is b. Static

Explanation: A body at rest is said to be in static equilibrium.

Ques. If a rigid body is in equilibrium, the sum of the moments on the body is

1. -1
2. 0
3. 1
4. None of the above

Ans. The correct answer is b. 0

Explanation: The sum of the moments on the rigid body under equilibrium is zero.

---

Short Answers Questions [2 Marks Questions]

Ques. Define the equilibrium of a rigid body.

Ans. A body is said to be in equilibrium if it is at rest or moving with uniform velocity. No net force or net torque acts on the body if it is in equilibrium.

Ques. What are some examples of static equilibrium?

Ans. Examples of static equilibrium are

• A book resting on a table
• A ladder leaning against the wall
• A ball placed at the bottom of a bowl
• A hanging picture on the wall

Ques. What is the difference between static and dynamic equilibrium?

Ans. A body under static equilibrium experiences no net force or torque i.e. it will remain in its position until and unless an external force is applied to change its state of rest. For example, a book resting on a table.

Dynamic equilibrium refers to the state of a body when it is moving with constant velocity and no net force and torque acting on it. For example, the motion of the moon around the Earth.

Read Also:

---

Long Answers Questions [3 Marks Questions]

Ques. Define stable, unstable, and neutral equilibrium.

Ans. Stable Equilibrium: An object is said to be in stable equilibrium when it is displaced from its original position it will return to that position by itself. For example, a ball placed in the bottom of a bowl is in stable equilibrium.

Unstable Equilibrium: An object is said to be in unstable equilibrium when it is displaced from its original position it will not return to that position by itself. For example, a pencil standing on its tip is in unstable equilibrium.

Neutral Equilibrium: An object is said to be in neutral equilibrium when it is displaced from its original position it will remain in that position. For example, a ball resting on a flat surface is in neutral equilibrium.

Ques. A uniform ladder of length 5 m and mass 100 kg is in equilibrium between the vertical smooth wall and rough horizontal surface. The ladder makes an angle of 45° with the horizontal surface. Find the minimum friction coefficient between the floor and ladder for this equilibrium.

Ans. Given

• Length of the ladder, l = 5 m
• Mass of the ladder, m = 100 kg

Let N₁ be the normal reaction between the ladder and the vertical wall and N₂ be the normal reaction between the ladder and the horizontal surface. Also, let f be the force of friction between the ladder and the horizontal surface.

Since the ladder is in equilibrium, the sum of all forces is equal to zero.

Applying the condition of horizontal equilibrium, we get

N₁ = f

Applying the condition of vertical equilibrium, we get

N₂ = mg

Also, we have f = μN₂

⇒ f = μ mg

⇒ N₁ = μ mg

Taking torque about the point of contact between the ladder and the horizontal surface

mg l/2 cos θ - N₁ l sin θ = 0

⇒ mg l/2 cos θ - μ mg l sin θ = 0

⇒ μ = 1/2 cot θ

Given θ = 45°, therefore,

μ = 1/2

Ques. Define mechanical equilibrium. What are the conditions for an object to be in a state of static equilibrium?

Ans. Mechanical equilibrium is defined as the state of a body when it is at rest or moving with uniform velocity i.e. no net force or net torque acting on it.

The conditions for an object to be in a state of static equilibrium is

• The vector sum of all the external forces acting on an object must be zero.
• The vector sum of all the external torques acting on an object about any point must be zero.

---

Very Long Answers Questions [5 Marks Questions]

Ques. A uniform ladder of length 8 m weighing 100 N leans against a smooth vertical wall. The ladder makes an angle of 60° with the horizontal floor. A person weighing 600 N stands on the ladder at 2 meters from the bottom. Find the force exerted by the wall on the ladder. Also, find the reaction force at the point of contact between the ladder and the floor.

Ans. We will use the torque equilibrium condition to find the force exerted on the ladder by the wall. Torque (τ) is given by the product of force and perpendicular distance from the point of rotation.

Let T be the force exerted by the wall on the ladder, and R be the reaction force at the point of contact between the ladder and the floor.

Taking moments about the point of contact with the floor:

(600 N) × (2 m) = T × (8 m) × cos(60°)

1200 Nm = 8T × 0.5

T = 1200 N / 4

T = 300 N

Now, using the condition for vertical equilibrium:

T + R = Weight of the ladder

300 N + R = 100 N

R = 100 N - 300 N

R = -200 N 

The negative sign indicates an upward force.

So, the force exerted by the wall on the ladder is 300 N, and the reaction force at the point of contact between the ladder and the floor is 200 N upward.

Ques. A rectangular wooden block of density 800 kg m^-3 having dimensions 4 x 2 x 1 m floats in water. What is the mass of the block and the minimum force required to submerge the block completely in water? Also, determine the buoyant force acting on the block when it is partially submerged to a depth of 0.6 m.

Ans. Given

• Length, l = 4 m
• Width, w = 2 m
• Height, h= 1 m

Volume of the block = l × w × h = 4 × 2 × 1 = 8 m³

Also, given that the density of wood = 800 kg/m³

Mass of the block = Density × Volume = 800 kg/m³ × 8 m³ = 6400 kg

The buoyant force acting on the block is equal to its weight when the block is fully submerged. 

The buoyant force is given by Archimedes' principle i.e.

Buoyant force = Volume of fluid displaced × Density of fluid × g

Therefore, Buoyant force = 8 m³ × 1000 kg/m³ × 9.8 m/s² = 78400 N

Hence, the minimum force required to submerge the block completely in water is 78400 N.

When the block is partially submerged to a depth of 0.6 m, then the volume of fluid displaced is given by 

V = 4 × 2 × 0.6 = 4.8 m³.

Buoyant force = 4.8 m³ × 1000 kg/m³ × 9.8 m/s² = 47040 N

Ques. A uniform rod AB of length 6 m and weight 200 N is supported horizontally by two vertical strings attached to its ends. The angle between the rod and one of the strings is 30°, and with the other string is 60°. Calculate the tension in each string.

Ans. Let T₁ and T₂ be the tensions in the two strings attached to the rod. 

Also let R₁ and R₂ be the reaction forces at points A and B, respectively.

Using the condition for horizontal equilibrium, we have

T₁ × cos30° + T₂ × cos60° = 0

⇒ √3/2 T₁ + T₂/2 = 0

⇒ T₂ = - √3/2 T₁ …(i)

The rod is in equilibrium, therefore the sum of the vertical forces must be equal to its weight. Using the condition for vertical equilibrium, we have

T₁ × sin30° + T₂ × sin60° - 200 = 0

T₁/2 + √3/2 T₂ - 200 = 0 …(ii)

On substituting equation (i) in the above equation, we get

T₁/2 + (√3/2)(- √3/2 T₁) = 200

⇒ T₁/2 - 3/4 T₁ = 200

⇒ T₁/4 = 200

⇒ T₁ = 800 N

Substituting T₁ = 800 N in equation (ii), we get

T₂ = - 400√3 N

---

Previous Year Questions

1. Extraction of metal from the ore cassiterite involves...[JEE Advanced 2011]
2. Commonly used vectors for human genome sequencing are...[NEET UG 2014]
3. Interfascicular cambium and cork cambium are formed due to​..
4. Pneumotaxic centre is present in​...[UP CPMT 2007]
5. Reaction of HBr with propene in the presence of peroxide gives….[NEET UG 2004]
6. Assuming the expression for the pressure exerted by the gas on the walls of the container, it can be shown that pressure is...[MHT CET 2016]
7. Which among the following is the strongest acid?...[TS EAMCET 2017]
8. Isopropyl alcohol on oxidation forms​..
9. A vector is not changed if​..
10. Which of the following arrangements does not represent the correct order of the property stated against it?...[JEE Main 2013]

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates

---

Check-Out: