Mechanics Practice Questions — JAMB UTME — Page 5

Question 81Question →

If two objects of unequal mass experience the same magnitude of net force for the same duration of time, the lighter object will undergo a greater change in linear momentum than the heavier object.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Both objects undergo the exact same change in linear momentum because impulse depends only on the force applied and the time duration, not on the mass of the object.

The statement is false because the change in momentum (

\Delta p

) is determined solely by the impulse applied (

F \Delta t

). Since the net force and time duration are identical for both objects, the change in momentum is the same for both, regardless of mass differences.

Step-by-Step Solution

1

State the relationship between force, time, and momentum change using Newton's Second Law.

Impulse

J = F \Delta t = \Delta p

, where

F

is the net force,

\Delta t

is the duration, and

\Delta p

is the change in linear momentum.

The impulse-momentum theorem directly links the net force acting over time to the resulting change in momentum.

2

Evaluate the given problem constraints for both the lighter mass (

m_1

) and heavier mass (

m_2

).

F_1 = F_2 = F

and

\Delta t_1 = \Delta t_2 = \Delta t

.

Both objects experience equal forces for equal durations.

3

Compare the resulting changes in linear momentum.

\Delta p_1 = F \Delta t

and

\Delta p_2 = F \Delta t

, so

\Delta p_1 = \Delta p_2

.

Because mass

m

does not alter the product

F \Delta t

, both objects experience identical momentum changes.

Key Concept

Impulse-Momentum Theorem and Newton's Second Law of Motion

Question 82Question →

A sledge of mass $8.0\text{ kg}$ sliding on ice with an initial velocity of $15\text{ m s}^{-1}$ enters a rough patch that exerts a constant retarding force of $24\text{ N}$ . Calculate the time, in seconds, required for the sledge to come to a complete stop.

Show answer & explanation

Answer: 5

Answer

The time required for the sledge to come to a complete stop is

5.0\text{ s}

.

By Newton's second law in terms of momentum, the rate of change of momentum is equal to the applied net force (

F = \frac{\Delta p}{\Delta t}

). Rearranging gives

\Delta t = \frac{m(v - u)}{F}

. Substituting

m = 8.0\text{ kg}

,

u = 15\text{ m s}^{-1}

,

v = 0\text{ m s}^{-1}

, and retarding force

F = -24\text{ N}

yields

\Delta t = \frac{8.0 \times (0 - 15)}{-24} = 5.0\text{ s}

.

Step-by-Step Solution

1

Determine the change in linear momentum of the sledge.

The change in linear momentum is

\Delta p = m(v - u) = 8.0\text{ kg} \times (0\text{ m s}^{-1} - 15\text{ m s}^{-1}) = -120\text{ kg m s}^{-1}

.

Linear momentum is defined as the product of mass and velocity.

2

Apply the impulse-momentum theorem to determine the time duration.

\Delta t = \frac{\Delta p}{F} = \frac{-120\text{ kg m s}^{-1}}{-24\text{ N}} = 5.0\text{ s}

.

Impulse delivered by a net force over a time interval equals the change in linear momentum.

Key Concept

Newton's Second Law and Impulse-Momentum Theorem

Question 83Question →

A spring with a force constant of $400\text{ N m}^{-1}$ is compressed by $0.1\text{ m}$ from its uncompressed length. When released, it launches a block of mass $0.16\text{ kg}$ along a smooth horizontal surface. What is the speed of the block immediately after leaving the spring?

5.0\text{ m s}^{-1}

2.5\text{ m s}^{-1}

12.5\text{ m s}^{-1}

25.0\text{ m s}^{-1}

Show answer & explanation

Answer: $5.0\text{ m s}^{-1}$

Answer

The speed of the block immediately after leaving the spring is

5.0\text{ m s}^{-1}

.

According to the principle of conservation of mechanical energy, the elastic potential energy stored in the compressed spring (

E_p = \frac{1}{2} k x^2

) is completely converted into kinetic energy (

E_k = \frac{1}{2} m v^2

) when the spring is released on a frictionless surface. Substituting

k = 400\text{ N m}^{-1}

,

x = 0.1\text{ m}

, and

m = 0.16\text{ kg}

yields

2.0\text{ J} = 0.08 v^2

, which solves to

v = 5.0\text{ m s}^{-1}

.

Step-by-Step Solution

1

Calculate the elastic potential energy stored in the compressed spring.

E_p = \frac{1}{2} k x^2 = \frac{1}{2} \times 400 \times (0.1)^2 = 200 \times 0.01 = 2.0\text{ J}

Work done in compressing the spring is stored as elastic potential energy.

2

Apply conservation of mechanical energy.

E_k = E_p \implies \frac{1}{2} m v^2 = 2.0\text{ J}

Since the surface is smooth, all potential energy transforms into kinetic energy.

3

Solve for the velocity

v

of the block.

\frac{1}{2} \times 0.16 \times v^2 = 2.0 \implies 0.08 v^2 = 2.0 \implies v^2 = 25 \implies v = 5.0\text{ m s}^{-1}

Isolate

v

by dividing by

0.08

and taking the square root.

Key Concept

Conservation of Mechanical Energy (Elastic Potential Energy to Kinetic Energy)

Question 84Question →

A crate is pulled along a smooth horizontal floor by a constant force of $80\text{ N}$ applied at an angle of $60^\circ$ to the horizontal. If the crate moves through a displacement of $15\text{ m}$ along the floor, what is the work done by the applied force?

600\text{ J}

1039\text{ J}

1200\text{ J}

300\text{ J}

Show answer & explanation

Answer: $600\text{ J}$

Answer

600\text{ J}

The work done by a constant force applied at an angle

\theta

to the direction of motion is given by

W = F s \cos\theta

. Substituting

F = 80\text{ N}

,

s = 15\text{ m}

, and

\cos(60^\circ) = 0.5

gives

W = 80 \times 15 \times 0.5 = 600\text{ J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Force

F = 80\text{ N}

, displacement

s = 15\text{ m}

, angle

\theta = 60^\circ

Extract parameters required for the work formula.

2

Apply the work formula for a force at an angle

W = F s \cos\theta

Only the component of force parallel to the displacement does work on the object.

3

Substitute the values and compute the result

W = 80 \times 15 \times \cos(60^\circ) = 1200 \times 0.5 = 600\text{ J}

Calculates the effective work done by the applied force.

Key Concept

Work Done by a Constant Force at an Angle

Estimated Time:1m 0s

Question 85Question →

An electric motor with an efficiency of $80\%$ is used to pull a $100\text{ kg}$ object up a smooth incline inclined at $30^\circ$ to the horizontal at a constant speed of $2\text{ m s}^{-1}$ . Taking acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what is the electrical power input to the motor, in watts?

Show answer & explanation

Answer: 1250

Answer

1250 W

The force needed to move the mass up the smooth incline at constant speed is the parallel component of weight,

F = mg \sin(30^\circ) = 500\text{ N}

. The useful power output is

P_{\text{out}} = Fv = 500 \times 2 = 1000\text{ W}

. Dividing by the efficiency of

0.80

yields the electrical power input

P_{\text{in}} = 1250\text{ W}

.

Step-by-Step Solution

1

Determine the force required along the inclined plane.

F = mg \sin(30^\circ) = 100\text{ kg} \times 10\text{ m s}^{-2} \times 0.5 = 500\text{ N}

At constant velocity, the applied force balances the component of weight parallel to the incline.

2

Calculate the useful power output delivered by the motor.

P_{\text{out}} = F \times v = 500\text{ N} \times 2\text{ m s}^{-1} = 1000\text{ W}

Mechanical power output is the product of pulling force and constant speed.

3

Calculate the total electrical power input required.

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{1000\text{ W}}{0.80} = 1250\text{ W}

Efficiency is defined as the ratio of useful power output to total power input.

Key Concept

Mechanical power on inclined planes and system efficiency

Question 86Question →

A water pump raises $600\text{ kg}$ of water through a vertical height of $20\text{ m}$ in $50\text{ s}$ . If the efficiency of the pump is $80\%$ , what is the electrical power input required to operate the pump? (Take $g = 10\text{ m s}^{-2}$ )

3.0\text{ kW}

2.4\text{ kW}

1.92\text{ kW}

120.0\text{ kW}

Show answer & explanation

Answer: $3.0\text{ kW}$

Answer

3.0\text{ kW}

The correct answer is

3.0\text{ kW}

. Raising

600\text{ kg}

of water by

20\text{ m}

requires

120,000\text{ J}

of gravitational potential energy. Doing this in

50\text{ s}

requires an output power of

2,400\text{ W}

(

2.4\text{ kW}

). Since the pump operates at

80\%

efficiency, the input power must be

\frac{2.4\text{ kW}}{0.80} = 3.0\text{ kW}

.

Step-by-Step Solution

1

Calculate the useful work output needed to elevate the water

W_{out} = mgh = 600\text{ kg} \times 10\text{ m s}^{-2} \times 20\text{ m} = 120,000\text{ J}

Work done against gravity equals the gravitational potential energy gained

2

Calculate the useful output power of the pump

P_{out} = \frac{W_{out}}{t} = \frac{120,000\text{ J}}{50\text{ s}} = 2,400\text{ W} = 2.4\text{ kW}

Power is defined as the rate at which work is performed

3

Calculate the required electrical power input using efficiency

P_{in} = \frac{P_{out}}{\text{Efficiency}} = \frac{2.4\text{ kW}}{0.80} = 3.0\text{ kW}

Efficiency is the ratio of useful output power to total input power

Key Concept

Work done against gravity, power, and mechanical/electrical efficiency

Estimated Time:1m 30s

Question 87Question →

A constant horizontal force acts on a body of mass $10\text{ kg}$ , accelerating it from rest to a speed of $12\text{ m s}^{-1}$ in a time of $4\text{ s}$ along a smooth horizontal surface. What is the average power delivered by the force during this time interval?

Show answer & explanation

Answer: 180

Answer

The average power delivered by the force during the 4-second interval is

180\text{ W}

.

By the work-energy theorem, the total work done by the constant force equals the gain in kinetic energy:

W = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times 144 = 720\text{ J}

. The average power is the rate at which work is performed over time:

P = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

.

Step-by-Step Solution

1

Calculate the final kinetic energy acquired by the body.

E_k = \frac{1}{2} m v^2 = \frac{1}{2} (10\text{ kg})(12\text{ m s}^{-1})^2 = 720\text{ J}

.

Since the body starts from rest on a smooth surface, all work done by the net force goes into increasing its kinetic energy.

2

Divide the total work done by the elapsed time to find the average power.

P_{\text{avg}} = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

.

Average power is defined as the rate of doing work over a given time interval.

Key Concept

Work-Energy Theorem and Average Power

Question 88Question →

A hydraulic press consists of a small effort piston of radius $2\text{ cm}$ and a large load piston of radius $8\text{ cm}$ . When an effort force of $50\text{ N}$ is applied to the small piston, it successfully lifts a load of $600\text{ N}$ placed on the large piston. What is the efficiency of this hydraulic press?

75\%

133\%

33\%

25\%

Show answer & explanation

Answer: $75\%$

Answer

The efficiency of the hydraulic press is

75\%

.

The correct answer is

75\%

. The mechanical advantage is

MA = \frac{600}{50} = 12

, and the velocity ratio is

VR = \left(\frac{8}{2}\right)^2 = 16

. Taking the ratio

\frac{MA}{VR} \times 100\%

yields

\frac{12}{16} \times 100\% = 75\%

.

Step-by-Step Solution

1

Calculate the Mechanical Advantage (MA) of the hydraulic press.

MA = \frac{\text{Load}}{\text{Effort}} = \frac{600\text{ N}}{50\text{ N}} = 12

Mechanical advantage measures the force multiplication factor of a machine.

2

Calculate the Velocity Ratio (VR) of the hydraulic press using the piston radii.

VR = \frac{\text{Area of load piston}}{\text{Area of effort piston}} = \frac{\pi R^2}{\pi r^2} = \left(\frac{8\text{ cm}}{2\text{ cm}}\right)^2 = 4^2 = 16

For a hydraulic press, velocity ratio equals the ratio of the cross-sectional areas of the pistons, which simplifies to the square of the radii ratio.

3

Compute the efficiency of the machine.

\eta = \frac{MA}{VR} \times 100\% = \frac{12}{16} \times 100\% = 75\%

Efficiency is defined as the ratio of Mechanical Advantage to Velocity Ratio expressed as a percentage.

Key Concept

Mechanical Advantage, Velocity Ratio, and Efficiency of a Hydraulic Press

Estimated Time:1m 30s

Question 89Question →

A block and tackle system consisting of $5$ pulleys is used to raise a load of $200\text{ N}$ through a vertical height of $4\text{ m}$ . If the efficiency of the system is $80\%$ , what is the work done against friction, in joules, during the lifting process?

Show answer & explanation

Answer: 200

Answer

The work done against friction during the lifting process is

200\text{ J}

.

The useful work output is achieved by lifting the

200\text{ N}

load through a vertical height of

4\text{ m}

, yielding

W_{\text{out}} = 200 \times 4 = 800\text{ J}

. Given an efficiency of

80\%

(

0.80

), the total work input required from the effort force is

W_{\text{in}} = \frac{800}{0.80} = 1000\text{ J}

. The energy lost to overcome frictional resistance in the pulleys equals the total work input minus useful work output:

1000\text{ J} - 800\text{ J} = 200\text{ J}

.

Step-by-Step Solution

1

Calculate useful work output

W_{\text{out}} = 800\text{ J}

Useful work output is the energy required to raise the load through the specified height (

W_{\text{out}} = \text{Load} \times \text{height}

).

2

Determine total work input

W_{\text{in}} = 1000\text{ J}

The total work input is calculated from the efficiency formula:

\text{Efficiency} = \frac{W_{\text{out}}}{W_{\text{in}}}

.

3

Compute work done against friction

W_{\text{friction}} = 200\text{ J}

The work lost overcoming friction is the difference between total work input and useful work output (

W_{\text{in}} - W_{\text{out}}

).

Key Concept

Work and Efficiency in Pulley Systems

Estimated Time:1m 30s

Question 90Question →

A machine with a velocity ratio of $6$ is used to raise a load of $540\text{ N}$ by applying an effort of $120\text{ N}$ . What is the efficiency of the machine?

75\%

133.3\%

22.2\%

450\%

Show answer & explanation

Answer: $75\%$

Answer

The efficiency of the machine is

75\%

.

The correct option is

75\%

. Mechanical Advantage is found by dividing the load (

540\text{ N}

) by the effort (

120\text{ N}

), giving

4.5

. Dividing this mechanical advantage by the given velocity ratio (

6

) and expressing as a percentage yields

\frac{4.5}{6} \times 100\% = 75\%

.

Step-by-Step Solution

1

Calculate Mechanical Advantage (MA)

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{540\text{ N}}{120\text{ N}} = 4.5

Mechanical advantage measures the force magnification produced by the machine.

2

Calculate Efficiency (\eta)

\eta = \left(\frac{\text{MA}}{\text{VR}}\right) \times 100\% = \left(\frac{4.5}{6}\right) \times 100\% = 75\%

Efficiency is the ratio of Mechanical Advantage to Velocity Ratio multiplied by 100%.

Key Concept

Efficiency of a Simple Machine

Estimated Time:1m 0s

Question 91Question →

An inclined plane is set at an angle of $30^\circ$ to the horizontal. An effort of $320\text{ N}$ applied parallel to the plane is used to push a load of $480\text{ N}$ up the incline at a constant speed. What is the efficiency of the inclined plane expressed as a percentage?

Show answer & explanation

Answer: 75%; 75; 75 percent

Answer

The efficiency of the inclined plane is

75\%

.

For an inclined plane inclined at

30^\circ

to the horizontal, the velocity ratio (VR) is given by

\text{VR} = \frac{1}{\sin 30^\circ} = 2

. The mechanical advantage (MA) is

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{480\text{ N}}{320\text{ N}} = 1.5

. Dividing MA by VR and multiplying by

100\%

yields an efficiency of

\left(\frac{1.5}{2}\right) \times 100\% = 75\%

.

Step-by-Step Solution

1

Determine the Velocity Ratio (VR) of the inclined plane from its angle of inclination

VR = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2

For an inclined plane with inclination angle \theta, the velocity ratio is equal to \frac{1}{\sin \theta}.

2

Calculate the Mechanical Advantage (MA)

MA = \frac{\text{Load}}{\text{Effort}} = \frac{480\text{ N}}{320\text{ N}} = 1.5

Mechanical advantage is defined as the ratio of load to effort force.

3

Calculate the efficiency of the machine

\text{Efficiency} = \frac{\text{MA}}{\text{VR}} \times 100\% = \frac{1.5}{2} \times 100\% = 75\%

Efficiency is the ratio of mechanical advantage to velocity ratio expressed as a percentage.

Key Concept

Efficiency of an Inclined Plane

Question 92Question →

A light rigid bar $PQ$ of length $5.0\text{ m}$ is hinged at end $P$ and held horizontally. An upward force of $40\text{ N}$ is applied at end $Q$ at an angle of $30^\circ$ to the bar. To keep the bar in horizontal equilibrium, a vertical downward force $F$ is applied at a distance of $2.0\text{ m}$ from $P$ . What is the magnitude of the force $F$ ?

20\text{ N}

50\text{ N}

86.6\text{ N}

100\text{ N}

Show answer & explanation

Answer: $50\text{ N}$

Answer

The magnitude of the force

F

required to maintain equilibrium is

50\text{ N}

.

The force of

40\text{ N}

applied at an angle of

30^\circ

to the bar has a perpendicular component of

40\sin 30^\circ = 20\text{ N}

. The counterclockwise moment about pivot

P

is

20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}

. For the bar to remain in equilibrium, the clockwise moment created by

F

must equal

100\text{ N}\cdot\text{m}

, giving

F \times 2.0\text{ m} = 100\text{ N}\cdot\text{m}

, which yields

F = 50\text{ N}

.

Step-by-Step Solution

1

Calculate the perpendicular component of the force applied at end

Q

.

F_{\perp} = 40\text{ N} \times \sin(30^\circ) = 40 \times 0.5 = 20\text{ N}

Only the force component perpendicular to the bar produces a moment about the pivot

P

.

2

Calculate the counterclockwise moment produced by the force at

Q

about pivot

P

.

\text{Moment}_{Q} = 20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}

Moment is defined as perpendicular force multiplied by distance from the pivot.

3

Apply the Principle of Moments about pivot

P

to find force

F

.

F \times 2.0\text{ m} = 100\text{ N}\cdot\text{m} \implies F = \frac{100}{2.0} = 50\text{ N}

For rotational equilibrium, total clockwise moment about

P

must equal total counterclockwise moment about

P

.

Key Concept

Principle of Moments and Rotational Equilibrium with Forces at an Angle

Question 93Question →

A uniform horizontal wooden beam $AB$ of length $4.0\text{ m}$ and mass $8.0\text{ kg}$ is supported on a pivot placed $1.0\text{ m}$ from end $A$ . A mass of $5.0\text{ kg}$ is suspended from end $B$ . What is the magnitude of the downward vertical force $F$ in newtons that must be applied at end $A$ to keep the beam in horizontal equilibrium? (Take $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 230

Answer

The magnitude of the downward force required at end A to maintain horizontal equilibrium is 230 N.

Taking moments about the pivot, the downward force F at end A produces an anticlockwise moment of F × 1.0 m. This balances the clockwise moments produced by the weight of the beam (80 N × 1.0 m) and the load at end B (50 N × 3.0 m). Equating anticlockwise and clockwise moments gives F × 1.0 = 80 + 150 = 230 N.

Step-by-Step Solution

1

Determine the weights and distance of each force from the pivot point

Weight of beam = 80 N acting at 1.0 m right of pivot; weight at B = 50 N acting at 3.0 m right of pivot; force F acts at 1.0 m left of pivot.

The center of gravity of a uniform 4.0 m beam is at its midpoint (2.0 m from end A).

2

Set up the moment equilibrium equation about the pivot

F × 1.0 = (80 × 1.0) + (50 × 3.0)

For static rotational equilibrium, total anticlockwise moments equal total clockwise moments.

3

Solve for the force magnitude F

F = 230 N

Summing clockwise moments yields 80 + 150 = 230 N m, which divided by 1.0 m gives F = 230 N.

Key Concept

Principle of Moments and Static Equilibrium

Estimated Time:1m 30s

Question 94Question →

A uniform horizontal beam $XY$ of length $3.0\text{ m}$ and weight $80\text{ N}$ is hinged at end $X$ to a vertical wall. It is held in horizontal equilibrium by a light cable attached to end $Y$ that makes an angle of $30^\circ$ with the beam. What is the tension in the cable?

40\text{ N}

80\text{ N}

160\text{ N}

240\text{ N}

Show answer & explanation

Answer: $80\text{ N}$

Answer

The tension in the cable is

80\text{ N}

.

Taking moments about the hinge at end

X

, the clockwise moment due to the weight of the beam acting at its midpoint (

1.5\text{ m}

) is

80\text{ N} \times 1.5\text{ m} = 120\text{ N}\cdot\text{m}

. The counterclockwise moment provided by the cable tension

T

at end

Y

(

3.0\text{ m}

) is

T \sin(30^\circ) \times 3.0\text{ m} = 1.5 T\text{ N}\cdot\text{m}

. Setting clockwise moments equal to counterclockwise moments gives

1.5 T = 120

, yielding

T = 80\text{ N}

.

Step-by-Step Solution

1

Identify the center of gravity and calculate the clockwise moment about the hinge at end

X

.

Since the beam is uniform, its weight of

80\text{ N}

acts at its center of gravity, which is at a distance of

1.5\text{ m}

from

X

. Clockwise moment =

80\text{ N} \times 1.5\text{ m} = 120\text{ N}\cdot\text{m}

.

For a uniform body, weight acts precisely at the midpoint.

2

Determine the counterclockwise moment exerted by the cable tension

T

about point

X

.

The perpendicular component of the tension is

T \sin(30^\circ)

. Counterclockwise moment =

T \sin(30^\circ) \times 3.0\text{ m} = 1.5 T\text{ N}\cdot\text{m}

.

Only the component of force perpendicular to the beam produces a moment about the pivot.

3

Apply the principle of moments for rotational equilibrium to solve for

T

.

1.5 T = 120 \implies T = \frac{120}{1.5} = 80\text{ N}

.

For equilibrium, total counterclockwise moments about any pivot must equal total clockwise moments.

Key Concept

Principle of Moments and Rotational Equilibrium

Estimated Time:1m 30s

Question 95Question →

A uniform horizontal plank $AB$ of length $5.0\text{ m}$ and mass $40\text{ kg}$ rests on two smooth supports, $C$ and $D$ , located $1.0\text{ m}$ from end $A$ and $1.0\text{ m}$ from end $B$ respectively. What is the maximum mass, in kilograms, of an object that can be placed at end $A$ without causing the plank to tilt?

Show answer & explanation

Answer: 60

Answer

The maximum mass that can be placed at end A without tilting the plank is 60 kg.

Just before tilting, the plank rotates around support C, causing the normal reaction at support D to drop to zero. Equating the anticlockwise moment of the added mass at end A about support C (

(m \cdot g) \times 1.0\text{ m}

) to the clockwise moment of the plank's weight about support C (

(40 \cdot g) \times 1.5\text{ m}

) yields

m = 60\text{ kg}

.

Step-by-Step Solution

1

Identify the tipping condition and pivot point

Support C acts as the pivot; the reaction at support D becomes zero (

R_D = 0

).

When extra weight is added at end A, the plank rotates about C and lifts off support D.

2

Calculate perpendicular distances from the pivot C

Distance to added mass

m

=

1.0\text{ m}

; Distance to plank's center of mass =

2.5\text{ m} - 1.0\text{ m} = 1.5\text{ m}

.

The weight of a uniform beam acts at its midpoint (2.5 m from either end).

3

Equate clockwise and counterclockwise moments about C

m \times g \times 1.0\text{ m} = 40\text{ kg} \times g \times 1.5\text{ m}

, giving

m = 60\text{ kg}

.

For the plank to remain balanced just before tipping, total anticlockwise moment must equal total clockwise moment.

Key Concept

Rotational Equilibrium and Tilting of Rigid Bodies

Estimated Time:1m 30s

Question 96Question →

A uniform rigid bar $AB$ of length $2.0\text{ m}$ and weight $120\text{ N}$ is hinged smoothly at end $A$ to a vertical post. The bar is held in equilibrium at an angle of $60^\circ$ above the horizontal by a force $F$ applied at end $B$ acting perpendicular to the bar. What is the magnitude of the force $F$ ?

30\text{ N}

60\text{ N}

120\text{ N}

15\text{ N}

Show answer & explanation

Answer: $30\text{ N}$

Answer

The magnitude of the force required to keep the bar in equilibrium is

30\text{ N}

.

To maintain rotational equilibrium, the clockwise moment created by the weight of the bar about hinge

A

must equal the counterclockwise moment created by force

F

. The weight of

120\text{ N}

acts at the bar's midpoint (

1.0\text{ m}

from

A

), and its perpendicular distance to the vertical line of action is

1.0 \cos(60^\circ) = 0.5\text{ m}

. Thus, the clockwise moment is

120 \times 0.5 = 60\text{ N}\cdot\text{m}

. Since force

F

acts perpendicularly at the end of the

2.0\text{ m}

bar, its moment is

F \times 2.0

. Setting

2.0 F = 60

gives

F = 30\text{ N}

.

Step-by-Step Solution

1

Identify the center of gravity and the position of applied forces.

For a uniform bar of length

L = 2.0\text{ m}

, its weight

W = 120\text{ N}

acts vertically downward at its center of gravity, which is at a distance of

1.0\text{ m}

from hinge

A

.

The weight of a uniform body acts through its midpoint.

2

Determine the perpendicular distance for each force relative to the pivot at

A

.

Perpendicular distance for weight:

d_W = 1.0\text{ m} \times \cos(60^\circ) = 0.5\text{ m}

. Perpendicular distance for force

F

:

d_F = 2.0\text{ m}

(since

F

is perpendicular to the bar).

The moment of a force is defined as the product of the force magnitude and the perpendicular distance from the pivot to the line of action of the force.

3

Apply the Principle of Moments about the pivot

A

.

\sum M_A = 0 \implies F \times 2.0\text{ m} = 120\text{ N} \times 0.5\text{ m} \implies 2.0 F = 60 \implies F = 30\text{ N}

.

For rotational equilibrium, the total counterclockwise moment about any pivot must equal the total clockwise moment.

Key Concept

Principle of Moments and Rotational Equilibrium

Question 97Question →

A simple pendulum of length $L$ with a bob of mass $m$ has a period of oscillation of $2.0\text{ s}$ . If the mass of the bob is increased to $4m$ and the length of the pendulum string is increased to $4L$ , what is the new period of oscillation?

2.0\text{ s}

4.0\text{ s}

8.0\text{ s}

16.0\text{ s}

Show answer & explanation

Answer: $4.0\text{ s}$

Answer

The new period of oscillation is

4.0\text{ s}

.

The period of a simple pendulum undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{L}{g}}

. The mass of the bob does not enter the period formula, meaning changes to bob mass have zero effect on the period. When the length of the pendulum is quadrupled (

L' = 4L

), the new period becomes

T' = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

. Since the original period was

2.0\text{ s}

, the new period is

2 \times 2.0\text{ s} = 4.0\text{ s}

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum.

The period is given by

T = 2\pi \sqrt{\frac{L}{g}}

.

The period of simple harmonic motion for a simple pendulum depends only on the length of the string

L

and the acceleration due to gravity

g

, and is independent of the mass of the bob

m

.

2

Substitute the scaled values into the formula to find the new period

T'

.

T' = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

.

Taking the square root of

4L

factors out a multiplier of

\sqrt{4} = 2

, while the change in mass from

m

to

4m

has no effect on the period.

3

Calculate the numerical value of the new period.

T' = 2 \times 2.0\text{ s} = 4.0\text{ s}

.

Multiplying the initial period of

2.0\text{ s}

by

2

yields

4.0\text{ s}

.

Key Concept

Mass Independence and Length Relationship of a Simple Pendulum

Question 98Question →

A uniform cylindrical rod of length $20\text{ cm}$ floats vertically in liquid $X$ of density $800\text{ kg/m}^3$ with $15\text{ cm}$ of its length submerged. When transferred to liquid $Y$ , it floats vertically with $12\text{ cm}$ of its length submerged. What is the density of liquid $Y$ ?

640\text{ kg/m}^3

1000\text{ kg/m}^3

1333\text{ kg/m}^3

500\text{ kg/m}^3

Show answer & explanation

Answer: $1000\text{ kg/m}^3$

Answer

The density of liquid

Y

is

1000\text{ kg/m}^3

.

According to the Law of Flotation, a floating body displaces its own weight of liquid. Therefore,

h_X \rho_X = h_Y \rho_Y

. Substituting

15\text{ cm} \times 800\text{ kg/m}^3 = 12\text{ cm} \times \rho_Y

yields

\rho_Y = 1000\text{ kg/m}^3

.

Step-by-Step Solution

1

Apply the Law of Flotation for a floating body of uniform cross-sectional area

A

.

Weight of rod

W = \text{Upthrust} = A \cdot h_{\text{submerged}} \cdot \rho_{\text{liquid}} \cdot g

.

A floating body displaces a weight of fluid equal to its own total weight.

2

Equate the upthrust in liquid

X

to the upthrust in liquid

Y

.

A \cdot h_X \cdot \rho_X \cdot g = A \cdot h_Y \cdot \rho_Y \cdot g \implies h_X \cdot \rho_X = h_Y \cdot \rho_Y

.

Since the rod is identical and floating freely in both liquids, its weight

W

remains unchanged.

3

Substitute the known values (

h_X = 15\text{ cm}

,

\rho_X = 800\text{ kg/m}^3

,

h_Y = 12\text{ cm}

) and solve for

\rho_Y

.

\rho_Y = \frac{h_X \cdot \rho_X}{h_Y} = \frac{15 \times 800}{12} = 1000\text{ kg/m}^3

.

Rearranging the linear equation yields the density of liquid

Y

.

Key Concept

Law of Flotation and Hydrometer Principle

Estimated Time:1m 15s

Question 99Question →

In Newton's law of universal gravitation, the gravitational force $F$ between two point masses $m_1$ and $m_2$ separated by a distance $r$ is expressed as $F = G \frac{m_1 m_2}{r^2}$ , where $G$ is the universal gravitational constant. What is the dimensional formula of $G$ in terms of mass ( $\text{M}$ ), length ( $\text{L}$ ), and time ( $\text{T}$ )?

\text{M}^{-1}\text{L}^3\text{T}^{-2}

\text{M}^{1}\text{L}^3\text{T}^{-2}

\text{M}^{-1}\text{L}^2\text{T}^{-2}

\text{M}^{-1}\text{L}^3\text{T}^{-1}

Show answer & explanation

Answer: $\text{M}^{-1}\text{L}^3\text{T}^{-2}$

Answer

\text{M}^{-1}\text{L}^3\text{T}^{-2}

The dimensional formula of the universal gravitational constant

G

is derived by expressing

G = \frac{F r^2}{m_1 m_2}

. Substituting the basic dimensions of force (

\text{M}\text{L}\text{T}^{-2}

), radius squared (

\text{L}^2

), and mass squared (

\text{M}^2

) gives

\frac{(\text{M}\text{L}\text{T}^{-2})(\text{L}^2)}{\text{M}^2} = \text{M}^{-1}\text{L}^3\text{T}^{-2}

.

Step-by-Step Solution

1

Make

G

the subject of the formula

G = \frac{F \cdot r^2}{m_1 \cdot m_2}

To analyze the dimensions of

G

, isolate it on one side of the equation.

2

Substitute the fundamental dimensions for force, distance, and mass

[F] = \text{M}\text{L}\text{T}^{-2}, [r^2] = \text{L}^2, [m_1 m_2] = \text{M}^2

Force is mass times acceleration, giving dimensions

\text{M}\text{L}\text{T}^{-2}

, while distance squared gives

\text{L}^2

and the product of two masses gives

\text{M}^2

.

3

Simplify the dimensional expression

[G] = \frac{(\text{M}\text{L}\text{T}^{-2}) \cdot \text{L}^2}{\text{M}^2} = \text{M}^{1-2} \text{L}^{1+2} \text{T}^{-2} = \text{M}^{-1}\text{L}^3\text{T}^{-2}

Apply laws of indices for fundamental dimensions

\text{M}

,

\text{L}

, and

\text{T}

.

Key Concept

Dimensional Analysis of Physical Constants

Estimated Time:1m 15s

Question 100Question →

A ball is thrown vertically upwards with an initial velocity of $30\text{ m/s}$ from the edge of a cliff that is $35\text{ m}$ above ground level. Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the total time elapsed before the ball hits the ground?

3.0\text{ s}

4.0\text{ s}

7.0\text{ s}

5.0\text{ s}

Show answer & explanation

Answer: $7.0\text{ s}$

Answer

The total time elapsed before the ball hits the ground is

7.0\text{ s}

.

Using the displacement equation

s = ut - \frac{1}{2}gt^2

with downward taken as negative, the net displacement when the ball reaches the ground is

-35\text{ m}

. Setting up the equation:

-35 = 30t - 5t^2

, which simplifies to

t^2 - 6t - 7 = 0

. Factoring gives

(t - 7)(t + 1) = 0

, yielding

t = 7.0\text{ s}

.

Step-by-Step Solution

1

Calculate the time taken (

t_1

) to reach maximum height

t_1 = \frac{u}{g} = \frac{30}{10} = 3.0\text{ s}

At maximum height, the final vertical velocity is

0\text{ m/s}

.

2

Calculate the maximum height (

h_{max}

) reached above the cliff

h_{max} = \frac{u^2}{2g} = \frac{30^2}{2(10)} = 45\text{ m}

Using the kinematic equation

v^2 = u^2 - 2gh

.

3

Determine total height above ground and time (

t_2

) to fall to the ground

Total height

H = 35\text{ m} + 45\text{ m} = 80\text{ m}

.

t_2 = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2(80)}{10}} = 4.0\text{ s}

The ball falls from rest from a peak height of

80\text{ m}

.

4

Calculate the total time of flight

t_{total} = t_1 + t_2 = 3.0\text{ s} + 4.0\text{ s} = 7.0\text{ s}

The complete journey consists of ascending to the peak and descending to the ground.

Key Concept

Kinematics of Vertical Motion Under Gravity

Estimated Time:1m 30s

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