Mechanics Practice Questions — JAMB UTME — Page 6

Question 101Question →

The speed $v$ of a longitudinal wave propagating through a gas depends on the pressure $P$ of the gas and its density $\rho$ according to the dimensional relationship $v = C P^x \rho^y$ , where $C$ is a dimensionless constant. Using dimensional analysis, what is the numerical value of $x - y$ ?

Show answer & explanation

Answer: 1

Answer

The numerical value of

x - y

is 1.0.

By applying the principle of dimensional homogeneity, the exponents are determined as

x = 0.5

(for pressure) and

y = -0.5

(for density). Thus,

x - y = 0.5 - (-0.5) = 1.0

.

Step-by-Step Solution

1

Determine the dimensions of speed, pressure, and density.

[v] = [L T^{-1}]

,

[P] = [M L^{-1} T^{-2}]

, and

[\rho] = [M L^{-3}]

.

Dimensional homogeneity requires expressed physical quantities to be broken down into fundamental dimensions (

M

,

L

,

T

).

2

Substitute dimensions into the relationship

v = C P^x \rho^y

.

[L T^{-1}] = [M L^{-1} T^{-2}]^x \, [M L^{-3}]^y = M^{x+y} \, L^{-x-3y} \, T^{-2x}

.

This establishes a system of algebraic equations by equating exponents of corresponding fundamental dimensions.

3

Solve for exponents

x

and

y

.

From time

T

:

-2x = -1 \implies x = 0.5

. From mass

M

:

x + y = 0 \implies y = -0.5

.

Equating the powers of fundamental dimensions on both sides yields the values of

x

and

y

.

4

Calculate the required expression

(x - y)

.

x - y = 0.5 - (-0.5) = 1.0

.

Subtracting negative

0.5

from

0.5

results in

1.0

.

Key Concept

Dimensional Analysis and Homogeneity

Question 102Question →

A beaker containing water of density $1000\text{ kg/m}^3$ rests on a digital weighing scale, giving an initial reading of $1.50\text{ kg}$ . A solid aluminum block of mass $0.80\text{ kg}$ and density $2500\text{ kg/m}^3$ is suspended from a string and completely immersed in the water without touching the bottom or sides of the beaker. What is the new reading on the digital weighing scale, in kilograms? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 1.82

Answer

The new reading on the digital weighing scale is

1.82\text{ kg}

.

When the aluminum block is fully submerged in the water, it displaces a volume of water equal to its own volume (

V = \frac{0.80}{2500} = 3.2 \times 10^{-4}\text{ m}^3

). The mass of this displaced water is

m_{\text{water}} = 1000 \times 3.2 \times 10^{-4} = 0.32\text{ kg}

. The upthrust exerted by the water upward on the block is equal to the weight of the displaced water (

3.2\text{ N}

). By Newton's Third Law, the block exerts an equal and opposite downward reaction force (

3.2\text{ N}

) on the water. This extra downward force adds an equivalent mass of

0.32\text{ kg}

to the digital scale reading, making the new reading

1.50\text{ kg} + 0.32\text{ kg} = 1.82\text{ kg}

.

Step-by-Step Solution

1

Calculate the volume of the submerged block

Volume

V = 3.2 \times 10^{-4}\text{ m}^3

The volume of fluid displaced by a completely submerged body equals the volume of the body itself.

2

Find the mass of the displaced water

Mass of displaced water

m_{\text{water}} = 0.32\text{ kg}

According to Archimedes' principle, the upthrust equals the weight of the displaced fluid, which corresponds to a displaced mass of

\rho_{\text{water}} V

.

3

Apply Newton's Third Law to determine the change in scale reading

Scale reading increase

\Delta m = 0.32\text{ kg}

The fluid exerts an upward buoyant force on the block, so by Newton's Third Law, the block exerts an equal downward reaction force on the fluid, transferring an effective weight equal to the upthrust onto the scale.

4

Compute the total new scale reading

New scale reading

= 1.82\text{ kg}

Sum the initial mass reading of the beaker system (

1.50\text{ kg}

) and the mass of the displaced water (

0.32\text{ kg}

).

Key Concept

Apparent weight transfer, Archimedes' principle, and Newton's Third Law

Question 103Question →

A particle undergoing simple harmonic motion moves with an angular frequency of $4\text{ rad/s}$ and an amplitude of $0.5\text{ m}$ . What is the maximum speed of the particle in $\text{m/s}$ ?

Show answer & explanation

Answer: 2

Answer

The maximum speed of the particle is

2.0\text{ m/s}

.

The magnitude of velocity in simple harmonic motion varies with displacement

x

according to

v = \omega \sqrt{A^2 - x^2}

. The speed reaches its maximum value when the particle passes through the equilibrium position (

x = 0

), giving

v_{\text{max}} = \omega A

. Substituting

\omega = 4\text{ rad/s}

and

A = 0.5\text{ m}

gives

v_{\text{max}} = 4 \times 0.5 = 2.0\text{ m/s}

.

Step-by-Step Solution

1

Identify the given physical parameters.

\omega = 4\text{ rad/s}

and

A = 0.5\text{ m}

These values define the speed profile of the simple harmonic oscillator.

2

Apply the SHM formula for maximum speed.

v_{\text{max}} = \omega A

Peak speed occurs at the equilibrium position where displacement is zero.

3

Substitute the values to calculate the maximum speed.

v_{\text{max}} = 4 \times 0.5 = 2.0\text{ m/s}

Multiplying angular frequency by amplitude yields the maximum linear velocity.

Key Concept

Maximum speed in Simple Harmonic Motion

Question 104Question →

A diver is swimming at a depth of $3.5\text{ m}$ below the surface of a freshwater lake. If the density of water is $1000\text{ kg/m}^3$ and the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the gauge pressure exerted on the diver in pascals?

Show answer & explanation

Answer: 35000

Answer

35000 Pa

The gauge pressure exerted by a static column of fluid is given by

P = h \rho g

. Using the values

h = 3.5\text{ m}

,

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

, and

g = 10\text{ m/s}^2

, the pressure is

P = 3.5 \times 1000 \times 10 = 35000\text{ Pa}

.

Step-by-Step Solution

1

Identify the formula for liquid hydrostatic pressure

P = h \rho g

Gauge pressure at a depth

h

in a static fluid depends on depth, fluid density, and gravitational field strength.

2

Substitute the given numerical values

P = 3.5 \times 1000 \times 10

Substitute depth

h = 3.5\text{ m}

, density

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

, and

g = 10\text{ m/s}^2

.

3

Perform the multiplication to determine the pressure

35000\text{ Pa}

Complete the calculation to get the pressure in SI units (Pascals).

Key Concept

Hydrostatic Pressure in Static Fluids

Estimated Time:45s

Question 105Question →

A solid uniform cylinder of height $0.20\text{ m}$ and cross-sectional area $5.0 \times 10^{-3}\text{ m}^2$ floats vertically at the boundary between oil of density $800\text{ kg/m}^3$ and water of density $1000\text{ kg/m}^3$ . If a height of $0.08\text{ m}$ of the cylinder extends into the water layer while the remaining upper portion is completely covered by the oil layer, what is the mass of the cylinder in kilograms?

Show answer & explanation

Answer: 0.88

Answer

The mass of the cylinder is

0.88\text{ kg}

.

According to the Law of Flotation, a floating body displaces its own weight of fluid. When floating at the interface of two immiscible liquids, the total mass of the body equals the sum of the masses of the displaced liquids. Displaced water mass is

\rho_w A h_w = 0.40\text{ kg}

and displaced oil mass is

\rho_o A h_o = 0.48\text{ kg}

, giving a total cylinder mass of

0.88\text{ kg}

.

Step-by-Step Solution

1

Find the height of the cylinder submerged in the oil layer.

h_o = 0.20\text{ m} - 0.08\text{ m} = 0.12\text{ m}

The total cylinder height is

0.20\text{ m}

, and

0.08\text{ m}

is submerged in water.

2

Calculate the volumes of water and oil displaced by the cylinder.

V_w = 5.0 \times 10^{-3} \times 0.08 = 4.0 \times 10^{-4}\text{ m}^3

;

V_o = 5.0 \times 10^{-3} \times 0.12 = 6.0 \times 10^{-4}\text{ m}^3

Volume displaced in each fluid equals cross-sectional area multiplied by the submerged height in that fluid.

3

Calculate the mass of the floating cylinder using the Law of Flotation.

m = \rho_w V_w + \rho_o V_o = (1000 \times 4.0 \times 10^{-4}) + (800 \times 6.0 \times 10^{-4}) = 0.40\text{ kg} + 0.48\text{ kg} = 0.88\text{ kg}

For a floating object in static equilibrium, its mass equals the total mass of the fluids displaced by its submerged parts.

Key Concept

Law of Flotation in Layered Liquids

Estimated Time:1m 30s

Question 106Question →

A car is traveling along a straight horizontal road at a constant speed of $20\text{ m/s}$ . The driver suddenly spots an obstacle ahead and takes $0.5\text{ s}$ to react before applying the brakes. Once the brakes are applied, the car decelerates uniformly at a rate of $4\text{ m/s}^2$ until coming to a complete stop. What is the total distance traveled by the car from the moment the driver spots the obstacle until the car stops completely?

50\text{ m}

60\text{ m}

110\text{ m}

40\text{ m}

Show answer & explanation

Answer: $60\text{ m}$

Answer

The total distance traveled by the car is

60\text{ m}

.

The motion consists of two distinct stages: first, constant speed motion during the

0.5\text{ s}

reaction time giving

s_1 = 20 \times 0.5 = 10\text{ m}

; second, uniform deceleration from

20\text{ m/s}

to rest over

s_2 = \frac{u^2}{2a} = \frac{400}{8} = 50\text{ m}

. Adding both stages yields a total distance of

60\text{ m}

.

Step-by-Step Solution

1

Calculate the reaction distance traveled at constant speed before braking.

s_1 = v \times t = 20\text{ m/s} \times 0.5\text{ s} = 10\text{ m}

.

During the reaction time, acceleration is zero, so distance equals speed multiplied by time.

2

Calculate the braking distance using the third equation of motion.

Using

v^2 = u^2 + 2as

:

0 = (20)^2 + 2(-4)s_2 \implies 8s_2 = 400 \implies s_2 = 50\text{ m}

.

The car decelerates from

u = 20\text{ m/s}

to

v = 0\text{ m/s}

at

a = -4\text{ m/s}^2

.

3

Sum the reaction distance and the braking distance to obtain the total stopping distance.

s_{\text{total}} = s_1 + s_2 = 10\text{ m} + 50\text{ m} = 60\text{ m}

.

Total distance is the sum of distances covered in both stages of motion.

Key Concept

Multi-stage linear motion combining constant velocity reaction distance and uniform deceleration braking distance.

Estimated Time:1m 30s

Question 107Question →

The critical velocity $v_c$ of a fluid flowing through a cylindrical pipe of diameter $D$ depends on the dynamic viscosity $\eta$ of the fluid, its density $\rho$ , and the pipe diameter $D$ according to the empirical relationship $v_c = R_e \eta^x \rho^y D^z$ , where $R_e$ is the dimensionless Reynolds number. Using dimensional analysis, calculate the numerical value of the sum of the exponents $x + y + z$ .

Show answer & explanation

Answer: -1

Answer

The sum of the exponents

x + y + z

is

-1

.

Using the principle of dimensional homogeneity, the dimensions of both sides of the formula

v_c = R_e \eta^x \rho^y D^z

must be equal. Equating the powers of Mass, Length, and Time yields

x = 1

,

y = -1

, and

z = -1

. Summing these three values gives

1 + (-1) + (-1) = -1

.

Step-by-Step Solution

1

Determine the dimensions of all physical quantities involved

Critical velocity

[v_c] = \text{L T}^{-1}

, dynamic viscosity

[\eta] = \text{M L}^{-1} \text{T}^{-1}

, fluid density

[\rho] = \text{M L}^{-3}

, and diameter

[D] = \text{L}

. The Reynolds number

R_e

is dimensionless.

Dimensional analysis requires replacing physical quantities with their base SI dimensions.

2

Formulate the dimensional homogeneity equation

\text{M}^0 \text{L}^1 \text{T}^{-1} = (\text{M L}^{-1} \text{T}^{-1})^x (\text{M L}^{-3})^y (\text{L})^z = \text{M}^{x+y} \text{L}^{-x-3y+z} \text{T}^{-x}.

By the principle of dimensional homogeneity, the total exponent of each fundamental dimension must match on both sides of the equation.

3

Solve the system of simultaneous linear equations for

x

,

y

, and

z

From

\text{T}

:

-x = -1 \implies x = 1

.
From

\text{M}

:

x + y = 0 \implies y = -1

.
From

\text{L}

:

-x - 3y + z = 1 \implies -1 + 3 + z = 1 \implies z = -1

.

Equating powers of fundamental quantities yields explicit values for each dimensional power.

4

Calculate the target sum

x + y + z

x + y + z = 1 + (-1) + (-1) = -1.

Combining the calculated exponents gives the required numerical value.

Key Concept

Principle of Dimensional Homogeneity and Derivation of Physical Formulas

Question 108Question →

The frequency of oscillation $f$ of a small liquid droplet executing spherical oscillations depends on the surface tension $\gamma$ of the liquid, its mass density $\rho$ , and the radius $r$ of the droplet according to the relationship $f = k \gamma^a \rho^b r^c$ , where $k$ is a dimensionless constant. Which of the following represents the correct value of the exponent $c$ ?

-\frac{3}{2}

-\frac{1}{2}

\frac{1}{2}

\frac{3}{2}

Show answer & explanation

Answer: $-\frac{3}{2}$

Answer

The correct value of the exponent

c

is

-\frac{3}{2}

.

By writing the dimensional equation

[M^0 L^0 T^{-1}] = [M T^{-2}]^a [M L^{-3}]^b [L]^c

, we solve for the exponents:

-2a = -1

gives

a = 1/2

,

a + b = 0

gives

b = -1/2

, and

-3b + c = 0

gives

3/2 + c = 0 \implies c = -3/2

. Thus, the option equal to

-3/2

is correct.

Step-by-Step Solution

1

Express each physical quantity in terms of its base dimensions

[M]

,

[L]

, and

[T]

.

Frequency

f = [T^{-1}]

, Surface tension

\gamma = \frac{\text{Force}}{\text{Length}} = [M T^{-2}]

, Density

\rho = [M L^{-3}]

, and Radius

r = [L]

.

Dimensional analysis requires reducing derived physical quantities to their base units.

2

Substitute dimensions into the given formula

f = k \gamma^a \rho^b r^c

.

[M^0 L^0 T^{-1}] = [M T^{-2}]^a [M L^{-3}]^b [L]^c = M^{a+b} L^{-3b+c} T^{-2a}

.

The principle of dimensional homogeneity requires both sides of a physical equation to have matching dimensions.

3

Equate exponents for each base dimension

M

,

L

, and

T

.

For

T

:

-2a = -1 \implies a = \frac{1}{2}

. For

M

:

a + b = 0 \implies b = -a = -\frac{1}{2}

. For

L

:

-3b + c = 0 \implies -3\left(-\frac{1}{2}\right) + c = 0

.

Matching powers across orthogonal base dimensions provides a system of linear equations.

4

Solve for the target exponent

c

.

\frac{3}{2} + c = 0 \implies c = -\frac{3}{2}.

Subtracting

3/2

from both sides gives the exact value of exponent

c

.

Key Concept

Dimensional Homogeneity and Dimensional Analysis

Estimated Time:2m 0s

Question 109Question →

The physical quantity work is defined as the product of force and displacement. What are the dimensional exponents $a$ , $b$ , and $c$ for mass, length, and time respectively in the dimensional formula for work, $[M^a L^b T^c]$ ?

1, 2, -2

1, 1, -2

1, 2, -1

1, -3, 0

Show answer & explanation

Answer: $1, 2, -2$

Answer

The dimensional exponents for mass, length, and time in the expression for work are

a = 1

,

b = 2

, and

c = -2

.

Work is calculated as force multiplied by displacement. Since force has dimensions

[M L T^{-2}]

and displacement has dimension

[L]

, the resulting dimensional formula for work is

[M^1 L^2 T^{-2}]

. The exponents of

M

,

L

, and

T

are therefore

1

,

2

, and

-2

respectively.

Step-by-Step Solution

1

Express the dimensions of force using base quantities

\text{Force} = \text{mass} \times \text{acceleration} = [M] \times [L T^{-2}] = [M L T^{-2}]

Acceleration has the dimensions of length divided by time squared.

2

Multiply force dimensions by displacement dimension to find the dimensions of work

\text{Work} = \text{Force} \times \text{Displacement} = [M L T^{-2}] \times [L] = [M^1 L^2 T^{-2}]

Displacement is a measure of length, contributing an additional factor of

[L]

.

3

Extract the exponents

a

,

b

, and

c

a = 1, b = 2, c = -2

Comparing

[M^a L^b T^c]

to

[M^1 L^2 T^{-2}]

gives the values of

a

,

b

, and

c

.

Key Concept

Dimensional analysis of work

Question 110Question →

A car moves $80\text{ km}$ due East along a straight highway and then turns to travel $60\text{ km}$ due North. What is the magnitude of the displacement of the car from its starting position?

100\text{ km}

140\text{ km}

20\text{ km}

70\text{ km}

Show answer & explanation

Answer: $100\text{ km}$

Answer

The magnitude of the displacement is

100\text{ km}

.

Because displacement is a vector quantity, perpendicular components must be combined vectorially using the Pythagorean theorem rather than scalar addition. The calculation

\sqrt{80^2 + 60^2} = 100\text{ km}

correctly yields the magnitude of the net displacement vector.

Step-by-Step Solution

1

Identify the vector components and their directions.

Eastward displacement

x = 80\text{ km}

, Northward displacement

y = 60\text{ km}

. The directions are mutually perpendicular (

90^\circ

to each other).

Displacement is a vector quantity, so direction must be accounted for when combining components.

2

Apply the Pythagorean theorem to calculate the resultant vector magnitude.

R = \sqrt{x^2 + y^2} = \sqrt{80^2 + 60^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\text{ km}

.

When two vector components act at right angles (

90^\circ

), the magnitude of their resultant is given by the hypotenuse of the right-angled triangle formed by the vectors.

Key Concept

Vector Addition of Perpendicular Components

Question 111Question →

The acoustic intensity $S$ (defined as power per unit area) of a sound wave propagating through a medium of density $\rho$ at speed $v$ is given by the empirical relationship $S = k A^x \omega^2 \rho v^w$ , where $A$ is the wave displacement amplitude, $\omega$ is the angular frequency, and $k$ is a dimensionless constant. Using the principles of dimensional analysis, calculate the numerical value of the exponent $x$ .

Show answer & explanation

Answer: 2

Answer

The numerical value of the exponent

x

is 2.

By applying the principle of dimensional homogeneity, the dimensions of intensity

[S] = M T^{-3}

are equated to

[A]^x [\omega]^2 [\rho] [v]^w = M L^{x - 3 + w} T^{-2 - w}

. Equating time exponents yields

-3 = -2 - w \implies w = 1

. Equating length exponents yields

0 = x - 3 + w \implies x = 2

.

Step-by-Step Solution

1

Determine the fundamental dimensions of acoustic intensity

S

[S] = \frac{[\text{Power}]}{[\text{Area}]} = \frac{M L^2 T^{-3}}{L^2} = M L^0 T^{-3}

Intensity is defined as power delivered per unit surface area perpendicular to the direction of propagation.

2

Write the dimensional formulas for all variables in the given equation

S = k A^x \omega^2 \rho^1 v^w

[A] = L

,

[\omega] = T^{-1}

,

[\rho] = M L^{-3}

,

[v] = L T^{-1}

Each physical quantity must be resolved into fundamental SI dimensions of Mass (

M

), Length (

L

), and Time (

T

).

3

Formulate the dimensional balance equation

M^1 L^0 T^{-3} = L^x \cdot T^{-2} \cdot M^1 L^{-3} \cdot L^w T^{-w} = M^1 L^{x - 3 + w} T^{-2 - w}

For physical validity, the dimensions on both sides of an equation must be identical (principle of dimensional homogeneity).

4

Equate the exponents of Time (

T

) to solve for

w

-3 = -2 - w \implies w = 1

The power of

T

on the left side must equal the sum of powers of

T

on the right side.

5

Equate the exponents of Length (

L

) to find

x

0 = x - 3 + w \implies 0 = x - 3 + 1 \implies x = 2

Substituting

w = 1

into the length exponent balance yields the value of

x

.

Key Concept

Principle of Dimensional Homogeneity

Estimated Time:2m 0s

Question 112Question →

The couple per unit twist $C$ (torque per unit angle of twist) of a solid wire of length $L$ , radius $r$ , and shear modulus $\eta$ is modeled by the equation:

C = \frac{\pi \eta r^x}{2 L}

Using dimensional analysis, determine the numerical value of the exponent $x$ .

Show answer & explanation

Answer: 4

Answer

The numerical value of the exponent x is 4.

Applying the principle of dimensional homogeneity requires the dimensions of couple per unit twist

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

to equal the dimensions of

\frac{\eta r^x}{L}

, which simplifies to

[\text{M L}^{x-2} \text{T}^{-2}]

. Equating exponents of length gives

2 = x - 2

, yielding

x = 4

.

Step-by-Step Solution

1

Determine the dimensional formula of couple per unit twist

C

.

[C] = \text{M L}^2 \text{T}^{-2}

Couple (torque) is force multiplied by perpendicular distance, which has dimensions

[\text{M L T}^{-2}][\text{L}] = [\text{M L}^2 \text{T}^{-2}]

. The angle of twist (in radians) is dimensionless.

2

Determine the dimensional formula of shear modulus

\eta

.

[\eta] = \text{M L}^{-1} \text{T}^{-2}

Shear modulus is defined as shear stress divided by shear strain. Stress has dimensions of force per unit area

[\text{M L T}^{-2}]/[\text{L}^2] = [\text{M L}^{-1} \text{T}^{-2}]

, while strain is dimensionless.

3

Set up the dimensional equation for the relation

C = \frac{\pi \eta r^x}{2 L}

.

[\text{M L}^2 \text{T}^{-2}] = \frac{[\text{M L}^{-1} \text{T}^{-2}][\text{L}]^x}{[\text{L}]} = [\text{M L}^{x-2} \text{T}^{-2}]

Pure numerical constants such as

\pi

and

2

are dimensionless. Length

L

and radius

r

both have dimension

[\text{L}]

.

4

Equate the exponents of length

\text{L}

on both sides of the dimensional equation.

x = 4

Comparing powers of

\text{L}

on both sides gives

2 = x - 2

, which solves to

x = 4

.

Key Concept

Dimensional Homogeneity in Mechanics

Question 113Question →

The tensile stress $\sigma$ on a solid wire subjected to a stretching force $F$ is defined as force per unit cross-sectional area, while the fractional change in length is the tensile strain $\epsilon$ . If Young's modulus $Y$ of the material is given by $Y = \frac{\sigma}{\epsilon}$ , which of the following is the SI unit of Young's modulus expressed in fundamental SI base units?

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

\text{kg}\cdot\text{m}\cdot\text{s}^{-2}

\text{kg}\cdot\text{m}^{-2}\cdot\text{s}^{-2}

\text{kg}\cdot\text{m}^{2}\cdot\text{s}^{-2}

Show answer & explanation

Answer: $\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$

Answer

\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

Young's modulus is calculated as tensile stress divided by tensile strain. Since tensile strain is the ratio of change in length to original length, it has no units. Therefore, the SI unit of Young's modulus is identical to that of stress. Stress is force divided by area:

\frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}

.

Step-by-Step Solution

1

Determine the dimensions of the force component

Force

F = ma \implies [F] = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}

Mass has fundamental unit

\text{kg}

and acceleration has derived unit

\text{m}\cdot\text{s}^{-2}

.

2

Determine the SI base units of tensile stress

\sigma

[\sigma] = \frac{[F]}{\text{Area}} = \frac{\text{kg}\cdot\text{m}\cdot\text{s}^{-2}}{\text{m}^2} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$

Stress is defined as force divided by cross-sectional area (

A = \text{m}^2

).

3

Evaluate the unit of Young's modulus

Y

[Y] = \frac{[\sigma]}{[\epsilon]} = \frac{\text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}}{1} = \text{kg}\cdot\text{m}^{-1}\cdot\text{s}^{-2}$

Strain

\epsilon = \frac{\Delta L}{L}

is the ratio of two lengths and is dimensionless.

Key Concept

Derivation of SI base units for mechanical moduli from physical definitions

Estimated Time:1m 15s

Question 114Question →

Match each physical scenario involving scalar and vector quantities on the left with its corresponding resultant value or component magnitude on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: (1) The particle's net displacement corresponds to 13 m; (2) The resultant of two equal forces of magnitude F at 60 degrees corresponds to F√3; (3) The vertical component of a 40 N force inclined at 60 degrees to the vertical corresponds to 20 N; (4) The resultant of perpendicular forces of 8 N and 15 N corresponds to 17 N.

Each scenario correctly applies vector algebra: 2D displacement resolution yields a 5-12-13 right triangle; the parallelogram rule for equal forces at 60 degrees produces F√3; resolving a force adjacent to the vertical axis uses cos(60°) to give 20 N; and perpendicular 8 N and 15 N forces synthesize to a 17 N resultant using the Pythagorean theorem.

Step-by-Step Solution

1

Calculate net displacement for Item 1

Net x-component:

10\text{ m} - 5\text{ m} = 5\text{ m}

East. Net y-component:

12\text{ m}

North. Magnitude

R = \sqrt{5^2 + 12^2} = 13\text{ m}

.

Displacements along parallel lines subtract scalar-wise, and perpendicular components combine via the Pythagorean theorem.

2

Determine the resultant of two equal forces at 60 degrees for Item 2

R = \sqrt{F^2 + F^2 + 2(F)(F)\cos(60^\circ)} = \sqrt{2F^2 + 2F^2(0.5)} = \sqrt{3F^2} = F\sqrt{3}

.

Applying the parallelogram law of vector addition.

3

Resolve the force vector along the vertical direction for Item 3

F_{\text{vertical}} = F \cos(\theta_{\text{vertical}}) = 40 \cos(60^\circ) = 40 \times 0.5 = 20\text{ N}

.

The component adjacent to the reference angle uses the cosine function.

4

Compute resultant magnitude of orthogonal forces for Item 4

R = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\text{ N}

.

Vectors at right angles sum directly using Pythagorean synthesis.

Key Concept

Vector resolution, component synthesis, and parallelogram law of vector addition

Question 115Question →

The centripetal acceleration $a$ of a particle moving in a circular path depends on its linear speed $v$ and the radius $r$ of the path according to the formula $a = k v^x r^y$ , where $k$ is a dimensionless constant. Using dimensional analysis, what is the numerical value of the product $x \cdot y$ ?

Show answer & explanation

Answer: -2

Answer

The numerical value of the product

x \cdot y

is

-2

.

By dimensional analysis, centripetal acceleration has dimensions

[a] = \text{L T}^{-2}

, velocity

[v] = \text{L T}^{-1}

, and radius

[r] = \text{L}

. Substituting into

a = k v^x r^y

gives

\text{L T}^{-2} = \text{L}^{x+y} \text{T}^{-x}

. Equating exponents of time yields

-x = -2 \implies x = 2

. Equating exponents of length gives

x + y = 1 \implies 2 + y = 1 \implies y = -1

. Consequently,

x \cdot y = 2 \cdot (-1) = -2

.

Step-by-Step Solution

1

Identify the base dimensions for each physical quantity.

[a] = \text{M}^0 \text{L}^1 \text{T}^{-2}

,

[v] = \text{M}^0 \text{L}^1 \text{T}^{-1}

, and

[r] = \text{M}^0 \text{L}^1 \text{T}^0

.

Dimensional analysis requires substituting fundamental dimensions of mass, length, and time.

2

Set up the dimensional homogeneity equation.

\text{L}^1 \text{T}^{-2} = (\text{L T}^{-1})^x \cdot (\text{L})^y = \text{L}^{x+y} \text{T}^{-x}.

Since

k

is dimensionless, the net dimensions on both sides of the equation must be identical.

3

Solve for exponents

x

and

y

by equating powers of corresponding base units.

For \text{T}:

-x = -2 \implies x = 2

. For \text{L}:

x + y = 1 \implies 2 + y = 1 \implies y = -1

.

Equating coefficients of identical base dimensions gives a system of linear equations.

4

Multiply the derived values of

x

and

y

.

x \cdot y = 2 \cdot (-1) = -2

.

The question asks specifically for the product of exponents

x

and

y

.

Key Concept

Dimensional Homogeneity and Exponent Analysis

Question 116Question →

An electric train accelerates uniformly from rest at a rate of $2\text{ m/s}^2$ and then immediately decelerates uniformly at $4\text{ m/s}^2$ until it comes to a complete stop. If the total distance covered by the train during this entire motion is $600\text{ m}$ , what is the total time taken for the journey?

20\text{ s}

25\text{ s}

30\text{ s}

40\text{ s}

Show answer & explanation

Answer: $30\text{ s}$

Answer

The total time taken for the journey is

30\text{ s}

.

For motion from rest back to rest, the velocity-time graph forms a triangle of total base

T

and height

v_{\text{max}}

. Acceleration time is

t_1 = v_{\text{max}}/2

and deceleration time is

t_2 = v_{\text{max}}/4 = 0.5 t_1

. The total displacement is

s = \frac{1}{2} (2) t_1^2 + \frac{1}{2} (4) t_2^2 = t_1^2 + 2 (0.5 t_1)^2 = 1.5 t_1^2 = 600\text{ m}

, yielding

t_1 = 20\text{ s}

and

t_2 = 10\text{ s}

. Summing these gives a total journey time of

30\text{ s}

.

Step-by-Step Solution

1

Relate maximum velocity to time in each stage.

Let

v_{\text{max}}

be the maximum velocity. Acceleration time

t_1 = \frac{v_{\text{max}}}{2}

and deceleration time

t_2 = \frac{v_{\text{max}}}{4}

.

Using

v = u + at

from rest to

v_{\text{max}}

and from

v_{\text{max}}

to rest.

2

Express total time

T

in terms of maximum velocity.

T = t_1 + t_2 = \frac{v_{\text{max}}}{2} + \frac{v_{\text{max}}}{4} = \frac{3}{4} v_{\text{max}}

, which gives

v_{\text{max}} = \frac{4}{3} T

.

Total duration is the sum of individual phase durations.

3

Set up distance equation from the area of the velocity-time graph.

\text{Total distance } s = \frac{1}{2} \times T \times v_{\text{max}} = \frac{1}{2} \times T \times \frac{4}{3} T = \frac{2}{3} T^2 = 600\text{ m}.

Area under a velocity-time triangle equals total displacement.

4

Solve for total time

T

.

T^2 = 600 \times \frac{3}{2} = 900 \implies T = 30\text{ s}

.

Taking the square root gives the total travel time.

Key Concept

Multi-stage linear motion with uniform acceleration and deceleration

Question 117Question →

An electric scooter starts from rest and accelerates uniformly at a rate of $3\text{ m/s}^2$ for a time of $6\text{ s}$ . What is the total distance, in meters, traveled by the scooter during this period?

Show answer & explanation

Answer: 54

Answer

The total distance traveled by the scooter is

54\text{ m}

.

Using the second equation of linear motion

s = ut + \frac{1}{2}at^2

with

u = 0\text{ m/s}

,

a = 3\text{ m/s}^2

, and

t = 6\text{ s}

yields

s = 0 + \frac{1}{2}(3)(36) = 54\text{ m}

.

Step-by-Step Solution

1

Identify known variables from the stem

Initial velocity

u = 0\text{ m/s}

, acceleration

a = 3\text{ m/s}^2

, time interval

t = 6\text{ s}

Extracting given values provides the foundation for selecting the correct kinematic equation.

2

Select the appropriate equation of linear motion

s = ut + \frac{1}{2}at^2

This formula directly relates displacement

s

to initial velocity

u

, constant acceleration

a

, and time

t

.

3

Calculate the numerical displacement

s = (0\text{ m/s})(6\text{ s}) + \frac{1}{2}(3\text{ m/s}^2)(6\text{ s})^2 = 0 + \frac{1}{2}(3)(36) = 54\text{ m}

Evaluating the mathematical expression yields the final total distance.

Key Concept

Uniform Linear Acceleration

Question 118Question →

Three coplanar forces act simultaneously at a point $O$ . The first force of magnitude $10\text{ N}$ acts due East, and the second force of magnitude $10\text{ N}$ acts at an angle of $60^\circ$ North of East. If a third force keeps the system in static equilibrium, what is the magnitude of this third force?

10\sqrt{3}\text{ N}

20\text{ N}

10\sqrt{2}\text{ N}

10\text{ N}

Show answer & explanation

Answer: $10\sqrt{3}\text{ N}$

Answer

The magnitude of the third force required for equilibrium is

10\sqrt{3}\text{ N}

.

Resolving the forces along the horizontal (East) and vertical (North) axes gives components of

15\text{ N}

and

5\sqrt{3}\text{ N}

respectively. The magnitude of the resultant force is

\sqrt{15^2 + (5\sqrt{3})^2} = \sqrt{300} = 10\sqrt{3}\text{ N}

. Since the third force balances the system, its magnitude must equal that of the resultant, which is

10\sqrt{3}\text{ N}

.

Step-by-Step Solution

1

Resolve the two given forces into horizontal (

x

) and vertical (

y

) Cartesian components.

F_{1x} = 10\text{ N}

,

F_{1y} = 0\text{ N}

;

F_{2x} = 10 \cos(60^\circ) = 5\text{ N}

,

F_{2y} = 10 \sin(60^\circ) = 5\sqrt{3}\text{ N}

.

Vector addition requires breaking non-orthogonal vectors into perpendicular component directions.

2

Calculate the total horizontal (

R_x

) and vertical (

R_y

) components of the resultant of the first two forces.

R_x = 10 + 5 = 15\text{ N}

,

R_y = 0 + 5\sqrt{3} = 5\sqrt{3}\text{ N}

.

Adding aligned components yields the net component along each axis.

3

Compute the magnitude of the resultant force

R

.

R = \sqrt{R_x^2 + R_y^2} = \sqrt{15^2 + (5\sqrt{3})^2} = \sqrt{225 + 75} = \sqrt{300} = 10\sqrt{3}\text{ N}

.

Applying Pythagoras' theorem to orthogonal components determines the net magnitude.

4

Determine the magnitude of the equilibrant (third force).

Equibrant magnitude

= R = 10\sqrt{3}\text{ N}

.

For static equilibrium, the third force must be equal in magnitude and opposite in direction to the resultant of the first two forces.

Key Concept

Vector Addition and Equilibrium of Forces

Question 119Question →

A motorcycle traveling at a constant speed of $18\text{ m/s}$ passes a landmark. Exactly $4.0\text{ s}$ after passing the landmark, the rider accelerates uniformly at a rate of $2.5\text{ m/s}^2$ until reaching a speed of $28\text{ m/s}$ . What is the total distance, in meters, traveled by the motorcycle from the instant it passed the landmark to the moment it reaches $28\text{ m/s}$ ?

Show answer & explanation

Answer: 164

Answer

The total distance traveled by the motorcycle is

164\text{ m}

.

The total distance is obtained by finding the displacement during the

4.0\text{ s}

of constant speed at

18\text{ m/s}

(

72\text{ m}

) and adding the displacement during uniform acceleration from

18\text{ m/s}

to

28\text{ m/s}

at

2.5\text{ m/s}^2

(

92\text{ m}

), giving a sum of

164\text{ m}

.

Step-by-Step Solution

1

Calculate the distance covered during the initial constant speed stage.

The distance covered in the first

4.0\text{ s}

is

72\text{ m}

.

At a constant velocity

v = 18\text{ m/s}

, the distance

s_1 = v \times t = 18 \times 4.0 = 72\text{ m}

.

2

Calculate the distance covered during the accelerated motion stage.

The distance covered while accelerating from

18\text{ m/s}

to

28\text{ m/s}

is

92\text{ m}

.

Using the kinematic equation

v^2 = u^2 + 2as_2

, substitute

u = 18\text{ m/s}

,

v = 28\text{ m/s}

, and

a = 2.5\text{ m/s}^2

to get

28^2 = 18^2 + 2(2.5)s_2 \implies 784 = 324 + 5s_2 \implies s_2 = 92\text{ m}

.

3

Find the total distance traveled.

Total distance

s_{\text{total}} = 164\text{ m}

.

The total distance is the sum of the distance covered during the constant speed period (

72\text{ m}

) and during uniform acceleration (

92\text{ m}

).

Key Concept

Kinematics equations for multi-stage motion combining uniform speed and uniform acceleration.

Question 120Question →

A traffic officer on a stationary motorcycle spots a car moving past at a constant speed of $30\text{ m/s}$ . The officer takes $2\text{ s}$ of reaction time before accelerating uniformly at $4\text{ m/s}^2$ to a maximum cruise speed of $40\text{ m/s}$ , after which the motorcycle continues at this constant speed. What is the total distance traveled by the motorcycle from its initial position to the point where it catches up with the car?

360\text{ m}

510\text{ m}

600\text{ m}

840\text{ m}

Show answer & explanation

Answer: $840\text{ m}$

Answer

840\text{ m}

The correct answer is

840\text{ m}

. Over the first

2\text{ s}

, the motorcycle is stationary while the car covers

60\text{ m}

. Over the next

10\text{ s}

, the motorcycle accelerates to

40\text{ m/s}

, covering

200\text{ m}

, while the car travels another

300\text{ m}

(totaling

360\text{ m}

). To close the remaining

160\text{ m}

gap at a relative speed of

10\text{ m/s}

requires an additional

16\text{ s}

. The total elapsed time of

28\text{ s}

yields a total catch-up distance of

30\text{ m/s} \times 28\text{ s} = 840\text{ m}

.

Step-by-Step Solution

1

Calculate car position and motorcycle position at the end of the reaction time period (

t = 2\text{ s}

).

During the reaction time

t_r = 2\text{ s}

, the motorcycle remains stationary (

s_m = 0\text{ m}

). The car travels a distance

s_c = 30\text{ m/s} \times 2\text{ s} = 60\text{ m}

.

The motorcycle does not begin accelerating until after the officer's reaction time elapses.

2

Determine the time and distance required for the motorcycle to reach its maximum speed of

40\text{ m/s}

.

Time to reach maximum speed:

t_{acc} = \frac{v_{max} - u}{a} = \frac{40 - 0}{4} = 10\text{ s}

. Distance during acceleration:

s_{acc} = \frac{v_{max}^2 - u^2}{2a} = \frac{40^2 - 0}{2(4)} = 200\text{ m}

.

The motorcycle accelerates uniformly from rest until reaching its capped maximum velocity.

3

Calculate total positions at

t_{total1} = 2\text{ s} + 10\text{ s} = 12\text{ s}

from the instant the car passed.

Motorcycle position:

x_m(12) = 200\text{ m}

. Car position:

x_c(12) = 30\text{ m/s} \times 12\text{ s} = 360\text{ m}

. Distance gap remaining:

\Delta x = 360\text{ m} - 200\text{ m} = 160\text{ m}

.

Comparing positions at

12\text{ s}

establishes the remaining distance gap to be closed during the constant speed phase.

4

Calculate the time required during the constant-speed phase to close the remaining gap and find the total meeting distance.

Relative speed:

v_{rel} = 40\text{ m/s} - 30\text{ m/s} = 10\text{ m/s}

. Additional time needed:

\Delta t = \frac{160\text{ m}}{10\text{ m/s}} = 16\text{ s}

. Total elapsed time:

t = 12\text{ s} + 16\text{ s} = 28\text{ s}

. Catch-up distance:

s_{total} = 30\text{ m/s} \times 28\text{ s} = 840\text{ m}

.

With both vehicles moving at constant speeds, the relative velocity determines how quickly the remaining separation is eliminated.

Key Concept

Multi-stage linear motion involving delayed reaction time, uniform acceleration, and constant speed pursuit.

Estimated Time:3m 0s

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