Physical Quantities, Units and Dimensions Alıştırma Soruları — JAMB UTME

Soru 1Soru →

The aerodynamic drag force $F$ acting on an object moving through a fluid of density $\rho$ with cross-sectional area $A$ at speed $v$ is modeled by the equation $F = \frac{1}{2} C_d \rho^a A^b v^c$ , where $C_d$ is a dimensionless constant. Using dimensional analysis, what is the numerical value of the exponent $c$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The numerical value of the exponent

c

is 2.

By applying the principle of dimensional homogeneity, the base dimension of time on the left side is

\text{T}^{-2}

(from force

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

). On the right side, the only quantity containing time is velocity

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

, raised to power

c

, giving

\text{T}^{-c}

. Equating the exponents gives

-2 = -c

, so

c = 2

.

Adım Adım Çözüm

1

Identify the base dimensions of each physical quantity in the given equation.

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

,

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

,

[A] = \text{L}^2

, and

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

.

C_d

is dimensionless (

[C_d] = 1

).

Dimensional homogeneity requires both sides of a physical equation to have identical base dimensions.

2

Substitute the base dimensions into the formula

F = \frac{1}{2} C_d \rho^a A^b v^c

and simplify.

\text{M L T}^{-2} = (\text{M L}^{-3})^a (\text{L}^2)^b (\text{L T}^{-1})^c = \text{M}^a \text{L}^{-3a + 2b + c} \text{T}^{-c}.

Combining powers of base dimensions allows direct comparison of corresponding exponents.

3

Equate the exponent of time (T) on both sides of the dimensional equation.

-2 = -c \implies c = 2.

The exponent of T on the left side is -2, which must equal the exponent of T on the right side (-c).

Anahtar Kavram

Principle of Dimensional Homogeneity

Tahmini Süre:1m 15s

Soru 2Soru →

The physical quantity impulse is defined as the product of force and time. Which of the following physical quantities has the same dimensions as impulse?

Linear momentum

Work done

Power

Force

Cevabı ve açıklamayı göster

Cevap: Linear momentum

Cevap

Linear momentum

Linear momentum is calculated as mass times velocity, which yields the base dimensions

[M L T^{-1}]

. This is identical to impulse, which is force times time (

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

).

Adım Adım Çözüm

1

Determine the dimensions of impulse.

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

Force has base dimensions

[M L T^{-2}]

and time has dimension

[T]

.

2

Determine the dimensions of linear momentum.

\text{Linear Momentum} = \text{Mass} \times \text{Velocity} = [M][L T^{-1}] = [M L T^{-1}]

Mass has base dimension

[M]

and velocity has dimensions

[L T^{-1}]

.

3

Compare the dimensions of impulse and linear momentum.

Both quantities have identical dimensions of

[M L T^{-1}]

.

By the impulse-momentum theorem, impulse equals change in momentum.

Anahtar Kavram

Dimensional analysis of physical quantities

Soru 3Soru →

The mechanical power $P$ dissipated by a circular disc of radius $R$ rotating at an angular velocity $\omega$ in a fluid of density $\rho$ is given by the relation $P = k \cdot \rho^a \cdot \omega^b \cdot R^c$ , where $k$ is a dimensionless constant. Using dimensional analysis, which of the following represents the correct values of the exponents $a$ , $b$ , and $c$ ?

a = 1, b = 3, c = 5

a = 1, b = 2, c = 4

a = 1, b = 3, c = 2

a = 1, b = 1, c = 5

Cevabı ve açıklamayı göster

Cevap: $a = 1, b = 3, c = 5$

Cevap

a = 1, b = 3, c = 5

By writing the dimensions of power as

[M L^2 T^{-3}]

, density as

[M L^{-3}]

, angular velocity as

[T^{-1}]

, and radius as

[L]

, dimensional homogeneity requires that

M L^2 T^{-3} = M^a L^{-3a + c} T^{-b}

. Equating powers gives

a = 1

,

b = 3

, and

c = 5

.

Adım Adım Çözüm

1

Express the base dimensions for each physical quantity

Power

[P] = M L^2 T^{-3}

, Density

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

, Angular velocity

[\omega] = T^{-1}

, and Radius

[R] = L

.

Dimensional analysis requires decomposing derived physical quantities into fundamental dimensions of mass (

M

), length (

L

), and time (

T

).

2

Substitute dimensions into the given equation

P = k \cdot \rho^a \cdot \omega^b \cdot R^c

M L^2 T^{-3} = (M L^{-3})^a \cdot (T^{-1})^b \cdot (L)^c = M^a \cdot L^{-3a + c} \cdot T^{-b}

.

The constant

k

is dimensionless, so its dimension is 1.

3

Equate the powers of

M

,

L

, and

T

on both sides of the equation

For mass

M

:

a = 1

. For time

T

:

-b = -3 \implies b = 3

. For length

L

:

-3a + c = 2 \implies -3(1) + c = 2 \implies c = 5

.

According to the principle of dimensional homogeneity, powers of fundamental dimensions must be equal on both sides of a physically correct equation.

Anahtar Kavram

Dimensional Homogeneity and Formula Derivation

Tahmini Süre:2m 0s

Soru 4Soru →

The volume flow rate

Q

of a viscous liquid flowing through a pipe of radius

r

under a pressure gradient

\frac{\Delta P}{l}

is modeled by the equation:

Q = k \eta^x r^y \left(\frac{\Delta P}{l}\right)^z

where

\eta

is the coefficient of dynamic viscosity and

k

is a dimensionless constant. What are the values of the exponents

x

,

y

, and

z

respectively?

x = -1, y = 4, z = 1

x = 1, y = 4, z = -1

x = -1, y = 3, z = 1

x = 1, y = 2, z = 1

Cevabı ve açıklamayı göster

Cevap: $x = -1, y = 4, z = 1$

Cevap

x = -1

,

y = 4

, and

z = 1

The dimensional representation of volume flow rate is

[L^3 T^{-1}]

, dynamic viscosity is

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

, radius is

[L]

, and pressure gradient is

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

. Equating the powers of

M

,

L

, and

T

gives

x + z = 0

,

-x - 2z = -1

, and

-x + y - 2z = 3

. Solving these simultaneously yields

x = -1

,

y = 4

, and

z = 1

.

Adım Adım Çözüm

1

Determine the fundamental dimensions of each physical quantity

Volume flow rate

Q = \frac{\text{Volume}}{\text{Time}} = [L^3 T^{-1}]

;
Dynamic viscosity

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

;
Radius

r = [L]

;
Pressure gradient

\frac{\Delta P}{l} = \frac{\text{Pressure}}{\text{Length}} = \frac{[M L^{-1} T^{-2}]}{[L]} = [M L^{-2} T^{-2}]

.

Correct base dimensions are required for dimensional analysis.

2

Set up the dimensional equation by substituting the base dimensions into the formula

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

.

The principle of dimensional homogeneity requires both sides of the equation to have matching exponents for

M

,

L

, and

T

.

3

Equate exponents for

M

,

T

, and

L

to form algebraic equations

For

M

:

x + z = 0 \implies z = -x

For

T

:

-x - 2z = -1

For

L

:

-x + y - 2z = 3

.

This creates a linear system of equations for the exponents

x

,

y

, and

z

.

4

Solve the system of linear equations

Substituting

z = -x

into the

T

equation:

-x - 2(-x) = -1 \implies x = -1

.
Hence

z = -(-1) = 1

.
Substituting

x = -1

and

z = 1

into the

L

equation:

-(-1) + y - 2(1) = 3 \implies 1 + y - 2 = 3 \implies y = 4

.

Yields the unique set of exponents

x = -1, y = 4, z = 1

.

Anahtar Kavram

Dimensional Analysis and Homogeneity

Tahmini Süre:2m 0s

Soru 5Soru →

The speed $v$ of a transverse wave traveling along a stretched string under tension $T$ with mass per unit length $\mu$ is given by $v = k T^x \mu^y$ , where $k$ is a dimensionless constant. What is the numerical value of the exponent $x$ ?

Cevabı ve açıklamayı göster

Cevap: 0.5

Cevap

The numerical value of the exponent

x

is 0.5.

Applying the principle of dimensional homogeneity, the dimensions on both sides must match. Speed

[v] = L T^{-1}

, tension force

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

, and mass per unit length

[\mu] = M L^{-1}

. Substituting these into

v = k T^x \mu^y

yields

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

. Comparing the exponents of

T

gives

-2x = -1

, leading to

x = 0.5

.

Adım Adım Çözüm

1

Determine the dimensions of speed

v

, tension force

T

, and linear density

\mu

.

[v] = L T^{-1}

,

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

,

[\mu] = M L^{-1}

Tension is a force (

F=ma

) with dimensions

[M L T^{-2}]

, and

\mu

is mass per unit length (

m/l

) with dimensions

[M L^{-1}]

.

2

Substitute the dimensional formulas into the equation

v = k T^x \mu^y

and collect powers of base dimensions.

M^0 L^1 T^{-1} = M^{x+y} L^{x-y} T^{-2x}

Combining exponents for base dimensions

M

,

L

, and

T

allows applying the principle of dimensional homogeneity.

3

Equate the exponent of

T

on both sides to solve for

x

.

-2x = -1 \implies x = 0.5

The exponent of

T

on the left side is

-1

and on the right side is

-2x

.

Anahtar Kavram

Dimensional Analysis and Determination of Exponents

Tahmini Süre:1m 30s

Soru 6Soru →

The viscous drag force $F$ acting on a spherical body moving through a fluid at speed $v$ is given by Stokes' law, $F = 6 \pi \eta r v$ , where $r$ is the radius of the sphere and $\eta$ is the coefficient of viscosity. What is the dimensional formula of the coefficient of viscosity $\eta$ ?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $\text{M L}^{-1} \text{T}^{-1}$

Cevap

The dimensional formula of the coefficient of viscosity is

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

.

Isolating the coefficient of viscosity from Stokes' law yields

\eta = \frac{F}{6\pi r v}

. Substituting the dimensions

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

,

[r] = \text{L}

, and

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

gives

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

.

Adım Adım Çözüm

1

Identify the dimensional formulas for force, radius, and velocity.

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

,

[r] = \text{L}

, and

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

.

Base physical quantities must be represented by their fundamental dimensions.

2

Rearrange Stokes' law to express

\eta

in terms of the other variables, treating

6\pi

as a dimensionless constant.

[\eta] = \frac{[F]}{[r][v]}

.

Numerical constants do not carry physical dimensions.

3

Substitute the base dimensions into the formula and simplify using index rules.

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

.

Subtracting denominator indices from numerator indices yields the final dimensional expression.

Anahtar Kavram

Dimensional analysis of physical constants and equations

Soru 7Soru →

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}

Cevabı ve açıklamayı göster

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

Cevap

\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}

.

Adım Adım Çözüm

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}

.

Anahtar Kavram

Dimensional Analysis of Physical Constants

Tahmini Süre:1m 15s

Soru 8Soru →

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 $x - y$ ?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Dimensional Analysis and Homogeneity

Soru 9Soru →

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 $x + y + z$ .

Cevabı ve açıklamayı göster

Cevap: -1

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Principle of Dimensional Homogeneity and Derivation of Physical Formulas

Soru 10Soru →

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}

Cevabı ve açıklamayı göster

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

Cevap

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.

Adım Adım Çözüm

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

.

Anahtar Kavram

Dimensional Homogeneity and Dimensional Analysis

Tahmini Süre:2m 0s

Soru 11Soru →

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

Cevabı ve açıklamayı göster

Cevap: $1, 2, -2$

Cevap

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.

Adım Adım Çözüm

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

.

Anahtar Kavram

Dimensional analysis of work

Soru 12Soru →

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 $x$ .

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Principle of Dimensional Homogeneity

Tahmini Süre:2m 0s

Soru 13Soru →

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 $x$ .

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Cevap: 4

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Dimensional Homogeneity in Mechanics

Soru 14Soru →

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}

Cevabı ve açıklamayı göster

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

Cevap

\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}

.

Adım Adım Çözüm

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.

Anahtar Kavram

Derivation of SI base units for mechanical moduli from physical definitions

Tahmini Süre:1m 15s

Soru 15Soru →

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 $x \cdot y$ ?

Cevabı ve açıklamayı göster

Cevap: -2

Cevap

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

.

Adım Adım Çözüm

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

.

Anahtar Kavram

Dimensional Homogeneity and Exponent Analysis

Soru 16Soru →

The energy density $u$ (defined as energy per unit volume) stored in an electrostatic field is related to the permittivity of free space $\epsilon_0$ and the electric field strength $E$ by the dimensional formula $u = k \epsilon_0^x E^y$ , where $k$ is a dimensionless constant. What is the value of the numerical exponent $y$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of the exponent

y

is 2.

By writing the dimensions of energy density

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

, permittivity

[M^{-1} L^{-3} T^4 I^2]

, and electric field strength

[M L T^{-3} I^{-1}]

, equating powers of electric current

I

yields

2x - y = 0

(or

y = 2x

). Substituting this into the equation for powers of mass

M

,

-x + y = 1

, yields

-x + 2x = 1

, so

x = 1

and

y = 2

.

Adım Adım Çözüm

1

Derive the dimensional formulas for energy density

u

, permittivity

\epsilon_0

, and electric field

E

.

[u] = M L^{-1} T^{-2}, [\epsilon_0] = M^{-1} L^{-3} T^4 I^2, [E] = M L T^{-3} I^{-1}.

Expressing quantities in terms of base dimensions (M, L, T, I) is required for dimensional homogeneity.

2

Form the dimensional equation

u = k \epsilon_0^x E^y

and combine powers.

M L^{-1} T^{-2} = M^{-x+y} L^{-3x+y} T^{4x-3y} I^{2x-y}.

Applies the principle of dimensional consistency across the formula.

3

Equate corresponding powers of base dimensions to set up equations for

x

and

y

.

For I: 2x - y = 0; for M: -x + y = 1.

Base unit exponents on both sides of a physically valid equation must match.

4

Solve the algebraic equations for the unknown exponent

y

.

x = 1, y = 2.

Substituting y = 2x into -x + y = 1 directly gives x = 1 and y = 2.

Anahtar Kavram

Dimensional Analysis and Dimensional Homogeneity

Tahmini Süre:1m 30s

Soru 17Soru →

The torque $\tau$ required to rotate a thin flat disk of radius $r$ at a constant angular velocity $\omega$ in a fluid of dynamic viscosity $\eta$ is expressed by the dimensional formula $\tau = k \eta^x \omega^y r^z$ , where $k$ is a dimensionless constant. What is the value of the sum of the exponents $x + y + z$ ?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The sum of the exponents

x + y + z

is 5.

By substituting the base dimensions into

\tau = k \eta^x \omega^y r^z

, we get

M L^2 T^{-2} = (M L^{-1} T^{-1})^x (T^{-1})^y L^z = M^x L^{-x+z} T^{-x-y}

. Equating exponents of

M

gives

x = 1

. Equating exponents of

T

gives

-1 - y = -2 \implies y = 1

. Equating exponents of

L

gives

-1 + z = 2 \implies z = 3

. Thus,

x + y + z = 1 + 1 + 3 = 5

.

Adım Adım Çözüm

1

Determine the dimensions of torque, dynamic viscosity, angular velocity, and radius in base mechanical dimensions (M, L, T).

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

,

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

,

[\omega] = T^{-1}

, and

[r] = L

.

Dimensional analysis requires converting all parameters into base dimensions.

2

Apply the principle of dimensional homogeneity to set up exponential equations for each base dimension.

M^1 L^2 T^{-2} = M^x L^{-x+z} T^{-x-y}

.

Both sides of a physically valid equation must share identical net dimensions.

3

Solve for each exponent individually by comparing indices.

x = 1

,

y = 1

,

z = 3

.

Matching powers of M yields

x=1

, matching powers of T yields

y=1

, and matching powers of L yields

z=3

.

4

Sum the three calculated exponent values.

1 + 1 + 3 = 5

.

The question asks specifically for the value of

x + y + z

.

Anahtar Kavram

Dimensional analysis and dimensional homogeneity

Tahmini Süre:1m 30s

Soru 18Soru →

The resistive force $F$ experienced by a small sphere of radius $r$ moving at velocity $v$ through a viscous fluid is given by Stokes' law: $F = 6\pi \eta r v$ , where $\eta$ is the coefficient of viscosity. What is the dimensional formula of $\eta$ ?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $\text{M L}^{-1}\text{T}^{-1}$

Cevap

The dimensional formula of the coefficient of viscosity

\eta

is

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

.

Rearranging Stokes' law gives

\eta = \frac{F}{6\pi r v}

. Since the constant

6\pi

has no dimensions, substituting

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

,

[r] = \text{L}

, and

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

yields

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

.

Adım Adım Çözüm

1

Isolate the coefficient of viscosity

\eta

from Stokes' formula

\eta = \frac{F}{6\pi r v}

Numerical constants like

6\pi

are dimensionless, so

[\eta] = \frac{[F]}{[r][v]}

.

2

Substitute the fundamental dimensions for force, radius, and velocity

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

,

[r] = \text{L}

, and

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

Force is mass times acceleration, radius is length, and velocity is displacement per unit time.

3

Perform exponent simplification for like base dimensions

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

Subtract indices of denominator dimensions from those in the numerator.

Anahtar Kavram

Dimensional Analysis of Viscosity

Alternatif Yöntem

Alternatively, unit derivation can be used: the SI unit of viscosity is

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

. Replacing Newtons with fundamental SI units (

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

) gives

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

, which corresponds directly to

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

.

Tahmini Süre:1m 0s

Physical Quantities, Units and Dimensions

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Alternatif Yöntem