All practice questions

Question 7761Question →

In Bohr's atomic model of the hydrogen atom, the radius of the ground state orbit ( $n = 1$ ) is $0.053\text{ nm}$ . According to de Broglie's condition for stationary electron orbits, what is the de Broglie wavelength of the electron in its second excited state? Express your answer in nanometers ( $\text{nm}$ ).

Show answer & explanation

Answer: 1

Answer

The de Broglie wavelength of the electron in its second excited state is

1.00\text{ nm}

.

The second excited state corresponds to the quantum number

n = 3

. According to Bohr's atomic model, the radius of the

n

-th orbit is given by

r_n = n^2 r_1

, yielding

r_3 = 3^2 \times 0.053\text{ nm} = 0.477\text{ nm}

. De Broglie explained Bohr's angular momentum quantization by showing that an integral number of electron matter-waves must fit around the orbital circumference:

2\pi r_n = n \lambda_n

. Solving for

\lambda_3

gives

\lambda_3 = \frac{2\pi r_3}{3} = 2\pi \times 3 \times 0.053\text{ nm} \approx 1.00\text{ nm}

.

Step-by-Step Solution

1

Identify the principal quantum number for the specified energy state

n = 3

The ground state corresponds to

n = 1

, the first excited state to

n = 2

, and the second excited state to

n = 3

.

2

Calculate the radius of the third stationary orbit

r_3 = 0.477\text{ nm}

In Bohr's model, the radius of the

n

-th orbit is proportional to

n^2

, so

r_3 = 3^2 \times 0.053\text{ nm} = 9 \times 0.053\text{ nm} = 0.477\text{ nm}

.

3

Apply de Broglie's standing wave condition for stationary orbits

\lambda_3 = \frac{2\pi r_3}{3}

De Broglie postulated that a stationary orbit contains an integral number of electron de Broglie wavelengths around its circumference:

2\pi r_n = n \lambda_n

.

4

Substitute values to compute the wavelength

\lambda_3 = 1.00\text{ nm}

\lambda_3 = \frac{2 \times 3.1416 \times 0.477\text{ nm}}{3} = 2\pi \times 3 \times 0.053\text{ nm} \approx 0.9992\text{ nm} \approx 1.00\text{ nm}

.

Key Concept

De Broglie Standing Wave Quantization in Bohr Atomic Model

Question 7762Question →

A ray of light traveling inside a transparent medium strikes the boundary with air. If the critical angle for total internal reflection at this boundary is $30^\circ$ , what is the refractive index of the medium?

2.00

0.50

0.87

1.15

Show answer & explanation

Answer: $2.00$

Answer

The refractive index of the medium is

2.00

.

For light traveling from a medium into air, the critical angle

C

is related to the refractive index

n

by

n = \frac{1}{\sin C}

. Substituting

C = 30^\circ

gives

n = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2.00

.

Step-by-Step Solution

1

Identify the relationship between critical angle

C

and refractive index

n

The formula for light passing into air is

n = \frac{1}{\sin C}

.

Total internal reflection occurs when light travels from a denser medium to a less dense medium (air,

n_{air} = 1

) at an angle greater than the critical angle.

2

Substitute the given critical angle

C = 30^\circ

into the formula

\sin 30^\circ = 0.5

, so

n = \frac{1}{0.5} = 2.00

.

Taking the reciprocal of

\sin 30^\circ

gives the correct refractive index.

Key Concept

Critical Angle and Refractive Index

Question 7763Question →

The pressure $P$ of a given mass of gas varies directly as its absolute temperature $T$ and inversely as its volume $V$ . Given that $P = 50\text{ kPa}$ when $T = 300\text{ K}$ and $V = 10\text{ m}^3$ , what is the value of $P$ when $T = 360\text{ K}$ and $V = 8\text{ m}^3$ ?

48\text{ kPa}

60\text{ kPa}

75\text{ kPa}

90\text{ kPa}

Show answer & explanation

Answer: $75\text{ kPa}$

Answer

75\text{ kPa}

The relationship is governed by

P = \frac{kT}{V}

. Substituting

P = 50

,

T = 300

, and

V = 10

yields

k = \frac{5}{3}

. Using

k = \frac{5}{3}

with

T = 360

and

V = 8

gives

P = \frac{(5/3) \times 360}{8} = 75\text{ kPa}

.

Step-by-Step Solution

1

Set up the general formula for joint and inverse variation.

P = \frac{kT}{V}

, where

k

is the constant of variation.

Pressure varies directly as temperature

T

and inversely as volume

V

.

2

Substitute initial conditions to determine

k

.

50 = \frac{k \times 300}{10} \implies 50 = 30k \implies k = \frac{5}{3}

.

The initial values

P = 50\text{ kPa}

,

T = 300\text{ K}

, and

V = 10\text{ m}^3

allow solving for

k

.

3

Calculate the new pressure with updated temperature and volume values.

P = \frac{\frac{5}{3} \times 360}{8} = \frac{600}{8} = 75\text{ kPa}

.

Substitute

k = \frac{5}{3}

,

T = 360\text{ K}

, and

V = 8\text{ m}^3

into the variation formula.

Key Concept

Joint and inverse variation in algebraic relationships

Estimated Time:1m 30s

Question 7764Question →

Let $A$ be a $3 \times 3$ square matrix such that $\det(A) > 0$ . If the matrix satisfies the property $\det(3A^{-1}) = \det(A^T A)$ , what is the value of $\det(A)$ ?

Show answer & explanation

Answer: 3

Answer

The value of

\det(A)

is 3.

Using fundamental determinant identities for a

3 \times 3

matrix (

n = 3

):

\det(3A^{-1}) = 3^3 \det(A^{-1}) = \frac{27}{\det(A)}

and

\det(A^T A) = \det(A^T)\det(A) = (\det(A))^2

. Equating them yields

\frac{27}{\det(A)} = (\det(A))^2 \implies (\det(A))^3 = 27

, which yields

\det(A) = 3

.

Step-by-Step Solution

1

Express

\det(3A^{-1})

in terms of

\det(A)

using determinant scaling and inverse properties.

\det(3A^{-1}) = 3^3 \det(A^{-1}) = \frac{27}{\det(A)}.

For an

n \times n

matrix

M

, scaling by constant

k

gives

\det(kM) = k^n \det(M)

, and

\det(M^{-1}) = \frac{1}{\det(M)}

.

2

Express

\det(A^T A)

in terms of

\det(A)

using transpose and multiplication properties.

\det(A^T A) = \det(A^T)\det(A) = (\det(A))^2.

The determinant of a product is the product of determinants, and

\det(A^T) = \det(A)

.

3

Set the two simplified expressions equal to each other and solve for

\det(A)

.

\frac{27}{\det(A)} = (\det(A))^2 \implies (\det(A))^3 = 27 \implies \det(A) = 3.

Taking the cube root of both sides gives the unique real value since

\det(A) > 0

.

Key Concept

Properties of Determinants (Scalar Multiplication, Transpose, Inverse, and Matrix Products)

Estimated Time:1m 30s

Question 7765Question →

Three capacitors, each of capacitance $12\text{ }\mu\text{F}$ , are arranged such that two of them are connected in parallel, and this combination is connected in series with the third capacitor. If the entire network is connected across a $20\text{ V}$ d.c. power supply, what is the total energy stored in the network?

0.8 \times 10^{-3}\text{ J}

1.6 \times 10^{-3}\text{ J}

3.6 \times 10^{-3}\text{ J}

7.2 \times 10^{-3}\text{ J}

Show answer & explanation

Answer: $1.6 \times 10^{-3}\text{ J}$

Answer

The total energy stored in the network is

1.6 \times 10^{-3}\text{ J}

.

Combining two

12\text{ }\mu\text{F}

capacitors in parallel gives a parallel capacitance of

24\text{ }\mu\text{F}

. Connecting this combination in series with the third

12\text{ }\mu\text{F}

capacitor results in an equivalent network capacitance of

C_{\text{eq}} = \frac{24 \times 12}{24 + 12} = 8\text{ }\mu\text{F}

. Substituting this into the energy formula

E = \frac{1}{2} C_{\text{eq}} V^2

yields

E = \frac{1}{2} \times 8 \times 10^{-6} \times 400 = 1.6 \times 10^{-3}\text{ J}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance of the parallel branch.

C_p = 12\text{ }\mu\text{F} + 12\text{ }\mu\text{F} = 24\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate the total equivalent capacitance of the network.

C_{\text{eq}} = \frac{C_p \times C_3}{C_p + C_3} = \frac{24 \times 12}{24 + 12} = \frac{288}{36} = 8\text{ }\mu\text{F} = 8 \times 10^{-6}\text{ F}

The parallel combination is connected in series with the third capacitor.

3

Calculate the total electrical energy stored in the combination.

E = \frac{1}{2} C_{\text{eq}} V^2 = \frac{1}{2} \times (8 \times 10^{-6}\text{ F}) \times (20\text{ V})^2 = 4 \times 10^{-6} \times 400 = 1.6 \times 10^{-3}\text{ J}

The formula for energy stored in a capacitor network is

E = \frac{1}{2} C V^2

.

Key Concept

Mixed capacitor networks and energy storage

Estimated Time:1m 30s

Question 7766Question →

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 7767Question →

Two electrostatic forces of magnitude $3.0\text{ N}$ and $4.0\text{ N}$ act on a small test charge at right angles ( $90^\circ$ ) to each other. What is the magnitude of the net electrostatic force acting on the charge?

7.0\text{ N}

5.0\text{ N}

1.0\text{ N}

12.0\text{ N}

Show answer & explanation

Answer: $5.0\text{ N}$

Answer

5.0\text{ N}

The net electrostatic force is found by vector addition. Since the two forces are perpendicular (

90^\circ

), the magnitude of their resultant is given by

\sqrt{F_1^2 + F_2^2} = \sqrt{3.0^2 + 4.0^2} = 5.0\text{ N}

.

Step-by-Step Solution

1

Identify the nature of force as a vector quantity.

Electrostatic forces must be combined using vector addition rather than simple scalar addition.

The forces act at right angles (

90^\circ

) to each other.

2

Apply the Pythagorean theorem to calculate the magnitude of the resultant net force

F_{\text{net}}

.

F_{\text{net}} = \sqrt{F_1^2 + F_2^2} = \sqrt{3.0^2 + 4.0^2} = \sqrt{9 + 16} = \sqrt{25} = 5.0\text{ N}

.

For perpendicular vectors, the resultant is the hypotenuse of a right-angled triangle formed by the vector components.

Key Concept

Vector Addition of Electrostatic Forces

Question 7768Question →

Match each statistical chart term or parameter in Column A with its corresponding definition or formula in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Class boundary matches the exact continuous limit of a class interval; Frequency density matches the quotient of class frequency and class width; Cumulative frequency matches the running total of frequencies plotted on an ogive; Pie chart sector angle matches the central angle formula involving multiplication by 360 degrees.

Each chart parameter in Column A directly corresponds to its standard mathematical definition, formula, or geometric representation in Column B.

Step-by-Step Solution

1

Identify the statistical definition for continuous class intervals in histograms.

Class boundaries remove gaps between non-overlapping class limits.

Histograms require continuous real class boundaries on the horizontal axis.

2

Recall the formula for histogram bar height when class intervals vary.

Frequency density =

\frac{\text{Frequency}}{\text{Class Width}}

.

Bar area must remain proportional to frequency.

3

Determine the parameter used to plot an ogive curve.

Cumulative frequency tracks accumulated totals across upper class boundaries.

An ogive represents cumulative distribution.

4

Identify the angular calculation for circular charts.

Sector angle =

\frac{\text{Class Frequency}}{\text{Total Frequency}} \times 360^\circ

.

A complete pie chart represents 360 degrees.

Key Concept

Data Representation and Chart Properties

Question 7769Question →

Given that $(x - 3)$ is a factor of the polynomial $P(x) = x^3 + kx^2 + mx - 6$ and that dividing $P(x)$ by $(x + 1)$ leaves a remainder of $-12$ , what is the value of $k + m$ ?

-1

-7

-5

5

Show answer & explanation

Answer: $-1$

Answer

The value of

k + m

is

-1

.

Applying the Factor Theorem

P(3) = 0

gives

3k + m = -7

, and applying the Remainder Theorem

P(-1) = -12

gives

k - m = -5

. Solving these simultaneous linear equations gives

k = -3

and

m = 2

, which sums to

k + m = -1

.

Step-by-Step Solution

1

Apply the Factor Theorem for the linear factor

(x - 3)

3k + m = -7

By the Factor Theorem, if

(x - 3)

is a factor, then

P(3) = 0

. Substituting

x = 3

gives

3^3 + k(3)^2 + m(3) - 6 = 0 \implies 27 + 9k + 3m - 6 = 0 \implies 9k + 3m = -21 \implies 3k + m = -7

.

2

Apply the Remainder Theorem for the divisor

(x + 1)

k - m = -5

By the Remainder Theorem, dividing

P(x)

by

(x + 1)

gives remainder

P(-1) = -12

. Substituting

x = -1

gives

(-1)^3 + k(-1)^2 + m(-1) - 6 = -12 \implies -1 + k - m - 6 = -12 \implies k - m - 7 = -12 \implies k - m = -5

.

3

Solve the system of linear equations for

k

and

m

k = -3

and

m = 2

Adding the two equations

(3k + m) + (k - m) = -7 + (-5)

gives

4k = -12 \implies k = -3

. Substituting

k = -3

into

k - m = -5

gives

-3 - m = -5 \implies m = 2

.

4

Calculate the target value

k + m

k + m = -1

Summing

k = -3

and

m = 2

yields

k + m = -3 + 2 = -1

.

Key Concept

Factor and Remainder Theorems

Estimated Time:1m 30s

Question 7770Question →

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 7771Question →

Two point charges, $q_1 = +2.0 \times 10^{-6}\text{ C}$ and $q_2 = -2.0 \times 10^{-6}\text{ C}$ , are fixed in a vacuum separated by a distance of $0.20\text{ m}$ . What is the magnitude of the net electric field intensity at the midpoint along the line joining the two charges? (Take Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ )

0.0\text{ N C}^{-1}

9.0 \times 10^5\text{ N C}^{-1}

1.8 \times 10^6\text{ N C}^{-1}

3.6 \times 10^6\text{ N C}^{-1}

Show answer & explanation

Answer: $3.6 \times 10^6\text{ N C}^{-1}$

Answer

3.6 \times 10^6\text{ N C}^{-1}

The correct answer is

3.6 \times 10^6\text{ N C}^{-1}

. At the midpoint (

r = 0.10\text{ m}

), the electric field due to the positive charge points towards the negative charge, and the field due to the negative charge also points towards the negative charge. Summing both equal field magnitudes of

1.8 \times 10^6\text{ N C}^{-1}

yields

3.6 \times 10^6\text{ N C}^{-1}

.

Step-by-Step Solution

1

Determine the distance from each point charge to the midpoint.

The distance

r = \frac{0.20\text{ m}}{2} = 0.10\text{ m}

.

The midpoint divides the total separation distance equally.

2

Calculate the magnitude of the electric field intensity

E_1

created by the positive charge

q_1

at the midpoint.

E_1 = \frac{k |q_1|}{r^2} = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.10)^2} = 1.8 \times 10^6\text{ N C}^{-1}

directed away from

q_1

(towards

q_2

).

Electric field vectors point away from positive charges.

3

Calculate the magnitude of the electric field intensity

E_2

created by the negative charge

q_2

at the midpoint.

E_2 = \frac{k |q_2|}{r^2} = \frac{9.0 \times 10^9 \times 2.0 \times 10^{-6}}{(0.10)^2} = 1.8 \times 10^6\text{ N C}^{-1}

directed towards

q_2

.

Electric field vectors point towards negative charges.

4

Combine the electric field vectors vectorially to find the net field intensity.

E_{\text{net}} = E_1 + E_2 = 1.8 \times 10^6 + 1.8 \times 10^6 = 3.6 \times 10^6\text{ N C}^{-1}

.

Because both

E_1

and

E_2

point in the exact same direction (towards

q_2

), their magnitudes add directly.

Key Concept

Superposition Principle of Electric Fields

Question 7772Question →

In an electric circuit, two capacitors $C_1 = 12\text{ }\mu\text{F}$ and $C_2 = 6\text{ }\mu\text{F}$ are connected in series. This series combination is then connected in parallel with a third capacitor $C_3$ of unknown value. When a direct-current potential difference of $100\text{ V}$ is applied across the entire network, the total electrostatic energy stored in the circuit is $100\text{ mJ}$ . What is the capacitance of $C_3$ in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 16

Answer

The capacitance of

C_3

is

16\text{ }\mu\text{F}

.

First, the series combination of

12\text{ }\mu\text{F}

and

6\text{ }\mu\text{F}

yields an equivalent branch capacitance of

4\text{ }\mu\text{F}

. Second, using

E = \frac{1}{2} C_{eq} V^2

with

E = 0.100\text{ J}

and

V = 100\text{ V}

gives a total circuit equivalent capacitance of

20\text{ }\mu\text{F}

. Finally, subtracting the branch capacitance from the total parallel equivalent capacitance gives

C_3 = 20\text{ }\mu\text{F} - 4\text{ }\mu\text{F} = 16\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the effective capacitance of the series branch containing

C_1

and

C_2

C_{12} = 4\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_{12}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{12} + \frac{1}{6} = \frac{3}{12} \implies C_{12} = 4\text{ }\mu\text{F}

.

2

Determine the total equivalent capacitance

C_{eq}

of the circuit using the given stored energy and voltage

C_{eq} = 20\text{ }\mu\text{F}

Energy stored in a capacitor network is

E = \frac{1}{2} C_{eq} V^2

. Rearranging gives

C_{eq} = \frac{2E}{V^2} = \frac{2 \times 0.100\text{ J}}{(100\text{ V})^2} = 20 \times 10^{-6}\text{ F} = 20\text{ }\mu\text{F}

.

3

Calculate the unknown capacitance

C_3

from the parallel combination formula

C_3 = 16\text{ }\mu\text{F}

Because the branch

C_{12}

and

C_3

are in parallel,

C_{eq} = C_{12} + C_3 \implies 20\text{ }\mu\text{F} = 4\text{ }\mu\text{F} + C_3 \implies C_3 = 16\text{ }\mu\text{F}

.

Key Concept

Series and parallel combinations of capacitors combined with electrostatic energy storage

Estimated Time:2m 0s

Question 7773Question →

A water wave with a frequency of $20\text{ Hz}$ and a wavelength of $0.60\text{ m}$ in deep water enters a shallow region where its speed becomes $8.0\text{ m s}^{-1}$ . What is the wavelength of the wave in the shallow region?

0.40\text{ m}

0.60\text{ m}

0.90\text{ m}

1.50\text{ m}

Show answer & explanation

Answer: $0.40\text{ m}$

Answer

The wavelength of the wave in the shallow region is

0.40\text{ m}

.

When a wave passes from one medium into another, its frequency remains constant because frequency depends only on the source. Applying the wave equation

v = f\lambda

to the shallow region gives

\lambda = \frac{v}{f} = \frac{8.0\text{ m s}^{-1}}{20\text{ Hz}} = 0.40\text{ m}

.

Step-by-Step Solution

1

Identify the constant parameter during wave refraction.

The frequency of the wave remains

f = 20\text{ Hz}

.

Frequency is determined entirely by the source producing the wave and does not change upon entering a new medium.

2

Calculate the new wavelength using the wave equation

v = f\lambda

.

$\lambda = \frac{v}{f} = \frac{8.0\text{ m s}^{-1}}{20\text{ Hz}} = 0.40\text{ m}$.

Dividing the wave speed in the new medium by the constant frequency yields the wavelength in that medium.

Key Concept

Constancy of wave frequency during refraction across media boundaries

Question 7774Question →

Match each physical quantity listed on the left with its corresponding SI unit expressed strictly in terms of fundamental (base) units on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electric permittivity of free space maps to

\text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2

, Magnetic permeability of free space maps to

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

, Thermal conductivity maps to

\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}

, and Specific heat capacity maps to

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

.

Each physical quantity is matched to its exact fundamental unit expression obtained by substituting basic formulas into SI base units (

\text{kg}

,

\text{m}

,

\text{s}

,

\text{A}

,

\text{K}

). Electric permittivity resolves to

\text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2

, magnetic permeability resolves to

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

, thermal conductivity resolves to

\text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}

, and specific heat capacity resolves to

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

.

Step-by-Step Solution

1

Derive the base SI units for Electric permittivity of free space (

\varepsilon_0

).

From Coulomb's Law,

F = \frac{q_1 q_2}{4\pi \varepsilon_0 r^2} \implies \varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}

. Expressing terms in SI fundamental units: charge

q = \text{A}\cdot\text{s}

, force

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

, distance

r = \text{m}

. Thus,

[\varepsilon_0] = \frac{(\text{A}\cdot\text{s})^2}{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}^2} = \text{kg}^{-1}\cdot\text{m}^{-3}\cdot\text{s}^4\cdot\text{A}^2

.

Relating derived electromagnetic quantities to fundamental SI units via governing physical equations.

2

Derive the base SI units for Magnetic permeability of free space (

\mu_0

).

From the force per unit length between parallel current-carrying conductors,

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r} \implies \mu_0 = \frac{2\pi F r}{I_1 I_2 L}

. Units:

[\mu_0] = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2})\cdot\text{m}}{\text{A}^2\cdot\text{m}} = \text{kg}\cdot\text{m}\cdot\text{s}^{-2}\cdot\text{A}^{-2}

.

Using Ampère's force law to solve for magnetic permeability in base units.

3

Derive the base SI units for Thermal conductivity (

k

).

From Fourier's Law of Heat Conduction,

\frac{Q}{t} = k A \frac{\Delta T}{\Delta x} \implies k = \frac{Q \cdot \Delta x}{t A \Delta T}

. Heat energy

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

, time

t = \text{s}

, area

A = \text{m}^2

, thickness

\Delta x = \text{m}

, temperature difference

\Delta T = \text{K}

. Thus,

[k] = \frac{(\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2})\cdot\text{m}}{\text{s}\cdot\text{m}^2\cdot\text{K}} = \text{kg}\cdot\text{m}\cdot\text{s}^{-3}\cdot\text{K}^{-1}

.

Connecting thermal conduction equations to fundamental mechanics and thermodynamic units.

4

Derive the base SI units for Specific heat capacity (

c

).

From

Q = m c \Delta T \implies c = \frac{Q}{m \Delta T}

. Units:

[c] = \frac{\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2}}{\text{kg}\cdot\text{K}} = \text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}

.

Applying the defining equation of heat capacity to reduce the unit to fundamental base units.

Key Concept

Derivation of complex derived SI units from fundamental base units using fundamental physical laws.

Question 7775Question →

A trader deposited $\text{₦}40,000$ in a microfinance bank offering compound interest at the rate of $10\%$ per annum, compounded annually. What is the total interest earned by the trader at the end of $2$ years?

\text{₦}8,000

\text{₦}8,400

\text{₦}40,400

\text{₦}48,400

Show answer & explanation

Answer: $\text{₦}8,400$

Answer

The total interest earned by the trader at the end of

2

years is

\text{₦}8,400

.

Using the compound interest formula

A = P(1 + r)^n

, the total accumulated amount after 2 years is

40,000 \times (1.1)^2 = \text{₦}48,400

. Subtracting the original principal of

\text{₦}40,000

gives the compound interest earned:

48,400 - 40,000 = \text{₦}8,400

.

Step-by-Step Solution

1

Calculate the total accumulated amount (

A

) using the compound interest formula

A = P\left(1 + \frac{R}{100}\right)^n

.

A = 40,000\left(1 + \frac{10}{100}\right)^2 = 40,000 \times (1.1)^2 = 40,000 \times 1.21 = \text{₦}48,400

.

The compound interest formula yields the total value of the investment after 2 years.

2

Subtract the initial principal (

P

) from the total accumulated amount (

A

) to find the compound interest (

I

).

I = A - P = 48,400 - 40,000 = \text{₦}8,400

.

Interest earned is the difference between total final amount and initial principal.

Key Concept

Calculation of compound interest vs total accumulated amount

Question 7776Question →

Which of the following is the set of real values of $x$ that satisfies the inequality $\frac{3 - x}{4} \le \frac{2x + 5}{3}$ ?

x \ge -1

x \le -1

x \ge 1

x \le 1

Show answer & explanation

Answer: $x \ge -1$

Answer

The set of real values of

x

that satisfies the inequality is

x \ge -1

.

Multiplying through by 12 gives

9 - 3x \le 8x + 20

. Grouping terms results in

-11x \le 11

. Dividing by

-11

requires reversing the inequality sign from

\le

to

\ge

, giving the solution

x \ge -1

.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the inequality by the lowest common multiple, 12.

3(3 - x) \le 4(2x + 5)

Eliminating fractions simplifies the algebraic expression.

2

Expand both sides by distributing the multipliers.

9 - 3x \le 8x + 20

Prepares terms for grouping variables on one side and constants on the other.

3

Collect all terms containing

x

on the left side and constant terms on the right side.

-3x - 8x \le 20 - 9 \implies -11x \le 11

Isolates the linear variable term.

4

Divide both sides by

-11

and flip the inequality sign.

x \ge -1

Dividing or multiplying an inequality by a negative number reverses the direction of the inequality sign.

Key Concept

Linear Inequalities and Reversing Inequality Sign on Division by Negative Numbers

Question 7777Question →

A student obtained a mean score of $62$ in $5$ subjects. After the score of a $6\text{th}$ subject was added, the overall mean score became $65$ . What is the score obtained in the $6\text{th}$ subject?

Show answer & explanation

Answer: 80

Answer

The score obtained in the 6th subject is 80.

The sum of scores for the first 5 subjects is

5 \times 62 = 310

. With the 6th subject included, the total sum of scores becomes

6 \times 65 = 390

. The score of the 6th subject is the difference between these two totals:

390 - 310 = 80

.

Step-by-Step Solution

1

Calculate the total score of the initial 5 subjects

310

The sum of scores for ungrouped data is equal to the number of data items multiplied by the mean: 5 × 62 = 310.

2

Calculate the total score of all 6 subjects after including the new score

390

The new total score is obtained by multiplying the new count of subjects by the new mean: 6 × 65 = 390.

3

Determine the score of the 6th subject

80

The difference between the total score of 6 subjects and the total score of 5 subjects gives the score of the 6th subject: 390 - 310 = 80.

Key Concept

Calculating a missing data value given the mean of ungrouped data before and after addition

Question 7778Question →

A brass container with an initial capacity of $500\text{ cm}^3$ at $20^\circ\text{C}$ is filled completely with ethanol. When the container and its contents are uniformly heated to $70^\circ\text{C}$ , a volume of $12\text{ cm}^3$ of ethanol overflows from the container. Given that the linear expansivity of brass is $2.0 \times 10^{-5}\text{ K}^{-1}$ , calculate the real cubic expansivity of ethanol in units of $10^{-4}\text{ K}^{-1}$ .

Show answer & explanation

Answer: 5.4

Answer

The real cubic expansivity of ethanol is

5.4 \times 10^{-4}\text{ K}^{-1}

, which corresponds to a value of

5.4

in units of

10^{-4}\text{ K}^{-1}

.

The real volume increase of a liquid consists of the apparent expansion (observed overflow volume) plus the volume expansion of the container itself. By finding the apparent cubic expansivity

\gamma_a = \frac{12}{500 \times 50} = 4.8 \times 10^{-4}\text{ K}^{-1}

and adding the vessel's cubic expansivity

\gamma_v = 3 \times 2.0 \times 10^{-5} = 0.6 \times 10^{-4}\text{ K}^{-1}

, we obtain the real cubic expansivity

\gamma_r = 5.4 \times 10^{-4}\text{ K}^{-1}

, which yields

5.4

in the required units.

Step-by-Step Solution

1

Determine the temperature increase of the system

\Delta T = 70^\circ\text{C} - 20^\circ\text{C} = 50\text{ K}

The thermal expansion is driven by the change in temperature.

2

Calculate the apparent cubic expansivity of ethanol (\gamma_a)

\gamma_a = \frac{\Delta V_{overflow}}{V_0 \cdot \Delta T} = \frac{12\text{ cm}^3}{500\text{ cm}^3 \times 50\text{ K}} = 4.8 \times 10^{-4}\text{ K}^{-1}

The overflow volume represents the apparent volume increase of the liquid relative to the expanding vessel.

3

Calculate the cubic expansivity of the brass vessel (\gamma_v)

\gamma_v = 3 \times \alpha_{brass} = 3 \times 2.0 \times 10^{-5}\text{ K}^{-1} = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}

Cubic expansivity of a solid vessel is three times its linear expansivity.

4

Compute the real cubic expansivity of ethanol (\gamma_r)

\gamma_r = \gamma_a + \gamma_v = 4.8 \times 10^{-4}\text{ K}^{-1} + 0.6 \times 10^{-4}\text{ K}^{-1} = 5.4 \times 10^{-4}\text{ K}^{-1}

The real expansion of a liquid is the sum of its apparent expansion and the expansion of the containing vessel.

Key Concept

Relationship between Real and Apparent Cubic Expansivity of Liquids

Question 7779Question →

A neutral insulated conductor loses $5.0 \times 10^{13}$ electrons during an electrostatics experiment. What is the magnitude of the net electric charge, in microcoulombs ( $\mu\text{C}$ ), acquired by the conductor? (Take elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ )

Show answer & explanation

Answer: 8

Answer

The magnitude of the net electric charge acquired by the conductor is

8\text{ }\mu\text{C}

.

According to the principle of charge quantization, the total electric charge

Q

is calculated using

Q = n e

. Multiplying

5.0 \times 10^{13}

electrons by

1.6 \times 10^{-19}\text{ C}

yields

8.0 \times 10^{-6}\text{ C}

, which converts to

8\text{ }\mu\text{C}

.

Step-by-Step Solution

1

Identify the relevant formula for quantization of charge.

The net charge acquired is given by

Q = n e

.

Electric charge is quantized, so the total charge magnitude equals the number of transferred electrons multiplied by the magnitude of charge on a single electron.

2

Calculate the magnitude of charge in Coulombs.

Q = (5.0 \times 10^{13}) \times (1.6 \times 10^{-19}\text{ C}) = 8.0 \times 10^{-6}\text{ C}

.

Multiplying the quantity of removed electrons by the elementary charge gives total charge in Coulombs.

3

Convert the calculated value from Coulombs to microcoulombs.

8.0 \times 10^{-6}\text{ C} = 8\text{ }\mu\text{C}

.

Since

1\text{ }\mu\text{C} = 10^{-6}\text{ C}

, dividing

8.0 \times 10^{-6}

by

10^{-6}

yields 8.

Key Concept

Quantization of Electric Charge

Question 7780Question →

During a head-on collision in Rutherford's $\alpha$ -particle scattering experiment, an $\alpha$ -particle of initial speed $v$ approaches a stationary heavy nucleus. The distance of closest approach achieved by the $\alpha$ -particle is $r_0$ . If the initial speed of the $\alpha$ -particle is increased to $2v$ , what will be the new distance of closest approach in terms of $r_0$ ?

\frac{r_0}{4}

\frac{r_0}{2}

2r_0

4r_0

Show answer & explanation

Answer: $\frac{r_0}{4}$

Answer

The new distance of closest approach will be

\frac{r_0}{4}

.

At the distance of closest approach, the entire initial kinetic energy of the

\alpha

-particle is converted into electric potential energy:

E_k = \frac{1}{2}mv^2 = \frac{k_e q_1 q_2}{r}

. Rearranging for

r

gives

r = \frac{2k_e q_1 q_2}{m v^2}

, showing that

r

is inversely proportional to

v^2

. When the initial speed is doubled to

2v

, the kinetic energy increases by a factor of

2^2 = 4

. Consequently, the distance of closest approach is reduced to one-fourth of its initial value,

\frac{r_0}{4}

.

Step-by-Step Solution

1

Apply conservation of mechanical energy at the distance of closest approach.

Initial kinetic energy equals electrostatic potential energy at distance

r_0

:

\frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_0}

.

At the distance of closest approach, the

\alpha

-particle momentarily stops, converting all kinetic energy into electric potential energy.

2

Express the distance of closest approach

r_0

in terms of initial speed

v

.

r_0 = \frac{2 q_1 q_2}{4\pi\varepsilon_0 m v^2} \propto \frac{1}{v^2}

.

Rearranging the energy conservation equation demonstrates that distance of closest approach is inversely proportional to

v^2

.

3

Substitute the new speed

v' = 2v

into the proportion.

r' \propto \frac{1}{(2v)^2} = \frac{1}{4v^2} = \frac{r_0}{4}

.

Doubling the speed quadruples the kinetic energy, reducing the distance required to bring the particle to rest by a factor of 4.

Key Concept

Distance of Closest Approach in Rutherford Scattering

Estimated Time:2m 0s

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept