All practice questions

Question 7861Question →

A resistor of resistance $15\,\Omega$ is connected across a direct-current source. If a steady current of $0.80\,\text{A}$ flows through the resistor, what is the potential difference across its terminals?

12\,\text{V}

18.75\,\text{V}

0.053\,\text{V}

14.2\,\text{V}

Show answer & explanation

Answer: $12\,\text{V}$

Answer

The potential difference across the terminals of the resistor is

12\,\text{V}

.

According to Ohm's law, the potential difference

V

across a resistor is equal to the product of the electric current

I

passing through it and its resistance

R

(

V = I \times R

). Substituting

I = 0.80\,\text{A}

and

R = 15\,\Omega

yields

V = 12\,\text{V}

.

Step-by-Step Solution

1

Identify the given physical quantities

Resistance

R = 15\,\Omega

and electric current

I = 0.80\,\text{A}

.

These are the given parameters needed to find potential difference.

2

Apply Ohm's Law formula for potential difference

V = I \times R

Ohm's law states that potential difference is directly proportional to current for a ohmic resistor of constant resistance.

3

Substitute the values and calculate

V = 0.80\,\text{A} \times 15\,\Omega = 12\,\text{V}

Multiplying current by resistance yields potential difference in volts.

Key Concept

Ohm's Law (

V = IR

)

Question 7862Question →

A progressive wave traveling through a primary medium is represented by the displacement equation $y = 0.04 \sin\left(100\pi t - 2.5\pi x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. If the wave propagates into a secondary medium where its speed increases by $20\%$ , what is the wavelength of the wave in the secondary medium?

0.96\text{ m}

0.67\text{ m}

1.50\text{ m}

0.16\text{ m}

Show answer & explanation

Answer: $0.96\text{ m}$

Answer

The wavelength of the wave in the secondary medium is

0.96\text{ m}

.

Comparing

y = 0.04 \sin(100\pi t - 2.5\pi x)

to the standard form

y = A \sin(\omega t - k x)

, the wave number is

k = 2.5\pi\text{ rad/m}

. The initial wavelength is

\lambda_1 = \frac{2\pi}{2.5\pi} = 0.8\text{ m}

. When a wave moves to a new medium, its frequency stays constant, making wavelength directly proportional to wave speed (

v \propto \lambda

). An increase of

20\%

in wave speed means the new wavelength is

\lambda_2 = 0.8 \times 1.20 = 0.96\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency

\omega

and wave number

k

from the wave equation.

From

y = A \sin(\omega t - k x)

, we identify

\omega = 100\pi\text{ rad/s}

and

k = 2.5\pi\text{ rad/m}

.

Standard wave equation parameters directly define the wave's spatial and temporal frequencies.

2

Calculate the wavelength

\lambda_1

in the initial medium.

$\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{2.5\pi} = 0.8\text{ m}$.

Wavelength is inversely related to the wave number

k

by

\lambda = \frac{2\pi}{k}

.

3

Apply the boundary conditions of wave refraction across media.

Frequency

f

remains constant across boundary; speed

v

and wavelength

\lambda

scale proportionally.

The frequency of a wave is determined solely by the source and does not change upon entering a new medium.

4

Determine the new wavelength

\lambda_2

in the secondary medium.

$\lambda_2 = \lambda_1 \times (1 + 0.20) = 0.8\text{ m} \times 1.20 = 0.96\text{ m}$.

Since

v = f \lambda

and

f

is constant,

\frac{v_2}{v_1} = \frac{\lambda_2}{\lambda_1} = 1.20

.

Key Concept

Invariance of Wave Frequency across Media and Wave Equation Parameter Extraction

Estimated Time:2m 0s

Question 7863Question →

A resistance thermometer has a resistance of $4.0\,\Omega$ at $0\,^\circ\text{C}$ and $6.0\,\Omega$ at $100\,^\circ\text{C}$ . The thermometer is connected in series with a fixed $10.0\,\Omega$ resistor across a $24.0\,\text{V}$ DC power source having an internal resistance of $1.0\,\Omega$ . If a steady current of $1.2\,\text{A}$ flows through the circuit, what is the temperature of the thermometer's environment?

250\,^\circ\text{C}

300\,^\circ\text{C}

450\,^\circ\text{C}

350\,^\circ\text{C}

Show answer & explanation

Answer: $250\,^\circ\text{C}$

Answer

The temperature of the thermometer environment is

250\,^\circ\text{C}

.

By applying the complete circuit equation

E = I(R_\theta + R_{\text{fixed}} + r)

, the total circuit resistance is found to be

20.0\,\Omega

. Subtracting the fixed resistance of

10.0\,\Omega

and the cell internal resistance of

1.0\,\Omega

yields the thermometer resistance

R_\theta = 9.0\,\Omega

. Substituting this into the thermometric relation

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100\,^\circ\text{C}

gives

\frac{9.0 - 4.0}{6.0 - 4.0} \times 100 = 250\,^\circ\text{C}

.

Step-by-Step Solution

1

Calculate total circuit resistance using Ohm's Law and internal resistance equation

R_{\text{total}} = \frac{E}{I} = \frac{24.0\,\text{V}}{1.2\,\text{A}} = 20.0\,\Omega

The electromotive force (e.m.f) of the source equals total current multiplied by total resistance including internal resistance.

2

Determine the resistance of the thermometer at the unknown temperature (

R_\theta

)

R_\theta = R_{\text{total}} - R_{\text{fixed}} - r = 20.0\,\Omega - 10.0\,\Omega - 1.0\,\Omega = 9.0\,\Omega

The circuit components are in series, so total resistance is the sum of external resistances and internal resistance.

3

Apply the linear resistance thermometer temperature scale formula

$\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100\,^\circ\text{C} = \frac{9.0 - 4.0}{6.0 - 4.0} \times 100 = \frac{5.0}{2.0} \times 100 = 250\,^\circ\text{C}$

Resistance varies linearly with temperature between the ice point (

0\,^\circ\text{C}

) and steam point (

100\,^\circ\text{C}

).

Key Concept

Integration of Ohm's Law, internal resistance of a cell, and resistance thermometry

Estimated Time:2m 0s

Question 7864Question →

Find the real value of $x$ that satisfies the logarithmic equation $\log_5(x^2 - 4) - \log_5(x - 2) = 2$ .

Show answer & explanation

Answer: 23

Answer

The value of

x

that satisfies the equation is 23.

Applying the logarithmic quotient rule gives

\log_5 \left(\frac{x^2 - 4}{x - 2}\right) = 2

. Factoring the numerator gives

\log_5 \left(\frac{(x - 2)(x + 2)}{x - 2}\right) = \log_5(x + 2) = 2

. Converting to exponential form yields

x + 2 = 5^2 = 25

, which simplifies to

x = 23

.

Step-by-Step Solution

1

Apply the logarithmic quotient rule

\log_5\left(\frac{x^2 - 4}{x - 2}\right) = 2

The difference of two logarithms of the same base equals the logarithm of their quotient.

2

Factor the numerator and simplify the expression

\log_5(x + 2) = 2

Factoring

x^2 - 4

into

(x-2)(x+2)

allows canceling the common factor

(x-2)

in the denominator.

3

Convert the logarithmic equation into exponential form

x + 2 = 5^2 = 25

By definition,

\log_b(y) = c

is equivalent to

b^c = y

.

4

Solve the linear equation for

x

x = 23

Subtracting 2 from both sides gives

x = 23

.

Key Concept

Logarithmic Quotient Rule and Logarithm-to-Exponent Conversion

Question 7865Question →

The speed of sound $v$ in a gas depends on the gas pressure $P$ and density $\rho$ according to the relation $v = k P^a \rho^b$ , where $k$ is a dimensionless constant. What is the numerical value of the exponent $a$ ?

Show answer & explanation

Answer: 0.5

Answer

The numerical value of the exponent

a

is

0.5

.

Applying dimensional homogeneity to the relation

v = k P^a \rho^b

yields

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

. Equating the powers of time

T

gives

-2a = -1

, which simplifies to

a = 0.5

.

Step-by-Step Solution

1

Determine the base dimensions of all physical quantities in the relationship

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

,

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

, and

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

Physical quantities must be expressed in fundamental dimensions (

M, L, T

) to apply dimensional analysis.

2

Formulate the dimensional balance equation

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

By the principle of dimensional homogeneity, the total dimensions on the left side must equal those on the right side.

3

Equate exponents of

T

to solve for

a

-2a = -1 \implies a = 0.5

Comparing powers of time

T

directly isolates the variable

a

.

Key Concept

Dimensional Homogeneity and Derivation of Exponents

Question 7866Question →

For a normal good, the substitution effect and the income effect operate in opposite directions following a change in the price of the good.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. For a normal good, the substitution effect and the income effect operate in the same direction.

The statement is false because for any normal good, a price decrease creates a substitution effect and an income effect that both work in the same direction to increase total quantity demanded.

Step-by-Step Solution

1

Examine the substitution effect direction upon a price decrease.

The substitution effect always induces a consumer to purchase more of a good when its relative price falls.

The good becomes relatively less expensive compared to alternative substitute goods.

2

Examine the income effect direction for a normal good.

The increase in real purchasing power leads to an increase in the quantity demanded of the normal good.

By definition, the demand for a normal good rises as real income increases.

3

Combine both effects to determine their directional relationship.

Both the substitution effect and the income effect push quantity demanded in the same upward direction.

Because both effects act constructively together, they do not oppose each other for normal goods.

Key Concept

Directional interaction of income and substitution effects for normal goods

Question 7867Question →

During an investigation on electrical circuits, a student measures electric current, potential difference, time, and thermodynamic temperature. Which of the measured physical quantities is a derived quantity?

Electric current

Potential difference

Time

Thermodynamic temperature

Show answer & explanation

Answer: Potential difference

Answer

Potential difference

Potential difference is a derived physical quantity because it is derived from work (energy) and electric charge (

V = \frac{W}{Q} = \frac{W}{I \cdot t}

), unlike electric current, time, and thermodynamic temperature which are fundamental SI quantities.

Step-by-Step Solution

1

Identify fundamental (base) physical quantities

The seven SI fundamental quantities are mass, length, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.

Fundamental quantities are basic quantities that do not depend on any other physical quantity.

2

Classify the given quantities

Electric current, time, and thermodynamic temperature are base quantities. Potential difference (

V = \frac{W}{Q}

) is derived from work and electric charge.

Derived quantities are physical quantities defined by combining fundamental quantities.

Key Concept

Fundamental and Derived Quantities

Estimated Time:1m 0s

Question 7868Question →

In a sports academy of 120 athletes, 70 play football, 60 play basketball, and 50 play tennis. If 10 athletes play none of these three sports and 15 athletes play all three sports, how many athletes play exactly two of these sports?

Show answer & explanation

Answer: 40

Answer

40 athletes play exactly two of the sports.

The correct answer is 40. Subtracting the 10 athletes who play no sports from the total of 120 leaves 110 athletes playing at least one sport. Using inclusion-exclusion, the sum of pairwise intersections is

S_2 = 70 + 60 + 50 + 15 - 110 = 85

. Since

S_2

contains the region of all three sports counted three times, subtracting

3 \times 15 = 45

gives 40 athletes who play exactly two sports.

Step-by-Step Solution

1

Determine the cardinality of the union of all three sets

n(F \cup B \cup T) = 120 - 10 = 110

Athletes who play none of the three sports are excluded from the total universal set.

2

Apply the Principle of Inclusion-Exclusion for three sets to find the sum of 2-set intersections

S_2 = n(F) + n(B) + n(T) + n(F \cap B \cap T) - n(F \cup B \cup T) = 70 + 60 + 50 + 15 - 110 = 85

The formula relates the total union, individual set cardinalities, pairwise intersections, and triple intersection.

3

Subtract three times the triple intersection from

S_2

to isolate regions corresponding to exactly two sports

Exactly two sports =

S_2 - 3 \times n(F \cap B \cap T) = 85 - 3(15) = 40

Each pairwise intersection sum

S_2

includes the triple intersection region three times.

Key Concept

Three-set inclusion-exclusion principle and region cardinality decomposition

Estimated Time:1m 30s

Question 7869Question →

In Rutherford's $\alpha$ -particle scattering experiment, most of the $\alpha$ -particles passed straight through the gold foil with negligible deflection, while a very small fraction was scattered through angles greater than $90^\circ$ . Which fundamental deduction about atomic structure was directly established by these rare, large-angle scatterings?

The positive charge and nearly all the atomic mass are concentrated in an extremely small, dense central region called the nucleus.

Electrons revolve around the positive center in discrete, non-radiating stationary orbits.

The positive charge is distributed uniformly throughout a sphere of atomic dimensions containing embedded electrons.

Orbiting electrons continuously radiate energy, causing them to gradually spiral inward toward the center.

Show answer & explanation

Answer: The positive charge and nearly all the atomic mass are concentrated in an extremely small, dense central region called the nucleus.

Answer

The positive charge and nearly all the atomic mass are concentrated in an extremely small, dense central region called the nucleus.

Large-angle deflection (

>90^\circ

) requires an immense repulsive Coulomb force, which can only occur if the entire positive charge and virtually all atomic mass are concentrated in a tiny central region (the nucleus). If positive charge were spread out over the entire atomic volume, the maximum electric field would be far too weak to turn back high-velocity alpha particles.

Step-by-Step Solution

1

Analyze the experimental observations from Rutherford's alpha-scattering experiment.

Most

\alpha

-particles pass undeflected (indicating mostly empty space), while a tiny fraction scatter at angles

>90^\circ

.

Understanding the physical cause of large-angle electrostatic deflection.

2

Relate electrostatic repulsive force to mass and charge distribution.

To turn around a fast-moving positive

\alpha

-particle (

He^{2+}

), it must experience a intense Coulomb repulsion

F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}

at very small distance

r

.

A diffuse charge distribution (Thomson model) yields weak electric fields that cannot cause wide-angle scattering.

3

Identify the structural conclusion drawn by Rutherford.

The entire positive charge and mass must reside in a massive, concentrated center (the nucleus).

Directly matches the core physical takeaway of the nuclear atomic model.

Key Concept

Rutherford Nuclear Model and Alpha Scattering Deduction

Question 7870Question →

According to Newton's law of universal gravitation, the gravitational force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2}$ . Which of the following correctly expresses the derived SI unit of the universal gravitational constant, $G$ , in terms of fundamental (base) units?

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

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

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

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

Show answer & explanation

Answer: $\text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}$

Answer

The SI unit of the universal gravitational constant

G

expressed in fundamental base units is

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

.

The expression

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

is correct because rearranging

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

yields

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

. Replacing

F

with its base equivalent

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

,

r

with

\text{m}

, and

m_1, m_2

with

\text{kg}

gives

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

.

Step-by-Step Solution

1

Rearrange the gravitational formula to solve for

G

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

To express the unit of

G

, we need it in terms of quantities with known units.

2

Substitute the SI unit for force in fundamental base units

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

Force is mass times acceleration (

F = ma

), so its base unit is

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

.

3

Substitute all base units into the formula for

G

\text{Unit of } G = \frac{(\text{kg}\cdot\text{m}\cdot\text{s}^{-2}) \cdot \text{m}^2}{\text{kg} \cdot \text{kg}} = \frac{\text{kg}\cdot\text{m}^3\cdot\text{s}^{-2}}{\text{kg}^2}

Combining length terms (

\text{m} \cdot \text{m}^2 = \text{m}^3

) and mass terms in the denominator.

4

Simplify the mass exponent using laws of indices

\text{Unit of } G = \text{kg}^{1 - 2}\cdot\text{m}^3\cdot\text{s}^{-2} = \text{kg}^{-1}\cdot\text{m}^3\cdot\text{s}^{-2}

Dividing by

\text{kg}^2

subtracts 2 from the mass exponent.

Key Concept

Derivation of Derived Units in Base SI Units

Estimated Time:1m 15s

Question 7871Question →

The rate of change of a function $f(x)$ with respect to $x$ is defined by $f'(x) = 3x^2 - 4x + 6\sin(3x)$ . If $f(0) = 7$ , determine the value of the constant of integration, $C$ .

Show answer & explanation

Answer: 9

Answer

The constant of integration

C

is equal to 9.

Integrating

f'(x) = 3x^2 - 4x + 6\sin(3x)

gives

f(x) = x^3 - 2x^2 - 2\cos(3x) + C

. Substituting

x = 0

and

f(0) = 7

leads to

7 = 0 - 0 - 2(1) + C

, which simplifies to

C = 9

.

Step-by-Step Solution

1

Integrate the rate of change function

f'(x) = 3x^2 - 4x + 6\sin(3x)

with respect to

x

.

f(x) = x^3 - 2x^2 - 2\cos(3x) + C

Using the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

and trigonometric integration rule

\int \sin(kx) dx = -\frac{1}{k}\cos(kx)

.

2

Substitute the initial condition

x = 0

and

f(0) = 7

into the expression for

f(x)

.

7 = (0)^3 - 2(0)^2 - 2\cos(3 \cdot 0) + C

The curve passes through

x = 0

with value

y = 7

.

3

Evaluate the trigonometric term at zero and solve for

C

.

7 = -2(1) + C \implies C = 9

Since

\cos(0) = 1

, the expression simplifies to

7 = -2 + C

, yielding

C = 9

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Boundary Conditions

Question 7872Question →

A sound transmitter and a projectile launcher are co-located at a distance of $210\text{ m}$ directly in front of a tall, flat vertical cliff. At time $t = 0\text{ s}$ , a projectile is launched directly away from the cliff at a constant speed of $60\text{ m s}^{-1}$ , while a sound pulse is emitted simultaneously towards the cliff. The sound wave reflects off the cliff face and travels back to overtake the moving projectile. Assuming the speed of sound in air is $340\text{ m s}^{-1}$ , calculate the distance from the cliff face, in meters, to the position where the reflected sound wave intercepts the projectile.

Show answer & explanation

Answer: 300

Answer

The distance of the projectile from the cliff face at the instant of interception is

300\text{ m}

.

The correct calculation accounts for both the two-part path of the sound wave (forward to cliff + back to projectile) and the displacement of the projectile moving away from the cliff over the same time interval, yielding an interception distance of

300\text{ m}

from the cliff.

Step-by-Step Solution

1

Set up the total distance expression for the sound wave from the cliff face

Total sound path =

210\text{ m} + x

The sound pulse must travel

210\text{ m}

forward to hit the cliff face, plus an additional distance

x

away from the cliff face after reflection to reach the projectile.

2

Set up the distance expression for the projectile from its starting point

Projectile path =

x - 210\text{ m}

The projectile starts

210\text{ m}

away from the cliff and moves farther away to position

x

.

3

Equate the time elapsed for both sound propagation and projectile movement

\frac{210 + x}{340} = \frac{x - 210}{60}

Both events happen simultaneously over the exact same time interval

t

.

4

Solve the linear equation for

x

x = 300\text{ m}

Cross-multiplying yields

60(210 + x) = 340(x - 210)

, which simplifies to

28x = 8400

, giving

x = 300\text{ m}

.

Key Concept

Echo reflection path combined with relative linear kinematics

Question 7873Question →

A progressive wave traveling along a stretched string is represented by the equation $y = 0.02 \sin(120\pi t - 3\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the velocity of the wave in $\text{m s}^{-1}$ ?

Show answer & explanation

Answer: 40

Answer

The velocity of the wave is

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

.

The standard progressive wave equation is

y = A \sin(\omega t - kx)

, where

\omega

is the angular frequency and

k

is the wave number (wave vector). Comparing the given equation

y = 0.02 \sin(120\pi t - 3\pi x)

with the standard form yields

\omega = 120\pi \text{ rad s}^{-1}

and

k = 3\pi \text{ m}^{-1}

. The velocity of propagation of the wave is given by

v = \frac{\omega}{k} = \frac{120\pi}{3\pi} = 40 \text{ m s}^{-1}

.

Step-by-Step Solution

1

Compare given equation with the standard progressive wave equation

Matching

y = 0.02 \sin(120\pi t - 3\pi x)

to

y = A \sin(\omega t - kx)

gives

\omega = 120\pi \text{ rad s}^{-1}

and

k = 3\pi \text{ rad m}^{-1}

.

Direct parameter identification from the wave function gives the angular frequency and wave number.

2

Calculate wave velocity

Wave velocity

v = \frac{\omega}{k} = \frac{120\pi}{3\pi} = 40 \text{ m s}^{-1}

.

The ratio of angular frequency to wave number equals the phase velocity of the wave.

Key Concept

Extracting wave parameters (angular frequency and wave number) from the mathematical wave equation to determine wave velocity.

Question 7874Question →

A particle is projected from ground level at an angle to the horizontal. At its highest point of trajectory, which of the following statements correctly describes its velocity and acceleration?

The vertical component of velocity is zero, the horizontal component of velocity is non-zero, and the acceleration is directed downward.

Both the total velocity and the acceleration of the particle are zero.

The total velocity of the particle is zero, but its acceleration is directed downward.

The vertical component of velocity is at its maximum value, and the acceleration is zero.

Show answer & explanation

Answer: The vertical component of velocity is zero, the horizontal component of velocity is non-zero, and the acceleration is directed downward.

Answer

At the highest point of projectile motion, the vertical velocity component is zero, the horizontal velocity component is non-zero, and acceleration due to gravity acts downward.

At the peak of a projectile trajectory, vertical upward velocity is fully decelerated by gravity to zero. However, horizontal velocity is maintained because there is no horizontal acceleration. Acceleration due to gravity continues to act constantly downward.

Step-by-Step Solution

1

Analyze vertical velocity at the highest point

v_y = 0

At maximum height, the upward vertical component of motion reduces to zero before the particle begins descending.

2

Analyze horizontal velocity throughout motion

v_x = u \cos\theta \neq 0

In the absence of air resistance, no horizontal force acts on the projectile, keeping horizontal velocity constant throughout flight.

3

Determine the direction of acceleration

a = g

acting vertically downward

Gravity is the only force acting on a projectile in motion, providing continuous downward acceleration.

Key Concept

Velocity components and acceleration at the apex of projectile motion

Estimated Time:45s

Question 7875Question →

In an X-ray tube operated at a fixed accelerating potential, replacing the target anode with a metal of higher atomic number decreases the cut-off (minimum) wavelength $\lambda_{\text{min}}$ of the continuous X-ray spectrum.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because, by the Duane-Hunt law (

eV = \frac{hc}{\lambda_{\text{min}}}

), the minimum cut-off wavelength depends strictly on the accelerating potential

V

applied across the tube and fundamental physical constants. It is completely independent of the target material's atomic number

Z

.

Step-by-Step Solution

1

Identify the relationship governing the minimum (cut-off) wavelength of continuous X-rays.

By the Duane-Hunt law, the maximum energy of an emitted X-ray photon equals the total kinetic energy of an accelerating electron:

E_{\text{max}} = eV = \frac{hc}{\lambda_{\text{min}}}

.

The continuous X-ray spectrum is produced via Bremsstrahlung (braking radiation) when high-speed electrons are decelerated by the electric fields of target nuclei.

2

Express the minimum wavelength

\lambda_{\text{min}}

in terms of operational variables.

Rearranging the equation yields

\lambda_{\text{min}} = \frac{hc}{eV}

, where

h

is Planck's constant,

c

is the speed of light,

e

is the elementary charge, and

V

is the tube potential.

This formula establishes that

\lambda_{\text{min}}

depends exclusively on the accelerating voltage

V

and physical constants.

3

Analyze the effect of altering the target material's atomic number

Z

at constant voltage

V

.

Changing the atomic number

Z

shifts characteristic spectral line wavelengths (via Moseley's law) and increases overall Bremsstrahlung intensity, but leaves

\lambda_{\text{min}}

entirely unchanged.

Since

Z

does not enter the Duane-Hunt expression for

\lambda_{\text{min}}

, varying

Z

has zero effect on the shortest emitted wavelength.

Key Concept

Duane-Hunt Law and Cut-off Wavelength Independence from Target Material

Question 7876Question →

An X-ray tube operates at an initial potential difference of $V_1$ . When the potential difference is increased by $20.0\text{ kV}$ , the minimum cutoff wavelength of the emitted X-rays is reduced to one-third of its initial value. What is the initial operating potential difference $V_1$ of the X-ray tube?

10.0\text{ kV}

30.0\text{ kV}

40.0\text{ kV}

5.0\text{ kV}

Show answer & explanation

Answer: $10.0\text{ kV}$

Answer

The initial operating potential difference of the tube is

10.0\text{ kV}

.

The Duane-Hunt law states that

\lambda_{\text{min}} = \frac{hc}{eV}

. Because minimum wavelength is inversely proportional to potential difference, reducing the minimum wavelength to one-third requires the operating potential to increase by a factor of three (

V_2 = 3V_1

). Given that

V_2 = V_1 + 20.0\text{ kV}

, setting

V_1 + 20.0\text{ kV} = 3V_1

gives

2V_1 = 20.0\text{ kV}

, which solves to

V_1 = 10.0\text{ kV}

.

Step-by-Step Solution

1

Relate the minimum cutoff wavelength to potential difference using the Duane-Hunt law.

\lambda_{\text{min}} = \frac{hc}{eV}

, which implies

\lambda_{\text{min}} \propto \frac{1}{V}

.

The maximum photon energy equals the kinetic energy of the incident electrons.

2

Set up the ratio between initial and final conditions.

Since

\lambda_2 = \frac{1}{3}\lambda_1

, the new voltage must be three times the initial voltage:

V_2 = 3V_1

.

Cutoff wavelength is inversely proportional to accelerating voltage.

3

Substitute

V_2 = V_1 + 20.0\text{ kV}

into the equation

V_2 = 3V_1

and solve for

V_1

.

V_1 + 20.0\text{ kV} = 3V_1 \implies 2V_1 = 20.0\text{ kV} \implies V_1 = 10.0\text{ kV}

.

Solving the linear algebraic equation yields the initial potential difference.

Key Concept

Duane-Hunt Law and Inverse Relationship between Minimum Wavelength and Accelerating Potential

Estimated Time:2m 0s

Question 7877Question →

The sum of the first $4$ terms of a geometric progression (GP) with a common ratio of $2$ is $45$ . What is the $6^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 96

Answer

The

6^{\text{th}}

term of the geometric progression is

96

.

Using the sum formula

S_4 = \frac{a(2^4 - 1)}{2 - 1} = 45

gives

15a = 45

, so the first term

a

is

3

. Substituting

a = 3

and

r = 2

into the term formula

T_6 = a r^5

gives

3 \times 32 = 96

.

Step-by-Step Solution

1

Express the sum of the first 4 terms using the GP sum formula to find the first term

a

.

Setting up

45 = \frac{a(2^4 - 1)}{2 - 1}

yields

15a = 45

, so

a = 3

.

The sum formula

S_n = \frac{a(r^n - 1)}{r - 1}

allows us to isolate the unknown initial term

a

when

S_4

and

r

are given.

2

Calculate the

6^{\text{th}}

term

T_6

using the

n^{\text{th}}

term formula

T_n = a r^{n-1}

.

T_6 = 3 \times 2^{6-1} = 3 \times 32 = 96

.

The exponent for the common ratio in the

n^{\text{th}}

term formula is

n - 1

, giving

5

as the exponent.

Key Concept

Sum and

n^{\text{th}}

term of a Geometric Progression

Question 7878Question →

What is the value of the definite integral $\int_{0}^{\frac{\pi}{2}} (3\sin x + 2\cos x) \, dx$ ?

5

2

-1

-5

Show answer & explanation

Answer: $5$

Answer

The value of the definite integral is

5

.

The antiderivative of

3\sin x + 2\cos x

is

-3\cos x + 2\sin x

. Evaluating this expression at the upper limit

x = \frac{\pi}{2}

gives

2

, and at the lower limit

x = 0

gives

-3

. Applying the Fundamental Theorem of Calculus gives

2 - (-3) = 5

.

Step-by-Step Solution

1

Find the indefinite integral of

3\sin x + 2\cos x

\int (3\sin x + 2\cos x) \, dx = -3\cos x + 2\sin x

The antiderivative of

\sin x

is

-\cos x

and the antiderivative of

\cos x

is

\sin x

.

2

Evaluate the antiderivative at the upper limit

x = \frac{\pi}{2}

-3\cos\left(\frac{\pi}{2}\right) + 2\sin\left(\frac{\pi}{2}\right) = -3(0) + 2(1) = 2

Substitute

x = \frac{\pi}{2}

into the antiderivative expression.

3

Evaluate the antiderivative at the lower limit

x = 0

-3\cos(0) + 2\sin(0) = -3(1) + 2(0) = -3

Substitute

x = 0

into the antiderivative expression.

4

Subtract the lower limit value from the upper limit value

2 - (-3) = 2 + 3 = 5

By the Fundamental Theorem of Calculus,

\int_{a}^{b} f(x)\,dx = F(b) - F(a)

.

Key Concept

Definite Integration of Trigonometric Functions

Estimated Time:1m 30s

Question 7879Question →

The sum of the first $n$ terms of an arithmetic progression (A.P.) is given by $S_n = 2n^2 + 3n$ . What is the $5^{\text{th}}$ term of the progression?

21

65

44

25

Show answer & explanation

Answer: $21$

Answer

The

5^{\text{th}}

term of the arithmetic progression is

21

.

The

n^{\text{th}}

term of a sequence can be determined from its sum formula using

T_n = S_n - S_{n-1}

. Evaluating

S_5 = 2(5)^2 + 3(5) = 65

and

S_4 = 2(4)^2 + 3(4) = 44

, the difference

T_5 = 65 - 44 = 21

gives the correct value of the

5^{\text{th}}

term.

Step-by-Step Solution

1

Calculate the sum of the first 5 terms (

S_5

)

S_5 = 2(5)^2 + 3(5) = 2(25) + 15 = 65

To find the sum up to the

5^{\text{th}}

term using the given formula

S_n = 2n^2 + 3n

.

2

Calculate the sum of the first 4 terms (

S_4

)

S_4 = 2(4)^2 + 3(4) = 2(16) + 12 = 44

To find the cumulative total of all terms prior to the

5^{\text{th}}

term.

3

Subtract

S_4

from

S_5

to isolate the

5^{\text{th}}

term (

T_5

)

T_5 = S_5 - S_4 = 65 - 44 = 21

The

n^{\text{th}}

term of any sequence is given by the relation

T_n = S_n - S_{n-1}

.

Key Concept

Relationship between the sum of

n

terms (

S_n

) and the

n^{\text{th}}

term (

T_n

) in an Arithmetic Progression

Estimated Time:1m 30s

Question 7880Question →

The energy of an electron in the first excited state of a hydrogen atom is $-3.4\text{ eV}$ . What is the minimum energy, in joules ( $\text{J}$ ), required to completely remove the electron from this state to ionize the atom? ( $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 5.44e-19

Answer

The minimum energy required to ionize the atom from its first excited state is

5.44 \times 10^{-19}\text{ J}

.

Ionization energy is defined as the minimum energy necessary to completely remove an electron from its bound atomic energy level to infinity (

E_{\infty} = 0\text{ eV}

). For an electron at

E = -3.4\text{ eV}

, the required energy change is

\Delta E = 0 - (-3.4\text{ eV}) = 3.4\text{ eV}

. Converting this to joules using

1\text{ eV} = 1.6 \times 10^{-19}\text{ J}

gives

3.4 \times 1.6 \times 10^{-19} = 5.44 \times 10^{-19}\text{ J}

.

Step-by-Step Solution

1

Determine the energy required for ionization in electron-volts (eV)

\Delta E = E_{\infty} - E_2 = 0\text{ eV} - (-3.4\text{ eV}) = 3.4\text{ eV}

Ionization requires supplying sufficient energy to raise the electron from its bound energy state (

E_2 = -3.4\text{ eV}

) to the ionization limit where it is free (

E_{\infty} = 0\text{ eV}

).

2

Convert the ionization energy from electron-volts to joules

E = 3.4\text{ eV} \times 1.6 \times 10^{-19}\text{ J/eV} = 5.44 \times 10^{-19}\text{ J}

To convert energy from electron-volts (eV) to joules (J), multiply the value in eV by the elementary charge conversion factor

1.6 \times 10^{-19}\text{ J/eV}

.

Key Concept

Ionization Energy and Energy Level Transitions

Estimated Time:1m 30s

Answer

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