All practice questions

Question 4441Question →

Monochromatic light with a photon energy of $4.5\text{ eV}$ strikes the surface of a sodium plate inside a vacuum tube. If the work function of sodium is $2.1\text{ eV}$ , what is the maximum kinetic energy of the emitted photoelectrons in electron-volts ( $\text{eV}$ )?

Show answer & explanation

Answer: 2.4

Answer

The maximum kinetic energy of the emitted photoelectrons is 2.4 eV.

According to Einstein's photoelectric theory, energy is conserved such that the energy of an incident photon (

E

) is partly used to liberate an electron from the metal surface (work function

W_0

) and the remaining energy appears as the maximum kinetic energy (

K_{\text{max}}

) of the photoelectron. Subtracting

2.1\text{ eV}

from

4.5\text{ eV}

gives

2.4\text{ eV}

.

Step-by-Step Solution

1

Identify the given values from the problem statement

Photon energy

E = 4.5\text{ eV}

, Work function

W_0 = 2.1\text{ eV}

Establishing known variables helps determine the correct photoelectric relation to apply.

2

Apply Einstein's photoelectric equation

K_{\text{max}} = E - W_0

The maximum kinetic energy of photoelectrons equals the excess energy of incident photons after overcoming the work function.

3

Calculate the value

K_{\text{max}} = 4.5\text{ eV} - 2.1\text{ eV} = 2.4\text{ eV}

Subtracting the work function from the photon energy yields the final numerical answer.

Key Concept

Einstein's Photoelectric Equation and Energy Conservation

Question 4442Question →

A uniform string of length $0.50\text{ m}$ and mass $2.0\text{ g}$ is fixed at both ends under a tension of $90\text{ N}$ . When vibrating, the second harmonic of this string resonates with the first overtone of an air column in a pipe closed at one end. Assuming the speed of sound in air is $340\text{ m/s}$ , calculate the length of the pipe in meters.

Show answer & explanation

Answer: 0.85

Answer

The length of the pipe is

0.85\text{ m}

.

The wave speed on the string is computed from tension and linear mass density as $150\text{ m/s}$, yielding a second harmonic frequency of $300\text{ Hz}$. Equating this to the first overtone (third harmonic) frequency formula of a closed air pipe, $f = \frac{3v_{air}}{4L_p}$, yields an exact pipe length of $0.85\text{ m}$.

Step-by-Step Solution

1

Calculate the linear mass density ($\mu$) of the string

$\mu = \frac{m}{L_s} = \frac{0.0020\text{ kg}}{0.50\text{ m}} = 4.0 \times 10^{-3}\text{ kg/m}$

Mass must be converted to kilograms before determining mass per unit length.

2

Determine the wave speed ($v_s$) along the stretched string

$v_s = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{90\text{ N}}{4.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{22500} = 150\text{ m/s}$

The velocity of a transverse wave on a string depends on tension and linear mass density.

3

Calculate the second harmonic frequency of the string ($f_{2,s}$)

$f_{2,s} = \frac{v_s}{L_s} = \frac{150\text{ m/s}}{0.50\text{ m}} = 300\text{ Hz}$

The fundamental frequency is $f_{1,s} = \frac{v_s}{2L_s} = 150\text{ Hz}$, so the second harmonic is twice the fundamental frequency.

4

Set up the resonance equation for the first overtone of a closed pipe

$f_{3,p} = \frac{3 v_{air}}{4 L_p} = 300\text{ Hz}$

A pipe closed at one end produces only odd harmonics, so the first overtone is the 3rd harmonic.

5

Solve for the length of the closed pipe ($L_p$)

$L_p = \frac{3 \times 340\text{ m/s}}{4 \times 300\text{ Hz}} = \frac{1020}{1200} = 0.85\text{ m}$

Rearranging the frequency formula gives the required air column length.

Key Concept

Coupled resonance between standing waves on strings and air columns in closed pipes

Question 4443Question →

If $y = (x^2 + 1)(2x - 3)^3$ , find the value of $\frac{dy}{dx}$ at $x = 2$ .

Show answer & explanation

Answer: 34

Answer

The value of

\frac{dy}{dx}

at

x = 2

is

34

.

Applying the product rule

\frac{dy}{dx} = u'v + uv'

alongside the chain rule gives

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

. Evaluating at

x = 2

yields

4(1) + 30(1) = 34

.

Step-by-Step Solution

1

Identify component functions for the product rule

Let

u(x) = x^2 + 1

and

v(x) = (2x - 3)^3

.

The function

y

is a product of two differentiable functions.

2

Differentiate each component function

u'(x) = 2x

and

v'(x) = 3(2x - 3)^2 \cdot 2 = 6(2x - 3)^2

.

The power rule gives

u'(x)

and the chain rule gives

v'(x)

by multiplying by the derivative of the inner function

(2x - 3)

.

3

Apply the product rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

.

To find the general derivative of a product of functions.

4

Evaluate the derivative at

x = 2

\frac{dy}{dx}\Big|_{x=2} = 2(2)(2(2) - 3)^3 + 6(2^2 + 1)(2(2) - 3)^2 = 4(1) + 30(1) = 34

.

To calculate the specific numerical value of the derivative at

x = 2

.

Key Concept

Product and Chain Rules of Differentiation

Estimated Time:1m 30s

Question 4444Question →

A $3\text{ }\mu\text{F}$ capacitor and a $6\text{ }\mu\text{F}$ capacitor are connected in series across a direct current voltage source. What is the total equivalent capacitance of the combination, in microfarads ( $\mu\text{F}$ )?

Show answer & explanation

Answer: 2

Answer

The total equivalent capacitance of the combination is

2\text{ }\mu\text{F}

.

For capacitors connected in series, the reciprocal of the total equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. Substituting

3\text{ }\mu\text{F}

and

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

gives

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

, which yields an equivalent capacitance of

2\text{ }\mu\text{F}

.

Step-by-Step Solution

1

State the formula for equivalent capacitance of two capacitors in series

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

Capacitors connected in series combine reciprocally, unlike resistors connected in series.

2

Substitute the values of

C_1

and

C_2

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

Find a common denominator and add the fractions.

3

Calculate the reciprocal to determine

C_{eq}

C_{eq} = 2\text{ }\mu\text{F}

Inverting

\frac{1}{2}

yields the total equivalent capacitance.

Key Concept

Equivalent Capacitance in Series

Question 4445Question →

A body of mass $2\text{ kg}$ is projected vertically upward from ground level with an initial speed of $30\text{ m s}^{-1}$ . During its entire flight, it experiences a constant resistive force due to air resistance of $5\text{ N}$ . Taking $g = 10\text{ m s}^{-2}$ , calculate the kinetic energy of the body in Joules when it returns to the ground level.

Show answer & explanation

Answer: 540

Answer

The kinetic energy of the body when it returns to ground level is

540\text{ J}

.

At launch, the body possesses an initial kinetic energy of

E_i = \frac{1}{2}(2)(30)^2 = 900\text{ J}

. During ascent, both gravity (

20\text{ N}

) and air resistance (

5\text{ N}

) retard the motion, giving a net downward force of

25\text{ N}

and a deceleration of

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

. The maximum height reached is

h = \frac{30^2}{2(12.5)} = 36\text{ m}

. Since the body travels up and down, total distance covered is

72\text{ m}

. Non-conservative work done against air resistance is

W = 5\text{ N} \times 72\text{ m} = 360\text{ J}

. Therefore, the remaining kinetic energy upon returning to the ground is

900\text{ J} - 360\text{ J} = 540\text{ J}

.

Step-by-Step Solution

1

Calculate initial kinetic energy of launch

E_i = 900\text{ J}

Kinetic energy is given by

\frac{1}{2}m u^2 = \frac{1}{2}(2\text{ kg})(30\text{ m s}^{-1})^2 = 900\text{ J}

.

2

Determine maximum height reached during ascent

h = 36\text{ m}

Net upward retarding force

F = mg + F_{\text{air}} = 20 + 5 = 25\text{ N}

, yielding a deceleration

a = 12.5\text{ m s}^{-2}

. Using

0 = u^2 - 2ah

,

h = \frac{900}{25} = 36\text{ m}

.

3

Calculate energy lost to air resistance over total trajectory

W_{\text{air}} = 360\text{ J}

Air resistance acts continuously over both ascent and descent (total distance

2h = 72\text{ m}

). Work dissipated

= F_{\text{air}} \times 2h = 5 \times 72 = 360\text{ J}

.

4

Subtract non-conservative work loss from initial mechanical energy

E_f = 540\text{ J}

By mechanical energy balance, final kinetic energy

E_f = E_i - W_{\text{air}} = 900 - 360 = 540\text{ J}

.

Key Concept

Work-Energy Theorem and Mechanical Energy Dissipation by Non-Conservative Forces

Question 4446Question →

The sum of the interior angles of a regular polygon is $1440^\circ$ . What is the measure, in degrees, of one exterior angle of this polygon?

Show answer & explanation

Answer: 36

Answer

The measure of one exterior angle of the polygon is

36^\circ

.

The sum of the interior angles of an

n

-sided polygon is given by

(n-2) \times 180^\circ

. Setting

(n-2) \times 180^\circ = 1440^\circ

gives

n-2 = 8

, so the polygon has

n = 10

sides (a decagon). The measure of each exterior angle of a regular polygon is

\frac{360^\circ}{n} = \frac{360^\circ}{10} = 36^\circ

.

Step-by-Step Solution

1

Set up the equation for the sum of interior angles of an

n

-sided polygon.

(n - 2) \times 180^\circ = 1440^\circ

The sum of interior angles of any convex

n

-sided polygon is

(n - 2) \times 180^\circ

.

2

Solve for

n

, the number of sides.

n - 2 = \frac{1440}{180} = 8 \implies n = 10

Dividing the interior angle sum by

180^\circ

gives

n - 2

.

3

Calculate the measure of one exterior angle.

Exterior angle

= \frac{360^\circ}{10} = 36^\circ

The sum of exterior angles of any convex polygon is

360^\circ

, so each exterior angle of a regular polygon with

n

sides is

\frac{360^\circ}{n}

.

Key Concept

Relationship between interior angle sum, number of sides, and exterior angles of a regular polygon.

Question 4447Question →

In a Coolidge X-ray tube, electrons are accelerated from rest through an unknown potential difference $V$ towards a target metal. If the shortest cutoff wavelength of the resulting continuous X-ray spectrum is recorded as $0.0414\text{ nm}$ , what is the operating potential difference $V$ of the tube?

(Take Planck's constant $h = 6.63 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.00 \times 10^8\text{ m s}^{-1}$ , elementary charge $e = 1.60 \times 10^{-19}\text{ C}$ )

30.0\text{ kV}

15.0\text{ kV}

300\text{ V}

60.0\text{ kV}

Show answer & explanation

Answer: $30.0\text{ kV}$

Answer

The operating potential difference

V

of the tube is

30.0\text{ kV}

.

According to the Duane-Hunt law for X-ray emission, the shortest wavelength

\lambda_{\min}

corresponds to the maximum kinetic energy acquired by an electron accelerated through a voltage

V

. The mathematical relationship is

e V = \frac{h c}{\lambda_{\min}}

. Substituting

h = 6.63 \times 10^{-34}\text{ J s}

,

c = 3.00 \times 10^8\text{ m s}^{-1}

,

e = 1.60 \times 10^{-19}\text{ C}

, and

\lambda_{\min} = 4.14 \times 10^{-11}\text{ m}

gives

V = \frac{1.989 \times 10^{-25}}{6.624 \times 10^{-30}} = 30.0\text{ kV}

.

Step-by-Step Solution

1

Convert the given minimum wavelength from nanometers to meters

\lambda_{\min} = 0.0414\text{ nm} = 0.0414 \times 10^{-9}\text{ m} = 4.14 \times 10^{-11}\text{ m}

SI unit consistency requires length to be in meters.

2

Apply the Duane-Hunt law for continuous X-ray production

E_{\max} = e V = h f_{\max} = \frac{h c}{\lambda_{\min}}

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

3

Rearrange the equation to solve for the accelerating voltage

V

V = \frac{h c}{e \lambda_{\min}}

Isolating the operating potential difference

V

on one side of the equation.

4

Substitute the physical constants and calculate

V

V = \frac{(6.63 \times 10^{-34}\text{ J s}) (3.00 \times 10^8\text{ m s}^{-1})}{(1.60 \times 10^{-19}\text{ C}) (4.14 \times 10^{-11}\text{ m})} = \frac{1.989 \times 10^{-25}}{6.624 \times 10^{-30}} = 30,012\text{ V} \approx 30.0\text{ kV}

Performing numeric evaluation gives the operating potential difference in kilovolts.

Key Concept

Duane-Hunt Law and Cutoff Wavelength in X-ray Tubes

Estimated Time:2m 0s

Question 4448Question →

In a telecommunications network, two independent relay switches, $R_1$ and $R_2$ , operate during a data transmission. The probability that switch $R_1$ functions successfully is $0.80$ , and the probability that switch $R_2$ functions successfully is $0.75$ . What is the probability that at least one of the two switches functions successfully during the transmission?

Show answer & explanation

Answer: 0.95

Answer

The probability that at least one switch functions successfully is 0.95.

Since the two switches operate independently, the probability of both failing is the product of their individual failure probabilities:

(1 - 0.80) \times (1 - 0.75) = 0.20 \times 0.25 = 0.05

. Therefore, the probability that at least one switch functions successfully is

1 - 0.05 = 0.95

. Alternatively, using the addition law for independent events:

P(R_1 \cup R_2) = P(R_1) + P(R_2) - P(R_1 \cap R_2) = 0.80 + 0.75 - (0.80 \times 0.75) = 1.55 - 0.60 = 0.95

.

Step-by-Step Solution

1

Determine the probabilities of individual switch failure.

P(R_1') = 0.20 and P(R_2') = 0.25

The probability of an event failing is the complement of its success probability: P(E') = 1 - P(E).

2

Compute the probability that both switches fail simultaneously.

P(R_1' ∩ R_2') = 0.20 × 0.25 = 0.05

Since the switches operate independently, the multiplication law for independent events applies: P(A ∩ B) = P(A) × P(B).

3

Calculate the probability that at least one switch functions successfully.

P(at least one) = 1 - 0.05 = 0.95

The complement of 'neither switch functioning' is 'at least one switch functioning'.

Key Concept

Probability laws for independent compound events and the complement rule

Estimated Time:1m 30s

Question 4449Question →

The $3^{\text{rd}}$ term of an arithmetic progression (AP) is $10$ and the $7^{\text{th}}$ term is $22$ . What is the sum of the first $12$ terms of the progression?

228

246

264

444

Show answer & explanation

Answer: $246$

Answer

The sum of the first

12

terms of the arithmetic progression is

246

.

Using the

n^{\text{th}}

term formula

T_n = a + (n-1)d

, the equations

a + 2d = 10

and

a + 6d = 22

yield

d = 3

and

a = 4

. Substituting these values into

S_{12} = \frac{12}{2}[2(4) + 11(3)]

gives

6 \times 41 = 246

.

Step-by-Step Solution

1

Set up simultaneous equations using the

n^{\text{th}}

term formula

T_n = a + (n-1)d

.

a + 2d = 10

and

a + 6d = 22

.

The

3^{\text{rd}}

term corresponds to

n=3

and the

7^{\text{th}}

term corresponds to

n=7

.

2

Subtract the first equation from the second to find the common difference

d

.

4d = 12 \implies d = 3

.

Subtracting eliminates the first term

a

.

3

Substitute

d = 3

back into the first equation to find the first term

a

.

a + 2(3) = 10 \implies a = 4

.

Determining the first term is necessary to calculate the sum.

4

Calculate the sum of the first

12

terms using

S_n = \frac{n}{2}[2a + (n-1)d]

.

S_{12} = \frac{12}{2}[2(4) + (12-1)(3)] = 6[8 + 33] = 6(41) = 246

.

Applying the AP sum formula with

n = 12

,

a = 4

, and

d = 3

.

Key Concept

Arithmetic Progression: Finding common difference, first term, and sum of terms

Estimated Time:1m 30s

Question 4450Question →

Given that $y = \frac{3x - 1}{(x + 2)^2}$ , what is the value of $\frac{dy}{dx}$ at $x = 1$ ?

\frac{5}{27}

\frac{13}{9}

\frac{25}{81}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{5}{27}$

Answer

The value of

\frac{dy}{dx}

at

x = 1

is

\frac{5}{27}

.

Applying the quotient rule

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

with

u = 3x - 1

and

v = (x + 2)^2

yields

\frac{dy}{dx} = \frac{8 - 3x}{(x + 2)^3}

. Substituting

x = 1

results in

\frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{27}

.

Step-by-Step Solution

1

Identify the numerator

u(x)

and denominator

v(x)

for the quotient rule.

Let

u = 3x - 1

and

v = (x + 2)^2

.

The given function

y = \frac{u}{v}

requires the quotient rule

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

.

2

Differentiate

u(x)

and

v(x)

with respect to

x

.

u' = 3

and, using the chain rule,

v' = 2(x + 2)(1) = 2(x + 2)

.

The derivative of the inner term

(x + 2)

is

1

, giving

v' = 2(x + 2)

.

3

Substitute

u, v, u', v'

into the quotient rule formula and simplify.

\frac{dy}{dx} = \frac{3(x + 2)^2 - (3x - 1) \cdot 2(x + 2)}{(x + 2)^4} = \frac{(x + 2)[3(x + 2) - 2(3x - 1)]}{(x + 2)^4} = \frac{3x + 6 - 6x + 2}{(x + 2)^3} = \frac{8 - 3x}{(x + 2)^3}

.

Factoring out

(x + 2)

simplifies the algebraic expression.

4

Evaluate the derivative at

x = 1

.

\frac{dy}{dx}\Big|_{x=1} = \frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{3^3} = \frac{5}{27}

.

Substituting

x = 1

gives the final numerical derivative value.

Key Concept

Quotient and Chain Rules of Differentiation

Question 4451Question →

A point charge $q_1 = +9.0 \times 10^{-9}\text{ C}$ is fixed at the origin ( $x = 0\text{ m}$ ), and a second point charge $q_2 = -4.0 \times 10^{-9}\text{ C}$ is fixed on the x-axis at $x = 0.5\text{ m}$ . At what position $x$ (in meters) along the x-axis is the net electric field intensity equal to zero?

Show answer & explanation

Answer: 1.5

Answer

The net electric field intensity is zero at

x = 1.5\text{ m}

.

The correct position is

x = 1.5\text{ m}

. At this point, the electric field from

+q_1

points in the

+x

direction with magnitude

E_1 = \frac{9.0 \times 10^9 \times 9.0 \times 10^{-9}}{1.5^2} = 36\text{ N C}^{-1}

, and the electric field from

-q_2

points in the

-x

direction with magnitude

E_2 = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-9}}{1.0^2} = 36\text{ N C}^{-1}

. The two vectors are equal in magnitude and opposite in direction, yielding a net electric field of zero.

Step-by-Step Solution

1

Determine the physical region where electric fields can cancel

The point of zero field lies to the right of

q_2

, i.e.,

x > 0.5\text{ m}

.

Between the charges, the fields due to

+q_1

and

-q_2

point in the same direction (+x). To the left of

q_1

,

q_1

is both larger in magnitude and closer, so

E_1 > E_2

everywhere. Hence, balance can only occur to the right of the smaller magnitude charge

q_2

.

2

Set up the condition for equal electric field magnitudes

\frac{k |q_1|}{x^2} = \frac{k |q_2|}{(x - 0.5)^2}

For the net field to be zero, the vector sum of

E_1

and

E_2

must equal zero, meaning their magnitudes must be equal.

3

Substitute values and simplify the algebraic equation

\frac{9.0 \times 10^{-9}}{x^2} = \frac{4.0 \times 10^{-9}}{(x - 0.5)^2} \implies \frac{9}{x^2} = \frac{4}{(x - 0.5)^2}

Coulomb's constant

k

and the power factor

10^{-9}

cancel from both sides.

4

Take square root on both sides to solve for x

\frac{3}{x} = \frac{2}{x - 0.5} \implies 3(x - 0.5) = 2x \implies x = 1.5\text{ m}

Taking the principal square root reduces the quadratic relation to a simple linear equation.

Key Concept

Electric Field Superposition and Zero Field Condition for Point Charges

Question 4452Question →

In prose fiction, the literary device known as deus ex machina resolves a narrative's central conflict through logical, pre-established character development rather than an unexpected external intervention.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false because deus ex machina relies on an artificial or unexpected external intervention rather than logical, character-driven resolution.

The statement is false. Deus ex machina refers to a plot device wherein a seemingly unsolvable conflict is abruptly resolved by an unexpected, artificial, or improbable intervention, rather than emerging organically from internal character development or cause-and-effect plot progression.

Step-by-Step Solution

1

Define the narrative device deus ex machina in prose structure.

Deus ex machina (literally 'god from the machine') is a structural plot resolution wherein a hopeless scenario is suddenly resolved by an improbable external force or contrived event.

Establishing the precise definition of the device is necessary to evaluate the claim.

2

Compare the definition with the claim made in the statement.

The statement asserts that deus ex machina operates through 'logical, pre-established character development', which describes organic plot resolution rather than deus ex machina.

Evaluating whether the statement aligns with established literary terminology reveals the falsehood.

3

Formulate the conclusion.

Because deus ex machina is defined by its artificial and unearned nature, attributing organic character-driven resolution to it is incorrect.

Organic resolution contrasts directly with the contrived nature of deus ex machina.

Key Concept

Deus ex Machina in Plot Resolution

Question 4453Question →

A vessel of negligible heat capacity contains $0.20\text{ kg}$ of a liquid at its boiling point of $100^\circ\text{C}$ . An electric heating element rated at $500\text{ W}$ is immersed in the liquid and switched on for $3.0\text{ minutes}$ . If $0.040\text{ kg}$ of the liquid is vaporized during this period, what is the specific latent heat of vaporization of the liquid?

2.25 \times 10^6\text{ J kg}^{-1}

2.25 \times 10^4\text{ J kg}^{-1}

4.50 \times 10^5\text{ J kg}^{-1}

1.25 \times 10^4\text{ J kg}^{-1}

Show answer & explanation

Answer: $2.25 \times 10^6\text{ J kg}^{-1}$

Answer

The specific latent heat of vaporization of the liquid is

2.25 \times 10^6\text{ J kg}^{-1}

.

The electrical energy supplied by the heater in

180\text{ seconds}

is

Q = 500 \times 180 = 90,000\text{ J}

. Since phase change occurs at constant boiling temperature (

100^\circ\text{C}

), the heat required to vaporize

0.040\text{ kg}

of liquid is given by

Q = m L_v

. Solving for

L_v

yields

\frac{90,000\text{ J}}{0.040\text{ kg}} = 2.25 \times 10^6\text{ J kg}^{-1}

.

Step-by-Step Solution

1

Convert the time duration into seconds

t = 3.0\text{ minutes} \times 60\text{ s/min} = 180\text{ s}

SI units require time to be in seconds when calculating electrical energy.

2

Calculate total electrical thermal energy supplied by the heater

Q = P \times t = 500\text{ W} \times 180\text{ s} = 90,000\text{ J}

Thermal energy produced equals power multiplied by time elapsed.

3

Apply the latent heat formula using the mass of liquid vaporized

L_v = \frac{Q}{m} = \frac{90,000\text{ J}}{0.040\text{ kg}} = 2,250,000\text{ J kg}^{-1} = 2.25 \times 10^6\text{ J kg}^{-1}

During vaporization at constant boiling temperature, heat energy supplied goes entirely into changing state.

Key Concept

Specific Latent Heat of Vaporization

Question 4454Question →

For any real object positioned in front of a convex spherical mirror, the image formed is always virtual, upright, and diminished, located between the pole and the principal focus.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true. A convex mirror always produces a virtual, erect, and diminished image positioned between the pole and the focus for any real object placed in front of it.

The statement accurately expresses the fundamental optics rule for convex mirrors: reflecting surface geometry causes incident parallel rays to diverge, producing a virtual, erect, and diminished image located between the mirror's pole and principal focus for all real object positions.

Step-by-Step Solution

1

Analyze ray tracing principles for a convex mirror.

Ray 1, traveling parallel to the principal axis, reflects such that it appears to originate from the principal focus

F

behind the mirror. Ray 2, directed toward the center of curvature

C

, reflects back along its original path.

Tracing these rays determines where their virtual extensions intersect.

2

Determine the location and characteristics of the virtual intersection.

The backward extensions of the reflected rays intersect behind the reflecting surface between the pole

P

and the principal focus

F

.

Since the light rays diverge and only their extensions meet, the image formed is virtual, erect, and diminished.

Key Concept

Image characteristics of convex spherical mirrors

Question 4455Question →

A metallic element $M$ forms two distinct oxides. Quantitative analysis shows that the first oxide contains $20.0\%$ oxygen by mass and has the empirical formula $MO$ . The second oxide contains $11.1\%$ oxygen by mass. Based on the Law of Multiple Proportions, what is the empirical formula of the second oxide?

M_2O

MO_2

M_2O_3

M_3O_4

Show answer & explanation

Answer: $M_2O$

Answer

M_2O

According to the Law of Multiple Proportions, when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. In the first oxide (

MO

),

20.0\text{ g}

of oxygen combines with

80.0\text{ g}

of metal

M

, giving a ratio of

4.0\text{ g } M

per

1.0\text{ g } O

. In the second oxide,

11.1\text{ g}

of oxygen combines with

88.9\text{ g}

of

M

, giving a ratio of

8.0\text{ g } M

per

1.0\text{ g } O

. Comparing the mass of metal combining with

1.0\text{ g}

of oxygen gives

4.0 : 8.0 = 1 : 2

. Since the first oxide has

1

atom of

M

per atom of

O

, the second oxide contains

2

atoms of

M

per atom of

O

, yielding the formula

M_2O

.

Step-by-Step Solution

1

Calculate the mass ratio of metal

M

to oxygen

O

in the first oxide

In

100\text{ g}

of the first oxide, mass of

O = 20.0\text{ g}

and mass of

M = 80.0\text{ g}

. Mass ratio

M : O = \frac{80.0\text{ g } M}{20.0\text{ g } O} = 4.0\text{ g } M / \text{g } O

.

Determining the mass of metal that combines with a fixed unit mass (

1.0\text{ g}

) of oxygen in the first compound.

2

Calculate the mass ratio of metal

M

to oxygen

O

in the second oxide

In

100\text{ g}

of the second oxide, mass of

O = 11.1\text{ g}

and mass of

M = 88.9\text{ g}

. Mass ratio

M : O = \frac{88.9\text{ g } M}{11.1\text{ g } O} = 8.01\text{ g } M / \text{g } O \approx 8.0\text{ g } M / \text{g } O

.

Determining the mass of metal that combines with a fixed unit mass (

1.0\text{ g}

) of oxygen in the second compound.

3

Apply the Law of Multiple Proportions to compare the masses of metal

M

combining with fixed oxygen

\text{Ratio of masses of } M = 4.0 : 8.0 = 1 : 2.

The Law of Multiple Proportions states that the masses of one element combining with a fixed mass of another are in simple whole-number ratios.

4

Deduce the empirical formula of the second oxide

Since the first oxide (

MO

) has

1

atom of

M

per atom of

O

, an oxide with twice the mass of

M

per atom of

O

must have

2

atoms of

M

per atom of

O

, giving the empirical formula

M_2O

.

Relating the atomic ratio of the second oxide directly to the known formula of the first oxide.

Key Concept

Law of Multiple Proportions

Question 4456Question →

Which of the following emissions resulting from natural radioactivity passes through a strong electric field without undergoing any deflection?

Gamma rays (

\gamma

-rays)

Alpha particles (

\alpha

-particles)

Beta-minus particles (

\beta^-

-particles)

Positrons (

\beta^+

-particles)

Show answer & explanation

Answer: Gamma rays ( $\gamma$ -rays)

Answer

Gamma rays (

\gamma

-rays) carry no electrical charge, so they pass through an electric field without any deflection.

Gamma rays consist of neutral electromagnetic radiation (photons). Because their charge is zero, they experience no electrostatic force in an electric field and continue in a straight, undeflected path.

Step-by-Step Solution

1

Identify the nature and electric charge of each type of radioactive emission.

Alpha particles have a charge of

+2e

, beta particles have a charge of

-e

or

+e

, and gamma rays are high-energy photons with zero net electric charge.

An electric field exerts an electrostatic force

F = qE

only on particles that carry a non-zero electric charge

q

.

2

Determine which emission experiences zero electrostatic force.

Since

q = 0

for gamma rays,

F = 0

, meaning gamma rays experience no deflection.

Uncharged radiation travels in a straight line unaffected by electrostatic or magnetic fields.

Key Concept

Electric Field Deflection of Radioactive Emissions

Estimated Time:45s

Question 4457Question →

Which of the following statements correctly distinguishes a chemical compound from a mixture?

A compound consists of elements combined in a fixed ratio by mass, whereas a mixture contains components in variable proportions.

A compound can be separated into its constituent elements using physical methods, whereas a mixture requires chemical reactions for separation.

A compound melts or boils over a wide range of temperatures, whereas a mixture exhibits a sharp melting point.

The chemical properties of a compound are an average of its constituent elements, whereas a mixture shows completely unique chemical properties.

Show answer & explanation

Answer: A compound consists of elements combined in a fixed ratio by mass, whereas a mixture contains components in variable proportions.

Answer

A compound consists of elements combined in a fixed ratio by mass, whereas a mixture contains components in variable proportions.

The correct statement identifies that chemical compounds are pure substances formed by chemical reactions where constituent elements combine in fixed, definite proportions by mass. In contrast, mixtures are formed by physical blending where the constituent components can be present in any variable proportion.

Step-by-Step Solution

1

Analyze the fundamental definition of a chemical compound regarding composition.

Compounds are formed by chemical combination of elements in definite, fixed mass ratios (Law of Definite Proportions).

Chemical bonds form between constituent atoms in fixed stoichiometric ratios.

2

Analyze the composition and properties of a mixture.

Mixtures consist of substances physically blended together in variable mass ratios, retaining their individual identities.

No chemical bonding occurs between distinct components of a mixture.

3

Evaluate the given options against these core criteria.

The statement highlighting fixed mass proportion for compounds versus variable composition for mixtures is the accurate distinction.

Physical separation techniques apply to mixtures, sharp melting points characterize pure compounds, and compound properties differ entirely from constituent elements.

Key Concept

Distinction Between Elements, Compounds, and Mixtures

Estimated Time:1m 0s

Question 4458Question →

An open pipe of physical length $0.58\text{ m}$ and a pipe closed at one end emit sound at the same frequency when both are vibrating in their first overtone mode. If the end correction at each open end is $0.01\text{ m}$ , what is the physical length of the closed pipe?

0.29\text{ m}

0.43\text{ m}

0.44\text{ m}

0.45\text{ m}

Show answer & explanation

Answer: $0.44\text{ m}$

Answer

The physical length of the closed pipe is

0.44\text{ m}

.

The effective length of an open pipe with two open ends is

L_{1,\text{eff}} = 0.58 + 2(0.01) = 0.60\text{ m}

. The first overtone for an open pipe is its second harmonic (

n=2

), giving

f = \frac{v}{0.60}

. For a pipe closed at one end, the first overtone is its third harmonic (

m=3

), giving

f = \frac{3v}{4L_{2,\text{eff}}}

. Equating frequencies gives

L_{2,\text{eff}} = 0.45\text{ m}

. Subtracting the single end correction (

e = 0.01\text{ m}

) yields the physical length

0.44\text{ m}

.

Step-by-Step Solution

1

Calculate the effective length and first overtone frequency of the open pipe.

L_{1,\text{eff}} = L_1 + 2e = 0.58\text{ m} + 2(0.01\text{ m}) = 0.60\text{ m}

, so

f_{\text{open,1st overtone}} = \frac{2v}{2L_{1,\text{eff}}} = \frac{v}{0.60}

.

An open pipe has two open ends, so end correction is applied at both ends (

2e

). Its first overtone corresponds to the second harmonic (

n=2

).

2

Express the first overtone frequency of the closed pipe in terms of its effective length.

f_{\text{closed,1st overtone}} = \frac{3v}{4L_{2,\text{eff}}}

.

A pipe closed at one end produces only odd harmonics (

m=1, 3, 5, \dots

). The first overtone corresponds to the third harmonic (

m=3

).

3

Equate the two frequencies and solve for the effective length of the closed pipe.

\frac{v}{0.60} = \frac{3v}{4L_{2,\text{eff}}} \implies 4L_{2,\text{eff}} = 1.80\text{ m} \implies L_{2,\text{eff}} = 0.45\text{ m}

.

Both pipes emit sound at the same frequency in their first overtone modes.

4

Calculate the physical length of the closed pipe.

L_2 = L_{2,\text{eff}} - e = 0.45\text{ m} - 0.01\text{ m} = 0.44\text{ m}

.

A closed pipe has only one open end, so its effective length is

L_2 + e

.

Key Concept

Standing Waves and End Correction in Open and Closed Organ Pipes

Question 4459Question →

When air containing unsaturated water vapour is cooled at constant atmospheric pressure without adding or removing moisture, the saturated vapour pressure of water decreases while the actual partial vapour pressure of water remains constant until the dew point is reached.

True

False

Show answer & explanation

Answer: True

Answer

True. Saturated vapour pressure decreases as temperature falls, while actual partial vapour pressure remains constant under constant total pressure until condensation starts at the dew point.

The statement is correct because saturated vapour pressure is a temperature-dependent property that decreases as air cools. So long as total pressure remains unchanged and no water vapour is added or removed, the actual partial vapour pressure of the water vapour stays constant until the dew point is reached, at which point the air becomes saturated (

100\%

relative humidity) and condensation begins.

Step-by-Step Solution

1

Analyze the temperature dependence of saturated vapour pressure (SVP).

SVP decreases as temperature drops.

SVP is determined by the kinetic energy of water molecules escaping into vapour at dynamic equilibrium, which decreases with decreasing temperature.

2

Determine the behavior of the actual partial vapour pressure of water during cooling at constant total pressure.

Actual partial vapour pressure remains constant prior to condensation.

Dalton's law dictates that partial pressure depends on the mole fraction of water vapour and total pressure; since no water vapour is added or removed, actual partial vapour pressure does not change.

3

Evaluate the condition for reaching the dew point.

At the dew point, SVP drops to equal the constant actual partial vapour pressure, achieving 100% relative humidity.

Relative humidity is defined as

\text{R.H.} = \frac{P_{\text{actual}}}{P_{\text{SVP}}} \times 100\%

; as

P_{\text{SVP}}

decreases towards

P_{\text{actual}}

,

\text{R.H.}

increases to

100\%

.

Key Concept

Saturated vs. actual vapour pressure, temperature dependence of SVP, and dew point

Question 4460Question →

Given the function $y = \frac{2x - 3}{(x^2 + 1)^2}$ , find the numerical value of $\frac{dy}{dx}$ at $x = 1$ .

Show answer & explanation

Answer: 1

Answer

The numerical value of

\frac{dy}{dx}

at

x = 1

is 1.

Applying the quotient rule

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

alongside the chain rule for the denominator yields

\frac{dy}{dx} = \frac{2(x^2 + 1)^2 - (2x - 3) \cdot 4x(x^2 + 1)}{(x^2 + 1)^4}

. Substituting

x = 1

evaluates to

\frac{2(4) - (-1)(8)}{16} = \frac{16}{16} = 1

.

Step-by-Step Solution

1

Identify the numerator and denominator functions

Let

u(x) = 2x - 3

and

v(x) = (x^2 + 1)^2

.

The given function is structured as a quotient

y = \frac{u}{v}

, requiring the quotient rule.

2

Find the derivatives

u'(x)

and

v'(x)

u'(x) = 2

and

v'(x) = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)

.

Differentiating

u(x)

follows standard polynomial rules;

v(x)

requires the chain rule.

3

Evaluate

u(1), u'(1), v(1),

and

v'(1)

at

x = 1

u(1) = -1

,

u'(1) = 2

,

v(1) = 4

, and

v'(1) = 8

.

Evaluating components before substitution simplifies the arithmetic.

4

Apply the quotient rule formula

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

at

x = 1

\frac{dy}{dx} = \frac{(2)(4) - (-1)(8)}{4^2} = \frac{8 + 8}{16} = 1.

Substitute the calculated component values into the quotient rule formula.

Key Concept

Quotient Rule and Chain Rule of Differentiation

Estimated Time:1m 30s

Answer

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