All practice questions

Question 7161Question →

A laboratory mixture contains insoluble chalk powder (calcium carbonate) suspended in water. Which physical separation technique is most appropriate for collecting the chalk as a residue and the clear water as a filtrate?

Filtration

Evaporation to dryness

Crystallization

Fractional distillation

Show answer & explanation

Answer: Filtration

Answer

Filtration

Filtration is the standard method for separating an insoluble solid from a liquid. The filter paper acts as a mechanical sieve; insoluble chalk particles remain on top as the residue, while pure liquid water passes through as the filtrate.

Step-by-Step Solution

1

Identify the physical state and solubility of the components in the mixture.

Chalk (calcium carbonate) is an insoluble solid suspended in liquid water.

Choosing the correct separation technique depends on whether the solute is soluble or insoluble in the solvent.

2

Select the separation technique designed for insoluble solid-liquid mixtures.

Filtration is selected because the filter paper pores trap the insoluble chalk particles (residue) while letting water molecules pass through (filtrate).

Techniques like evaporation or distillation are intended for soluble solutes or miscible liquids, whereas filtration directly separates suspended particles.

Key Concept

Separation of insoluble solids from liquids using filtration

Estimated Time:45s

Question 7162Question →

A sample of nitrogen gas enclosed in a constant-volume container exerts a pressure of $2.50\text{ atm}$ at a temperature of $-23^\circ\text{C}$ . To what temperature, in degrees Celsius ( $^\circ\text{C}$ ), must the gas be heated so that its pressure increases to $4.00\text{ atm}$ ?

Show answer & explanation

Answer: 127

Answer

The gas must be heated to

127^\circ\text{C}

.

According to Gay-Lussac's Pressure Law, for a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature (

P \propto T

). Converting

-23^\circ\text{C}

to Kelvin gives

250\text{ K}

. Solving

\frac{2.50}{250} = \frac{4.00}{T_2}

gives

T_2 = 400\text{ K}

. Converting back to Celsius (

400 - 273

) yields the correct temperature of

127^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to the thermodynamic temperature scale (Kelvin).

T_1 = -23 + 273 = 250\text{ K}

Gas laws strictly require temperature to be expressed in Kelvin.

2

Use Gay-Lussac's Pressure Law equation

\frac{P_1}{T_1} = \frac{P_2}{T_2}

to solve for

T_2

.

T_2 = \frac{P_2 \times T_1}{P_1} = \frac{4.00 \times 250}{2.50} = 400\text{ K}

At constant volume, the pressure of a given mass of gas is directly proportional to its absolute temperature.

3

Convert the calculated temperature

T_2

back to degrees Celsius.

t_2 = 400 - 273 = 127^\circ\text{C}

The question explicitly requests the final temperature in degrees Celsius.

Key Concept

Pressure Law (Gay-Lussac's Law)

Question 7163Question →

An uncatalyzed exothermic reaction has an activation energy for the forward reaction of $110\text{ kJ mol}^{-1}$ and an overall enthalpy change ( $\Delta H$ ) of $-70\text{ kJ mol}^{-1}$ . If a catalyst is added that lowers the activation energy of the forward reaction by $35\text{ kJ mol}^{-1}$ , what is the activation energy of the reverse catalyzed reaction?

145\text{ kJ mol}^{-1}

75\text{ kJ mol}^{-1}

40\text{ kJ mol}^{-1}

215\text{ kJ mol}^{-1}

Show answer & explanation

Answer: $145\text{ kJ mol}^{-1}$

Answer

145\text{ kJ mol}^{-1}

For an exothermic reaction, the reverse activation energy is greater than the forward activation energy by

|\Delta H|

. The uncatalyzed reverse activation energy is

110 - (-70) = 180\text{ kJ mol}^{-1}

. Since a catalyst reduces both forward and reverse activation energy barriers by the exact same amount (

35\text{ kJ mol}^{-1}

), the reverse catalyzed activation energy is

180 - 35 = 145\text{ kJ mol}^{-1}

. Alternatively, using the catalyzed forward barrier of

75\text{ kJ mol}^{-1}

,

E_{a,\text{rev, cat}} = 75 - (-70) = 145\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Calculate the activation energy for the forward catalyzed reaction.

E_{a,\text{fwd, cat}} = 110\text{ kJ mol}^{-1} - 35\text{ kJ mol}^{-1} = 75\text{ kJ mol}^{-1}

A catalyst lowers the energy barrier for the forward reaction by the specified amount.

2

Relate the forward activation energy, reverse activation energy, and enthalpy change.

\Delta H = E_{a,\text{fwd, cat}} - E_{a,\text{rev, cat}}

The difference between forward and reverse activation energy yields the reaction enthalpy change.

3

Solve for the reverse catalyzed activation energy.

E_{a,\text{rev, cat}} = 75\text{ kJ mol}^{-1} - (-70\text{ kJ mol}^{-1}) = 145\text{ kJ mol}^{-1}

Subtracting a negative enthalpy change adds its absolute value to the forward activation energy.

Key Concept

Effect of Catalysts on Forward and Reverse Activation Energies in Energy Profile Diagrams

Estimated Time:1m 30s

Question 7164Question →

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

Show answer & explanation

Answer: 1

Answer

The numerical value of

x - y

is 1.0.

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

x = 0.5

(for pressure) and

y = -0.5

(for density). Thus,

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

.

Step-by-Step Solution

1

Determine the dimensions of speed, pressure, and density.

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

,

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

, and

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

.

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

M

,

L

,

T

).

2

Substitute dimensions into the relationship

v = C P^x \rho^y

.

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

.

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

3

Solve for exponents

x

and

y

.

From time

T

:

-2x = -1 \implies x = 0.5

. From mass

M

:

x + y = 0 \implies y = -0.5

.

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

x

and

y

.

4

Calculate the required expression

(x - y)

.

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

.

Subtracting negative

0.5

from

0.5

results in

1.0

.

Key Concept

Dimensional Analysis and Homogeneity

Question 7165Question →

A naturally occurring sample of boron consists of two stable isotopes, $^{10}\text{B}$ and $^{11}\text{B}$ . If the relative percentage abundance of $^{10}\text{B}$ is $20.0\%$ and that of $^{11}\text{B}$ is $80.0\%$ , what is the relative atomic mass of boron?

Show answer & explanation

Answer: 10.8

Answer

The relative atomic mass of boron is

10.8

.

The relative atomic mass of an element is defined as the weighted average mass of its naturally occurring isotopes relative to

\frac{1}{12}\text{th}

the mass of a carbon-12 atom. Applying the formula

\text{RAM} = \frac{\sum (\text{isotopic mass} \times \% \text{ abundance})}{100}

, we get

\frac{(10 \times 20) + (11 \times 80)}{100} = \frac{200 + 880}{100} = 10.8

.

Step-by-Step Solution

1

Determine the mass contribution of the

^{10}\text{B}

isotope

10 \times 0.20 = 2.0

The weighted contribution of an isotope is its mass multiplied by its fractional abundance.

2

Determine the mass contribution of the

^{11}\text{B}

isotope

11 \times 0.80 = 8.8

The weighted contribution of the second isotope is calculated using its percentage abundance.

3

Sum the weighted contributions to find the relative atomic mass

2.0 + 8.8 = 10.8

The relative atomic mass of an element is the weighted average mass of all naturally occurring isotopes relative to carbon-12.

Key Concept

Calculation of Relative Atomic Mass from Isotopic Abundances

Question 7166Question →

Two sound waves of frequencies $440\text{ Hz}$ and $445\text{ Hz}$ travel through air and superpose to produce beats. What is the resulting beat frequency, in hertz?

Show answer & explanation

Answer: 5

Answer

The beat frequency produced by the superposition of the two sound waves is

5\text{ Hz}

.

When two waves of slightly different frequencies interfere, periodic variations in sound intensity called beats occur. The number of beats heard per second is equal to the absolute difference between the frequencies of the two superposing waves:

f_{\text{beat}} = |f_2 - f_1| = |445\text{ Hz} - 440\text{ Hz}| = 5\text{ Hz}

.

Step-by-Step Solution

1

Identify the frequencies of the interfering waves

f_1 = 440\text{ Hz}

and

f_2 = 445\text{ Hz}

Beat frequency is determined by the absolute difference between the individual wave frequencies.

2

Subtract the lower frequency from the higher frequency to find the beat frequency

f_{\text{beat}} = |445 - 440| = 5\text{ Hz}

The rate of periodic intensity variation (beat frequency) is governed by

f_{\text{beat}} = |f_2 - f_1|

.

Key Concept

Beat Frequency and Wave Superposition

Question 7167Question →

In a cathode-ray experiment, a beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field intensity between the deflection plates is $4.0 \times 10^{4}\text{ V m}^{-1}$ and the magnetic flux density is $2.0 \times 10^{-3}\text{ T}$ , what is the velocity of the electrons in the cathode-ray beam?

2.0 \times 10^{7}\text{ m s}^{-1}

5.0 \times 10^{-8}\text{ m s}^{-1}

8.0 \times 10^{1}\text{ m s}^{-1}

6.0 \times 10^{7}\text{ m s}^{-1}

Show answer & explanation

Answer: $2.0 \times 10^{7}\text{ m s}^{-1}$

Answer

The velocity of the electrons in the cathode-ray beam is

2.0 \times 10^{7}\text{ m s}^{-1}

.

For cathode rays passing undeflected through perpendicular electric (

E

) and magnetic (

B

) fields, the electric force

eE

pulling electrons toward the positive plate is exactly balanced by the magnetic force

evB

. Equating

eE = evB

leads directly to

v = \frac{E}{B}

. Substituting

E = 4.0 \times 10^{4}\text{ V m}^{-1}

and

B = 2.0 \times 10^{-3}\text{ T}

gives

v = \frac{4.0 \times 10^{4}}{2.0 \times 10^{-3}} = 2.0 \times 10^{7}\text{ m s}^{-1}

.

Step-by-Step Solution

1

Identify the equilibrium condition for an undeflected electron beam in crossed fields.

The electric force

F_e = eE

balances the magnetic Lorentz force

F_b = evB

, so

eE = evB

.

When cathode rays pass undeflected, the net transverse force acting on each electron is zero.

2

Rearrange the equilibrium equation to solve for velocity

v

.

v = \frac{E}{B}

Canceling the elementary charge

e

from both sides isolates electron speed as a function of field strengths.

3

Substitute the given values into the velocity equation.

v = \frac{4.0 \times 10^{4}\text{ V m}^{-1}}{2.0 \times 10^{-3}\text{ T}} = 2.0 \times 10^{7}\text{ m s}^{-1}

Executing the division yields the velocity of cathode rays.

Key Concept

Velocity selector principle in crossed electric and magnetic fields (J.J. Thomson cathode ray velocity determination)

Question 7168Question →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & x \neq 3 \\ k + 2, & x = 3 \end{cases}

If

f(x)

is continuous at

x = 3

, what is the value of the constant

k

?

3

5

7

-2

Show answer & explanation

Answer: $5$

Answer

The constant value is

k = 5

.

For the function to be continuous at

x = 3

, the limit as

x \to 3

must equal the value of the function at

x = 3

, which is

f(3) = k + 2

. Factoring the numerator gives

2x^2 - 5x - 3 = (2x + 1)(x - 3)

. Canceling the common factor

(x - 3)

leaves

\lim_{x \to 3}(2x + 1) = 7

. Setting

k + 2 = 7

yields

k = 5

.

Step-by-Step Solution

1

Evaluate the limit of

f(x)

as

x

approaches

3

\lim_{x \to 3} \frac{2x^2 - 5x - 3}{x - 3} = \lim_{x \to 3} \frac{(2x + 1)(x - 3)}{x - 3} = \lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7

Direct substitution yields the indeterminate form

\frac{0}{0}

, so the numerator must be factored to cancel the common term

(x - 3)

.

2

Apply the definition of continuity at

x = 3

f(3) = \lim_{x \to 3} f(x) \implies k + 2 = 7

For a function to be continuous at a point

x = c

, the function value

f(c)

must equal the limit of

f(x)

as

x \to c

.

3

Solve for the constant

k

k = 7 - 2 = 5

Subtract

2

from both sides of the equation.

Key Concept

Continuity of a Piecewise Function at a Point

Question 7169Question →

A sample of pure ammonia gas ( $\text{NH}_3$ ) has a mass of $3.4\text{ g}$ . What is the total number of atoms contained in this sample? $[\text{N} = 14, \text{H} = 1, N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

4.816 \times 10^{23}

1.204 \times 10^{23}

3.612 \times 10^{23}

2.408 \times 10^{23}

Show answer & explanation

Answer: $4.816 \times 10^{23}$

Answer

4.816 \times 10^{23}

atoms

The molar mass of ammonia (

\text{NH}_3

) is

17\text{ g/mol}

. A mass of

3.4\text{ g}

corresponds to

0.2\text{ moles}

of

\text{NH}_3

molecules, which contains

1.204 \times 10^{23}

molecules. Because one molecule of

\text{NH}_3

consists of

4

atoms (

1

nitrogen atom and

3

hydrogen atoms), multiplying

1.204 \times 10^{23}

by

4

gives

4.816 \times 10^{23}

total atoms.

Step-by-Step Solution

1

Calculate the molar mass of ammonia (

\text{NH}_3

)

\text{Molar Mass} = 14 + 3(1) = 17\text{ g/mol}

Molar mass is required to convert sample mass into moles.

2

Calculate the number of moles of

\text{NH}_3

\text{Moles} = \frac{3.4\text{ g}}{17\text{ g/mol}} = 0.2\text{ mol}

Dividing given mass by molar mass yields the mole quantity.

3

Determine the number of molecules of

\text{NH}_3

\text{Molecules} = 0.2\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 1.204 \times 10^{23}\text{ molecules}

Multiplying moles by Avogadro's constant gives total molecules.

4

Calculate the total number of individual atoms

\text{Total atoms} = 1.204 \times 10^{23} \times 4 = 4.816 \times 10^{23}\text{ atoms}

Each

\text{NH}_3

molecule contains 1 nitrogen atom and 3 hydrogen atoms (4 atoms in total).

Key Concept

Converting mass to number of constituent atoms using Avogadro's constant and molecular stoichiometry.

Question 7170Question →

What is the $x$ -intercept of the normal line to the curve $y = x^2 - 4x + 5$ at the point where $x = 3$ ?

2

7

-1

-7

Show answer & explanation

Answer: 7

Answer

The

x

-intercept of the normal line is 7.

Differentiating

y = x^2 - 4x + 5

gives

\frac{dy}{dx} = 2x - 4

. Substituting

x = 3

yields a tangent gradient of

2

, so the normal gradient is

-\frac{1}{2}

. The curve passes through

(3, 2)

at

x = 3

. The equation of the normal is

y - 2 = -\frac{1}{2}(x - 3)

, which simplifies to

x + 2y - 7 = 0

. Setting

y = 0

gives

x = 7

.

Step-by-Step Solution

1

Find the

y

-coordinate corresponding to

x = 3

y = 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2

. The point on the curve is

(3, 2)

.

The normal line passes through the specific point of tangency on the curve.

2

Calculate the gradient of the tangent and normal lines at

x = 3

\frac{dy}{dx} = 2x - 4

. At

x = 3

,

m_t = 2(3) - 4 = 2

. Therefore,

m_n = -\frac{1}{m_t} = -\frac{1}{2}

.

The normal line is perpendicular to the tangent line.

3

Determine the equation of the normal line

y - 2 = -\frac{1}{2}(x - 3) \implies 2(y - 2) = -(x - 3) \implies x + 2y - 7 = 0

.

Use the point-slope form

y - y_1 = m(x - x_1)

to write the linear equation.

4

Find the

x

-intercept of the normal line

Set

y = 0

:

x + 2(0) - 7 = 0 \implies x = 7

.

The

x

-intercept occurs where

y = 0

.

Key Concept

Tangents and Normals to Curves

Estimated Time:1m 30s

Question 7171Question →

A metal surface inside a vacuum cell is illuminated by incident photons each having an energy of $6.2\text{ eV}$ . If the maximum kinetic energy of the emitted photoelectrons is $2.4\text{ eV}$ , what is the work function of the metal in electron-volts ( $\text{eV}$ )?

Show answer & explanation

Answer: 3.8

Answer

The work function of the metal is

3.8\text{ eV}

.

According to Einstein's photoelectric equation, the total energy of an incident photon (

E = 6.2\text{ eV}

) equals the work function of the metal (

W_0

) plus the maximum kinetic energy of the ejected photoelectrons (

K_{\max} = 2.4\text{ eV}

). Subtracting the kinetic energy from the photon energy gives

W_0 = 6.2\text{ eV} - 2.4\text{ eV} = 3.8\text{ eV}

.

Step-by-Step Solution

1

State Einstein's photoelectric equation

E = W_0 + K_{\max}

Relates the incident photon energy to the metal work function and photoelectron kinetic energy.

2

Rearrange the equation to solve for the work function

W_0 = E - K_{\max}

Isolates the work function

W_0

on one side of the equation.

3

Substitute the given numerical values and compute

W_0 = 6.2\text{ eV} - 2.4\text{ eV} = 3.8\text{ eV}

Evaluates the difference to obtain the minimum energy needed to remove an electron.

Key Concept

Einstein's Photoelectric Equation and Work Function

Question 7172Question →

Power is defined as the rate at which work is done or energy is transferred. Which of the following expressions represents the correct dimensional formula for power?

M L^2 T^{-3}

M L^2 T^{-2}

M L T^{-3}

M L T^{-2}

Show answer & explanation

Answer: $M L^2 T^{-3}$

Answer

M L^2 T^{-3}

Power is defined as work done per unit time (

P = \frac{W}{t}

). The dimensional formula for work is

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

, and for time is

[t] = T

. Dividing work by time gives

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

.

Step-by-Step Solution

1

Write the fundamental definition of power in terms of work and time.

\text{Power} = \frac{\text{Work}}{\text{Time}}

By definition, power measures the rate of doing work.

2

Determine the dimensions of work.

[\text{Work}] = [\text{Force}] \times [\text{Distance}] = (M L T^{-2}) \times L = M L^2 T^{-2}

Force has dimensions of mass times acceleration (

M L T^{-2}

), and multiplying by distance (

L

) yields energy or work dimensions.

3

Divide the dimensions of work by the dimension of time (

T

).

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

Dividing by time increases the negative exponent of time from

-2

to

-3

.

Key Concept

Dimensional Formula for Power

Question 7173Question →

Match each experimental observation of cathode rays in a discharge tube with the corresponding physical property or characteristic it demonstrates.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Formation of a sharp shadow of a Maltese cross matches with Cathode rays travel in straight lines. 2. Rotation of a paddle wheel matches with Cathode rays possess particle mass and mechanical momentum. 3. Deflection toward a positively charged plate matches with Cathode rays carry a negative electrical charge. 4. Deflection perpendicular to a magnetic field matches with Cathode rays act as a current of moving charged particles obeying magnetic force laws.

Each experimental setup provides specific proof of a cathode ray property: sharp shadow formation proves straight-line motion; turning a paddle wheel demonstrates particle mass and momentum; attraction to a positive plate confirms negative charge; and deflection in a magnetic field confirms that the beam acts as moving electrical charges.

Step-by-Step Solution

1

Analyze the Maltese cross shadow experiment

Sharp shadows indicate straight-line propagation of rays from the cathode surface.

Light and particle beams traveling in straight lines produce sharp geometrical shadows of opaque obstructions.

2

Analyze the paddle wheel experiment

The paddle wheel rotates when struck by cathode rays, demonstrating kinetic energy and momentum transfer.

Mechanical rotation requires a force resulting from the momentum transfer of moving material particles.

3

Analyze electric field deflection

The ray path bends toward the positive anode plate.

Electrostatic attraction pulls negatively charged entities toward positive electric potentials.

4

Analyze magnetic field deflection

The beam bends laterally perpendicular to both the trajectory and the magnetic field vector.

Moving electrical charges experience a magnetic Lorentz force given by

F = qvB\sin\theta

.

Key Concept

Experimental evidence establishing the properties of cathode rays

Estimated Time:1m 30s

Question 7174Question →

What is the result of evaluating the indefinite integral $\int \left( (2x - 1)^2 - 6\cos(2x) \right) \, dx$ ?

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x)

\frac{4}{3}x^3 - 2x^2 + x + 3\sin(2x) + C

\frac{4}{3}x^3 - 2x^2 + x - 12\sin(2x) + C

Show answer & explanation

Answer: $\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C$

Answer

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

Expanding the squared term

(2x - 1)^2

yields

4x^2 - 4x + 1

. Integrating term by term using the power rule gives

\frac{4}{3}x^3 - 2x^2 + x

. The integral of

-6\cos(2x)

is

-\frac{6}{2}\sin(2x) = -3\sin(2x)

. Adding the arbitrary constant

C

results in

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

.

Step-by-Step Solution

1

Expand the squared binomial term inside the integral

(2x - 1)^2 = 4x^2 - 4x + 1

Expanding the expression allows for term-by-term integration using standard integration rules.

2

Integrate each term of the expanded polynomial

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

Apply the power rule for integration: \int x^n \, dx = \frac{x^{n+1}}{n+1}.

3

Integrate the trigonometric term

\int -6\cos(2x) \, dx = -6 \cdot \frac{1}{2}\sin(2x) = -3\sin(2x)

The standard integral of \cos(kx) is \frac{1}{k}\sin(kx).

4

Combine all integrated terms and append the constant of integration

\frac{4}{3}x^3 - 2x^2 + x - 3\sin(2x) + C

Indefinite integration requires an arbitrary constant of integration C.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 7175Question →

Which of the following sets contains all values of $x$ in the interval $0^\circ \le x \le 360^\circ$ that satisfy the trigonometric equation $\sqrt{3}\sin x + \cos x = 0$ ?

150^\circ \text{ and } 330^\circ

30^\circ \text{ and } 210^\circ

120^\circ \text{ and } 300^\circ

150^\circ \text{ only}

Show answer & explanation

Answer: $150^\circ \text{ and } 330^\circ$

Answer

150^\circ \text{ and } 330^\circ

The given equation

\sqrt{3}\sin x + \cos x = 0

simplifies to

\tan x = -\frac{1}{\sqrt{3}}

. Since tangent is negative in the second and fourth quadrants with a reference angle of

30^\circ

, the solutions in the domain

0^\circ \le x \le 360^\circ

are

180^\circ - 30^\circ = 150^\circ

and

360^\circ - 30^\circ = 330^\circ

.

Step-by-Step Solution

1

Rearrange the trigonometric equation into single ratio form

\sqrt{3}\sin x = -\cos x \implies \frac{\sin x}{\cos x} = -\frac{1}{\sqrt{3}} \implies \tan x = -\frac{1}{\sqrt{3}}

Dividing both sides by

\cos x

converts the sum of sine and cosine terms into a simple tangent equation.

2

Determine the reference angle

\text{Reference angle } \alpha = 30^\circ

The acute angle whose tangent is

\frac{1}{\sqrt{3}}

is

30^\circ

.

3

Identify the quadrants and find all solutions in

0^\circ \le x \le 360^\circ

x = 180^\circ - 30^\circ = 150^\circ

(Quadrant II) and

x = 360^\circ - 30^\circ = 330^\circ

(Quadrant IV)

The tangent function is negative in Quadrants II and IV.

Key Concept

Solving simple trigonometric equations by reducing to a basic ratio and finding all solutions within a given domain.

Estimated Time:1m 30s

Question 7176Question →

Match each physical quantity on the left with its correct fundamental SI base unit expression or fundamental status on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Thermodynamic temperature matches with 'Fundamental physical quantity measured in kelvin (K)'; Electric charge matches with 'Derived physical quantity expressed in base SI units as A·s'; Linear momentum matches with 'Derived physical quantity expressed in base SI units as kg·m·s⁻¹'; Power matches with 'Derived physical quantity expressed in base SI units as kg·m²·s⁻³'.

Each physical quantity is correctly paired with either its fundamental status or its base SI unit breakdown derived from core physics definitions.

Step-by-Step Solution

1

Identify fundamental quantities versus derived quantities.

Thermodynamic temperature is a basic fundamental quantity (unit:

\text{K}

). Electric current is fundamental (unit:

\text{A}

), but electric charge is derived (

Q = I t

).

Fundamental quantities cannot be defined in terms of other physical quantities.

2

Decompose Electric Charge into base units.

Since

Q = I \cdot t

, its unit is

\text{A}\cdot\text{s}

.

Current is measured in amperes and time in seconds.

3

Decompose Linear Momentum into base units.

p = m \cdot v \Rightarrow \text{unit} = \text{kg} \cdot \text{m}\cdot\text{s}^{-1}

.

Mass is in kilograms and velocity is in meters per second.

4

Decompose Power into base units.

P = \frac{W}{t} = \frac{F \cdot d}{t} = \frac{(m \cdot a) \cdot d}{t} \Rightarrow \frac{\text{kg} \cdot (\text{m}\cdot\text{s}^{-2}) \cdot \text{m}}{\text{s}} = \text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}

.

Power is work done per unit time.

Key Concept

Fundamental and Derived Quantities

Question 7177Question →

Increasing the intensity of incident monochromatic light of a fixed frequency above the threshold frequency increases the maximum kinetic energy of the photoelectrons emitted from a metal surface.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Increasing light intensity at constant frequency increases the number of emitted photoelectrons per second (photoelectric current), but leaves their maximum kinetic energy unchanged.

The statement is false. In quantum physics, increasing the intensity of monochromatic light increases the photon flux, which elevates the rate of photoelectron emission (photoelectric current). However, the maximum kinetic energy of each photoelectron depends solely on the energy of an individual photon (

E = hf

) minus the metal's work function (

W_0

), both of which remain unchanged when intensity is increased at fixed frequency.

Step-by-Step Solution

1

Identify the factors determining maximum kinetic energy in the photoelectric effect.

Maximum kinetic energy

K_{\text{max}}

is governed by Einstein's photoelectric equation

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

.

Energy transfer occurs on a one-photon-to-one-electron basis.

2

Determine the physical quantity affected by changing light intensity.

Intensity dictates the number of photons striking the metal surface per unit time.

Higher intensity means a greater photon flux, which increases the photoelectron emission rate.

3

Evaluate the statement.

The statement falsely attributes an increase in kinetic energy to an increase in intensity.

Since photon frequency

f

and work function

W_0

remain constant,

K_{\text{max}}

does not change.

Key Concept

Independence of photoelectron kinetic energy from light intensity

Question 7178Question →

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

Show answer & explanation

Answer: 1.82

Answer

The new reading on the digital weighing scale is

1.82\text{ kg}

.

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

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

). The mass of this displaced water is

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

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

3.2\text{ N}

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

3.2\text{ N}

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

0.32\text{ kg}

to the digital scale reading, making the new reading

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

.

Step-by-Step Solution

1

Calculate the volume of the submerged block

Volume

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

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

2

Find the mass of the displaced water

Mass of displaced water

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

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

\rho_{\text{water}} V

.

3

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

Scale reading increase

\Delta m = 0.32\text{ kg}

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

4

Compute the total new scale reading

New scale reading

= 1.82\text{ kg}

Sum the initial mass reading of the beaker system (

1.50\text{ kg}

) and the mass of the displaced water (

0.32\text{ kg}

).

Key Concept

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

Question 7179Question →

Match each kinetic theory concept on the left with its correct physical description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Temperature matches with the measure of average translational kinetic energy; Pressure matches with the average force per unit area exerted by colliding gas molecules on container walls; Root-mean-square speed matches with the square root of the mean of squared speeds.

Temperature measures average translational kinetic energy per particle. Pressure originates from force per unit area due to elastic wall collisions. Root-mean-square speed is the square root of the mean of squared molecular speeds.

Step-by-Step Solution

1

Identify the kinetic theory definition of Temperature

Temperature is directly proportional to the mean translational kinetic energy of the gas particles (

E_k \propto T

).

Absolute temperature reflects the average kinetic energy of molecular motion.

2

Identify the microscopic origin of Gas Pressure

Pressure is caused by molecular collisions with the container walls, transferring momentum and creating force per unit area.

Frequent elastic collisions of particles on container walls produce measurable pressure.

3

Identify the mathematical definition of Root-Mean-Square Speed

v_{rms} = \sqrt{\overline{v^2}}

, representing the square root of the average of squared molecular velocities.

This parameter represents the effective speed of gas particles relevant to thermal kinetic energy.

Key Concept

Kinetic Theory Interpretation of Gas Properties

Question 7180Question →

A convex mirror forms an upright image that is $\frac{1}{3}$ the size of an object. When the object is moved $20\text{ cm}$ further away from the mirror, the size of the image becomes $\frac{1}{5}$ the size of the object. What is the radius of curvature of the mirror?

20\text{ cm}

10\text{ cm}

40\text{ cm}

5\text{ cm}

Show answer & explanation

Answer: $20\text{ cm}$

Answer

The radius of curvature of the convex mirror is

20\text{ cm}

.

For a convex mirror, the focal length is negative (

f = -f_0

). The magnification formula

m = \frac{f}{f - u}

for a virtual upright image gives

m = \frac{-f_0}{-f_0 - u} = \frac{f_0}{f_0 + u}

. For

m_1 = \frac{1}{3}

, we get

u_1 = 2f_0

. For

m_2 = \frac{1}{5}

, we get

u_2 = 4f_0

. The object displacement is

u_2 - u_1 = 2f_0 = 20\text{ cm}

, which gives

f_0 = 10\text{ cm}

. The radius of curvature is

R = 2f_0 = 20\text{ cm}

, making the choice stating

20\text{ cm}

correct.

Step-by-Step Solution

1

Apply the magnification formula and sign convention for a convex mirror at the first position.

For a convex mirror,

f = -f_0

and the image is virtual (

v_1 = -v_{01}

). Magnification

m_1 = -\frac{v_1}{u_1} = \frac{v_{01}}{u_1} = \frac{1}{3}

, so

v_{01} = \frac{u_1}{3}

.

Convex mirrors always form virtual, upright, and diminished images.

2

Substitute

v_1

into the mirror equation for the first position to express

u_1

in terms of focal length magnitude

f_0

.

\frac{1}{-f_0} = \frac{1}{u_1} + \frac{1}{-v_{01}} = \frac{1}{u_1} - \frac{3}{u_1} = -\frac{2}{u_1} \implies u_1 = 2f_0

.

The mirror formula is

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

with cartesian sign conventions.

3

Repeat the mirror equation calculation for the second object position.

For

m_2 = \frac{1}{5}

,

v_{02} = \frac{u_2}{5}

. Substituting gives

\frac{1}{-f_0} = \frac{1}{u_2} - \frac{5}{u_2} = -\frac{4}{u_2} \implies u_2 = 4f_0

.

The second object position gives a magnification of

\frac{1}{5}

.

4

Use the known object shift distance to solve for

f_0

and radius of curvature

R

.

u_2 - u_1 = 20\text{ cm} \implies 4f_0 - 2f_0 = 20\text{ cm} \implies 2f_0 = 20\text{ cm} \implies f_0 = 10\text{ cm}

. Since

R = 2f_0

,

R = 20\text{ cm}

.

The distance between the two object positions is

20\text{ cm}

, and the radius of curvature of a spherical mirror is twice its focal length.

Key Concept

Spherical Mirror Formula and Sign Conventions for Convex Mirrors

Answer

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