All practice questions

Question 12881Question →

A magnetometer stationed at a field site records the horizontal component of the Earth's magnetic field as $36\ \mu\text{T}$ and the total magnetic field intensity as $45\ \mu\text{T}$ . What is the magnitude of the vertical component of the Earth's magnetic field, in $\mu\text{T}$ ?

Show answer & explanation

Answer: 27

Answer

The magnitude of the vertical component of the Earth's magnetic field is

27\ \mu\text{T}

.

The total magnetic field strength

B

is the hypotenuse of a right-angled triangle formed by the horizontal component

B_h

and vertical component

B_v

. By applying the Pythagorean relation

B^2 = B_h^2 + B_v^2

, substituting

B = 45\ \mu\text{T}

and

B_h = 36\ \mu\text{T}

yields

B_v = \sqrt{45^2 - 36^2} = \sqrt{729} = 27\ \mu\text{T}

.

Step-by-Step Solution

1

Relate total magnetic intensity to its orthogonal components

B^2 = B_h^2 + B_v^2

The total magnetic field vector of the Earth is the vector sum of mutually perpendicular horizontal and vertical components.

2

Isolate the vertical component variable

B_v = \sqrt{B^2 - B_h^2}

Applying the Pythagorean theorem allows direct calculation of the missing perpendicular side.

3

Substitute given values and compute numerical result

B_v = \sqrt{45^2 - 36^2} = \sqrt{2025 - 1296} = \sqrt{729} = 27\ \mu\text{T}

Evaluates the exact magnitude of the vertical component.

Key Concept

Orthogonal resolution of Earth's magnetic field components

Estimated Time:1m 15s

Question 12882Question →

Equal masses of zinc granules are added separately to beaker P containing $100\text{ cm}^3$ of $1.0\text{ mol dm}^{-3}\text{ H}_2\text{SO}_4(aq)$ and beaker Q containing $100\text{ cm}^3$ of $1.0\text{ mol dm}^{-3}\text{ HNO}_3(aq)$ at room temperature. Which of the following statements correctly identifies and explains the primary gaseous product liberated in each beaker?

Beaker P liberates hydrogen gas through typical acid-metal displacement, whereas beaker Q liberates oxides of nitrogen because trioxonitrate(V) acid acts as a strong oxidizing agent.

Both beakers liberate equal volumes of hydrogen gas because both acids are strong electrolytes that fully ionize in aqueous solution.

Beaker Q liberates twice the volume of hydrogen gas produced in beaker P because trioxonitrate(V) acid undergoes complete reduction to hydrogen ions.

Beaker P liberates sulfur(IV) oxide gas due to the dehydrating action of tetraoxosulfate(VI) acid, while beaker Q liberates hydrogen gas.

Show answer & explanation

Answer: Beaker P liberates hydrogen gas through typical acid-metal displacement, whereas beaker Q liberates oxides of nitrogen because trioxonitrate(V) acid acts as a strong oxidizing agent.

Answer

Beaker P liberates hydrogen gas through typical acid-metal displacement, whereas beaker Q liberates oxides of nitrogen because trioxonitrate(V) acid acts as a strong oxidizing agent.

Dilute tetraoxosulfate(VI) acid (

H_2SO_4

) reacts with active metals such as zinc via standard single-replacement to evolve hydrogen gas (

H_2

). Conversely, trioxonitrate(V) acid (

HNO_3

) is a strong oxidizing acid; the nitrate ions (

NO_3^-

) are preferentially reduced to oxides of nitrogen rather than hydrogen ions being reduced to hydrogen gas.

Step-by-Step Solution

1

Analyze the chemical reaction between zinc and dilute tetraoxosulfate(VI) acid in beaker P.

Zinc displaces hydrogen ions from dilute

H_2SO_4

to yield zinc tetraoxosulfate(VI) and hydrogen gas:

\text{Zn}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{H}_2(g)

.

Dilute tetraoxosulfate(VI) acid exhibits standard acid behavior with reactive metals above hydrogen in the electrochemical series.

2

Analyze the chemical reaction between zinc and dilute trioxonitrate(V) acid in beaker Q.

Nitrate ions (

NO_3^-

) act as powerful oxidizing agents, undergoing reduction to form nitrogen oxides (such as

N_2O

,

NO

, or

NO_2

) instead of liberating hydrogen gas.

Trioxonitrate(V) acid (

HNO_3

) is a strong oxidizing acid, so hydrogen ions are not reduced to

H_2(g)

during reaction with metals.

3

Synthesize the observations from both beakers to select the correct explanation.

Beaker P produces

H_2(g)

while beaker Q produces oxides of nitrogen.

The distinct chemical property of

HNO_3

as an oxidizing acid alters the gaseous product compared to typical mineral acids.

Key Concept

Oxidizing property of trioxonitrate(V) acid versus typical acid-metal displacement reactions

Estimated Time:2m 0s

Question 12883Question →

A glass window pane of thickness $4.0\text{ mm}$ and surface area $1.5\text{ m}^2$ maintains an inner surface temperature of $20^\circ\text{C}$ and an outer surface temperature of $5^\circ\text{C}$ . If the thermal conductivity of glass is $0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , what is the rate of heat transfer by conduction through the window?

4500\text{ W}

4.5\text{ W}

86,400\text{ W}

0.072\text{ W}

Show answer & explanation

Answer: $4500\text{ W}$

Answer

4500\text{ W}

(or

4.5\text{ kW}

)

The rate of conductive heat transfer is governed by Fourier's law of thermal conduction:

\frac{Q}{t} = \frac{k A (T_1 - T_2)}{d}

. Substituting

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = 1.5\text{ m}^2

, temperature difference

\Delta T = 15\text{ K}

, and thickness

d = 0.004\text{ m}

gives

\frac{Q}{t} = \frac{0.80 \times 1.5 \times 15}{0.004} = 4500\text{ W}

.

Step-by-Step Solution

1

Identify the given parameters and convert units to standard SI units

k = 0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = 1.5\text{ m}^2

,

\Delta T = 20^\circ\text{C} - 5^\circ\text{C} = 15\text{ K}

, and

d = 4.0\text{ mm} = 4.0 \times 10^{-3}\text{ m}

.

Thermal conductivity formulas require distance/thickness in meters.

2

Apply the law of thermal conduction formula

\frac{Q}{t} = \frac{k A \Delta T}{d}

The rate of heat transfer by conduction is directly proportional to thermal conductivity, surface area, and temperature difference, and inversely proportional to thickness.

3

Substitute the values and calculate the heat transfer rate

\frac{Q}{t} = \frac{0.80 \times 1.5 \times 15}{4.0 \times 10^{-3}} = \frac{18}{0.004} = 4500\text{ W}

Performing accurate arithmetic yields

4500\text{ Joules per second}

(Watts).

Key Concept

Rate of thermal conduction through a uniform slab

Question 12884Question →

A double-glazed window of total surface area $1.5\text{ m}^2$ consists of two glass panes, each of thickness $4.0\text{ mm}$ and thermal conductivity $0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , separated by a stagnant air gap of thickness $2.0\text{ mm}$ with thermal conductivity $0.025\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . If a steady-state temperature difference of $18.0^\circ\text{C}$ is maintained across the window's outer boundary surfaces, what is the rate of heat transfer through the window in watts?

Show answer & explanation

Answer: 300

Answer

The steady-state rate of heat transfer through the double-glazed window is

300\text{ W}

.

Heat conduction through composite layers in series is governed by the total thermal resistance. The thermal resistance per unit area of each layer is

r = \frac{d}{k}

. For two

4.0\text{ mm}

glass panes (

r = 0.005\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

each) and one

2.0\text{ mm}

air gap (

r = 0.080\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

), the total unit resistance is

r_{\text{total}} = 0.090\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

. Multiplying by the area

1.5\text{ m}^2

and temperature difference

18.0^\circ\text{C}

gives

\frac{Q}{t} = \frac{1.5 \times 18.0}{0.090} = 300\text{ W}

.

Step-by-Step Solution

1

Convert layer thicknesses to standard units (meters)

d_{\text{glass}} = 4.0\text{ mm} = 0.004\text{ m}

,

d_{\text{air}} = 2.0\text{ mm} = 0.002\text{ m}

SI units are required for calculations using thermal conductivity in

\text{W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

.

2

Calculate thermal resistance per unit area for each layer

r_{\text{glass}} = \frac{0.004}{0.80} = 0.005\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

,

r_{\text{air}} = \frac{0.002}{0.025} = 0.080\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

Thermal resistance per unit area is given by

r = \frac{d}{k}

.

3

Sum thermal resistances in series to obtain total unit resistance

r_{\text{total}} = r_{\text{glass1}} + r_{\text{air}} + r_{\text{glass2}} = 0.005 + 0.080 + 0.005 = 0.090\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

Heat flows sequentially through all three layers in series.

4

Calculate rate of heat flow across total window area

\frac{Q}{t} = \frac{A \cdot \Delta T}{r_{\text{total}}} = \frac{1.5 \cdot 18.0}{0.090} = 300\text{ W}

Rate of thermal conduction through composite series layers is

\frac{Q}{t} = \frac{\Delta T}{R_{\text{total}}}

where

R_{\text{total}} = \frac{r_{\text{total}}}{A}

.

Key Concept

Series thermal conduction through composite layers and thermal resistance

Question 12885Question →

Fill in the blanks with the correct numerical values based on volume stoichiometry at constant temperature and pressure.

Fill in the blanks below

When

40\text{ cm}^3

of nitrogen(II) oxide gas (

NO

) is reacted with

30\text{ cm}^3

of oxygen gas (

O_2

) according to the equation

2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}

, the volume of unreacted oxygen gas remaining is

\text{cm}^3

and the total volume of the resulting gaseous mixture is

\text{cm}^3

.

Show answer & explanation

Answer

The volume of unreacted oxygen gas remaining is

10\text{ cm}^3

and the total volume of the resulting gaseous mixture is

50\text{ cm}^3

.

According to Gay-Lussac's Law of Combining Volumes,

2\text{ volumes}

of

NO

react with

1\text{ volume}

of

O_2

to yield

2\text{ volumes}

of

NO_2

. Therefore,

40\text{ cm}^3

of

NO

reacts with

20\text{ cm}^3

of

O_2

, leaving

10\text{ cm}^3

of unreacted

O_2

(

30\text{ cm}^3 - 20\text{ cm}^3

). The reaction produces

40\text{ cm}^3

of

NO_2

gas. Summing the product volume (

40\text{ cm}^3

) and unreacted excess gas (

10\text{ cm}^3

) gives a total residual gas volume of

50\text{ cm}^3

.

Step-by-Step Solution

1

Determine combining volume ratios from the balanced chemical equation.

The mole/volume ratio is

2\text{ vol } NO : 1\text{ vol } O_2 : 2\text{ vol } NO_2

.

By Gay-Lussac's Law of Combining Volumes, gases at the same temperature and pressure react in volumes that bear simple whole-number ratios to one another and to gaseous products.

2

Identify the limiting reactant and calculate volumes consumed and produced.

40\text{ cm}^3

of

NO

requires

\frac{1}{2} \times 40 = 20\text{ cm}^3

of

O_2

, producing

40\text{ cm}^3

of

NO_2

.

NO

is completely used up.

40\text{ cm}^3

of

NO

is the limiting reactant because

30\text{ cm}^3

of

O_2

is available, which is more than the required

20\text{ cm}^3

.

3

Calculate remaining unreacted oxygen and total residual gas volume.

Excess

O_2 = 30\text{ cm}^3 - 20\text{ cm}^3 = 10\text{ cm}^3

. Total residual volume =

40\text{ cm}^3\,(NO_2) + 10\text{ cm}^3\,(O_2) = 50\text{ cm}^3

.

The total final volume equals the volume of gaseous product formed plus any unreacted excess gas.

Key Concept

Gay-Lussac's Law of Combining Volumes and Avogadro's Law

Estimated Time:1m 30s

Question 12886Question →

A student tested a sample of liquid ethanol in the laboratory. The literature boiling point of pure ethanol is $78.4^\circ\text{C}$ at $1\text{ atm}$ . When heated, the liquid sample began boiling at $76.2^\circ\text{C}$ and the temperature continuously rose until distillation finished at $81.5^\circ\text{C}$ . Which of the following conclusions best accounts for this experimental observation?

The sample is impure because a pure liquid boils at a sharp, constant temperature, whereas impurities broaden the boiling range.

The sample is pure because the average of the boiling range is approximately equal to the literature boiling point of pure ethanol.

The sample is pure because impurities only cause boiling to occur at a single lower fixed temperature.

The sample is impure because pure ethanol must boil at a temperature significantly higher than

85.0^\circ\text{C}

.

Show answer & explanation

Answer: The sample is impure because a pure liquid boils at a sharp, constant temperature, whereas impurities broaden the boiling range.

Answer

The sample is impure because a pure liquid boils at a sharp, constant temperature, whereas impurities broaden the boiling range.

A key criterion of chemical purity for any liquid is a sharp, fixed boiling point at constant pressure. When a liquid contains impurities, distillation occurs over a temperature interval (range) rather than at a single constant temperature.

Step-by-Step Solution

1

Recall the fundamental physical criteria for chemical purity.

Pure liquids boil at a sharp, constant temperature at standard pressure, whereas pure solids melt at a sharp temperature.

Physical constants such as boiling point and melting point are characteristic of pure substances.

2

Analyze the observed temperature profile during boiling.

The sample boils over a range from

76.2^\circ\text{C}

to

81.5^\circ\text{C}

rather than at a constant

78.4^\circ\text{C}

.

Impurities or mixed components alter phase change behavior, causing boiling to take place across a temperature interval.

3

Conclude the purity status of the ethanol sample.

The broad temperature range confirms that the ethanol sample is impure.

A constant boiling temperature is a required criterion for confirming liquid purity.

Key Concept

Criteria of Purity: Sharp Boiling and Melting Points

Question 12887Question →

A particle executing simple harmonic motion has a maximum speed of $3.0\text{ m/s}$ and a maximum acceleration of $12.0\text{ m/s}^2$ . What is the period of oscillation of the particle?

\frac{\pi}{4}\text{ s}

\frac{\pi}{2}\text{ s}

\pi\text{ s}

2\pi\text{ s}

Show answer & explanation

Answer: $\frac{\pi}{2}\text{ s}$

Answer

The period of oscillation of the particle is

\frac{\pi}{2}\text{ s}

.

For simple harmonic motion, maximum speed is

v_{\text{max}} = \omega A

and maximum acceleration is

a_{\text{max}} = \omega^2 A

. Dividing the maximum acceleration by the maximum speed gives

\omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{12.0}{3.0} = 4.0\text{ rad/s}

. Using the formula for the period

T = \frac{2\pi}{\omega}

, we find

T = \frac{2\pi}{4.0} = \frac{\pi}{2}\text{ s}

.

Step-by-Step Solution

1

Relate maximum speed and maximum acceleration to angular frequency

\omega = \frac{a_{\text{max}}}{v_{\text{max}}}

Since

v_{\text{max}} = \omega A

and

a_{\text{max}} = \omega^2 A

, dividing

a_{\text{max}}

by

v_{\text{max}}

eliminates the amplitude

A

and gives

\omega

.

2

Calculate the angular frequency

\omega

\omega = \frac{12.0\text{ m/s}^2}{3.0\text{ m/s}} = 4.0\text{ rad/s}

Substitute the given numerical values into the expression for angular frequency.

3

Calculate the period

T

T = \frac{2\pi}{\omega} = \frac{2\pi}{4.0} = \frac{\pi}{2}\text{ s}

The period of simple harmonic motion is related to angular frequency by

T = \frac{2\pi}{\omega}

.

Key Concept

Relationship between maximum velocity, maximum acceleration, angular frequency, and period in simple harmonic motion.

Estimated Time:1m 15s

Question 12888Question →

At a specific location on the Earth's surface, the horizontal component of the Earth's magnetic field is $3.2 \times 10^{-5}\text{ T}$ and the vertical component is also $3.2 \times 10^{-5}\text{ T}$ . What is the angle of dip at this location?

45^\circ

0^\circ

90^\circ

30^\circ

Show answer & explanation

Answer: $45^\circ$

Answer

The angle of dip at this location is

45^\circ

.

The angle of dip

\theta

is defined by

\tan\theta = \frac{B_v}{B_h}

. Since the vertical and horizontal components of Earth's magnetic field are equal, their ratio is equal to 1. The angle whose tangent is 1 is

45^\circ

.

Step-by-Step Solution

1

Identify the relationship between the horizontal component (

B_h

), vertical component (

B_v

), and angle of dip (

\theta

).

\tan\theta = \frac{B_v}{B_h}

The angle of dip

\theta

is the angle made by the Earth's total magnetic field vector with the horizontal.

2

Substitute the given values into the formula.

\tan\theta = \frac{3.2 \times 10^{-5}\text{ T}}{3.2 \times 10^{-5}\text{ T}} = 1

Both

B_v

and

B_h

are given as

3.2 \times 10^{-5}\text{ T}

.

3

Calculate the angle

\theta

.

\theta = \arctan(1) = 45^\circ

The inverse tangent of 1 is

45^\circ

.

Key Concept

Earth's Magnetic Field Components and Angle of Dip

Question 12889Question →

A major evolutionary advancement in the circulatory system of birds and mammals compared to amphibians is the complete separation of oxygenated and deoxygenated blood. Which structural feature of the heart is directly responsible for preventing the mixing of oxygenated and deoxygenated blood in mammals?

A completely divided four-chambered structure with an intact inter-ventricular septum

A three-chambered arrangement with a single undivided ventricle

A two-chambered configuration with one atrium and one ventricle

A three-chambered heart featuring a partially divided ventricle

Show answer & explanation

Answer: A completely divided four-chambered structure with an intact inter-ventricular septum

Answer

A completely divided four-chambered structure with an intact inter-ventricular septum is directly responsible for preventing the mixing of oxygenated and deoxygenated blood in mammals.

The mammalian heart has four distinct chambers (two atria and two ventricles). The muscular inter-ventricular septum completely divides the ventricular cavity into left and right sides, ensuring oxygenated blood from the lungs and deoxygenated blood from the body remain entirely separate.

Step-by-Step Solution

1

Analyze the circulatory requirement

Complete separation of oxygenated and deoxygenated blood requires double circulation where pulmonary and systemic circuits do not blend inside the heart.

Mixing of blood reduces oxygen delivery efficiency to metabolically active homoiothermic tissue.

2

Compare vertebrate heart chamber evolution

Fish have 2 chambers (single circulation); Amphibians have 3 chambers (2 atria, 1 ventricle, blood mixes); Reptiles have 3 chambers with an incomplete ventricular septum; Birds and Mammals have 4 distinct chambers.

The complete inter-ventricular septum in mammals physically isolates the right ventricle (pumping to lungs) from the left ventricle (pumping to body).

Key Concept

Vertebrate Comparative Heart Structure and Double Circulation

Question 12890Question →

A poultry breeder recorded two phenotypic traits in a flock of domestic chickens: comb shape (single, pea, rose, or walnut) and body weight at maturity. Which of the following statements correctly categorizes these two traits based on their pattern of variation and underlying genetic control?

Comb shape exhibits discontinuous variation controlled by major discrete genes, whereas body weight exhibits continuous variation influenced by polygenes and environmental factors.

Both comb shape and body weight exhibit continuous variation because both traits are influenced by environmental factors such as feed quality.

Comb shape demonstrates continuous variation due to multiple allele interactions, whereas body weight exhibits discontinuous variation due to single-gene inheritance.

Both comb shape and body weight demonstrate discontinuous variation because both traits fall into distinct phenotypic categories without intermediate forms.

Show answer & explanation

Answer: Comb shape exhibits discontinuous variation controlled by major discrete genes, whereas body weight exhibits continuous variation influenced by polygenes and environmental factors.

Answer

Comb shape exhibits discontinuous variation controlled by major discrete genes, whereas body weight exhibits continuous variation influenced by polygenes and environmental factors.

The option stating that comb shape exhibits discontinuous variation controlled by major discrete genes while body weight exhibits continuous variation influenced by polygenes and environmental factors is correct. Comb shape is a qualitative trait displaying clear-cut, non-overlapping phenotypes (discontinuous variation) under the control of specific gene loci. In contrast, body weight is a quantitative trait exhibiting a smooth gradient of intermediate phenotypes (continuous variation) governed by multiple additive genes (polygenes) interacting with nutrition and environment.

Step-by-Step Solution

1

Analyze the nature of comb shape phenotype distribution.

Comb shape presents clear, distinct, non-overlapping categories (single, pea, rose, walnut) without intermediate gradations.

Discontinuous variation is characterized by qualitative traits controlled by monogenic or oligogenic inheritance with little to no environmental effect.

2

Analyze the nature of body weight phenotype distribution.

Body weight presents a continuous spectrum of quantitative values that can be measured and plotted on a bell-shaped curve.

Continuous variation is characterized by polygenic inheritance where multiple additive genes interact alongside environmental influences.

3

Synthesize the correct classification.

Comb shape is a discontinuous trait, while body weight is a continuous trait.

This correctly pairs the phenotypic distribution pattern with its underlying genetic mechanism.

Key Concept

Continuous vs Discontinuous Variation and Genetic Control

Question 12891Question →

An element $X$ forms two distinct gaseous oxides, Oxide A and Oxide B. Quantitative analysis reveals that $14.0\text{ g}$ of $X$ combines with $16.0\text{ g}$ of oxygen in Oxide A, whereas $14.0\text{ g}$ of $X$ combines with $32.0\text{ g}$ of oxygen in Oxide B. What is the ratio of the masses of oxygen combining with a fixed mass of element $X$ , and which law of chemical combination does this illustrate?

1:2

; Law of Multiple Proportions

1:2

; Law of Definite Proportions

7:8

; Law of Conservation of Mass

2:1

; Law of Combining Volumes

Show answer & explanation

Answer: $1:2$ ; Law of Multiple Proportions

Answer

The ratio of oxygen masses is

1:2

, which illustrates the Law of Multiple Proportions.

For a constant mass of element

X

(

14.0\text{ g}

), the mass of oxygen in Oxide A (

16.0\text{ g}

) and Oxide B (

32.0\text{ g}

) forms a simple whole-number ratio of

16:32 = 1:2

. This directly satisfies John Dalton's Law of Multiple Proportions.

Step-by-Step Solution

1

Identify the fixed mass of element

X

in both compounds

Mass of element

X = 14.0\text{ g}

in both Oxide A and Oxide B

To apply the Law of Multiple Proportions, the mass of one element must be held constant

2

Determine the masses of oxygen combined with the fixed mass of element

X

Oxide A has

16.0\text{ g}

of oxygen, Oxide B has

32.0\text{ g}

of oxygen

These are the given masses of oxygen reacting with

14.0\text{ g}

of

X

3

Calculate the simple ratio between the masses of oxygen

\frac{16.0}{32.0} = \frac{1}{2} \text{ or } 1:2

Dividing both masses by the common factor

16.0\text{ g}

yields a simple whole-number ratio

4

Match the observed phenomenon to the appropriate chemical law

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

Key Concept

Law of Multiple Proportions

Estimated Time:1m 30s

Question 12892Question →

An $\alpha$ -particle (charge $+2e$ ) and a $\beta^-$ -particle (charge $-e$ ) emitted from a natural radioactive source enter a region of uniform electric field $E$ perpendicularly with equal initial kinetic energies. If $y_\alpha$ and $y_\beta$ represent the magnitudes of their transverse deflections after traveling the same horizontal distance through the field, what is the ratio $\frac{y_\beta}{y_\alpha}$ ?

\frac{1}{2}

2

\frac{m_\alpha}{2 m_\beta}

1

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The ratio of the transverse deflections

\frac{y_\beta}{y_\alpha}

is

\frac{1}{2}

.

For a charged particle entering a uniform electric field perpendicularly with kinetic energy

K

, the transverse deflection is

y = \frac{|q| E L^2}{4K}

. Because both particles enter with identical kinetic energies and travel the same horizontal distance

L

, the deflection is directly proportional to the magnitude of charge

|q|

and independent of mass. Since the

\beta^-

-particle has charge magnitude

e

and the

\alpha

-particle has charge magnitude

2e

, the ratio of their deflections is

\frac{e}{2e} = \frac{1}{2}

.

Step-by-Step Solution

1

Express initial horizontal velocity in terms of kinetic energy

K

and mass

m

.

v_x = \sqrt{\frac{2K}{m}}

, so time spent in the electric field of horizontal length

L

is

t = \frac{L}{v_x} = L \sqrt{\frac{m}{2K}}

.

Both particles enter horizontally with equal initial kinetic energy

K

.

2

Derive the formula for transverse deflection

y

in a uniform electric field

E

.

The transverse force is

F_y = |q|E

, giving acceleration

a_y = \frac{|q|E}{m}

. The deflection is

y = \frac{1}{2} a_y t^2 = \frac{1}{2} \left(\frac{|q|E}{m}\right) \left(\frac{L^2 m}{2K}\right) = \frac{|q|E L^2}{4K}

.

The particle mass

m

cancels out when time is expressed in terms of kinetic energy.

3

Calculate

y_\alpha

and

y_\beta

using their respective charge magnitudes

|q_\alpha| = 2e

and

|q_\beta| = e

.

y_\alpha = \frac{2e E L^2}{4K} = \frac{e E L^2}{2K}

and

y_\beta = \frac{e E L^2}{4K}

.

An

\alpha

-particle carries a charge of

+2e

while a

\beta^-

-particle carries a charge of

-e

.

4

Compute the ratio

\frac{y_\beta}{y_\alpha}

.

\frac{y_\beta}{y_\alpha} = \frac{\frac{e E L^2}{4K}}{\frac{e E L^2}{2K}} = \frac{1}{2}

.

Dividing

y_\beta

by

y_\alpha

cancels all terms except the ratio of charge magnitudes.

Key Concept

Deflection of charged radioactive emissions in a uniform electric field under equal kinetic energy

Question 12893Question →

A battery of electromotive force (e.m.f.) $24\,\text{V}$ and internal resistance $2\,\Omega$ is connected across a parallel network consisting of two resistors of resistances $6\,\Omega$ and $12\,\Omega$ . What is the total electrical energy, in Joules, dissipated in the external circuit during an operating time of $5\,\text{minutes}$ ?

Show answer & explanation

Answer: 19200

Answer

The total electrical energy dissipated in the external circuit over 5 minutes is

19,200\,\text{J}

.

To compute the external energy dissipated, first combine the parallel resistors to get an equivalent external resistance of

4\,\Omega

. Adding the

2\,\Omega

internal resistance yields a total circuit resistance of

6\,\Omega

, which draws

4\,\text{A}

of current from the

24\,\text{V}

battery. The power delivered to the external load is

P_{ext} = I^2 R_p = 4^2 \times 4 = 64\,\text{W}

. Multiplying this power by the time duration in seconds (

5 \times 60 = 300\,\text{s}

) yields

19,200\,\text{J}

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the external parallel circuit

R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\,\Omega

The two external resistors are connected in parallel.

2

Calculate the total current supplied by the battery

I = \frac{E}{R_p + r} = \frac{24}{4 + 2} = \frac{24}{6} = 4\,\text{A}

Ohm's law for a complete circuit accounts for both external resistance and internal resistance.

3

Determine the power dissipated exclusively in the external circuit

P_{ext} = I^2 R_p = (4)^2 \times 4 = 16 \times 4 = 64\,\text{W}

Electrical power dissipated across the external load depends on the total current squared times the equivalent external resistance.

4

Convert time from minutes to seconds and calculate total energy dissipated

t = 5 \times 60 = 300\,\text{s}

,

E = P_{ext} \times t = 64 \times 300 = 19,200\,\text{J}

Electrical energy is the product of power in Watts and time in seconds.

Key Concept

Electrical Energy and Power in Circuits with Internal Resistance

Question 12894Question →

A light ray enters the first face of a glass prism surrounded by air. The prism has a refracting angle of $75^\circ$ and a refractive index of $\sqrt{2}$ . Inside the prism, the ray strikes the second refracting face at the critical angle for total internal reflection. Calculate the angle of incidence, in degrees, at the first face.

Show answer & explanation

Answer: 45

Answer

The angle of incidence at the first face is

45^\circ

.

To find the angle of incidence at the first face, first determine the critical angle

C

at the second glass-air surface using

\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}

, which yields

C = 45^\circ

. Since the ray strikes the second face at this critical angle,

r_2 = 45^\circ

. Next, using the geometric relationship for a prism

A = r_1 + r_2

, the angle of refraction at the first surface is

r_1 = A - r_2 = 75^\circ - 45^\circ = 30^\circ

. Finally, applying Snell's law at the first face gives

\sin i = n \sin r_1 = \sqrt{2} \sin 30^\circ = \frac{\sqrt{2}}{2}

. Taking the inverse sine yields

i = 45^\circ

.

Step-by-Step Solution

1

Find the critical angle at the second face

Critical angle

C = 45^\circ

, so

r_2 = 45^\circ

Light travels from glass to air at the critical angle, so

\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}

.

2

Determine the angle of refraction at the first face

r_1 = 30^\circ

The apex angle of a prism satisfies

A = r_1 + r_2

, hence

r_1 = A - r_2 = 75^\circ - 45^\circ = 30^\circ

.

3

Apply Snell's Law at the entry boundary

\sin i = \frac{\sqrt{2}}{2}

Refraction at the first surface gives

\sin i = n \sin r_1 = \sqrt{2} \sin 30^\circ = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2}

.

4

Solve for the incident angle

i

i = 45^\circ

\arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

.

Key Concept

Refraction through a prism combined with total internal reflection critical angle condition

Question 12895Question →

In Nigeria, transitioning northward from the tropical rainforest into the Guinea savanna zone presents organisms with prolonged dry periods and frequent seasonal fires. Which of the following morphological adaptations is most characteristic of dominant tree species established in the Guinea savanna biome?

Thick, corky fire-resistant bark and deciduous leaves shed during the dry season

Broad evergreen leaves with drip tips and prominent supporting buttress roots

Specialized respiratory roots (pneumatophores) projecting above waterlogged muddy substrate

Complete reduction of leaves into rigid spines with thick succulent stems for water storage

Show answer & explanation

Answer: Thick, corky fire-resistant bark and deciduous leaves shed during the dry season

Answer

Thick, corky fire-resistant bark and deciduous leaves shed during the dry season

The correct option highlights the twin challenges of the Guinea savanna: dry season drought and seasonal fires. Thick corky bark provides insulation against heat damage from grass fires, and shedding leaves reduces transpirational water loss during dry months.

Step-by-Step Solution

1

Analyze the environmental stresses of the Guinea savanna biome

The Guinea savanna experiences a distinct dry season lasting 4 to 7 months accompanied by frequent grass fires.

Plant adaptations must target water conservation and thermal protection against fire.

2

Match environmental stresses with appropriate plant adaptations

Thick, corky bark insulates cambium cells from fire heat, while deciduous leaf-shedding stops transpirational loss during dry spells.

This combination ensures survival in tropical savanna climates.

Key Concept

Adaptive features of plants in Nigerian terrestrial biomes (Guinea Savanna)

Estimated Time:1m 0s

Question 12896Question →

An electric heater rated at $1.0\text{ kW}$ is immersed in $0.20\text{ kg}$ of water at its boiling point of $100^\circ\text{C}$ . Assuming all the energy supplied by the heater is used to vaporize the water and there is no heat loss to the surroundings, how long will it take for all the water to completely turn into steam? (Take the specific latent heat of vaporization of water $L_v = 2.25 \times 10^6\text{ J kg}^{-1}$ )

450\text{ s}

45,000\text{ s}

84\text{ s}

2250\text{ s}

Show answer & explanation

Answer: $450\text{ s}$

Answer

The total time required to completely vaporize the water is

450\text{ s}

.

At the boiling point of

100^\circ\text{C}

, all thermal energy supplied goes into changing the liquid water to steam without increasing its temperature. The required heat is

Q = m L_v = 0.20\text{ kg} \times (2.25 \times 10^6\text{ J kg}^{-1}) = 450,000\text{ J}

. Dividing energy by the heater power (

1000\text{ W}

) yields

450\text{ s}

.

Step-by-Step Solution

1

Calculate the total thermal energy required to vaporize the water at constant temperature.

Q = m L_v = 0.20\text{ kg} \times (2.25 \times 10^6\text{ J kg}^{-1}) = 4.5 \times 10^5\text{ J} = 450,000\text{ J}

Phase change at the boiling point occurs at constant temperature, so only latent heat is required.

2

Convert the power of the heater from kilowatts to watts.

P = 1.0\text{ kW} = 1000\text{ W} = 1000\text{ J s}^{-1}

Standard SI unit for electrical power is watts (joules per second).

3

Determine the time required using the relationship between energy, power, and time.

t = \frac{Q}{P} = \frac{450,000\text{ J}}{1000\text{ W}} = 450\text{ s}

Power is the rate of energy transfer (

P = \frac{Q}{t}

).

Key Concept

Latent heat of vaporization and electrical power heat transfer at constant temperature during a phase change.

Estimated Time:1m 30s

Question 12897Question →

A botanical survey identifies a flowering plant species characterized by brightly colored petals, sweet fragrance, and sugary nectar-secreting glands. Which of the following structural modifications would also be present in this flower to complement its mode of pollination?

Sticky, compact stigmas located inside the flower crown

Feathery stigmas hanging freely outside the floral envelope

Smooth, lightweight pollen grains produced in vast numbers

Pendulous stamens with versatile anthers exposed to air currents

Show answer & explanation

Answer: Sticky, compact stigmas located inside the flower crown

Answer

Sticky, compact stigmas located inside the flower crown

Bright petals, fragrance, and nectar are distinct adaptations for insect pollination. Flowers relying on insects feature sticky, compact stigmas located within the floral crown so that pollen from visiting insects readily adheres to the female reproductive organ.

Step-by-Step Solution

1

Identify the primary mode of pollination from the given floral traits.

Brightly colored petals, fragrance, and nectar glands indicate insect pollination (entomophily).

Visual attractions, scent, and food rewards are specialized mechanisms to attract animal vectors like insects.

2

Deduce corresponding structural adaptations required for entomophily.

The flower requires sticky or lobed stigmas positioned inside where visiting insects brush past them.

A sticky surface ensures pollen grains transferred from an insect's body adhere effectively.

3

Distinguish entomophilous adaptations from anemophilous features.

Features such as feathery stigmas, smooth light pollen, and versatile anthers are specialized for wind pollination.

Wind-pollinated flowers rely on air currents rather than animal vectors, requiring high exposure and aerodynamic pollen.

Key Concept

Floral Adaptations for Insect Pollination (Entomophily)

Estimated Time:1m 0s

Question 12898Question →

Match each chemical transformation or physical process on the left with its correct thermodynamic energy classification and enthalpy description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Sulfuric acid dissolution matches Exothermic with heat evolved; Calcium carbonate thermal decomposition matches Endothermic with thermal energy absorbed; Photosynthesis matches Endothermic with light energy absorbed; Ethanol combustion matches Exothermic with product bond formation releasing excess energy.

Each process is correctly linked to its thermodynamic behavior: acid dissolution and ethanol combustion release heat to surroundings (exothermic), while carbonate thermal decomposition and photosynthesis require net absorption of external energy (endothermic).

Step-by-Step Solution

1

Classify processes that release heat to the environment

The dissolution of concentrated acid and the combustion of ethanol raise the temperature of their surroundings, identifying them as exothermic reactions with negative enthalpy changes (

\Delta H < 0

).

Exothermic changes transfer thermal energy outward from the system.

2

Classify processes that require continuous energy absorption

The thermal breakdown of limestone and photosynthesis take in thermal or radiant energy from the outside, classifying them as endothermic reactions with positive enthalpy changes (

\Delta H > 0

).

Endothermic processes require net energy input to proceed.

Key Concept

Distinction between exothermic (

\Delta H < 0

, energy released) and endothermic (

\Delta H > 0

, energy absorbed) processes in physical and chemical systems.

Estimated Time:1m 30s

Question 12899Question →

An electric water pump with an efficiency of $80\%$ lifts $60\text{ kg}$ of water vertically through a height of $10\text{ m}$ in $1\text{ minute}$ . What is the input power of the pump in watts? (Take $g = 10\text{ m s}^{-2}$ )

Show answer & explanation

Answer: 125

Answer

The input power of the pump is

125\text{ W}

.

The total gravitational potential energy gained by

60\text{ kg}

of water lifted

10\text{ m}

is

W = mgh = 60 \times 10 \times 10 = 6000\text{ J}

. Performed over

60\text{ seconds}

, the useful power output is

100\text{ W}

. Accounting for an efficiency of

80\%

(

0.80

), the required input power is

P_{\text{in}} = \frac{100\text{ W}}{0.80} = 125\text{ W}

.

Step-by-Step Solution

1

Calculate the useful work done to lift the water

W = mgh = 60\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 6000\text{ J}

The useful work done equals the gravitational potential energy gained by the lifted mass of water.

2

Determine the useful output power of the pump

P_{\text{out}} = \frac{W}{t} = \frac{6000\text{ J}}{60\text{ s}} = 100\text{ W}

Power is the rate at which work is done, and

1\text{ minute}

must be converted to

60\text{ seconds}

.

3

Calculate the required input power using efficiency

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{100\text{ W}}{0.80} = 125\text{ W}

Efficiency is defined as

\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}

, so

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}}

.

Key Concept

Work, Power, and Efficiency of a Pump System

Estimated Time:1m 30s

Question 12900Question →

A sample of gas is enclosed in a cylinder fitted with a frictionless piston at a constant temperature of $27^\circ\text{C}$ . The initial pressure and volume of the gas are $1.2 \times 10^5\text{ N/m}^2$ and $600\text{ cm}^3$ respectively. If the pressure on the gas is increased to $3.6 \times 10^5\text{ N/m}^2$ , what is the final volume of the gas?

200\text{ cm}^3

1800\text{ cm}^3

5400\text{ cm}^3

432\text{ cm}^3

Show answer & explanation

Answer: $200\text{ cm}^3$

Answer

The final volume of the gas is

200\text{ cm}^3

.

According to Boyle's Law, the pressure and volume of a fixed mass of gas are inversely proportional at constant temperature (

P_1V_1 = P_2V_2

). Substituting the given values gives

(1.2 \times 10^5\text{ N/m}^2)(600\text{ cm}^3) = (3.6 \times 10^5\text{ N/m}^2)V_2

, which simplifies to

V_2 = 200\text{ cm}^3

.

Step-by-Step Solution

1

Identify the given quantities and state the relevant gas law formula.

P_1 = 1.2 \times 10^5\text{ N/m}^2

,

V_1 = 600\text{ cm}^3

,

P_2 = 3.6 \times 10^5\text{ N/m}^2

. Since temperature is constant, apply Boyle's Law:

P_1V_1 = P_2V_2

.

Boyle's Law states that for a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure.

2

Rearrange the equation to solve for the unknown final volume (

V_2

).

V_2 = \frac{P_1 V_1}{P_2}

Isolating

V_2

allows direct substitution of the known variables.

3

Substitute the values into the rearranged equation and compute the result.

V_2 = \frac{(1.2 \times 10^5\text{ N/m}^2) \times (600\text{ cm}^3)}{3.6 \times 10^5\text{ N/m}^2} = \frac{7.2 \times 10^7}{3.6 \times 10^5} = 200\text{ cm}^3

Carrying out the arithmetic yields the compressed gas volume.

Key Concept

Boyle's Law (

P_1V_1 = P_2V_2

at constant temperature)

Answer

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