All practice questions

Question 4401Question →

A cooperative society invested $\text{₦}80,000$ in a fixed deposit fund that pays compound interest at a rate of $10\%$ per annum compounded annually. What is the total compound interest earned by the cooperative society at the end of $2$ years?

\text{₦}16,000

\text{₦}16,800

\text{₦}96,000

\text{₦}96,800

Show answer & explanation

Answer: $\text{₦}16,800$

Answer

\text{₦}16,800

Using the compound interest formula for total amount,

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

, with principal

P = \text{₦}80,000

, rate

r = 10\%

, and time

n = 2

years:

A = 80000 \times (1.1)^2 = \text{₦}96,800

. The interest earned is the difference between the total amount and the principal:

I = 96800 - 80000 = \text{₦}16,800

.

Step-by-Step Solution

1

Calculate the total accumulated amount

A

using the compound interest formula

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

A = 80000 \times \left(1 + \frac{10}{100}\right)^2 = 80000 \times (1.1)^2 = 80000 \times 1.21 = \text{₦}96,800

The formula yields the full final balance after compound growth over 2 years.

2

Subtract the initial principal

P

from the total amount

A

to find the compound interest earned

I

I = A - P = 96800 - 80000 = \text{₦}16,800

Interest earned represents only the financial growth beyond the starting capital.

Key Concept

Compound Interest and Total Accumulated Amount

Question 4402Question →

Complete the sentence below by filling in the blank with the appropriate specialized medical term.

Fill in the blanks below

During a routine cardiovascular examination, the physician performed to listen to the internal sounds of the patient's heart and lungs using a stethoscope.

Show answer & explanation

Answer

auscultation

The word 'auscultation' is the precise medical register term for the act of listening to internal body sounds, particularly using a stethoscope to examine cardiac and respiratory functions.

Step-by-Step Solution

1

Analyze the context of the sentence to determine the relevant professional domain and action.

The setting is medical clinical examination, specifically the act of listening to internal bodily organs like the heart and lungs using a stethoscope.

Specialized registers require choosing the precise technical word assigned to a specific professional procedure.

2

Identify the precise medical register term for listening to organ sounds.

The correct technical term is 'auscultation'.

'Auscultation' is derived from Latin meaning 'to listen', and specifically denotes diagnostic listening to body sounds.

Key Concept

Occupational and Specialized Registers: Medical Terminology

Question 4403Question →

A container holds a large number of red, blue, and yellow counters. The theoretical probability of selecting a red counter at random is $\frac{2}{5}$ , and the theoretical probability of selecting a blue counter is $\frac{1}{3}$ . In an experiment where a counter is drawn and replaced $300$ times, a yellow counter is selected $72$ times. What is the absolute difference between the theoretical expected number of yellow counters and the experimental frequency of yellow counters obtained?

8

148

28

80

Show answer & explanation

Answer: $8$

Answer

The absolute difference between the theoretical expected number of yellow counters and the experimental frequency is

8

.

The theoretical probability of drawing a yellow counter is

1 - (\frac{2}{5} + \frac{1}{3}) = \frac{4}{15}

. For

300

trials, the theoretical expected count of yellow counters is

\frac{4}{15} \times 300 = 80

. Subtracting the experimental observed count of

72

from

80

gives an absolute difference of

8

.

Step-by-Step Solution

1

Calculate the theoretical probability of drawing a yellow counter.

P(\text{Yellow}) = 1 - \left(P(\text{Red}) + P(\text{Blue})\right) = 1 - \left(\frac{2}{5} + \frac{1}{3}\right) = 1 - \frac{11}{15} = \frac{4}{15}

The sum of probabilities of all mutually exclusive outcomes in the sample space must equal

1

.

2

Calculate the theoretical expected frequency of yellow counters in

300

trials.

\text{Expected frequency} = P(\text{Yellow}) \times \text{Total trials} = \frac{4}{15} \times 300 = 80

The expected number of occurrences of an outcome is the product of its theoretical probability and the number of trials.

3

Find the absolute difference between expected frequency and experimental frequency.

|80 - 72| = 8

The question asks for the difference between the theoretical expected value (

80

) and the actual experimental result (

72

).

Key Concept

Experimental and Theoretical Probability

Estimated Time:1m 30s

Question 4404Question →

A rectangular wire loop is pulled horizontally to the right out of a region containing a uniform magnetic field directed perpendicularly into the page. According to Lenz's law, what is the direction of the induced current in the loop and the direction of the resulting magnetic force acting on the loop?

Clockwise induced current; magnetic force directed to the left

Counter-clockwise induced current; magnetic force directed to the left

Clockwise induced current; magnetic force directed to the right

Counter-clockwise induced current; magnetic force directed to the right

Show answer & explanation

Answer: Clockwise induced current; magnetic force directed to the left

Answer

Clockwise induced current; magnetic force directed to the left

According to Lenz's law, an induced electric current flows in a direction such that its magnetic field opposes the change in magnetic flux that produced it. As the loop is pulled to the right, the magnetic flux pointing into the page decreases. The loop responds by inducing a current that creates additional magnetic field into the page to resist this decrease. By the right-hand rule, a current circulating clockwise produces a magnetic field into the page. Additionally, the magnetic force created on the loop must oppose the external motion, acting to the left.

Step-by-Step Solution

1

Determine the change in magnetic flux passing through the loop

As the rectangular loop is pulled to the right out of the magnetic field region, the magnetic flux directed into the page through the loop is decreasing.

Electromagnetic induction depends on the rate of change of magnetic flux.

2

Apply Lenz's law to determine the direction of the induced magnetic field and current

To oppose the decrease in inward flux, the induced current must create its own magnetic field directed into the page. By the right-hand grip rule, an inward induced field corresponds to a clockwise current flow.

Lenz's law states that the direction of an induced current always opposes the change in magnetic flux causing it.

3

Determine the direction of the net magnetic force on the loop

The induced magnetic force must oppose the mechanical motion pulling the loop to the right, so the net magnetic force acts to the left.

Lenz's law is a consequence of the conservation of energy, ensuring mechanical work must be done against electromagnetic forces.

Key Concept

Lenz's Law and Direction of Induced Current

Question 4405Question →

A rigid vessel of fixed volume contains an ideal gas at an initial pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Additional gas is pumped into the vessel until the total number of moles of gas is doubled. If the temperature of the gas increases to $87^\circ\text{C}$ , what is the final pressure of the gas in units of $10^5\text{ Pa}$ ?

Show answer & explanation

Answer: 3.6

Answer

The final pressure of the gas is

3.6 \times 10^5\text{ Pa}

(which is

3.6

in units of

10^5\text{ Pa}

).

Using the ideal gas equation

PV = nRT

at fixed volume, the ratio of final to initial pressure is given by

P_2/P_1 = (n_2/n_1) \times (T_2/T_1)

. Converting temperatures to Kelvin yields

T_1 = 300\text{ K}

and

T_2 = 360\text{ K}

. Given that the number of moles doubles (

n_2/n_1 = 2

), substituting the values gives

P_2 = 1.50 \times 10^5 \times 2 \times (360/300) = 3.60 \times 10^5\text{ Pa}

.

Step-by-Step Solution

1

Convert initial and final temperatures from Celsius to absolute temperature in Kelvin.

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 87 + 273 = 360\text{ K}

.

Gas laws require absolute temperatures in Kelvin for thermodynamic calculations.

2

Formulate the pressure relation from the ideal gas equation

PV = nRT

.

Since volume

V

and the universal gas constant

R

are constant,

P_2 = P_1 \times \frac{n_2}{n_1} \times \frac{T_2}{T_1}

.

Pressure is directly proportional to both the number of moles and the absolute temperature when volume is fixed.

3

Substitute the mole ratio

n_2/n_1 = 2

, initial pressure, and Kelvin temperatures to compute

P_2

.

P_2 = 1.50 \times 10^5 \times 2 \times \frac{360}{300} = 3.60 \times 10^5\text{ Pa}

.

Evaluates the final gas pressure in the requested numerical units.

Key Concept

Ideal Gas Equation and Variable Moles under Constant Volume

Question 4406Question →

An electric heater rated at $690\text{ W}$ is immersed in a thermally insulated container holding a mixture of $0.50\text{ kg}$ of ice and $0.50\text{ kg}$ of liquid water in equilibrium at $0^\circ\text{C}$ . The heater is operated for $5.0\text{ minutes}$ . Assuming negligible heat capacity for the container and no heat loss to the surroundings, what is the final equilibrium temperature of the mixture?
(Specific latent heat of fusion of ice $L_f = 3.3 \times 10^5\text{ J kg}^{-1}$ , specific heat capacity of water $c_w = 4200\text{ J kg}^{-1}\text{ K}^{-1}$ )

0^\circ\text{C}

10^\circ\text{C}

20^\circ\text{C}

5^\circ\text{C}

Show answer & explanation

Answer: $10^\circ\text{C}$

Answer

The final equilibrium temperature of the mixture is

10^\circ\text{C}

.

The total energy supplied by the

690\text{ W}

heater in

300\text{ s}

is

207,000\text{ J}

. Melting all

0.50\text{ kg}

of ice at

0^\circ\text{C}

requires

165,000\text{ J}

. The remaining

42,000\text{ J}

heats the combined

1.00\text{ kg}

of water (initial water + melted ice) through

\Delta T = \frac{42,000}{1.00 \times 4200} = 10^\circ\text{C}

, reaching a final equilibrium temperature of

10^\circ\text{C}

.

Step-by-Step Solution

1

Calculate the total thermal energy provided by the electric heater.

Q_{\text{total}} = P \times t = 690\text{ W} \times (5.0 \times 60\text{ s}) = 207,000\text{ J}

.

Power multiplied by time yields total energy supplied.

2

Determine the energy required to completely melt the

0.50\text{ kg}

of ice at

0^\circ\text{C}

.

Q_{\text{melt}} = m_{\text{ice}} \times L_f = 0.50\text{ kg} \times 330,000\text{ J kg}^{-1} = 165,000\text{ J}

.

Latent heat of fusion changes state from solid ice to liquid water at constant temperature

0^\circ\text{C}

.

3

Calculate the remaining thermal energy available to increase the temperature.

Q_{\text{rem}} = 207,000\text{ J} - 165,000\text{ J} = 42,000\text{ J}

.

After complete melting, excess energy goes into sensible heating.

4

Calculate the total mass of liquid water and the resulting temperature rise.

m_{\text{total}} = 0.50\text{ kg (melted ice)} + 0.50\text{ kg (initial water)} = 1.00\text{ kg}

.

\Delta T = \frac{Q_{\text{rem}}}{m_{\text{total}} \times c_w} = \frac{42,000\text{ J}}{1.00\text{ kg} \times 4200\text{ J kg}^{-1}\text{ K}^{-1}} = 10^\circ\text{C}

.

All water now absorbs energy to raise the temperature.

Key Concept

Phase changes occur at constant temperature (latent heat

Q = mL

). Once the phase change is complete, additional thermal energy increases temperature as sensible heat (

Q = mc\Delta T

) using the total combined mass.

Question 4407Question →

Two point charges of $+2.0 \times 10^{-6}\text{ C}$ and $+4.0 \times 10^{-6}\text{ C}$ are placed in a vacuum at a distance of $0.3\text{ m}$ apart. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electrostatic force exerted between them in Newtons?

Show answer & explanation

Answer: 0.8

Answer

The magnitude of the electrostatic force between the charges is 0.8 N.

According to Coulomb's Law, the force between two point charges is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance between them:

F = \frac{k q_1 q_2}{r^2}

. Substituting

q_1 = 2.0 \times 10^{-6}\text{ C}

,

q_2 = 4.0 \times 10^{-6}\text{ C}

,

r = 0.3\text{ m}

, and

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

into the formula gives

F = 0.8\text{ N}

.

Step-by-Step Solution

1

Identify known quantities from the problem statement

q_1 = 2.0 \times 10^{-6}\text{ C}

,

q_2 = 4.0 \times 10^{-6}\text{ C}

,

r = 0.3\text{ m}

,

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

Clear identification of parameters is required for substitution into Coulomb's Law.

2

Apply Coulomb's Law formula

F = \frac{k q_1 q_2}{r^2}

Coulomb's Law quantifies the electrostatic force between two stationary point charges.

3

Substitute values and perform arithmetic calculation

F = \frac{9.0 \times 10^9 \times (2.0 \times 10^{-6}) \times (4.0 \times 10^{-6})}{0.09} = \frac{0.072}{0.09} = 0.8\text{ N}

Squaring the separation distance

0.3\text{ m}

gives

0.09\text{ m}^2

, and evaluating the numerator gives

0.072\text{ N}\cdot\text{m}^2

.

Key Concept

Coulomb's Law

Question 4408Question →

A metal rod of length $0.4\text{ m}$ glides at a constant speed of $5.0\text{ m s}^{-1}$ to the right along horizontal parallel conducting rails placed in a uniform magnetic field of $0.5\text{ T}$ directed vertically into the page. The rails are connected at their left end by a $2.0\ \Omega$ resistor. What is the magnitude of the external force required to maintain the uniform speed of the rod, and in which direction does the induced magnetic force act on the rod?

0.10\text{ N}

, directed to the left

0.10\text{ N}

, directed to the right

0.50\text{ N}

, directed to the left

0.50\text{ N}

, directed to the right

Show answer & explanation

Answer: $0.10\text{ N}$ , directed to the left

Answer

0.10\text{ N}

, directed to the left

The motional e.m.f. generated across the moving conductor is

\mathcal{E} = BLv = 0.5 \times 0.4 \times 5.0 = 1.0\text{ V}

. The induced current is

I = \mathcal{E}/R = 1.0 / 2.0 = 0.5\text{ A}

. The magnetic force resisting the rod's motion is

F_B = BIL = 0.5 \times 0.5 \times 0.4 = 0.10\text{ N}

. By Lenz's law, this magnetic force acts to the left, opposing the motion to the right. To maintain uniform speed, an equal external force of

0.10\text{ N}

must be applied.

Step-by-Step Solution

1

Calculate the induced electromotive force (e.m.f.) across the moving rod

\mathcal{E} = B L v = 0.5\text{ T} \times 0.4\text{ m} \times 5.0\text{ m s}^{-1} = 1.0\text{ V}

Motional e.m.f. is produced when a conductor cuts magnetic flux lines at a perpendicular velocity.

2

Calculate the induced current flowing in the circuit

I = \frac{\mathcal{E}}{R} = \frac{1.0\text{ V}}{2.0\ \Omega} = 0.5\text{ A}

Ohm's law relates the induced e.m.f. and total circuit resistance.

3

Determine the magnitude of the magnetic force acting on the current-carrying rod

F_B = B I L = 0.5\text{ T} \times 0.5\text{ A} \times 0.4\text{ m} = 0.10\text{ N}

A magnetic field exerts a force on a straight conductor carrying current.

4

Determine the direction of the magnetic force using Lenz's law and Newton's first law

The magnetic force acts to the left (opposing motion to the right). An external force of equal magnitude (

0.10\text{ N}

) to the right is required to maintain constant speed.

Lenz's law states that induced effects always oppose the change causing them (motion to the right).

Key Concept

Motional Electromotive Force, Magnetic Force on a Conductor, and Lenz's Law

Question 4409Question →

An inclined plane of length $5\text{ m}$ is used to lift a load of $400\text{ N}$ through a vertical height of $1\text{ m}$ . If an effort of $100\text{ N}$ is applied parallel to the incline, what is the efficiency of the machine?

80\%

125\%

25\%

20\%

Show answer & explanation

Answer: $80\%$

Answer

The efficiency of the machine is

80\%

.

The correct answer of

80\%

is obtained by finding the velocity ratio (distance moved by effort divided by distance moved by load,

5 / 1 = 5

) and the mechanical advantage (load divided by effort,

400 / 100 = 4

), then calculating efficiency as

(\text{MA} / \text{VR}) \times 100\% = (4 / 5) \times 100\% = 80\%

.

Step-by-Step Solution

1

Calculate the Velocity Ratio (VR) of the inclined plane

\text{VR} = \frac{\text{Length of incline}}{\text{Height}} = \frac{5\text{ m}}{1\text{ m}} = 5

Velocity Ratio is the distance moved by the effort divided by the distance moved by the load.

2

Calculate the Mechanical Advantage (MA)

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{400\text{ N}}{100\text{ N}} = 4

Mechanical Advantage measures how many times a machine multiplies the applied force.

3

Calculate the Efficiency

\text{Efficiency} = \left(\frac{\text{MA}}{\text{VR}}\right) \times 100\% = \left(\frac{4}{5}\right) \times 100\% = 80\%

Efficiency is defined as the ratio of Mechanical Advantage to Velocity Ratio expressed as a percentage.

Key Concept

Efficiency of Simple Machines (Inclined Plane)

Estimated Time:45s

Question 4410Question →

A hypermetropic eye has its unassisted near point situated at $75\text{ cm}$ from the eye. Calculate the focal length, in $\text{cm}$ , of the converging spectacle lens needed to enable the person to read a book comfortably at a distance of $25\text{ cm}$ from the eye.

Show answer & explanation

Answer: 37.5

Answer

The focal length of the required converging lens is

37.5\text{ cm}

.

To correct hypermetropia, the spectacle lens must form a virtual image of an object located at the desired near point (

u = +25\text{ cm}

) at the eye's actual, unassisted near point (

v = -75\text{ cm}

). Substituting

u = +25\text{ cm}

and

v = -75\text{ cm}

into the thin lens formula

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

gives

\frac{1}{f} = \frac{1}{25} - \frac{1}{75} = \frac{2}{75}\text{ cm}^{-1}

. Taking the reciprocal yields

f = 37.5\text{ cm}

.

Step-by-Step Solution

1

Determine the object distance and image distance with sign conventions

u = +25\text{ cm}

and

v = -75\text{ cm}

.

The object is placed at the normal reading distance (

25\text{ cm}

), and the spectacle lens creates a virtual image on the same side of the lens at the person's near point (

75\text{ cm}

).

2

Set up the thin lens formula

\frac{1}{f} = \frac{1}{25} - \frac{1}{75}.

The thin lens formula relates the focal length to object and image distances.

3

Calculate the focal length

f

f = 37.5\text{ cm}.

Simplifying

\frac{3 - 1}{75} = \frac{2}{75}

and taking the reciprocal yields

f = \frac{75}{2} = 37.5\text{ cm}

.

Key Concept

Correction of hypermetropia (farsightedness) using thin lens formula with virtual image sign convention

Question 4411Question →

A variable force $F$ acts on a body moving along a straight horizontal path. The force increases linearly from $0\text{ N}$ at position $x = 0\text{ m}$ to $20\text{ N}$ at $x = 4\text{ m}$ , remains constant at $20\text{ N}$ from $x = 4\text{ m}$ to $x = 7\text{ m}$ , and then decreases linearly back to $0\text{ N}$ at $x = 10\text{ m}$ . What is the total work done by the force in Joules over the $10\text{ m}$ displacement?

Show answer & explanation

Answer: 130

Answer

130 J

The work done by a variable force is equal to the total area under the force-displacement (

F

-

x

) graph. The region under the graph forms a trapezoid bounded by parallel sides of length

10\text{ m}

(total displacement) and

3\text{ m}

(constant force interval), with a height of

20\text{ N}

. Calculating the area yields

W = \frac{1}{2}(10 + 3) \times 20 = 130\text{ J}

.

Step-by-Step Solution

1

Relate work done to the force-displacement graph

Work done

W

equals the total area under the

F

-

x

graph between

x = 0\text{ m}

and

x = 10\text{ m}

.

By definition,

W = \int F \, dx

, which corresponds to the geometric area under the force-displacement curve.

2

Calculate the geometric area under each section of the graph

First section (

0

to

4\text{ m}

): Triangle area =

\frac{1}{2} \times 4 \times 20 = 40\text{ J}

. Second section (

4

to

7\text{ m}

): Rectangle area =

(7 - 4) \times 20 = 60\text{ J}

. Third section (

7

to

10\text{ m}

): Triangle area =

\frac{1}{2} \times (10 - 7) \times 20 = 30\text{ J}

.

Breaking down a complex piecewise curve into simple geometric shapes allows straightforward area evaluation without calculus.

3

Sum the areas of all sections

Total Work

W = 40\text{ J} + 60\text{ J} + 30\text{ J} = 130\text{ J}

.

The total work done is the scalar sum of the work done across each contiguous segment of displacement.

Key Concept

Work Done by a Variable Force (Area under Force-Displacement Graph)

Question 4412Question →

Two marksmen, Kemi and Chidi, independently shoot at a target once. The probability that Kemi hits the target is $0.7$ , and the probability that Chidi hits the target is $0.6$ . What is the probability that at least one of them hits the target?

Show answer & explanation

Answer: 0.88

Answer

The probability that at least one marksman hits the target is 0.88.

Since the two shooting attempts are independent, the probability that both marksmen miss is

(1 - 0.7) \times (1 - 0.6) = 0.3 \times 0.4 = 0.12

. Subtracting this probability from 1 yields

1 - 0.12 = 0.88

, which represents the probability that at least one marksman hits the target.

Step-by-Step Solution

1

Find the probability of each event not occurring (missing the target).

P(\text{Kemi misses}) = 0.3

and

P(\text{Chidi misses}) = 0.4

The sum of an event's probability and its complement is always 1.

2

Determine the probability of both events failing simultaneously.

P(\text{both miss}) = 0.3 \times 0.4 = 0.12

For independent events

A

and

B

,

P(A' \cap B') = P(A') \times P(B')

.

3

Use the complement rule to find the probability of at least one success.

P(\text{at least one hits}) = 1 - 0.12 = 0.88

The event 'at least one hits' is the exact complement of 'neither hits'.

Key Concept

Compound probability laws for independent events and complement of combined events

Question 4413Question →

A uniform metal rod of length $0.50\text{ m}$ and cross-sectional area $2.0 \times 10^{-3}\text{ m}^2$ has a thermal conductivity of $400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . If a temperature difference of $50\text{ K}$ is maintained between its ends, what is the rate of heat flow through the rod?

80\text{ W}

40\text{ W}

20\text{ W}

160\text{ W}

Show answer & explanation

Answer: $80\text{ W}$

Answer

The rate of heat flow through the rod is

80\text{ W}

.

The rate of heat transfer by conduction is calculated using Fourier's law,

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

. Substituting

k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = 2.0 \times 10^{-3}\text{ m}^2

,

\Delta T = 50\text{ K}

, and

d = 0.50\text{ m}

yields

80\text{ W}

.

Step-by-Step Solution

1

Identify the given parameters for thermal conduction.

Thermal conductivity

k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

, cross-sectional area

A = 2.0 \times 10^{-3}\text{ m}^2

, temperature difference

\Delta T = 50\text{ K}

, and length

d = 0.50\text{ m}

.

These parameters form the components of Fourier's law of thermal conduction.

2

Apply the conduction rate formula

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

.

\frac{Q}{t} = \frac{400 \times (2.0 \times 10^{-3}) \times 50}{0.50}

Heat transfer per unit time depends directly on conductivity, area, and temperature difference, and inversely on thickness/length.

3

Calculate the numerical value.

\frac{Q}{t} = \frac{40}{0.50} = 80\text{ W}

Dividing the product of the numerator terms (

40\text{ J/s}\cdot\text{m}

) by length (

0.50\text{ m}

) yields the rate of energy transfer.

Key Concept

Thermal Conduction Rate (Fourier's Law)

Estimated Time:45s

Question 4414Question →

A crate on a smooth surface is simultaneously pulled by two horizontal forces. One force of magnitude $9.0\text{ N}$ acts towards the East, and another force of magnitude $12.0\text{ N}$ acts towards the North. What is the magnitude of the resultant force acting on the crate in newtons?

Show answer & explanation

Answer: 15

Answer

The magnitude of the resultant force acting on the crate is

15.0\text{ N}

.

Because the two pulling forces are perpendicular, their resultant magnitude is calculated using vector addition via the Pythagorean theorem:

R = \sqrt{9.0^2 + 12.0^2} = 15.0\text{ N}

.

Step-by-Step Solution

1

Determine the angle between the two given vectors

The forces are perpendicular (

90^\circ

).

East and North cardinal directions are orthogonal to each other.

2

Compute the resultant magnitude using vector synthesis

R = \sqrt{9.0^2 + 12.0^2} = \sqrt{81 + 144} = \sqrt{225} = 15.0\text{ N}

For perpendicular force vectors, the resultant magnitude is the hypotenuse of the right triangle formed by the vector components.

Key Concept

Vector addition of perpendicular forces using the Pythagorean theorem

Estimated Time:1m 15s

Question 4415Question →

Read the excerpt below carefully:

*During the tribunal hearing, the magistrate observed that the defendant's guilt was an open secret among the townspeople long before the trial commenced.*

Which figure of speech is expressed by the highlighted phrase, and what is its contextual meaning in the sentence?

An oxymoron indicating a fact that is officially unacknowledged yet known to almost everyone

A hyperbole suggesting that the defendant shouted a full confession publicly in the courtroom

A literal statement implying that confidential trial documents were left unlocked in public view

A metaphor describing a deeply mysterious occurrence that remains completely hidden from the public

Show answer & explanation

Answer: An oxymoron indicating a fact that is officially unacknowledged yet known to almost everyone

Answer

An oxymoron indicating a fact that is officially unacknowledged yet known to almost everyone

The phrase 'open secret' is an oxymoron because it combines two contradictory concepts: 'open' (publicly visible) and 'secret' (concealed). In English lexis and literary usage, this expression connotes a situation or piece of information that is officially kept quiet or unverified, yet is widely known and understood by the public.

Step-by-Step Solution

1

Identify the literary device used in the highlighted phrase 'open secret'.

The phrase places two contradictory terms side by side ('open', meaning public/unconcealed, and 'secret', meaning hidden/confidential). This juxtaposition of contradictory words forms an oxymoron.

Recognizing the structural relationship between opposing terms reveals the underlying figure of speech.

2

Analyze the contextual connotation of the phrase within the sentence.

In the context of the tribunal hearing, saying something is an 'open secret' means that while it is not formally announced or acknowledged, practically everyone in the community is already aware of it.

Figurative expressions must be interpreted according to their pragmatic contextual usage rather than literal word-for-word definitions.

Key Concept

Oxymoron and Figurative Connotation

Question 4416Question →

A sample of gas enclosed in a rigid container exerts a pressure of $3.0 \times 10^5 \text{ N m}^{-2}$ with a density of $0.40 \text{ kg m}^{-3}$ . If the gas is heated at constant volume until its pressure increases to $1.2 \times 10^6 \text{ N m}^{-2}$ , what is the final root-mean-square (r.m.s.) speed of the gas molecules?

3000 \text{ m s}^{-1}

6000 \text{ m s}^{-1}

1500 \text{ m s}^{-1}

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

Show answer & explanation

Answer: $3000 \text{ m s}^{-1}$

Answer

The final root-mean-square speed of the gas molecules is

3000 \text{ m s}^{-1}

.

Using the kinetic theory equation

P = \frac{1}{3}\rho v_{\text{rms}}^2

, the initial speed is

v_1 = \sqrt{\frac{3 \times 3.0 \times 10^5}{0.40}} = 1500 \text{ m s}^{-1}

. When heated at constant volume, density remains unchanged. The new pressure

1.2 \times 10^6 \text{ N m}^{-2}

is 4 times the initial pressure, so the new speed is

v_2 = \sqrt{4} \times v_1 = 2 \times 1500 \text{ m s}^{-1} = 3000 \text{ m s}^{-1}

.

Step-by-Step Solution

1

Express the relationship between pressure, density, and root-mean-square speed using kinetic theory.

P = \frac{1}{3}\rho v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}

According to the kinetic theory of gases, the pressure exerted by a gas is related to its density and molecular r.m.s. speed.

2

Calculate the initial root-mean-square speed

v_1

using initial pressure

P_1 = 3.0 \times 10^5 \text{ N m}^{-2}

and density

\rho = 0.40 \text{ kg m}^{-3}

.

v_1 = \sqrt{\frac{3 \times (3.0 \times 10^5)}{0.40}} = \sqrt{\frac{9.0 \times 10^5}{0.40}} = \sqrt{2.25 \times 10^6} = 1500 \text{ m s}^{-1}

Establishes the baseline molecular speed prior to heating.

3

Determine the final root-mean-square speed

v_2

after the pressure increases to

P_2 = 1.2 \times 10^6 \text{ N m}^{-2}

at constant volume.

v_2 = \sqrt{\frac{3 \times (1.2 \times 10^6)}{0.40}} = \sqrt{\frac{3.6 \times 10^6}{0.40}} = \sqrt{9.0 \times 10^6} = 3000 \text{ m s}^{-1}

Since the volume is rigid, density

\rho

remains constant while pressure increases due to heating.

Key Concept

Root-Mean-Square Speed and Pressure Relation in Kinetic Theory

Question 4417Question →

When a strong bar magnet is dropped vertically through a long, hollow copper tube, its downward acceleration is equal to the acceleration due to gravity ( $g$ ) because copper is a non-magnetic material.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. The magnet falls with an acceleration less than

g

because induced eddy currents in the copper tube create an upward magnetic force that opposes the motion.

The statement is false because the relative motion between the falling magnet and the conductive copper tube induces eddy currents. According to Lenz's law, these induced currents set up a magnetic field that opposes the falling magnet's motion, creating an upward retarding force that reduces the downward acceleration to a value less than

g

.

Step-by-Step Solution

1

Identify the physical interactions as the magnet falls through the tube.

The falling magnet creates a changing magnetic flux through the surrounding copper tube.

Relative motion between a magnetic field source and a conductor produces a time-varying magnetic flux in the conductor.

2

Apply Faraday's law of electromagnetic induction.

Electromotive force (e.m.f.) and circular eddy currents are induced in the conductive copper walls.

A changing magnetic flux induces electric currents in any closed conductive path.

3

Apply Lenz's law to determine the magnetic effect of the induced eddy currents.

The induced eddy currents produce a magnetic field that opposes the downward motion of the falling magnet, generating an upward magnetic force (

F_{\text{mag}}

).

Lenz's law dictates that an induced current always flows in a direction such that its magnetic field opposes the change causing it.

4

Analyze the net force and resulting acceleration.

The net downward force is

F_{\text{net}} = mg - F_{\text{mag}} < mg

, so the downward acceleration

a = g - \frac{F_{\text{mag}}}{m} < g

.

The presence of an upward magnetic force reduces the net downward acceleration below the free-fall value of

g

.

Key Concept

Lenz's Law and Eddy Currents in Conductors

Estimated Time:1m 0s

Question 4418Question →

A stretched string of length $0.50\text{ m}$ fixed at both ends vibrates in its third harmonic mode at a frequency of $450\text{ Hz}$ . What is the speed of the transverse wave along the string in $\text{m/s}$ ?

Show answer & explanation

Answer: 150

Answer

The speed of the transverse wave along the string is

150\text{ m/s}

.

For a string fixed at both ends, standing wave modes produce harmonics given by

f_n = \frac{n v}{2L}

. Given

n = 3

,

L = 0.50\text{ m}

, and

f_3 = 450\text{ Hz}

, substituting these into the equation yields

450 = \frac{3v}{2(0.50)} = 3v

, leading to

v = 150\text{ m/s}

.

Step-by-Step Solution

1

Identify the standing wave frequency equation for a string fixed at both ends.

The frequency of the

n

-th harmonic is

f_n = \frac{n v}{2L}

, where

n

is the harmonic number,

v

is the wave speed, and

L

is the string length.

Fixed ends require nodes at both boundaries, producing standing wave modes with wavelengths

\lambda_n = \frac{2L}{n}

.

2

Substitute the given physical quantities into the harmonic equation.

450 = \frac{3 \times v}{2 \times 0.50}

.

The question specifies

L = 0.50\text{ m}

, third harmonic mode (

n = 3

), and frequency

f_3 = 450\text{ Hz}

.

3

Solve for the wave speed

v

.

v = 150\text{ m/s}

.

Simplifying

2 \times 0.50 = 1.0

gives

3v = 450

, so

v = \frac{450}{3} = 150\text{ m/s}

.

Key Concept

Standing Waves and Harmonics in Vibrating Strings

Question 4419Question →

A uniform metallic wire of resistance $R$ is stretched uniformly until its radius decreases by $20\%$ . The stretched wire is subsequently cut into two equal halves, which are then connected in parallel across a constant potential difference $V$ . What is the ratio of the total electrical power dissipated in this parallel combination to the power dissipated by the original unstretched wire under the same potential difference?

1.64

0.61

2.56

0.41

Show answer & explanation

Answer: 1.64

Answer

The ratio of the total power dissipated in the parallel combination to the original power is 1.64

When a wire of initial resistance

R

is stretched so that its radius decreases by

20\%

, its new radius is

0.8r

. Because volume is conserved (

A_1 L_1 = A_2 L_2

), reducing the cross-sectional area to

0.64A

causes the length to increase to

L/0.64

. Consequently, the resistance scales inversely with the fourth power of the radius:

R_{\text{stretched}} = R / (0.8)^4 = R / 0.4096 = 2.4414R

. Cutting this wire into two equal pieces gives two resistors of

1.2207R

each. Connecting them in parallel yields an equivalent resistance

R_{\text{eq}} = 1.2207R / 2 = 0.61035R

. Power at constant voltage is

P = V^2/R

, so the new power is

P_{\text{new}} = V^2 / (0.61035R) = 1.64 (V^2/R) = 1.64 P_{\text{orig}}

.

Step-by-Step Solution

1

Determine the new resistance of the wire after stretching

R_{\text{stretched}} = \frac{R}{(0.8)^4} = \frac{R}{0.4096} \approx 2.4414 R

Since mass and density remain constant, volume

V_{\text{vol}} = A \cdot L

is conserved. Decreasing radius to

r_2 = 0.8 r_1

reduces area to

A_2 = 0.64 A_1

and increases length to

L_2 = L_1 / 0.64

. Resistance

R = \rho L / A \propto 1/r^4

.

2

Calculate the equivalent resistance of the two equal halves connected in parallel

R_{\text{eq}} = \frac{1}{4} R_{\text{stretched}} = \frac{2.4414 R}{4} \approx 0.61035 R

Cutting the stretched wire in half gives two pieces each of resistance

R_{\text{half}} = R_{\text{stretched}} / 2

. Connecting two identical resistors in parallel yields an equivalent resistance

R_{\text{eq}} = R_{\text{half}} / 2 = R_{\text{stretched}} / 4

.

3

Calculate the ratio of power dissipated across a constant potential difference V

\frac{P_{\text{new}}}{P_{\text{orig}}} = \frac{V^2 / R_{\text{eq}}}{V^2 / R} = \frac{R}{R_{\text{eq}}} = \frac{1}{0.61035} \approx 1.64

Electrical power dissipated at constant voltage is given by

P = V^2 / R

, which means power is inversely proportional to equivalent resistance.

Key Concept

Dependence of electrical resistance on conductor geometry under volume conservation, and power dissipation in parallel circuits.

Question 4420Question →

A thermometer calibrated on an arbitrary scale $X$ registers a lower fixed point of $-10^\circ\text{X}$ and an upper fixed point of $110^\circ\text{X}$ . What is the true temperature in degrees Celsius ( $^\circ\text{C}$ ) when this thermometer reads $20^\circ\text{X}$ ?

Show answer & explanation

Answer: 25

Answer

The true temperature on the Celsius scale is

25^\circ\text{C}

.

Using the relation

\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta}{100}

, substituting

X = 20

,

\text{LFP}_X = -10

, and

\text{UFP}_X = 110

gives

\frac{20 - (-10)}{110 - (-10)} = \frac{30}{120} = 0.25

. Multiplying

0.25

by

100

gives

25^\circ\text{C}

.

Step-by-Step Solution

1

Set up the linear relationship between the arbitrary temperature scale

X

and the Celsius scale

\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}

Thermometric properties vary linearly with temperature between fixed points.

2

Substitute the given numerical values into the formula

\frac{20 - (-10)}{110 - (-10)} = \frac{\theta - 0}{100 - 0} \implies \frac{30}{120} = \frac{\theta}{100}

The lower fixed point on scale

X

is

-10^\circ\text{X}

and the upper fixed point is

110^\circ\text{X}

.

3

Solve for the unknown temperature

\theta

in degrees Celsius

\theta = \frac{30}{120} \times 100 = 25^\circ\text{C}

Simplifying the fraction

\frac{30}{120}

yields

\frac{1}{4}

, and

\frac{1}{4} \times 100 = 25

.

Key Concept

Linear interpolation and conversion between thermometric temperature scales

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept