All practice questions

Question 1441Question →

At a specific location on Earth, the total magnetic field intensity is $40\ \mu\text{T}$ and the angle of dip is $60^\circ$ . What is the magnitude of the horizontal component of Earth's magnetic field at this location, in $\mu\text{T}$ ?

Show answer & explanation

Answer: 20

Answer

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

20\ \mu\text{T}

.

The horizontal component

B_h

of Earth's magnetic field is derived using

B_h = B \cos\theta

. Substituting

B = 40\ \mu\text{T}

and

\theta = 60^\circ

yields

B_h = 40 \times 0.5 = 20\ \mu\text{T}

.

Step-by-Step Solution

1

Recall the resolving formula for the horizontal component of Earth's magnetic field

B_h = B \cos\theta

The horizontal component is the vector projection of total field

B

onto the horizontal plane inclined at angle

\theta

.

2

Evaluate the cosine function for

60^\circ

\cos(60^\circ) = 0.5

Standard trigonometric value for

60^\circ

.

3

Multiply total magnetic field strength by

\cos(60^\circ)

B_h = 40 \times 0.5 = 20\ \mu\text{T}

Calculates the exact horizontal component magnitude.

Key Concept

Components of Earth's Magnetic Field

Question 1442Question →

An astronomical telescope operating in normal adjustment consists of an objective lens with a focal length of $80\text{ cm}$ and an eyepiece with a focal length of $4\text{ cm}$ . What is the magnitude of the angular magnification produced by the telescope?

Show answer & explanation

Answer: 20

Answer

The magnitude of the angular magnification produced by the telescope is 20.

The angular magnification

M

of an astronomical telescope in normal adjustment is defined as the ratio of the focal length of the objective lens

f_o

to the focal length of the eyepiece lens

f_e

, given by

M = \frac{f_o}{f_e}

. Substituting the given values

f_o = 80\text{ cm}

and

f_e = 4\text{ cm}

yields

M = \frac{80}{4} = 20

.

Step-by-Step Solution

1

Identify the given optical parameters and formula for angular magnification.

Objective focal length

f_o = 80\text{ cm}

, Eyepiece focal length

f_e = 4\text{ cm}

. Formula:

M = \frac{f_o}{f_e}

.

For an astronomical telescope in normal adjustment, the light rays emerge parallel, and the angular magnification is given by the ratio of the focal length of the objective lens to that of the eyepiece.

2

Calculate the angular magnification value.

M = \frac{80\text{ cm}}{4\text{ cm}} = 20

.

Dividing the focal length of the objective lens by the focal length of the eyepiece yields the dimensionless magnification ratio.

Key Concept

Angular Magnification of an Astronomical Telescope in Normal Adjustment

Question 1443Question →

A $600\text{ cm}^3$ sample of air contaminated with sulphur(IV) oxide ( $\text{SO}_2$ ) gas was passed through an excess aqueous solution of sodium hydroxide to absorb all the $\text{SO}_2$ , reducing the volume of the gas sample to $576\text{ cm}^3$ . The remaining gas mixture was then passed over excess heated copper turnings to remove oxygen gas, after which the unreacted gas volume measured $456\text{ cm}^3$ under the same conditions of temperature and pressure. What is the percentage by volume of sulphur(IV) oxide in the original air sample?

Show answer & explanation

Answer: 4

Answer

The percentage by volume of sulphur(IV) oxide in the original air sample is 4.0%.

Sodium hydroxide (

\text{NaOH}

) selectively absorbs acidic pollutant gases such as sulphur(IV) oxide (

\text{SO}_2

). The volume decrease from

600\text{ cm}^3

to

576\text{ cm}^3

indicates that

24\text{ cm}^3

of

\text{SO}_2

was absorbed. Dividing

24\text{ cm}^3

by the original sample volume of

600\text{ cm}^3

and multiplying by 100 gives

4.0\%

.

Step-by-Step Solution

1

Calculate the volume of sulphur(IV) oxide gas absorbed by the sodium hydroxide solution.

Volume of

\text{SO}_2 = 600\text{ cm}^3 - 576\text{ cm}^3 = 24\text{ cm}^3

.

Sodium hydroxide reacts with acidic oxide pollutants like

\text{SO}_2

, causing a reduction in gas volume equal to the volume of

\text{SO}_2

present.

2

Calculate the percentage composition by volume relative to the total initial sample.

\text{Percentage of } \text{SO}_2 = \left(\frac{24\text{ cm}^3}{600\text{ cm}^3}\right) \times 100\% = 4.0\%.

The volumetric percentage is determined by expressing the volume of the target gas component over the total volume of the original air sample.

Key Concept

Volumetric determination of air composition and gaseous pollutants.

Question 1444Question →

In an economy, the Central Bank issues a total of $\text{N}450\text{ billion}$ in currency. Out of this amount, commercial banks hold $\text{N}50\text{ billion}$ as vault cash in their tills. If the demand deposits held by the public in commercial banks total $\text{N}800\text{ billion}$ , what is the value of the narrow money supply ( $M_1$ ) in billions of Naira?

Show answer & explanation

Answer: 1200

Answer

The value of the narrow money supply (

M_1

) is 1200 billion Naira.

The narrow money supply (

M_1

) consists of currency in circulation outside commercial banks plus demand deposits. Currency in circulation is found by subtracting vault cash from total currency issued by the central bank (

\text{N}450\text{ billion} - \text{N}50\text{ billion} = \text{N}400\text{ billion}

). Adding demand deposits (

\text{N}800\text{ billion}

) yields a total narrow money supply of

\text{N}1200\text{ billion}

.

Step-by-Step Solution

1

Calculate currency in circulation outside commercial banks

400 billion Naira

Vault cash held inside commercial bank vaults is excluded from currency in circulation outside the banking system.

2

Calculate narrow money supply (

M_1

)

1200 billion Naira

Narrow money supply (

M_1

) is the sum of currency in circulation outside commercial banks and demand deposits.

Key Concept

Components and Calculation of Narrow Money Supply (

M_1

)

Question 1445Question →

An oxygen nucleus $^{16}_{8}\text{O}$ has a measured nuclear mass of $15.9906\text{ u}$ . Given that the mass of a proton is $1.0078\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , calculate the total binding energy of the nucleus in $\text{MeV}$ . (Take $1\text{ u} = 931.5\text{ MeV}$ )

Show answer & explanation

Answer: 131.71

Answer

The total binding energy of the oxygen nucleus is

131.71\text{ MeV}

.

Summing the masses of 8 individual protons and 8 individual neutrons gives

16.1320\text{ u}

. Subtracting the actual nuclear mass of

15.9906\text{ u}

leaves a mass defect of

0.1414\text{ u}

. Multiplying this mass defect by the equivalence constant

931.5\text{ MeV/u}

yields a total binding energy of

131.71\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total combined mass of the free constituent nucleons

Mass of 8 protons + 8 neutrons =

8(1.0078\text{ u}) + 8(1.0087\text{ u}) = 8.0624\text{ u} + 8.0696\text{ u} = 16.1320\text{ u}

Oxygen-16 has

Z = 8

protons and

A - Z = 16 - 8 = 8

neutrons.

2

Calculate the mass defect (

\Delta m

)

\Delta m = 16.1320\text{ u} - 15.9906\text{ u} = 0.1414\text{ u}

Mass defect is the difference between total constituent mass and actual nuclear mass.

3

Convert mass defect into total binding energy (

E_b

)

E_b = 0.1414\text{ u} \times 931.5\text{ MeV/u} = 131.7141\text{ MeV} \approx 131.71\text{ MeV}

Applying the energy equivalent factor of

931.5\text{ MeV}

per atomic mass unit.

Key Concept

Mass defect and nuclear binding energy equivalence

Question 1446Question →

A solid sample of lead with a mass of $0.80\text{ kg}$ is kept at its melting point of $327^\circ\text{C}$ . If $15,000\text{ J}$ of thermal energy is supplied to the sample, calculate the mass of lead, in kilograms, that remains in the solid state. (Take the specific latent heat of fusion of lead as $2.5 \times 10^4\text{ J/kg}$ ).

Show answer & explanation

Answer: 0.2

Answer

The mass of lead remaining in the solid state is

0.20\text{ kg}

.

Thermal energy

Q = 15,000\text{ J}

supplied to lead at its melting point melts a portion calculated by

m_{\text{melted}} = \frac{Q}{L_f} = \frac{15,000}{25,000} = 0.60\text{ kg}

. Subtracting this melted mass from the original

0.80\text{ kg}

yields

0.20\text{ kg}

of remaining solid lead.

Step-by-Step Solution

1

Calculate the mass of lead that melts.

Melted mass

m_{\text{melted}} = 0.60\text{ kg}

.

At the melting point, thermal energy supplied goes entirely into phase change without changing temperature:

Q = m_{\text{melted}} L_f

.

2

Determine the remaining mass of solid lead.

Remaining solid mass

m_{\text{solid}} = 0.20\text{ kg}

.

The un-melted portion equals the initial total mass minus the mass that has melted (

m_{\text{solid}} = m_{\text{total}} - m_{\text{melted}}

).

Key Concept

Latent Heat of Fusion and Phase Change

Question 1447Question →

A short bar magnet with magnetic dipole moment $1.6\text{ A}\cdot\text{m}^2$ is placed along the magnetic meridian with its north pole pointing towards the Earth's magnetic south pole. A neutral point is located on the axial line of the magnet at a distance of $0.2\text{ m}$ from its center. What is the magnitude of the horizontal component of the Earth's magnetic field at this location, in microtesla ( $\mu\text{T}$ )? (Take $\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}$ )

Show answer & explanation

Answer: 40

Answer

The magnitude of the horizontal component of Earth's magnetic field at this location is 40 μT.

At a neutral point, the horizontal component of Earth's magnetic field is equal in magnitude and opposite in direction to the magnetic field generated by the bar magnet. For a magnet aligned with its north pole pointing south, neutral points lie on its axial line at distance

d

. Using

B_h = \frac{\mu_0}{4\pi} \frac{2M}{d^3}

with

M = 1.6\text{ A}\cdot\text{m}^2

and

d = 0.2\text{ m}

yields

B_h = 4.0 \times 10^{-5}\text{ T} = 40\ \mu\text{T}

.

Step-by-Step Solution

1

Determine the condition for the neutral point

B_{\text{axial}} = B_h

When a magnet's north pole points south, its axial magnetic field opposes Earth's horizontal field, creating neutral points along the axis where the magnetic fields cancel out completely.

2

Apply the short bar magnet formula for field along the axial line

B_h = \frac{\mu_0}{4\pi} \frac{2M}{d^3}

The magnetic field produced at an axial point at distance

d

from the center of a short bar magnet of magnetic moment

M

is given by this formula.

3

Substitute the given numerical parameters

B_h = 10^{-7} \times \frac{2 \times 1.6}{(0.2)^3}

Substituting

M = 1.6\text{ A}\cdot\text{m}^2

,

d = 0.2\text{ m}

, and

\frac{\mu_0}{4\pi} = 10^{-7}\text{ T}\cdot\text{m}\cdot\text{A}^{-1}

into the field equation.

4

Calculate the magnitude of the horizontal field component in microtesla

B_h = 4.0 \times 10^{-5}\text{ T} = 40\ \mu\text{T}

Dividing

3.2 \times 10^{-7}

by

8 \times 10^{-3}

gives

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

, which converts to

40\ \mu\text{T}

.

Key Concept

Neutral points created by a bar magnet aligned with Earth's magnetic meridian

Question 1448Question →

Light of frequency $8.0 \times 10^{14}\text{ Hz}$ illuminates a photosensitive plate, causing photoelectrons to be emitted with a maximum kinetic energy of $1.2\text{ eV}$ . If the same plate is subsequently illuminated by light of frequency $1.2 \times 10^{15}\text{ Hz}$ , what is the stopping potential, in volts, needed to reduce the photoelectric current to zero? (Take $h = 6.6 \times 10^{-34}\text{ J s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 2.85

Answer

The stopping potential needed to reduce the photoelectric current to zero is 2.85 V.

Using Einstein's photoelectric equation

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

, the initial photon energy is

E_1 = h f_1 = \frac{6.6 \times 10^{-34} \times 8.0 \times 10^{14}}{1.6 \times 10^{-19}} = 3.3\text{ eV}

. Given

K_1 = 1.2\text{ eV}

, the work function of the metal is

W_0 = 3.3\text{ eV} - 1.2\text{ eV} = 2.1\text{ eV}

. For the second frequency

f_2 = 1.2 \times 10^{15}\text{ Hz}

, the photon energy is

E_2 = \frac{6.6 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}} = 4.95\text{ eV}

. The new maximum kinetic energy is

K_2 = 4.95\text{ eV} - 2.1\text{ eV} = 2.85\text{ eV}

. Since

e V_s = K_{\text{max}}

, the stopping potential required to reduce the current to zero is

2.85\text{ V}

.

Step-by-Step Solution

1

Calculate the photon energy

E_1

of the initial light in electron-volts

E_1 = \frac{6.6 \times 10^{-34} \times 8.0 \times 10^{14}}{1.6 \times 10^{-19}} = 3.3\text{ eV}

Photon energy is related to frequency by

E = h f

.

2

Determine the work function

W_0

of the photosensitive plate

W_0 = E_1 - K_1 = 3.3\text{ eV} - 1.2\text{ eV} = 2.1\text{ eV}

By Einstein's photoelectric equation,

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

.

3

Calculate the photon energy

E_2

for the second light frequency

E_2 = \frac{6.6 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}} = 4.95\text{ eV}

The energy of the second photon is calculated using

f_2 = 1.2 \times 10^{15}\text{ Hz}

.

4

Find the maximum kinetic energy

K_2

and corresponding stopping potential

V_s

K_2 = 4.95\text{ eV} - 2.1\text{ eV} = 2.85\text{ eV}

, giving

V_s = 2.85\text{ V}

The stopping potential in volts is numerical equal to the maximum kinetic energy expressed in electron-volts (

e V_s = K_{\text{max}}

).

Key Concept

Einstein's Photoelectric Equation and Stopping Potential

Estimated Time:2m 0s

Question 1449Question →

In a physics experiment to determine the acceleration due to gravity $g$ using a free-fall apparatus, the distance of fall is measured as $h = (2.00 \pm 0.04)\text{ m}$ and the duration of fall is measured as $t = (0.50 \pm 0.01)\text{ s}$ . Given that $g = \frac{2h}{t^2}$ , what is the percentage error in the calculated value of $g$ ?

Show answer & explanation

Answer: 6

Answer

The percentage error in the calculated value of

g

is 6%.

For a physical quantity defined by

g = \frac{2h}{t^2}

, the maximum percentage error is determined by adding the percentage error in

h

to twice the percentage error in

t

. The percentage error in

h

is

\frac{0.04}{2.00} \times 100\% = 2\%

and in

t

is

\frac{0.01}{0.50} \times 100\% = 2\%

. Therefore, the total percentage error in

g

is

2\% + 2(2\%) = 6\%

.

Step-by-Step Solution

1

Calculate the percentage error in the distance measurement

h

2%

The fractional error in height is

\frac{\Delta h}{h} = \frac{0.04}{2.00} = 0.02

, which corresponds to

2\%

.

2

Calculate the percentage error in the time measurement

t

2%

The fractional error in time is

\frac{\Delta t}{t} = \frac{0.01}{0.50} = 0.02

, which corresponds to

2\%

.

3

Apply the power-law error propagation rule to find the maximum percentage error in

g

6%

For

g = \frac{2h}{t^2}

, the total relative error is

\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t} = 2\% + 2(2\%) = 6\%

, since the exponent of

t

is 2.

Key Concept

Error propagation in physical formulas involving powers and quotients

Estimated Time:2m 0s

Question 1450Question →

A simple pendulum has a period of oscillation of $1.6\text{ s}$ on the surface of the Earth, where the acceleration due to gravity is $10.0\text{ m/s}^2$ . What is the period of oscillation of this pendulum when placed on a moon where the acceleration due to gravity is $2.5\text{ m/s}^2$ ?

Show answer & explanation

Answer: 3.2

Answer

The period of oscillation of the pendulum on the moon is

3.2\text{ s}

.

The period of a simple pendulum is given by

T = 2\pi \sqrt{\frac{l}{g}}

. Because the length

l

is constant, period is inversely proportional to the square root of acceleration due to gravity (

T \propto \frac{1}{\sqrt{g}}

). Reducing the local gravity from

10.0\text{ m/s}^2

to

2.5\text{ m/s}^2

decreases gravity by a factor of 4, which increases the period by a factor of

\sqrt{4} = 2

. Multiplying the initial period of

1.6\text{ s}

by 2 yields

3.2\text{ s}

.

Step-by-Step Solution

1

Relate the period of oscillation of a simple pendulum to gravitational acceleration.

The period formula is

T = 2\pi \sqrt{\frac{l}{g}}

, showing that

T

is inversely proportional to

\sqrt{g}

.

The length of the pendulum

l

remains unchanged.

2

Formulate a ratio comparing the pendulum's period on the moon to its period on Earth.

\frac{T_{moon}}{T_{earth}} = \sqrt{\frac{g_{earth}}{g_{moon}}}

Dividing the two equations cancels the constant terms

2\pi

and

\sqrt{l}

.

3

Substitute the known numerical values and solve for

T_{moon}

.

T_{moon} = 1.6 \times \sqrt{\frac{10.0}{2.5}} = 1.6 \times 2.0 = 3.2\text{ s}

The ratio of gravities is 4, whose square root is 2, doubling the initial period.

Key Concept

Dependence of Simple Pendulum Period on Gravitational Acceleration

Estimated Time:1m 30s

Question 1451Question →

A satellite communication system transmits an ultra-high frequency electromagnetic wave with a wavelength of $0.05\text{ m}$ . If the wave travels at a speed of $3.0 \times 10^{8}\text{ m/s}$ in a vacuum, what is the frequency of the transmission in gigahertz ( $\text{GHz}$ )?

Show answer & explanation

Answer: 6

Answer

The frequency of the electromagnetic transmission is

6\text{ GHz}

.

Using the wave equation

c = f \lambda

, the frequency is

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.05\text{ m}} = 6.0 \times 10^9\text{ Hz}

. Converting to gigahertz (

1\text{ GHz} = 10^9\text{ Hz}

) yields

6\text{ GHz}

.

Step-by-Step Solution

1

Identify the relationship between electromagnetic wave velocity, frequency, and wavelength.

c = f \lambda

, where

c = 3.0 \times 10^8\text{ m/s}

and

\lambda = 0.05\text{ m}

.

All electromagnetic waves travel at the speed of light

c

in a vacuum.

2

Rearrange the wave equation to isolate frequency (

f

).

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.05\text{ m}} = 6.0 \times 10^9\text{ Hz}

.

Dividing the wave speed by the wavelength yields the frequency in hertz.

3

Convert the frequency into gigahertz (

\text{GHz}

).

\frac{6.0 \times 10^9\text{ Hz}}{10^9\text{ Hz/GHz}} = 6\text{ GHz}

.

The prefix giga (G) denotes a factor of

10^9

.

Key Concept

Electromagnetic wave propagation equation (

c = f \lambda

) and unit conversion

Estimated Time:1m 15s

Question 1452Question →

In an experiment to determine the Young's modulus $Y$ of a metal wire using the formula $Y = \frac{4FL}{\pi d^2 e}$ , the force $F$ , length $L$ , diameter $d$ , and extension $e$ are measured with maximum percentage errors of $1.2\%$ , $0.8\%$ , $1.5\%$ , and $2.0\%$ respectively. What is the maximum percentage error in the calculated value of $Y$ ?

Show answer & explanation

Answer: 7

Answer

The maximum percentage error in the calculated value of Young's modulus

Y

is

7.0\%

.

The maximum percentage error is determined by adding the percentage error of each measured quantity, scaled by the absolute value of its power. For

Y = \frac{4FL}{\pi d^2 e}

, the calculation is

1.2\% + 0.8\% + 2(1.5\%) + 2.0\% = 7.0\%

.

Step-by-Step Solution

1

Set up the relative error propagation formula for the derived quantity

Y

.

\frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta e}{e}

Mathematical constants (

4

and

\pi

) carry no measurement error, and exponents multiply the fractional error of their respective variables.

2

Convert fractional errors into percentage errors by multiplying each term by

100\%

.

\text{Percentage Error}(Y) = \text{Percentage Error}(F) + \text{Percentage Error}(L) + 2 \times \text{Percentage Error}(d) + \text{Percentage Error}(e)

Percentage error is directly proportional to fractional error.

3

Substitute the given percentage errors into the formula and sum them up.

\text{Percentage Error}(Y) = 1.2\% + 0.8\% + 2(1.5\%) + 2.0\% = 7.0\%

To find the worst-case (maximum) error, all individual percentage contributions are added regardless of whether the variable appears in the numerator or denominator.

Key Concept

Error Propagation in Derived Physical Quantities

Question 1453Question →

A particle executes simple harmonic motion along a straight line with an angular frequency of $6.0\text{ rad/s}$ . What is the magnitude of the acceleration of the particle, in $\text{m/s}^2$ , when its displacement from the mean position is $0.50\text{ m}$ ?

Show answer & explanation

Answer: 18

Answer

18.0 m/s^2

The magnitude of acceleration in simple harmonic motion is calculated using

a = \omega^2 x

. Substituting

\omega = 6.0\text{ rad/s}

and

x = 0.50\text{ m}

gives

a = (6.0)^2 \times 0.50 = 36 \times 0.50 = 18.0\text{ m/s}^2

.

Step-by-Step Solution

1

Identify the formula relating acceleration to angular frequency and displacement in SHM.

a = \omega^2 x

In simple harmonic motion, the magnitude of acceleration is directly proportional to displacement from the equilibrium position.

2

Substitute the given physical values into the equation.

a = (6.0)^2 \times 0.50

The given values are angular frequency

\omega = 6.0\text{ rad/s}

and displacement

x = 0.50\text{ m}

.

3

Compute the numerical product.

a = 18.0\text{ m/s}^2

Squaring

6.0

yields

36

, and multiplying by

0.50

gives

18.0

.

Key Concept

Acceleration in Simple Harmonic Motion

Estimated Time:1m 0s

Question 1454Question →

A consumer derives a total utility of $85$ utils from consuming $5$ plates of rice. If the marginal utility derived from the $6^{\text{th}}$ plate is $7$ utils and the marginal utility from the $7^{\text{th}}$ plate is $3$ utils, what is the total utility derived from consuming $7$ plates of rice?

Show answer & explanation

Answer: 95

Answer

The total utility derived from consuming 7 plates of rice is 95 utils.

Total Utility (

TU

) is the cumulative sum of Marginal Utilities (

MU

) derived from each additional unit consumed. Starting from

TU_5 = 85

utils, consuming the

6^{\text{th}}

plate adds

7

utils to yield

TU_6 = 92

utils. Consuming the

7^{\text{th}}

plate adds another

3

utils, resulting in a total utility of

95

utils for

7

plates.

Step-by-Step Solution

1

Calculate the total utility after consuming 6 plates of rice

TU_6 = 92

utils

Total utility increases by the marginal utility of the

6^{\text{th}}

unit (

TU_6 = TU_5 + MU_6 = 85 + 7 = 92

utils).

2

Calculate the total utility after consuming 7 plates of rice

TU_7 = 95

utils

Total utility increases by the marginal utility of the

7^{\text{th}}

unit (

TU_7 = TU_6 + MU_7 = 92 + 3 = 95

utils).

Key Concept

Relationship between Total Utility and Marginal Utility under the Law of Diminishing Marginal Utility

Question 1455Question →

A concave mirror produces a real image that is $3$ times the size of an object placed in front of it. If the distance between the object and its image is $40\text{ cm}$ , what is the focal length of the mirror in $\text{cm}$ ?

Show answer & explanation

Answer: 15

Answer

The focal length of the concave mirror is

15\text{ cm}

.

Using the magnification relation

v = 3u

and the object-image separation of

40\text{ cm}

, we obtain

3u - u = 40\text{ cm}

, which yields

u = 20\text{ cm}

and

v = 60\text{ cm}

. Substituting these distances into the mirror equation

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

gives

\frac{1}{f} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}

, so

f = 15\text{ cm}

.

Step-by-Step Solution

1

Relate the image distance

v

to the object distance

u

using the linear magnification formula

v = 3u

Since the mirror forms a real, magnified image that is 3 times the size of the object, linear magnification

m = \frac{v}{u} = 3

.

2

Formulate an equation from the given object-to-image separation distance to solve for

u

and

v

u = 20\text{ cm}

and

v = 60\text{ cm}

The separation distance between the image and object is

v - u = 40\text{ cm}

. Substituting

v = 3u

yields

2u = 40\text{ cm} \implies u = 20\text{ cm}

and

v = 60\text{ cm}

.

3

Substitute the values of

u

and

v

into the mirror formula to compute the focal length

f

f = 15\text{ cm}

Applying

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}

gives

f = 15\text{ cm}

.

Key Concept

Linear magnification and mirror formula for concave mirrors

Question 1456Question →

In a macroeconomy, National Income is measured at $\text{₦1,200 million}$ . Economic records show social security contributions of $\text{₦85 million}$ , corporate profit taxes of $\text{₦120 million}$ , undistributed corporate profits of $\text{��95 million}$ , and government transfer payments of $\text{₦150 million}$ . Calculate the total Personal Income of the economy in millions of Naira.

Show answer & explanation

Answer: 1050

Answer

The total Personal Income is 1050 million Naira.

Personal Income is calculated by adjusting National Income: subtracting corporate profit taxes, undistributed corporate profits, and social security contributions, while adding government transfer payments. Thus,

\text{Personal Income} = 1200 - (85 + 120 + 95) + 150 = 1050

million Naira.

Step-by-Step Solution

1

Calculate total deductions from National Income

300 million Naira

Social security contributions, corporate profit taxes, and undistributed profits represent earned income that households do not directly receive.

2

Adjust National Income by subtracting deductions and adding transfer payments

1050 million Naira

Transfer payments (such as pensions and welfare) are received by individuals without rendering immediate productive services, so they are added to determine total Personal Income.

Key Concept

Derivation of Personal Income from National Income

Question 1457Question →

A lithium-6 nucleus $^{6}_{3}\text{Li}$ has a measured nuclear mass of $6.0151\text{ u}$ . Given that the mass of a proton is $1.0078\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , calculate the binding energy per nucleon of the nucleus in $\text{MeV}$ . (Take $1\text{ u} = 931\text{ MeV}$ )

Show answer & explanation

Answer: 5.34

Answer

The binding energy per nucleon of the

^{6}_{3}\text{Li}

nucleus is approximately

5.34\text{ MeV}

(or

5.34\text{ MeV/nucleon}

).

The total mass of 3 free protons and 3 free neutrons is

6.0495\text{ u}

. Subtracting the actual nuclear mass (

6.0151\text{ u}

) yields a mass defect of

0.0344\text{ u}

. Multiplying by

931\text{ MeV/u}

gives a total binding energy of

32.0264\text{ MeV}

. Dividing by the

6

nucleons in lithium-6 yields

5.34\text{ MeV}

per nucleon.

Step-by-Step Solution

1

Determine the number of protons and neutrons and compute the total constituent mass.

Protons

Z = 3

, Neutrons

N = 3

. Total nucleon mass

m_{\text{nucleons}} = 3(1.0078\text{ u}) + 3(1.0087\text{ u}) = 6.0495\text{ u}

.

Free nucleons have a combined mass greater than the bound nucleus.

2

Calculate the mass defect (

\Delta m

).

\Delta m = 6.0495\text{ u} - 6.0151\text{ u} = 0.0344\text{ u}

.

The difference between total constituent mass and measured nuclear mass represents the lost mass converted into binding energy.

3

Convert mass defect to total binding energy in

\text{MeV}

.

E_b = 0.0344\text{ u} \times 931\text{ MeV/u} = 32.0264\text{ MeV}

.

Using the equivalence

1\text{ u} = 931\text{ MeV}

.

4

Calculate binding energy per nucleon by dividing by mass number

A = 6

.

\frac{32.0264\text{ MeV}}{6} = 5.3377\text{ MeV} \approx 5.34\text{ MeV}.

Binding energy per nucleon measures the stability per particle in the nucleus.

Key Concept

Mass Defect and Binding Energy per Nucleon

Question 1458Question →

An ice block of mass $0.15\text{ kg}$ at $-10^\circ\text{C}$ absorbs thermal energy until it turns completely into liquid water at $0^\circ\text{C}$ . Taking the specific heat capacity of ice as $2100\text{ J kg}^{-1}\text{ K}^{-1}$ and the specific latent heat of fusion of ice as $3.36 \times 10^5\text{ J kg}^{-1}$ , what is the total thermal energy supplied in joules?

Show answer & explanation

Answer: 53550

Answer

The total thermal energy supplied is 53,550 J.

To transform sub-zero ice into liquid water at its melting point, thermal energy is absorbed in two distinct stages: warming the ice from

-10^\circ\text{C}

to

0^\circ\text{C}

(

Q_1 = 0.15 \times 2100 \times 10 = 3150\text{ J}

) and melting the ice completely at

0^\circ\text{C}

(

Q_2 = 0.15 \times 3.36 \times 10^5 = 50400\text{ J}

). Adding these yields a total energy of

53550\text{ J}

.

Step-by-Step Solution

1

Calculate the thermal energy required to warm the solid ice from

-10^\circ\text{C}

to its melting point (

0^\circ\text{C}

).

Q_1 = 3150\text{ J}

Before the phase transition can begin, sensible heat must be supplied to raise the ice temperature using

Q_1 = m c_{\text{ice}} \Delta T

.

2

Calculate the thermal energy required to melt the ice at constant temperature (

0^\circ\text{C}

).

Q_2 = 50400\text{ J}

Phase transformation at constant temperature requires latent heat of fusion using

Q_2 = m L_f

.

3

Sum the energy required for both stages to determine total energy.

Q_{\text{total}} = 53550\text{ J}

The total energy supplied is the sum of sensible warming heat (

Q_1

) and latent melting heat (

Q_2

).

Key Concept

Multi-step heat energy calculation involving sensible heat (

m c \Delta T

) and latent heat of fusion (

m L_f

).

Question 1459Question →

A radioactive parent nucleus of Thorium, $^{232}_{90}\text{Th}$ , undergoes a natural radioactive decay series by emitting a total of $6$ $\alpha$ -particles and $4$ $\beta^-$ -particles to form a stable daughter isotope of Lead ( $\text{Pb}$ ). Calculate the number of neutrons present in the nucleus of the resulting daughter isotope.

Show answer & explanation

Answer: 126

Answer

The resulting daughter nucleus contains 126 neutrons.

Emitting

6

\alpha

-particles reduces the mass number by

6 \times 4 = 24

units and the atomic number by

6 \times 2 = 12

units. Emitting

4

\beta^-

-particles leaves the mass number unchanged while increasing the atomic number by

4 \times 1 = 4

units. Consequently, the daughter nucleus has a mass number

A = 232 - 24 = 208

and an atomic number

Z = 90 - 12 + 4 = 82

. The number of neutrons is

N = A - Z = 208 - 82 = 126

.

Step-by-Step Solution

1

Calculate the mass number (

A

) of the daughter nucleus after all emissions

A = 232 - (6 \times 4) = 208

An alpha particle carries away 4 mass units (

^{4}_{2}\text{He}

), while a beta-minus particle carries 0 mass units (

^{0}_{-1}\text{e}

).

2

Calculate the atomic number (

Z

) of the daughter nucleus after all emissions

Z = 90 - (6 \times 2) + (4 \times 1) = 82

Each alpha decay reduces nuclear charge by 2, and each beta-minus decay increases nuclear charge by 1 due to neutron-to-proton conversion.

3

Compute the number of neutrons (

N

)

N = A - Z = 208 - 82 = 126

The number of neutrons in any nuclide is given by subtracting the atomic number (protons) from the mass number (nucleons).

Key Concept

Mass and Atomic Number Conservation in Radioactive Decay Chains

Estimated Time:2m 0s

Question 1460Question →

An agro-processing enterprise operating at full capacity can produce either $150\text{ bags}$ of cassava flour or $100\text{ bags}$ of garri per day using its fixed processing equipment. Assuming a constant rate of transformation between the two goods, what is the opportunity cost, in bags of cassava flour, of increasing garri production from $40\text{ bags}$ to $70\text{ bags}$ per day?

Show answer & explanation

Answer: 45

Answer

The opportunity cost of increasing garri production by 30 bags is 45 bags of cassava flour.

The opportunity cost of producing one additional bag of garri is

150 / 100 = 1.5\text{ bags}

of cassava flour. Increasing garri output from

40

to

70\text{ bags}

requires producing

30

additional bags. Consequently, the total opportunity cost is

30 \times 1.5 = 45\text{ bags}

of cassava flour foregone.

Step-by-Step Solution

1

Calculate the unit opportunity cost of garri

Opportunity cost of 1 bag of garri = 150 / 100 = 1.5 bags of cassava flour

Given full utilization of fixed resources, producing maximum cassava flour (150) versus maximum garri (100) establishes a constant trade-off ratio of 1.5.

2

Calculate the increase in garri production

70 - 40 = 30 bags of garri

The question specifies an expansion in garri output from 40 bags to 70 bags.

3

Calculate total foregone cassava flour

30 × 1.5 = 45 bags of cassava flour

Multiplying the extra garri produced by the unit opportunity cost yields the total foregone alternative.

Key Concept

Opportunity Cost Calculation

Answer

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