All practice questions

Question 1481Question →

During a natural radioactive decay series, a parent nucleus of Actinium-227 ( $^{227}_{89}\text{Ac}$ ) emits $5$ $\alpha$ -particles and $3$ $\beta^-$ -particles to reach a stable state. What is the atomic number ( $Z$ ) of the resulting stable daughter nucleus?

Show answer & explanation

Answer: 82

Answer

The atomic number of the resulting stable daughter nucleus is 82.

Each alpha particle emission decreases the nuclear charge (atomic number) by 2, so emitting 5 alpha particles reduces the atomic number by 10. Each beta-minus emission increases the nuclear charge by 1, so emitting 3 beta-minus particles increases the atomic number by 3. Starting with an initial atomic number of 89, the resulting atomic number is 89 - 10 + 3 = 82.

Step-by-Step Solution

1

Determine the atomic number reduction from alpha emissions.

5 alpha particles reduce the atomic number by 5 * 2 = 10.

An alpha particle (^4_2He) carries 2 protons, so each emission decreases Z by 2.

2

Determine the atomic number increase from beta-minus emissions.

3 beta-minus particles increase the atomic number by 3 * 1 = 3.

A beta-minus particle (^0_-1e) is emitted when a neutron converts to a proton, increasing Z by 1.

3

Apply the conservation of atomic number to find the final value.

Z_final = 89 - 10 + 3 = 82.

Subtracting the alpha contribution and adding the beta-minus contribution from the initial atomic number gives the daughter nucleus's atomic number.

Key Concept

Conservation of atomic number (charge) in natural radioactive decay chains

Question 1482Question →

A thermistor has an electrical resistance of $800\,\Omega$ at the melting point of ice ( $0^\circ\text{C}$ ) and $200\,\Omega$ at the boiling point of water ( $100^\circ\text{C}$ ). Assuming the thermometric property varies linearly with temperature, what is the temperature in degrees Celsius ( $^\circ\text{C}$ ) when the resistance of the thermistor is $500\,\Omega$ ?

Show answer & explanation

Answer: 50

Answer

The temperature corresponding to a resistance of

500\,\Omega

is

50^\circ\text{C}

.

Applying the standard thermometric interpolation relation

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}

with

R_0 = 800\,\Omega

,

R_{100} = 200\,\Omega

, and

R_\theta = 500\,\Omega

gives

\theta = \frac{500 - 800}{200 - 800} \times 100 = \frac{-300}{-600} \times 100 = 50^\circ\text{C}

.

Step-by-Step Solution

1

Identify the values of the thermometric property at the ice point and steam point.

R_0 = 800\,\Omega

and

R_{100} = 200\,\Omega

.

These established values represent the fixed points of

0^\circ\text{C}

and

100^\circ\text{C}

respectively.

2

Set up the linear relationship formula for temperature conversion on the Celsius scale.

$\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}$

The temperature scale is defined linearly between the two fixed calibration points.

3

Substitute the unknown resistance

R_\theta = 500\,\Omega

into the equation and compute the result.

$\theta = \frac{500 - 800}{200 - 800} \times 100^\circ\text{C} = \frac{-300}{-600} \times 100^\circ\text{C} = 50^\circ\text{C}$

Dividing the change from the lower fixed point by the total interval between fixed points gives the fraction of

100^\circ\text{C}

.

Key Concept

Linear relationship between a thermometric property and temperature

Question 1483Question →

A nation's annual foreign trade record shows visible merchandise exports of $\$450\text{ million}$ and visible merchandise imports of $\$310\text{ million}$ . If the net invisible trade balance for the same accounting year records a deficit of $\$65\text{ million}$ , what is the country's overall Current Account balance in millions of dollars?

Show answer & explanation

Answer: 75

Answer

The Current Account balance is $75 million (a surplus of 75 million dollars).

The Current Account balance is determined by adding the net invisible balance to the visible balance (Balance of Trade). The visible balance is

\$450\text{ million} - \$310\text{ million} = \$140\text{ million}

. Subtracting the invisible deficit of

\$65\text{ million}

results in a Current Account surplus of

\$75\text{ million}

.

Step-by-Step Solution

1

Calculate the Balance of Trade (visible balance)

450 million -

310 million = $140 million surplus

Balance of Trade includes only visible merchandise goods exported and imported.

2

Add the net invisible trade balance to the Balance of Trade

140 million + (-

65 million) = $75 million surplus

The Current Account balance aggregates visible trade (Balance of Trade) and net invisible transactions such as services, income, and unrequited transfers.

Key Concept

Current Account Balance Calculation

Question 1484Question →

A $2 \times 2$ matrix is given by $A = \begin{pmatrix} x + 2 & 3 \\ 4 & x - 2 \end{pmatrix}$ . If $A$ is a singular matrix, what is the positive value of $x$ ?

Show answer & explanation

Answer: 4

Answer

The positive value of

x

is

4

.

A matrix is singular when its determinant is zero. Expanding

\det(A) = (x+2)(x-2) - 12 = 0

yields

x^2 - 16 = 0

. Solving

x^2 = 16

gives

x = \pm 4

. The positive value is

4

.

Step-by-Step Solution

1

Apply the condition for a matrix to be singular

\det(A) = 0

By definition, a matrix is singular if and only if its determinant is zero.

2

Formulate the determinant equation for matrix

A

(x + 2)(x - 2) - (3)(4) = 0

The determinant of a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is calculated as

ad - bc

.

3

Expand and simplify the algebraic equation

x^2 - 4 - 12 = 0 \implies x^2 - 16 = 0

Expanding

(x+2)(x-2)

gives

x^2 - 4

, and subtracting

12

yields

x^2 - 16

.

4

Solve for

x

and select the positive root

x^2 = 16 \implies x = 4 \text{ or } x = -4 \implies x = 4

Taking the square root gives two roots, and the problem explicitly specifies the positive value.

Key Concept

Singular Matrices and Determinants

Estimated Time:1m 15s

Question 1485Question →

A cold-storage logistics firm acquired an industrial freezer under a hire purchase agreement. The cash price of the freezer is $\text{₦}1,500,000$ . The contract required an initial deposit of $20\%$ of the cash price and $10$ equal monthly installments of $\text{₦}140,000$ each. What is the total hire purchase charge (interest) paid by the firm on this transaction?

Show answer & explanation

Answer: 200000

Answer

The total hire purchase charge (interest) paid on the freezer is ₦200,000.

The total hire purchase price is determined by adding the 20% initial deposit (₦300,000) to the 10 monthly installments of ₦140,000 (₦1,400,000), giving ₦1,700,000. Subtracting the cash price of ₦1,500,000 gives the total hire purchase charge (interest) of ₦200,000.

Step-by-Step Solution

1

Calculate the initial deposit amount

₦300,000

The deposit is 20% of the cash price of ₦1,500,000 (0.20 × ₦1,500,000).

2

Calculate total installment payments

₦1,400,000

The hirer pays 10 equal installments of ₦140,000 each (10 × ₦140,000).

3

Determine the Total Hire Purchase Price

₦1,700,000

The total cost under hire purchase equals the deposit plus total installments (₦300,000 + ₦1,400,000).

4

Calculate the Hire Purchase Charge (Interest)

₦200,000

The extra cost/interest is the difference between the Total Hire Purchase Price and the Cash Price (₦1,700,000 - ₦1,500,000).

Key Concept

Hire Purchase Price and Interest Calculation

Question 1486Question →

In a given trading year, a country recorded visible merchandise exports of $\$520\text{ million}$ and visible merchandise imports of $\$380\text{ million}$ . What is the country's Balance of Trade in millions of dollars?

Show answer & explanation

Answer: 140

Answer

The country's Balance of Trade is 140 million dollars.

The Balance of Trade measures the difference between a country's visible exports and visible imports. Subtracting

\$380\text{ million}

from

\$520\text{ million}

gives a favorable Balance of Trade of

140\text{ million dollars}

.

Step-by-Step Solution

1

Identify visible exports and visible imports from the given data.

Visible exports =

520 million; Visible imports =

380 million.

Balance of Trade deals specifically with merchandise (visible) transactions.

2

Apply the Balance of Trade formula.

Balance of Trade = Visible Exports - Visible Imports.

Trade balance represents the net earnings from physical goods traded.

3

Calculate the difference.

520 million -

380 million = 140 million dollars.

Subtracting total merchandise imports from total merchandise exports yields a trade surplus of 140 million dollars.

Key Concept

Balance of Trade Calculation

Question 1487Question →

A trigonometric function is defined by $y = A \cos(B x) + C$ , where $A > 0$ and $B > 0$ . If the function has a maximum value of $7$ , a minimum value of $-3$ , and a period of $120^\circ$ , what is the value of $A + B + C$ ?

Show answer & explanation

Answer: 10

Answer

The value of

A + B + C

is 10.

Solving the system

A + C = 7

and

-A + C = -3

gives

C = 2

and

A = 5

. Using the period relationship

120^\circ = 360^\circ / B

yields

B = 3

. Adding these parameters together gives

5 + 3 + 2 = 10

.

Step-by-Step Solution

1

Calculate vertical shift C and amplitude A

C = 2, A = 5

For y = A cos(Bx) + C, Max = A + C and Min = -A + C. Thus C = (Max + Min)/2 = 2 and A = (Max - Min)/2 = 5.

2

Calculate period coefficient B

B = 3

The period T in degrees is given by 360° / B. With T = 120°, B = 360° / 120° = 3.

3

Compute the sum A + B + C

10

Sum the derived parameters: 5 + 3 + 2 = 10.

Key Concept

Deriving amplitude, period coefficient, and vertical shift from trigonometric graph properties

Question 1488Question →

In a meteorological study conducted over $120$ days in a coastal town, rain was observed on $45$ days. According to a theoretical weather model, the probability of rain on any given day during this period is $\frac{1}{3}$ . Calculate the absolute difference between the observed number of rainy days and the expected number of rainy days predicted by the theoretical model.

Show answer & explanation

Answer: 5

Answer

The absolute difference between the observed and expected number of rainy days is 5 days.

The expected number of rainy days based on theoretical probability is

\frac{1}{3} \times 120 = 40

days. The experimental observation recorded

45

rainy days. The absolute difference between the observed and expected number of rainy days is

|45 - 40| = 5

days.

Step-by-Step Solution

1

Calculate the theoretical expected number of rainy days

Expected rainy days = 40

Multiply the total number of trial days (120) by the theoretical probability of rain (1/3).

2

Identify the experimental (observed) number of rainy days

Observed rainy days = 45

The problem states that rain occurred on 45 days during the 120-day period.

3

Compute the positive absolute difference

Absolute difference = 5

Subtract the theoretical expected count (40) from the experimental observed count (45).

Key Concept

Difference Between Experimental Frequency and Theoretical Expectation

Question 1489Question →

A metal container with a volume of $200\text{ cm}^3$ at $30^\circ\text{C}$ is completely filled with an oil. The linear expansivity of the metal container is $2.0 \times 10^{-5}\text{ K}^{-1}$ and the real cubic expansivity of the oil is $6.0 \times 10^{-4}\text{ K}^{-1}$ . If the temperature of the system is raised to $80^\circ\text{C}$ , what volume of oil (in $\text{cm}^3$ ) will overflow?

Show answer & explanation

Answer: 5.4

Answer

The volume of oil that overflows is

5.4\text{ cm}^3

.

When a container filled to the brim with liquid is heated, both the liquid and container expand. The apparent cubic expansivity of the liquid is the difference between its real cubic expansivity and the volumetric expansivity of the vessel (3 * alpha). Subtracting 0.6 * 10^-4 K^-1 from 6.0 * 10^-4 K^-1 gives an apparent cubic expansivity of 5.4 * 10^-4 K^-1. Multiplying this by the initial volume of 200 cm^3 and the temperature rise of 50 K gives an overflow volume of 5.4 cm^3.

Step-by-Step Solution

1

Determine the temperature increase (\Delta T)

\Delta T = 80^\circ\text{C} - 30^\circ\text{C} = 50\text{ K}

Thermal expansion depends directly on the change in temperature.

2

Calculate the cubic expansivity of the container (\gamma_v)

\gamma_v = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1} = 0.6 \times 10^{-4}\text{ K}^{-1}

The volumetric expansivity of an isotropic solid container is three times its linear expansivity.

3

Calculate the apparent cubic expansivity of the liquid (\gamma_a)

\gamma_a = \gamma_r - \gamma_v = 6.0 \times 10^{-4}\text{ K}^{-1} - 0.6 \times 10^{-4}\text{ K}^{-1} = 5.4 \times 10^{-4}\text{ K}^{-1}

Apparent expansivity accounts for the simultaneous expansion of the vessel containing the liquid.

4

Calculate the volume of liquid that overflows (\Delta V)

\Delta V = V_0 \times \gamma_a \times \Delta T = 200\text{ cm}^3 \times 5.4 \times 10^{-4}\text{ K}^{-1} \times 50\text{ K} = 5.4\text{ cm}^3

The overflow volume equals the apparent expansion of the liquid volume.

Key Concept

Real and Apparent Expansion of Liquids

Question 1490Question →

A musical note played on an instrument produces a fundamental frequency of $440\text{ Hz}$ . Given that the speed of sound in air is $330\text{ m/s}$ , what is the wavelength, in meters, of the sound wave produced in air corresponding to its second overtone?

Show answer & explanation

Answer: 0.25

Answer

0.25 m

The fundamental frequency (

f_1 = 440\text{ Hz}

) is the first harmonic. The second overtone is the third harmonic, which has a frequency of

3 \times 440\text{ Hz} = 1320\text{ Hz}

. Substituting this into the wave speed equation

\lambda = \frac{v}{f}

gives

\lambda = \frac{330\text{ m/s}}{1320\text{ Hz}} = 0.25\text{ m}

.

Step-by-Step Solution

1

Determine the harmonic number for the second overtone

The second overtone is the third harmonic (

n = 3

)

Overtones are integer harmonics above the fundamental frequency (

n = 1

). Thus, the 1st overtone is

n = 2

and the 2nd overtone is

n = 3

.

2

Calculate the frequency of the second overtone

f_3 = 3 \times 440\text{ Hz} = 1320\text{ Hz}

The frequency of the

n

-th harmonic is

n

times the fundamental frequency.

3

Calculate the wavelength using the wave speed equation

\lambda = \frac{v}{f_3} = \frac{330\text{ m/s}}{1320\text{ Hz}} = 0.25\text{ m}

Wavelength is determined by dividing the speed of sound by the wave frequency.

Key Concept

Harmonics, Overtones, and Wave Equation

Question 1491Question →

If the matrix $P = \begin{pmatrix} k & 2 & 0 \\ 1 & 4 & 1 \\ 3 & 0 & 2 \end{pmatrix}$ has a determinant equal to $18$ , find the value of $k$ .

Show answer & explanation

Answer: 2

Answer

The value of

k

is

2

.

Expanding matrix

P

along the first row gives

\det(P) = k(8 - 0) - 2(2 - 3) + 0(0 - 12) = 8k + 2

. Setting

8k + 2 = 18

yields

8k = 16

, which simplifies to

k = 2

.

Step-by-Step Solution

1

Expand the matrix determinant along the first row.

\det(P) = k(8 - 0) - 2(2 - 3) + 0 = 8k + 2

Applying the cofactor expansion formula for a

3 \times 3

matrix across row 1.

2

Set the resulting determinant expression equal to the given value of 18.

8k + 2 = 18

The question specifies that the determinant of

P

is equal to

18

.

3

Solve the linear equation for

k

.

k = 2

Subtracting 2 from both sides yields

8k = 16

, and dividing by 8 gives

k = 2

.

Key Concept

Determinant of a

3 \times 3

Matrix

Question 1492Question →

A projectile is launched from ground level with an initial speed of $40\text{ m/s}$ at an angle $\theta$ to the horizontal such that $\sin\theta = 0.8$ and $\cos\theta = 0.6$ . Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the projectile in $\text{m/s}$ when it reaches a height of $35\text{ m}$ above the ground?

Show answer & explanation

Answer: 30

Answer

The speed of the projectile at a height of

35\text{ m}

above the ground is

30\text{ m/s}

.

The correct answer is

30\text{ m/s}

. The vertical component of velocity at height

35\text{ m}

is found using

v_y^2 = u_y^2 - 2gh = 32^2 - 2(10)(35) = 324

, giving

v_y = 18\text{ m/s}

. Since horizontal velocity component remains constant at

v_x = 24\text{ m/s}

, the overall speed magnitude is given by

v = \sqrt{v_x^2 + v_y^2} = \sqrt{24^2 + 18^2} = \sqrt{900} = 30\text{ m/s}

.

Step-by-Step Solution

1

Resolve initial velocity into horizontal and vertical components

u_x = 24\text{ m/s}

and

u_y = 32\text{ m/s}

Horizontal component

u_x = u\cos\theta = 40 \times 0.6 = 24\text{ m/s}

and vertical component

u_y = u\sin\theta = 40 \times 0.8 = 32\text{ m/s}

.

2

Calculate vertical velocity component at height

h = 35\text{ m}

v_y = 18\text{ m/s}

Using

v_y^2 = u_y^2 - 2gh

, we get

v_y^2 = 32^2 - 2(10)(35) = 1024 - 700 = 324

, yielding

v_y = 18\text{ m/s}

.

3

Calculate the magnitude of total velocity at height

h = 35\text{ m}

v = 30\text{ m/s}

Because air resistance is neglected, horizontal velocity remains constant (

v_x = u_x = 24\text{ m/s}

). Total speed is

v = \sqrt{v_x^2 + v_y^2} = \sqrt{24^2 + 18^2} = 30\text{ m/s}

.

Key Concept

Independence of perpendicular velocity components and calculation of instantaneous speed in projectile motion.

Question 1493Question →

Calculate the magnitude of the linear momentum, in $\text{kg m s}^{-1}$ , of a body of mass $6\text{ kg}$ moving at a constant velocity of $15\text{ m s}^{-1}$ .

Show answer & explanation

Answer: 90

Answer

The magnitude of the linear momentum is

90\text{ kg m s}^{-1}

.

Linear momentum (

p

) is the product of an object's mass (

m

) and its velocity (

v

). Substituting

m = 6\text{ kg}

and

v = 15\text{ m s}^{-1}

yields

p = 6 \times 15 = 90\text{ kg m s}^{-1}

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 6\text{ kg}

and velocity

v = 15\text{ m s}^{-1}

These are the parameters provided in the problem statement.

2

Apply the linear momentum formula

p = mv = 6\text{ kg} \times 15\text{ m s}^{-1} = 90\text{ kg m s}^{-1}

Linear momentum is defined as the product of an object's mass and its velocity.

Key Concept

Linear Momentum Definition

Estimated Time:30s

Question 1494Question →

What is the smallest positive angle $\theta$ , in degrees, that satisfies the trigonometric equation $\sqrt{3}\tan(3\theta) = 3$ ?

Show answer & explanation

Answer: 20

Answer

The smallest positive value of

\theta

is

20^\circ

.

Isolating

\tan(3\theta)

gives

\tan(3\theta) = \frac{3}{\sqrt{3}} = \sqrt{3}

. The principal acute angle whose tangent is

\sqrt{3}

is

60^\circ

. Equating

3\theta = 60^\circ

and solving for

\theta

yields

\theta = 20^\circ

.

Step-by-Step Solution

1

Isolate the trigonometric function

\tan(3\theta)

\tan(3\theta) = \sqrt{3}

Dividing both sides by \sqrt{3} isolates the tangent term and rationalizes the fraction to \sqrt{3}.

2

Determine the principal angle for

3\theta

3\theta = 60^\circ

The smallest positive angle whose tangent equals \sqrt{3} is 60^\circ.

3

Solve for

\theta

\theta = 20^\circ

Dividing the principal angle 60^\circ by the coefficient 3 gives the smallest positive angle \theta.

Key Concept

Solving trigonometric equations with multiple angle arguments

Question 1495Question →

A potentiometer wire of length $100\text{ cm}$ has a resistance of $5.0\ \Omega$ . It is connected in series with a driver cell of e.m.f. $3.0\text{ V}$ (having negligible internal resistance) and a protective series resistor $R_s$ . A balance point is obtained at $60.0\text{ cm}$ along the wire for a cell of e.m.f. $1.2\text{ V}$ . What is the value of the series resistor $R_s$ in ohms?

Show answer & explanation

Answer: 2.5

Answer

The resistance of the series resistor

R_s

is

2.5\ \Omega

.

At balance, the potential drop across the

60.0\text{ cm}

portion of the potentiometer wire equals the test cell e.m.f. (

1.2\text{ V}

). Since the

100\text{ cm}

wire has a total resistance of

5.0\ \Omega

, the

60.0\text{ cm}

section has a resistance of

3.0\ \Omega

. This requires a main circuit current of

I = \frac{1.2\text{ V}}{3.0\ \Omega} = 0.4\text{ A}

. The total driver circuit resistance is

R_{total} = \frac{3.0\text{ V}}{0.4\text{ A}} = 7.5\ \Omega

. Subtracting the wire's

5.0\ \Omega

resistance gives

R_s = 2.5\ \Omega

.

Step-by-Step Solution

1

Calculate the resistance of the balanced portion of the potentiometer wire.

R_{balance} = 5.0\ \Omega \times \frac{60.0\text{ cm}}{100\text{ cm}} = 3.0\ \Omega

The resistance of a uniform wire is directly proportional to its length.

2

Determine the current in the main potentiometer circuit.

I = \frac{1.2\text{ V}}{3.0\ \Omega} = 0.4\text{ A}

At the balance point, no current flows through the galvanometer branch, so the potential difference across the balance length equals the e.m.f. of the test cell.

3

Compute the total resistance of the driver circuit.

R_{total} = \frac{3.0\text{ V}}{0.4\text{ A}} = 7.5\ \Omega

According to Ohm's law, total resistance equals the total e.m.f. divided by the circuit current.

4

Calculate the value of the series resistor

R_s

.

R_s = 7.5\ \Omega - 5.0\ \Omega = 2.5\ \Omega

The total resistance is the sum of the potentiometer wire resistance and the series resistance.

Key Concept

Potentiometer balance condition and circuit analysis

Estimated Time:1m 30s

Question 1496Question →

An electric cart moving along a straight horizontal track at an initial speed of $10\text{ m/s}$ accelerates uniformly at $3\text{ m/s}^2$ for a duration of $4\text{ s}$ . Immediately after this phase, it applies its brakes and decelerates uniformly at $2\text{ m/s}^2$ until it comes to a complete stop. What is the total distance, in meters, traveled by the cart during the entire motion?

Show answer & explanation

Answer: 185

Answer

The total distance traveled by the cart during the entire motion is

185\text{ m}

.

The total distance is calculated by analyzing the two distinct stages of motion. In the first phase, traveling at an initial velocity of

10\text{ m/s}

with an acceleration of

3\text{ m/s}^2

for

4\text{ s}

yields a distance of

64\text{ m}

and a peak velocity of

22\text{ m/s}

. In the second phase, decelerating from

22\text{ m/s}

to rest at

2\text{ m/s}^2

requires a distance of

121\text{ m}

. Adding both distances (

64\text{ m} + 121\text{ m}

) yields the total distance of

185\text{ m}

.

Step-by-Step Solution

1

Determine the distance (

s_1

) and final velocity (

v_1

) during the uniform acceleration phase

s_1 = 64\text{ m}

and

v_1 = 22\text{ m/s}

Using kinematic formulas

s_1 = u_1 t_1 + \frac{1}{2} a_1 t_1^2

and

v_1 = u_1 + a_1 t_1

for constant acceleration.

2

Determine the stopping distance (

s_2

) during the uniform deceleration phase

s_2 = 121\text{ m}

Using

v_2^2 = v_1^2 - 2 a_2 s_2

with final speed

v_2 = 0\text{ m/s}

.

3

Sum the distances from both stages

S_{\text{total}} = 185\text{ m}

Total displacement for multi-stage motion along a straight line is the sum of displacements in each stage.

Key Concept

Multi-stage linear motion with constant acceleration and deceleration

Question 1497Question →

A micrometer screw gauge has a pitch of $0.5\text{ mm}$ and $50$ divisions on its circular thimble scale. When its anvil and spindle are fully brought together without any object, the zero mark of the thimble scale lies $4$ divisions past the main scale index line in the tightening direction (a positive zero error). When the wall thickness of a hollow metal pipe is measured with this micrometer, the main scale reading is $3.5\text{ mm}$ and the $32\text{nd}$ thimble division aligns with the index line. If the outer diameter of the pipe is separately measured to be $25.40\text{ mm}$ , what is the corrected inner diameter of the metal pipe in millimeters?

Show answer & explanation

Answer: 17.84

Answer

17.84 mm

The least count of the micrometer screw gauge is

\frac{0.5\text{ mm}}{50} = 0.01\text{ mm}

. The observed wall thickness is

3.5\text{ mm} + (32 \times 0.01\text{ mm}) = 3.82\text{ mm}

. Accounting for the positive zero error of

+0.04\text{ mm}

gives a true wall thickness of

3.82\text{ mm} - 0.04\text{ mm} = 3.78\text{ mm}

. Subtracting twice the wall thickness from the outer diameter yields the inner diameter:

25.40\text{ mm} - 2(3.78\text{ mm}) = 17.84\text{ mm}

.

Step-by-Step Solution

1

Calculate the least count (precision) of the micrometer screw gauge

Least count = 0.01 mm

Least count is defined as pitch divided by total circular scale divisions: 0.5 mm / 50 = 0.01 mm.

2

Determine the observed wall thickness from the instrument readings

Observed thickness = 3.82 mm

Observed reading = Main scale reading + (Thimble division * Least count) = 3.5 mm + (32 * 0.01 mm) = 3.82 mm.

3

Correct the observed thickness for positive zero error

Corrected wall thickness = 3.78 mm

Positive zero error (+0.04 mm) must be subtracted from the observed reading: 3.82 mm - 0.04 mm = 3.78 mm.

4

Calculate the corrected inner diameter of the hollow pipe

Inner diameter = 17.84 mm

The inner diameter equals the outer diameter minus twice the wall thickness: 25.40 mm - 2(3.78 mm) = 25.40 mm - 7.56 mm = 17.84 mm.

Key Concept

Measurement of length, micrometer zero error correction, and geometry of hollow cylinders

Estimated Time:2m 30s

Question 1498Question →

A rocket carries a payload of mass $25\text{ kg}$ resting on a spring balance. During its vertical ascent, the spring balance indicates a reading of $350\text{ N}$ . Taking the acceleration due to gravity as $g = 10\text{ m s}^{-2}$ , what is the magnitude of the upward acceleration of the rocket in $\text{m s}^{-2}$ ?

Show answer & explanation

Answer: 4

Answer

The upward acceleration of the rocket is

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

.

When an object of mass

m

accelerates upward at rate

a

, the scale must exert an upward force

R

that overcomes the weight

mg

and provides net acceleration

ma

, such that

R = m(g + a)

. Substituting

R = 350\text{ N}

,

m = 25\text{ kg}

, and

g = 10\text{ m s}^{-2}

yields

350 = 25(10 + a)

, which simplifies to

a = 4.0\text{ m s}^{-2}

.

Step-by-Step Solution

1

Formulate the equation of motion for apparent weight during upward acceleration

R - mg = ma \implies R = m(g + a)

The spring balance supports the mass against gravity while simultaneously providing the force for upward acceleration.

2

Substitute the given numerical parameters into the relation

350 = 25(10 + a)

The measured reading

R = 350\text{ N}

, mass

m = 25\text{ kg}

, and standard gravity

g = 10\text{ m s}^{-2}

are supplied.

3

Solve the algebraic equation for the rocket's acceleration

a

a = 4.0\text{ m s}^{-2}

Dividing

350\text{ N}

by

25\text{ kg}

gives an effective total acceleration of

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

; subtracting

g = 10\text{ m s}^{-2}

isolates the upward acceleration

a

.

Key Concept

Apparent Weight and Mass Measurement in Accelerating Frames

Question 1499Question →

The derived SI unit of electric capacitance, the farad ( $\text{F}$ ), can be expressed in terms of fundamental SI base units as $\text{kg}^a \cdot \text{m}^b \cdot \text{s}^c \cdot \text{A}^d$ . What is the numerical value of the sum of the exponents $a + b + c + d$ ?

Show answer & explanation

Answer: 3

Answer

The numerical value of the sum of the exponents

a + b + c + d

is 3.

The farad (

\text{F}

), when resolved into fundamental SI base units, is

\text{kg}^{-1} \cdot \text{m}^{-2} \cdot \text{s}^4 \cdot \text{A}^2

. Adding the exponents yields

a = -1

,

b = -2

,

c = 4

, and

d = 2

, giving a total sum of

(-1) + (-2) + 4 + 2 = 3

.

Step-by-Step Solution

1

Relate electric capacitance to fundamental physical quantities

Capacitance is defined as

C = \frac{Q}{V}

, where

Q

is electric charge and

V

is electric potential difference.

This fundamental relation connects capacitance to charge and energy per unit charge.

2

Express charge and potential difference in terms of SI base units

Electric charge

Q

has base units

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

. Potential difference

V = \frac{\text{Work}}{Q}

has base units

\frac{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}}{\text{A} \cdot \text{s}} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}

.

Work is force times distance (

\text{kg} \cdot \text{m} \cdot \text{s}^{-2} \times \text{m} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}

) and electric current is a fundamental base quantity.

3

Determine the base unit representation of the farad

The farad is expressed as

\frac{\text{A} \cdot \text{s}}{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-3} \cdot \text{A}^{-1}} = \text{kg}^{-1} \cdot \text{m}^{-2} \cdot \text{s}^4 \cdot \text{A}^2

.

Dividing the unit of charge by the unit of potential difference yields the derived base unit expression.

4

Calculate the sum of the exponents

a + b + c + d = (-1) + (-2) + 4 + 2 = 3

.

Summing the powers corresponding to kilograms, meters, seconds, and amperes gives the final scalar value.

Key Concept

Expressing derived SI units in terms of fundamental SI base units

Question 1500Question →

A uniform copper wire with a cross-sectional area of $2.0 \times 10^{-6}\,\text{m}^2$ carries a steady electric current of $4.0\,\text{A}$ . Given that the free electron density of copper is $2.5 \times 10^{28}\,\text{m}^{-3}$ and the elementary charge is $1.6 \times 10^{-19}\,\text{C}$ , calculate the drift velocity of the electrons in the wire in $\text{m/s}$ .

Show answer & explanation

Answer: 0.0005

Answer

The drift velocity of the electrons in the wire is

0.0005\,\text{m/s}

(or

5.0 \times 10^{-4}\,\text{m/s}

).

Applying the formula

I = n A e v_d

and solving for drift velocity yields

v_d = \frac{I}{n A e} = \frac{4.0}{(2.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})} = \frac{4.0}{8000} = 0.0005\,\text{m/s}

.

Step-by-Step Solution

1

State the formula relating electric current to microscopic drift velocity.

I = n A e v_d

Electric current is the net charge flowing per unit time across a conductor's cross-section.

2

Isolate the drift velocity

v_d

as the target variable.

v_d = \frac{I}{n A e}

Algebraic rearrangement places all known quantities on the right-hand side.

3

Substitute the numerical values into the equation.

v_d = \frac{4.0}{(2.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})}

Inserting

I = 4.0\,\text{A}

,

n = 2.5 \times 10^{28}\,\text{m}^{-3}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

, and

e = 1.6 \times 10^{-19}\,\text{C}

.

4

Evaluate the arithmetic computation.

v_d = 0.0005\,\text{m/s}

Dividing

4.0

by

8000

yields

0.0005\,\text{m/s}

.

Key Concept

Relationship between electric current, charge carrier density, and drift velocity

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