All practice questions

Question 7941Question →

A sports analyst recorded the number of points scored by a basketball player across five consecutive games as $12$ , $14$ , $15$ , $16$ , and $18$ . What is the mean deviation of these scores?

1.6

0.0

2.0

4.0

Show answer & explanation

Answer: $1.6$

Answer

The mean deviation of the scores is

1.6

.

The arithmetic mean of the five scores is

15

. The sum of the absolute differences between each score and the mean is

|12-15| + |14-15| + |15-15| + |16-15| + |18-15| = 3 + 1 + 0 + 1 + 3 = 8

. Dividing this total by the number of scores (

5

) gives

1.6

, which correctly represents the mean deviation.

Step-by-Step Solution

1

Calculate the mean (

\bar{x}

) of the data set.

\bar{x} = \frac{12 + 14 + 15 + 16 + 18}{5} = \frac{75}{5} = 15

The mean deviation measures dispersion relative to the arithmetic mean.

2

Compute the absolute deviations

|x - \bar{x}|

for each data point.

|12 - 15| = 3, \quad |14 - 15| = 1, \quad |15 - 15| = 0, \quad |16 - 15| = 1, \quad |18 - 15| = 3

Absolute values prevent positive and negative deviations from cancelling each other out.

3

Calculate the mean of the absolute deviations.

\text{Mean Deviation} = \frac{\sum |x - \bar{x}|}{n} = \frac{3 + 1 + 0 + 1 + 3}{5} = \frac{8}{5} = 1.6

Dividing the sum of absolute deviations by the number of observations yields the mean deviation.

Key Concept

Mean Deviation of Ungrouped Data

Estimated Time:1m 30s

Question 7942Question →

What is the maximum value of the curve $y = \frac{x}{x^2 + 4}$ ?

2

\frac{1}{4}

-\frac{1}{4}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{1}{4}$

Answer

The maximum value of the curve is

\frac{1}{4}

.

Differentiating

y = \frac{x}{x^2 + 4}

using the quotient rule gives

\frac{dy}{dx} = \frac{4 - x^2}{(x^2 + 4)^2}

. Setting

\frac{dy}{dx} = 0

yields critical points at

x = \pm 2

. Substituting

x = 2

into the original function gives

y = \frac{2}{8} = \frac{1}{4}

, which is the maximum value of the function.

Step-by-Step Solution

1

Differentiate

y = \frac{x}{x^2 + 4}

with respect to

x

using the quotient rule.

\frac{dy}{dx} = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

Stationary points occur where the first derivative equals zero.

2

Set

\frac{dy}{dx} = 0

to solve for the stationary points.

4 - x^2 = 0 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2

A rational expression equals zero when its numerator is zero.

3

Evaluate

y

at each critical point to determine the function values.

For

x = 2

:

y = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}

. For

x = -2

:

y = \frac{-2}{(-2)^2 + 4} = -\frac{1}{4}

.

The question asks for the maximum value of

y

on the curve.

4

Compare the stationary values to select the maximum.

The maximum value is

\frac{1}{4}

at

x = 2

.

Since

\frac{1}{4} > -\frac{1}{4}

,

x = 2

corresponds to the maximum point.

Key Concept

Stationary Points and Maxima/Minima of Rational Functions

Estimated Time:1m 30s

Question 7943Question →

The $5^{\text{th}}$ term of an arithmetic progression (AP) is $18$ and its $11^{\text{th}}$ term is $42$ . Calculate the value of the $20^{\text{th}}$ term of this progression.

Show answer & explanation

Answer: 78

Answer

78

Using the AP general term formula

T_n = a + (n-1)d

, the equations

a + 4d = 18

and

a + 10d = 42

yield

d = 4

and

a = 2

. Evaluating

T_{20} = 2 + 19(4)

yields

78

.

Step-by-Step Solution

1

Set up simultaneous equations using the general term formula

T_n = a + (n-1)d

a + 4d = 18

and

a + 10d = 42

Relate given terms to the first term

a

and common difference

d

.

2

Solve for the common difference

d

6d = 24 \implies d = 4

Subtracting the 5th term equation from the 11th term equation eliminates

a

.

3

Solve for the first term

a

a + 16 = 18 \implies a = 2

Substitute the value of

d

back into the first equation.

4

Calculate the 20th term

T_{20} = 2 + 19(4) = 78

Apply the nth term formula for

n = 20

.

Key Concept

Determining terms of an Arithmetic Progression using simultaneous equations

Estimated Time:1m 30s

Question 7944Question →

In Bohr's model of the hydrogen atom, the energy of an electron in a stationary orbit with principal quantum number $n$ is given by $E_n = -\frac{13.6}{n^2}\text{ eV}$ . What is the frequency of the photon emitted when an electron transitions from the $n = 4$ energy state to the $n = 2$ energy state? ( $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

6.15 \times 10^{14}\text{ Hz}

3.85 \times 10^{33}\text{ Hz}

2.05 \times 10^{14}\text{ Hz}

6.20 \times 10^{14}\text{ Hz}

Show answer & explanation

Answer: $6.15 \times 10^{14}\text{ Hz}$

Answer

6.15 \times 10^{14}\text{ Hz}

The energy of the emitted photon corresponds to the difference between the initial energy level (

n = 4

) and the final energy level (

n = 2

), giving

\Delta E = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}

. Converting

2.55\text{ eV}

to Joules yields

4.08 \times 10^{-19}\text{ J}

. Dividing this energy by Planck's constant (

6.63 \times 10^{-34}\text{ J}\cdot\text{s}

) produces a photon frequency of

6.15 \times 10^{14}\text{ Hz}

.

Step-by-Step Solution

1

Calculate the energy levels for

n = 4

and

n = 2

E_4 = -\frac{13.6}{4^2} = -0.85\text{ eV}

and

E_2 = -\frac{13.6}{2^2} = -3.40\text{ eV}

Electron energy in Bohr's model depends inversely on the square of the principal quantum number.

2

Find the energy of the emitted photon

\Delta E = E_4 - E_2 = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}

The energy of the emitted photon equals the energy lost during the downward transition between states.

3

Convert the photon energy from electron-volts to Joules

\Delta E = 2.55 \times 1.6 \times 10^{-19}\text{ J} = 4.08 \times 10^{-19}\text{ J}

Energy must be converted to SI units (Joules) before applying Planck's equation.

4

Calculate the photon frequency using

f = \frac{\Delta E}{h}

f = \frac{4.08 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}} \approx 6.15 \times 10^{14}\text{ Hz}

Photon energy and frequency are related by the Planck relation

E = h f

.

Key Concept

Bohr Model Energy Transitions and Photon Frequency

Question 7945Question →

In a quality control experiment, a technician tested $200$ light bulbs and found that $36$ were defective. According to the manufacturer's specification, the theoretical probability of a bulb being defective is $\frac{3}{20}$ . What is the difference between the theoretical probability and the experimental probability of selecting a non-defective light bulb?

\frac{3}{100}

\frac{67}{100}

\frac{33}{100}

\frac{17}{100}

Show answer & explanation

Answer: $\frac{3}{100}$

Answer

The correct answer is

\frac{3}{100}

(or

0.03

).

The experimental probability of obtaining a non-defective bulb is

1 - \frac{36}{200} = 0.82

. The theoretical probability of a non-defective bulb is

1 - 0.15 = 0.85

. Subtracting the two gives

0.85 - 0.82 = 0.03 = \frac{3}{100}

.

Step-by-Step Solution

1

Calculate the experimental probability of selecting a defective bulb

\text{P}_{\text{exp}}(\text{defective}) = \frac{36}{200} = 0.18

Relative frequency is determined by dividing the observed frequency of defective bulbs by the total number of trials.

2

Find the experimental probability of selecting a non-defective bulb

\text{P}_{\text{exp}}(\text{non-defective}) = 1 - 0.18 = 0.82

The sum of the probabilities of an event and its complement equals

1

.

3

Find the theoretical probability of selecting a non-defective bulb

\text{P}_{\text{theo}}(\text{non-defective}) = 1 - \frac{3}{20} = 1 - 0.15 = 0.85

Subtracting the given theoretical probability of a defective bulb from

1

yields the theoretical probability of a non-defective bulb.

4

Calculate the difference between the two probabilities of selecting a non-defective bulb

0.85 - 0.82 = 0.03 = \frac{3}{100}

Subtract the experimental probability from the theoretical probability.

Key Concept

Experimental probability measures relative frequency from observed outcomes, while theoretical probability is based on expected mathematical outcomes. Complementary probabilities fulfill

P(E') = 1 - P(E)

.

Estimated Time:1m 30s

Question 7946Question →

A potentiometer wire of length $100\text{ cm}$ has a resistance of $10\text{ }\Omega$ . It is connected in series with a driver cell of electromotive force $3.0\text{ V}$ and internal resistance $2.0\text{ }\Omega$ , alongside an external series resistor of $8.0\text{ }\Omega$ . A test cell of unknown electromotive force $E$ gives a balance point at a length of $60\text{ cm}$ from the zero end of the wire. What is the value of $E$ in volts?

Show answer & explanation

Answer: 0.9

Answer

The electromotive force of the test cell is

0.9\text{ V}

.

The e.m.f. of the test cell is balanced by the potential difference across a length of

60\text{ cm}

of the potentiometer wire. Accounting for the driver cell's internal resistance (

2.0\text{ }\Omega

), external series resistor (

8.0\text{ }\Omega

), and wire resistance (

10\text{ }\Omega

), the total resistance of the primary circuit is

20.0\text{ }\Omega

. This yields a primary current of

0.15\text{ A}

and a potential drop across the wire of

1.5\text{ V}

. The resulting potential gradient is

0.015\text{ V/cm}

, which when multiplied by the balance length of

60\text{ cm}

gives an e.m.f. of

0.9\text{ V}

.

Step-by-Step Solution

1

Calculate total resistance in the primary driver circuit

R_{total} = 10\text{ }\Omega + 2.0\text{ }\Omega + 8.0\text{ }\Omega = 20.0\text{ }\Omega

The driver cell's internal resistance, potentiometer wire, and external series resistor are connected in series.

2

Calculate the current flowing through the potentiometer wire

I = \frac{3.0\text{ V}}{20.0\text{ }\Omega} = 0.15\text{ A}

Apply Ohm's law to the complete primary circuit.

3

Find the voltage drop across the potentiometer wire

V_{wire} = 0.15\text{ A} \times 10\text{ }\Omega = 1.5\text{ V}

The potential difference across the wire depends on its resistance and the primary current.

4

Determine the potential gradient along the wire

k = \frac{1.5\text{ V}}{100\text{ cm}} = 0.015\text{ V/cm}

Potential gradient is the potential drop per unit length of the wire.

5

Calculate the e.m.f. of the unknown test cell

E = 0.015\text{ V/cm} \times 60\text{ cm} = 0.9\text{ V}

At the balance point, no current flows from the test cell, so its e.m.f. equals the potential drop across the balance length.

Key Concept

Potentiometer principle and potential gradient

Estimated Time:2m 0s

Question 7947Question →

An object has a mass of $12\text{ kg}$ on Earth, where the acceleration due to gravity is $10\text{ m s}^{-2}$ . What are the mass and weight of the object on the Moon, where the acceleration due to gravity is $1.6\text{ m s}^{-2}$ ?

12\text{ kg}

and

19.2\text{ N}

1.92\text{ kg}

and

19.2\text{ N}

12\text{ kg}

and

120\text{ N}

1.92\text{ kg}

and

120\text{ N}

Show answer & explanation

Answer: $12\text{ kg}$ and $19.2\text{ N}$

Answer

The mass on the Moon is

12\text{ kg}

and the weight on the Moon is

19.2\text{ N}

.

Mass is constant everywhere in the universe, so the object maintains a mass of

12\text{ kg}

on the Moon. Weight is the gravitational force exerted on the object, given by

W = mg

. On the Moon,

W = 12 \cdot 1.6 = 19.2\text{ N}

. Thus, the option specifying

12\text{ kg}

and

19.2\text{ N}

is correct.

Step-by-Step Solution

1

Determine the mass of the object on the Moon.

Mass

= 12\text{ kg}

.

Mass is a scalar physical quantity representing the quantity of matter in a body; it is constant and independent of location or gravitational field.

2

Calculate the weight of the object on the Moon using

W = m \cdot g_{\text{moon}}

.

W = 12\text{ kg} \times 1.6\text{ m s}^{-2} = 19.2\text{ N}

.

Weight is the force of gravitational attraction acting on a mass and varies directly with local gravitational field strength.

Key Concept

Mass is an intrinsic property that remains constant across different locations, whereas weight is a force that depends on local gravitational acceleration (

W = mg

).

Estimated Time:45s

Question 7948Question →

A right-angled triangle has a perimeter of $60\text{ cm}$ and a hypotenuse of length $25\text{ cm}$ . What is the area of the triangle in $\text{cm}^2$ ?

Show answer & explanation

Answer: 150

Answer

150

Let the perpendicular sides of the right-angled triangle be

a

and

b

, and the hypotenuse be

c = 25\text{ cm}

. From the perimeter,

a + b + 25 = 60

, so

a + b = 35\text{ cm}

. By the Pythagorean theorem,

a^2 + b^2 = 25^2 = 625

. Squaring both sides of

a + b = 35

yields

(a + b)^2 = a^2 + b^2 + 2ab = 35^2 = 1225

. Substituting

a^2 + b^2 = 625

gives

625 + 2ab = 1225 \implies 2ab = 600 \implies ab = 300

. The area of the right-angled triangle is

\frac{1}{2}ab = \frac{1}{2} \times 300 = 150\text{ cm}^2

.

Step-by-Step Solution

1

Determine the sum of the two legs from the perimeter

a + b = 35\text{ cm}

The perimeter of the triangle is

a + b + c = 60\text{ cm}

, where the hypotenuse

c = 25\text{ cm}

.

2

Use the Pythagorean theorem for the sum of squares of the legs

a^2 + b^2 = 625

In any right-angled triangle with hypotenuse

25\text{ cm}

,

a^2 + b^2 = 25^2 = 625

.

3

Expand

(a + b)^2

to find the product of the legs

ab

1225 = 625 + 2ab \implies 2ab = 600 \implies ab = 300

Using the algebraic identity

(a + b)^2 = a^2 + b^2 + 2ab

allows determining

ab

directly without solving for individual side lengths.

4

Calculate the area of the right-angled triangle

\text{Area} = 150\text{ cm}^2

The area of a right-angled triangle with perpendicular sides

a

and

b

is given by

\frac{1}{2}ab

.

Key Concept

Perimeter and Area of Right-Angled Triangles using Algebraic Identities

Question 7949Question →

Evaluate the algebraic limit:

\lim_{x \to 2} \frac{x^3 - 8}{x^2 + x - 6}

What is the value of this limit?

\frac{4}{5}

\frac{8}{5}

\frac{12}{5}

0

Show answer & explanation

Answer: $\frac{12}{5}$

Answer

The correct value of the limit is

\frac{12}{5}

.

Evaluating the limit by direct substitution gives the indeterminate form

\frac{0}{0}

. Factoring the numerator

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

and denominator

x^2 + x - 6 = (x - 2)(x + 3)

allows cancellation of

(x - 2)

. Evaluating

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

at

x = 2

yields

\frac{12}{5}

.

Step-by-Step Solution

1

Check for direct substitution

Substituting

x = 2

gives

\frac{2^3 - 8}{2^2 + 2 - 6} = \frac{0}{0}

, which is an indeterminate form.

Direct substitution results in

\frac{0}{0}

, requiring algebraic factorization.

2

Factor the numerator and the denominator

Numerator:

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

Denominator:

x^2 + x - 6 = (x - 2)(x + 3)

Use the difference of cubes formula

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

and quadratic factorization.

3

Cancel the common factor and compute the limit

\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 3)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 3} = \frac{2^2 + 2(2) + 4}{2 + 3} = \frac{12}{5}

Eliminating the factor

(x - 2)

removes the removable discontinuity at

x = 2

.

Key Concept

Limits of Indeterminate Forms (0/0) using Factorization

Question 7950Question →

A uniform metallic conductor of length $200\,\text{m}$ and cross-sectional area $2.0 \times 10^{-6}\,\text{m}^2$ has a resistivity of $1.6 \times 10^{-8}\,\Omega\cdot\text{m}$ at an initial temperature of $20\,^\circ\text{C}$ . The temperature coefficient of resistivity for the material is $5.0 \times 10^{-3}\,^\circ\text{C}^{-1}$ . If the operating temperature of the conductor increases to $120\,^\circ\text{C}$ while it is connected across a constant potential difference of $12\,\text{V}$ , what is the magnitude of the electric current flowing through the conductor in amperes?

Show answer & explanation

Answer: 5

Answer

The electric current flowing through the conductor is

5.0\,\text{A}

.

The temperature change of

100\,^\circ\text{C}

increases the resistivity of the material from

1.6 \times 10^{-8}\,\Omega\cdot\text{m}

to

2.4 \times 10^{-8}\,\Omega\cdot\text{m}

via

\rho = \rho_0(1 + \alpha \Delta T)

. Substituting this updated resistivity into

R = \frac{\rho L}{A}

gives a resistance of

2.4\,\Omega

. Applying Ohm's Law

I = \frac{V}{R}

with a potential difference of

12\,\text{V}

yields

5.0\,\text{A}

.

Step-by-Step Solution

1

Calculate the temperature difference

\Delta T = 100\,^\circ\text{C}

The temperature change relative to the reference temperature dictates the change in resistivity.

2

Calculate the resistivity at the final temperature

\rho = 2.4 \times 10^{-8}\,\Omega\cdot\text{m}

Resistivity depends on temperature according to

\rho = \rho_0(1 + \alpha \Delta T)

.

3

Calculate the total electrical resistance of the conductor

R = 2.4\,\Omega

Resistance is related to physical geometry and resistivity by

R = \frac{\rho L}{A}

.

4

Apply Ohm's law to solve for the current

I = 5.0\,\text{A}

Electric current is determined by potential difference divided by resistance (

I = V / R

).

Key Concept

Temperature dependence of resistivity and Ohm's Law

Estimated Time:2m 0s

Question 7951Question →

A sample of ideal gas in a syringe occupies a volume of $300\text{ cm}^3$ at a temperature of $27^\circ\text{C}$ . If the pressure of the gas remains constant, what is its volume in $\text{cm}^3$ when the temperature is raised to $127^\circ\text{C}$ ?

Show answer & explanation

Answer: 400

Answer

400 cm³

According to Charles's Law, at constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature (

V \propto T

). Converting the given temperatures to kelvins yields

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Applying

\frac{V_1}{T_1} = \frac{V_2}{T_2}

gives

V_2 = \frac{300 \times 400}{300} = 400\text{ cm}^3

.

Step-by-Step Solution

1

Convert the initial and final temperatures from degrees Celsius to kelvins.

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

and

T_2 = 127 + 273 = 400\text{ K}

Gas law calculations require absolute temperature measured on the kelvin scale.

2

Apply Charles's Law, which relates volume and absolute temperature at constant pressure.

\frac{V_1}{T_1} = \frac{V_2}{T_2}

The pressure of the gas is maintained constant.

3

Substitute the values into the equation and solve for the final volume

V_2

.

\frac{300\text{ cm}^3}{300\text{ K}} = \frac{V_2}{400\text{ K}} \implies V_2 = 400\text{ cm}^3

Cross-multiplying gives

V_2 = \frac{300 \times 400}{300} = 400\text{ cm}^3

.

Key Concept

Charles's Law

Question 7952Question →

What is the length of a simple pendulum that has a period of oscillation of $2.0\text{ s}$ at a location where the acceleration due to gravity is $g = \pi^2\text{ m/s}^2$ ?

0.5\text{ m}

1.0\text{ m}

2.0\text{ m}

4.0\text{ m}

Show answer & explanation

Answer: $1.0\text{ m}$

Answer

The length of the simple pendulum is

1.0\text{ m}

.

Using the period equation

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

, squaring both sides yields

T^2 = 4\pi^2 l / g

. Rearranging gives

l = \frac{T^2 g}{4\pi^2}

. Substituting

T = 2.0\text{ s}

and

g = \pi^2\text{ m/s}^2

yields

l = \frac{4 \pi^2}{4 \pi^2} = 1.0\text{ m}

.

Step-by-Step Solution

1

State the formula for the period of a simple pendulum.

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

This formula relates the oscillation period

T

, pendulum length

l

, and gravitational acceleration

g

.

2

Square both sides of the equation to solve for

l

.

T^2 = 4\pi^2 \left(\frac{l}{g}\right) \implies l = \frac{T^2 \cdot g}{4\pi^2}

Isolating

l

allows direct evaluation using the given numerical values.

3

Substitute

T = 2.0\text{ s}

and

g = \pi^2\text{ m/s}^2

into the expression.

l = \frac{(2.0)^2 \cdot \pi^2}{4\pi^2} = \frac{4\pi^2}{4\pi^2} = 1.0\text{ m}

Simplifying by canceling

\pi^2

and

4

gives the exact length.

Key Concept

Simple Pendulum Period and Length Relationship

Question 7953Question →

Two capacitors with capacitances $C_1 = 4.0\text{ }\mu\text{F}$ and $C_2 = 12.0\text{ }\mu\text{F}$ are connected in series across a $120.0\text{ V}$ d.c. power supply. After becoming fully charged, the capacitors are disconnected from the supply and reconnected in parallel with plates of like polarity connected together. What is the total electrostatic energy lost in the reconnection process?

5.4 \times 10^{-3}\text{ J}

1.62 \times 10^{-2}\text{ J}

1.76 \times 10^{-2}\text{ J}

2.16 \times 10^{-2}\text{ J}

Show answer & explanation

Answer: $5.4 \times 10^{-3}\text{ J}$

Answer

The total electrostatic energy lost in the reconnection process is

5.4 \times 10^{-3}\text{ J}

.

The correct answer is derived by finding the total initial energy stored in series (

2.16 \times 10^{-2}\text{ J}

), determining the total combined charge (

720\text{ }\mu\text{C}

) and parallel capacitance (

16.0\text{ }\mu\text{F}

) upon reconnection to find the final stored energy (

1.62 \times 10^{-2}\text{ J}

), and calculating the difference of

5.4 \times 10^{-3}\text{ J}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance and charge of the initial series arrangement.

C_s = \frac{C_1 C_2}{C_1 + C_2} = \frac{4.0 \times 12.0}{4.0 + 12.0} = 3.0\text{ }\mu\text{F}

, so charge on each capacitor

Q = C_s V = (3.0 \times 10^{-6}\text{ F})(120\text{ V}) = 360\text{ }\mu\text{C}

.

Capacitors in series store identical charge equal to the product of equivalent series capacitance and total applied voltage.

2

Calculate the initial total electrostatic energy stored.

E_i = \frac{1}{2} C_s V^2 = \frac{1}{2} (3.0 \times 10^{-6}\text{ F})(120\text{ V})^2 = 2.16 \times 10^{-2}\text{ J}

.

Initial stored energy is determined by the series combination connected across the supply voltage.

3

Determine total charge and equivalent capacitance after parallel reconnection.

Q_p = Q_1 + Q_2 = 360\text{ }\mu\text{C} + 360\text{ }\mu\text{C} = 720\text{ }\mu\text{C}

and

C_p = C_1 + C_2 = 4.0\text{ }\mu\text{F} + 12.0\text{ }\mu\text{F} = 16.0\text{ }\mu\text{F}

.

Connecting like-polarity plates aggregates the individual charges and sums the capacitances in parallel.

4

Calculate the final potential difference and final stored energy.

V_p = \frac{Q_p}{C_p} = \frac{720\text{ }\mu\text{C}}{16.0\text{ }\mu\text{F}} = 45\text{ V}

, so

E_f = \frac{1}{2} C_p V_p^2 = \frac{1}{2} (16.0 \times 10^{-6}\text{ F})(45\text{ V})^2 = 1.62 \times 10^{-2}\text{ J}

.

Charge redistributes until both capacitors reach a common potential difference

V_p

.

5

Calculate the energy lost during reconnection.

\Delta E = E_i - E_f = 2.16 \times 10^{-2}\text{ J} - 1.62 \times 10^{-2}\text{ J} = 5.4 \times 10^{-3}\text{ J}

.

The energy dissipated as heat and spark during charge redistribution is the difference between initial and final total energies.

Key Concept

Energy dissipation during charge sharing between reconnected capacitors

Question 7954Question →

A uniform conductor of length $4.0\,\text{m}$ and cross-sectional area $2.0 \times 10^{-7}\,\text{m}^2$ has an electrical resistance of $0.50\,\Omega$ . What is the electrical resistivity of the material of the conductor in units of $10^{-8}\,\Omega\cdot\text{m}$ ?

Show answer & explanation

Answer: 2.5

Answer

The electrical resistivity of the material is

2.5 \times 10^{-8}\,\Omega\cdot\text{m}

, which gives a numerical value of

2.5

in units of

10^{-8}\,\Omega\cdot\text{m}

.

The electrical resistance

R

of a conductor is given by

R = \frac{\rho L}{A}

, where

\rho

is the resistivity,

L

is the length, and

A

is the cross-sectional area. Rearranging to solve for resistivity yields

\rho = \frac{R \cdot A}{L}

. Substituting

R = 0.50\,\Omega

,

A = 2.0 \times 10^{-7}\,\text{m}^2

, and

L = 4.0\,\text{m}

gives

\rho = \frac{0.50 \times 2.0 \times 10^{-7}}{4.0} = 2.5 \times 10^{-8}\,\Omega\cdot\text{m}

. Expressed in units of

10^{-8}\,\Omega\cdot\text{m}

, the answer is

2.5

.

Step-by-Step Solution

1

Identify the relevant formula linking resistance, resistivity, length, and area.

R = \frac{\rho L}{A}

The resistance of a uniform conductor is directly proportional to its length and inversely proportional to its cross-sectional area.

2

Rearrange the equation to solve for resistivity

\rho

.

\rho = \frac{R \cdot A}{L}

Isolating

\rho

allows direct evaluation using the given quantitative values.

3

Substitute the known numerical values into the equation.

\rho = \frac{0.50 \times (2.0 \times 10^{-7})}{4.0} = 2.5 \times 10^{-8}\,\Omega\cdot\text{m}

Performing the algebraic calculation gives the resistivity in SI units.

Key Concept

Electrical Resistivity and Conductor Dimensions

Question 7955Question →

A ray of light traveling within a dense glass prism of refractive index $1.60$ strikes the boundary with a surrounding transparent liquid. If total internal reflection just occurs at an angle of incidence of $45.0^\circ$ in the glass, what is the refractive index of the liquid? (Take $\sin 45.0^\circ = 0.707$ )

1.13

2.26

0.44

1.60

Show answer & explanation

Answer: $1.13$

Answer

The refractive index of the liquid is

1.13

.

For light traveling from a denser medium (

n_{\text{glass}}

) to a rarer medium (

n_{\text{liquid}}

), the critical angle

\theta_c

is defined by

\sin \theta_c = \frac{n_{\text{liquid}}}{n_{\text{glass}}}

. Substituting

n_{\text{glass}} = 1.60

and

\sin 45.0^\circ = 0.707

gives

n_{\text{liquid}} = 1.60 \times 0.707 = 1.13

.

Step-by-Step Solution

1

Identify the given parameters and formula for total internal reflection

Refractive index of denser medium

n_{\text{glass}} = 1.60

, critical angle

\theta_c = 45.0^\circ

, and formula

\sin \theta_c = \frac{n_{\text{rarer}}}{n_{\text{denser}}}

.

Total internal reflection occurs at the critical angle when light travels from an optically denser medium to a less dense (rarer) medium.

2

Rearrange the equation to solve for the refractive index of the liquid (

n_{\text{liquid}}

)

n_{\text{liquid}} = n_{\text{glass}} \times \sin \theta_c

.

Multiplying both sides of the critical angle equation by

n_{\text{glass}}

isolates the target variable.

3

Substitute the values and calculate

n_{\text{liquid}}

n_{\text{liquid}} = 1.60 \times 0.707 = 1.1312 \approx 1.13

.

Carrying out the arithmetic yields the refractive index of the liquid.

Key Concept

Total Internal Reflection and Critical Angle

Question 7956Question →

Match each length measuring instrument listed on the left with its standard precision and appropriate physical measurement application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Metre rule matches precision

0.1\text{ cm}

for pendulum rod; Vernier caliper matches precision

0.01\text{ cm}

for tube diameters; Micrometer screw gauge matches precision

0.01\text{ mm}

for wire diameter; Flexible measuring tape matches precision

0.1\text{ cm}

for flexible/large distances.

Each instrument is matched according to its fundamental physical least count and specialized geometry: Metre rule measures rigid lengths to

0.1\text{ cm}

, Vernier caliper measures internal/external diameters to

0.01\text{ cm}

, Micrometer screw gauge measures fine dimensions to

0.01\text{ mm}

, and flexible measuring tape measures long or curved distances to

0.1\text{ cm}

.

Step-by-Step Solution

1

Identify the least count (precision) of each measuring instrument.

Metre rule:

0.1\text{ cm}

, Vernier caliper:

0.01\text{ cm}

, Micrometer screw gauge:

0.01\text{ mm}

, Measuring tape:

0.1\text{ cm}

.

Least count defines the minimum measurable dimension and precision limit for each tool.

2

Match each instrument to its specific physical design feature and suitable application.

Metre rule

\rightarrow

rigid straight measurements (pendulum rod); Vernier caliper

\rightarrow

internal/external diameters (jaws); Micrometer screw gauge

\rightarrow

very thin objects (wire diameter); Tape measure

\rightarrow

curved/large dimensions.

The mechanical structure of the instrument dictates its intended physical measurement domain.

Key Concept

Instrument least count and application domain in length measurement

Question 7957Question →

If a constant magnitude net force continuously acts on a moving body strictly perpendicular to its direction of motion, the magnitude of the body's linear momentum remains constant while its direction of motion continuously changes.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. A net force directed perpendicular to velocity changes only the direction of motion, keeping speed and the magnitude of linear momentum constant.

The statement is true because a force perpendicular to displacement does zero work, preserving kinetic energy and speed. Because speed is unchanged, the scalar magnitude of momentum remains constant while its vector direction curves continuously.

Step-by-Step Solution

1

Determine the work done by a perpendicular force.

The work done is

W = F \Delta s \cos(90^\circ) = 0\text{ J}

.

Force perpendicular to displacement performs no work on the object.

2

Relate work done to speed and magnitude of linear momentum.

Zero work means zero change in kinetic energy, so speed

v

is constant, making magnitude

p = mv

constant.

Linear momentum magnitude depends solely on mass and speed.

3

Evaluate the effect on momentum direction.

The force produces an acceleration vector perpendicular to velocity, changing the vector direction of linear momentum

\vec{p}

.

By Newton's Second Law (

\vec{F} = \frac{d\vec{p}}{dt}

), force dictates the rate of change of momentum vector.

Key Concept

Vector nature of linear momentum and perpendicular force action

Question 7958Question →

A solid metal sphere has an initial volume of $1000\text{ cm}^3$ at $20^\circ\text{C}$ . If it is heated to a final temperature of $70^\circ\text{C}$ and the linear expansivity of the metal is $2.0 \times 10^{-5}\text{ K}^{-1}$ , what is the increase in the volume of the sphere?

3.0\text{ cm}^3

1.0\text{ cm}^3

2.0\text{ cm}^3

4.2\text{ cm}^3

Show answer & explanation

Answer: $3.0\text{ cm}^3$

Answer

The increase in the volume of the sphere is

3.0\text{ cm}^3

.

The value of

3.0\text{ cm}^3

is correct because the volume expansion requires the cubical expansivity

\gamma = 3\alpha = 6.0 \times 10^{-5}\text{ K}^{-1}

. Multiplying this coefficient by the initial volume (

1000\text{ cm}^3

) and the temperature change (

50\text{ K}

) yields

\Delta V = 1000 \times 6.0 \times 10^{-5} \times 50 = 3.0\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the temperature change (

\Delta T

).

\Delta T = 70^\circ\text{C} - 20^\circ\text{C} = 50^\circ\text{C} = 50\text{ K}

Thermal expansion depends on the change in temperature rather than the initial or final temperature alone.

2

Determine the cubical (volume) expansivity (

\gamma

) from linear expansivity (

\alpha

).

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

For an isotropic solid, volume expands in three orthogonal dimensions, making cubical expansivity equal to three times linear expansivity.

3

Calculate the increase in volume (

\Delta V

).

\Delta V = V_1 \gamma \Delta T = 1000\text{ cm}^3 \times (6.0 \times 10^{-5}\text{ K}^{-1}) \times 50\text{ K} = 3.0\text{ cm}^3

Substitute initial volume, volume expansivity, and temperature change into the volume expansion formula.

Key Concept

Relationship between linear and volume expansivity (

\gamma = 3\alpha

) and application of the volume expansion formula.

Estimated Time:1m 30s

Question 7959Question →

A radioactive isotope has a half-life of $10\text{ hours}$ . If a sample initially contains $64\text{ g}$ of the isotope, what mass of the isotope has decayed after an elapsed time of $30\text{ hours}$ ?

8\text{ g}

16\text{ g}

48\text{ g}

56\text{ g}

Show answer & explanation

Answer: $56\text{ g}$

Answer

The mass of the isotope that has decayed after

30\text{ hours}

is

56\text{ g}

.

The correct answer is

56\text{ g}

. With a half-life of

10\text{ hours}

, an elapsed time of

30\text{ hours}

represents

3

half-lives. After

3

half-lives, the mass remaining undecayed is

64\text{ g} \times (1/2)^3 = 8\text{ g}

. Therefore, the mass that has decayed is

64\text{ g} - 8\text{ g} = 56\text{ g}

.

Step-by-Step Solution

1

Calculate the number of half-lives (

n

) that have elapsed.

n = \frac{t}{T_{1/2}} = \frac{30\text{ hours}}{10\text{ hours}} = 3\text{ half-lives}

Determining how many half-life intervals occur during the total elapsed time.

2

Calculate the mass of the sample remaining undecayed (

N

).

N = N_0 \left(\frac{1}{2}\right)^n = 64\text{ g} \times \left(\frac{1}{2}\right)^3 = 64\text{ g} \times \frac{1}{8} = 8\text{ g}

Using the radioactive decay formula to find the remaining undecayed mass.

3

Calculate the mass of the sample that has decayed (

N_{\text{decayed}}

).

N_{\text{decayed}} = N_0 - N = 64\text{ g} - 8\text{ g} = 56\text{ g}

Subtracting the remaining mass from the initial mass to find the amount decayed.

Key Concept

Radioactive Decay Law and Half-life

Estimated Time:45s

Question 7960Question →

A point $P(x, y)$ moves in the Cartesian plane such that it is always equidistant from the two fixed points $A(2, 1)$ and $B(6, 5)$ . If the locus of $P$ intersects the line $2x + y = 14$ at the point $(x_0, y_0)$ , what is the value of $x_0$ ?

Show answer & explanation

Answer: 7

Answer

The value of

x_0

is

7

.

The locus of points equidistant from

A(2, 1)

and

B(6, 5)

is the perpendicular bisector of line segment

AB

. The midpoint of

AB

is

(4, 3)

and its slope is

1

, giving the perpendicular bisector a slope of

-1

. The equation of this locus is

y - 3 = -1(x - 4)

, or

x + y = 7

. Subtracting

x + y = 7

from

2x + y = 14

directly gives

x_0 = 7

.

Step-by-Step Solution

1

Determine the equation of the locus of point P

The locus of P is the perpendicular bisector of segment AB, represented by

x + y = 7

.

The set of all points equidistant from two fixed points forms the perpendicular bisector of the line segment connecting those points.

2

Find the point of intersection with the line

2x + y = 14

Solving

x + y = 7

and

2x + y = 14

simultaneously gives

x_0 = 7

.

The intersection point of two geometric lines must satisfy both equations simultaneously.

Key Concept

Perpendicular Bisector Locus and Line Intersections

Answer

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