All practice questions

Question 7981Question →

A binary operation $\star$ defined on the set of real numbers $\mathbb{R}$ is given by $x \star y = \frac{x + y}{x - y + 1}$ for $x - y \neq -1$ . If $5 \star p = 3$ , what is the value of $p$ ?

Show answer & explanation

Answer: 3.25

Answer

The value of

p

is

3.25

(or

\frac{13}{4}

).

Applying the operation rule

x \star y = \frac{x + y}{x - y + 1}

with

x = 5

and

y = p

gives

\frac{5 + p}{6 - p} = 3

. Cross-multiplying gives

5 + p = 18 - 3p

, which simplifies to

4p = 13

or

p = 3.25

.

Step-by-Step Solution

1

Substitute

x = 5

and

y = p

into the given binary operation definition

\frac{5 + p}{5 - p + 1} = 3

This sets up the equation for the given condition

5 \star p = 3

.

2

Simplify the denominator in the algebraic fraction

\frac{5 + p}{6 - p} = 3

Combining the constants

5 + 1 = 6

simplifies the denominator.

3

Multiply both sides by

(6 - p)

and expand

5 + p = 18 - 3p

Eliminating the denominator allows linear terms in

p

to be collected.

4

Rearrange terms to solve for

p

4p = 13 \implies p = 3.25

Adding

3p

to both sides and subtracting

5

gives

4p = 13

.

Key Concept

Solving linear equations derived from non-commutative binary operations

Question 7982Question →

Which of the following combinations consists entirely of derived physical quantities?

Length, acceleration, and thermodynamic temperature

Volume, momentum, and electrical potential difference

Mass, velocity, and luminous intensity

Time, force, and electric current

Show answer & explanation

Answer: Volume, momentum, and electrical potential difference

Answer

The combination comprising volume, momentum, and electrical potential difference consists entirely of derived physical quantities.

The combination containing volume, momentum, and electrical potential difference consists entirely of derived physical quantities because volume depends on length (

L^3

), momentum depends on mass, length, and time (

M\cdot L\cdot T^{-1}

), and electrical potential difference depends on mass, length, time, and electric current (

M\cdot L^2\cdot T^{-3}\cdot I^{-1}

). None of these three are base/fundamental quantities.

Step-by-Step Solution

1

Identify the seven fundamental physical quantities in SI units.

The seven fundamental quantities are mass, length, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.

Fundamental quantities serve as the basic foundation from which all other physical quantities are defined.

2

Classify each physical quantity present in the given combinations.

Volume (

m^3

), momentum (

kg\cdot m/s

), and potential difference (

V

or

kg\cdot m^2/(A\cdot s^3)

) are all derived from fundamental units. Other combinations contain fundamental quantities such as length, mass, time, temperature, or electric current.

Derived quantities are physical quantities defined by mathematical combinations of fundamental quantities.

3

Select the option where every listed quantity is a derived quantity.

The group containing volume, momentum, and electrical potential difference is the only set where all items are derived quantities.

It fulfills the requirement of containing exclusively derived physical quantities.

Key Concept

Fundamental quantities are independent quantities defined by international standard protocols, whereas derived quantities are formed by mathematical combinations of fundamental quantities.

Question 7983Question →

An electric immersion heater with a resistance of $20\,\Omega$ carries a steady current of $3\,\text{A}$ . How much electrical energy is dissipated as heat by the heater in $10\,\text{seconds}$ ?

600\,\text{J}

1200\,\text{J}

1800\,\text{J}

180\,\text{J}

Show answer & explanation

Answer: $1800\,\text{J}$

Answer

The total electrical energy dissipated as heat by the heater is

1800\,\text{J}

.

According to Joule's law of heating, the electrical energy

E

converted into heat energy in a resistor is given by

E = I^2 R t

. Substituting

I = 3\,\text{A}

,

R = 20\,\Omega

, and

t = 10\,\text{s}

yields

E = (3)^2 \times 20 \times 10 = 9 \times 200 = 1800\,\text{J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Resistance

R = 20\,\Omega

, Current

I = 3\,\text{A}

, Time

t = 10\,\text{s}

Recognize the required values needed to compute heat energy dissipated in a resistor.

2

Select the appropriate formula for heat energy

E = I^2 R t

Joule's Law of Heating relates electrical energy, current, resistance, and time.

3

Substitute the known values into the equation and calculate

E = (3)^2 \times 20 \times 10 = 9 \times 20 \times 10 = 1800\,\text{J}

Perform arithmetic multiplication to find the final heat energy in Joules.

Key Concept

Joule's Law of Heating

Question 7984Question →

A block of mass $0.5\text{ kg}$ absorbs $4500\text{ J}$ of heat energy as its temperature rises from $25^\circ\text{C}$ to $45^\circ\text{C}$ . What is the heat capacity of the block?

450\text{ J K}^{-1}

225\text{ J K}^{-1}

112.5\text{ J K}^{-1}

225\text{ J kg}^{-1}\text{ K}^{-1}

Show answer & explanation

Answer: $225\text{ J K}^{-1}$

Answer

225\text{ J K}^{-1}

Heat capacity (

C

) is defined as thermal energy absorbed divided by temperature change:

C = \frac{Q}{\Delta T} = \frac{4500\text{ J}}{45\text{ K} - 25\text{ K}} = 225\text{ J K}^{-1}

.

Step-by-Step Solution

1

Calculate the temperature change (

\Delta T

).

\Delta T = 45^\circ\text{C} - 25^\circ\text{C} = 20^\circ\text{C} = 20\text{ K}

Heat capacity depends on the temperature change in Kelvin or degrees Celsius.

2

Apply the heat capacity formula

C = \frac{Q}{\Delta T}

.

C = \frac{4500\text{ J}}{20\text{ K}} = 225\text{ J K}^{-1}

Heat capacity (

C

) measures the heat required to raise the temperature of the entire body by

1\text{ K}

, regardless of mass.

Key Concept

Heat capacity (

C

) represents the energy required to change an entire object's temperature by one kelvin (

C = \frac{Q}{\Delta T}

), whereas specific heat capacity (

c

) is per unit mass (

c = \frac{Q}{m\Delta T}

).

Estimated Time:1m 0s

Question 7985Question →

For an object of constant mass moving at constant speed along a complete circular path, the net vector impulse imparted to the object over one full revolution is zero, even though a continuous centripetal force acts on the object throughout the motion.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because the impulse-momentum theorem states that net impulse equals the change in momentum (

\vec{J} = \Delta \vec{p}

). After one full revolution, the object's initial and final velocity vectors are identical, resulting in zero change in momentum.

The statement is correct because linear momentum is a vector quantity. Over a complete circular revolution, the initial and final velocity vectors are identical in both magnitude and direction, making the change in momentum—and therefore the net vector impulse—equal to zero.

Step-by-Step Solution

1

Recall the vector definition of impulse and the impulse-momentum theorem.

The net impulse

\vec{J}

acting on a body equals its change in linear momentum:

\vec{J} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i = m\vec{v}_f - m\vec{v}_i

.

Impulse is a vector quantity dependent on the initial and final states of momentum over the time interval.

2

Evaluate the velocity vector of the object after one complete circular revolution at constant speed.

Since the speed is constant and the trajectory completes a closed loop, the final velocity vector

\vec{v}_f

has the exact same magnitude and direction as the initial velocity vector

\vec{v}_i

.

A complete revolution returns the object to its starting point with its velocity pointing in the initial direction.

3

Calculate the change in momentum

\Delta \vec{p}

.

\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = m(0) = \vec{0}

. Thus, net impulse

\vec{J} = \vec{0}

.

Subtracting identical vectors yields zero vector magnitude.

Key Concept

Impulse-Momentum Theorem and Vector Nature of Linear Momentum

Estimated Time:1m 30s

Question 7986Question →

A battery with an electromotive force (e.m.f.) of $12.0\text{ V}$ and an internal resistance of $1.0\text{ }\Omega$ is connected across a series combination of a $2.0\text{ }\Omega$ resistor and a $3.0\text{ }\Omega$ resistor. What is the potential difference across the terminals of the battery?

10.0\text{ V}

12.0\text{ V}

2.0\text{ V}

6.0\text{ V}

Show answer & explanation

Answer: $10.0\text{ V}$

Answer

10.0\text{ V}

The total resistance of the circuit is the sum of the external load resistors and the internal resistance (

2.0 + 3.0 + 1.0 = 6.0\text{ }\Omega

). The total current is

I = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

. The potential difference across the terminals of the battery is the voltage across the external load

V = I \times R_{\text{ext}} = 2.0\text{ A} \times 5.0\text{ }\Omega = 10.0\text{ V}

(or

E - Ir = 12.0 - 2.0 = 10.0\text{ V}

).

Step-by-Step Solution

1

Calculate total external resistance in series

R_{\text{ext}} = 2.0\text{ }\Omega + 3.0\text{ }\Omega = 5.0\text{ }\Omega

Resistors in series add directly to give the total external load resistance.

2

Calculate total circuit resistance including internal resistance

R_{\text{total}} = R_{\text{ext}} + r = 5.0\text{ }\Omega + 1.0\text{ }\Omega = 6.0\text{ }\Omega

The cell's internal resistance is in series with the external circuit.

3

Determine the total current drawn from the battery

I = \frac{E}{R_{\text{total}}} = \frac{12.0\text{ V}}{6.0\text{ }\Omega} = 2.0\text{ A}

Applying Ohm's law to the entire circuit.

4

Calculate terminal potential difference across the battery

V = E - Ir = 12.0\text{ V} - (2.0\text{ A} \times 1.0\text{ }\Omega) = 10.0\text{ V}

Terminal voltage is the e.m.f. minus the lost volts across internal resistance (or equivalently

V = I \times R_{\text{ext}}

).

Key Concept

Terminal potential difference vs. electromotive force (e.m.f.) and lost volts

Estimated Time:1m 30s

Question 7987Question →

An electric crane lifts a load of $250\text{ kg}$ vertically upwards through a height of $12\text{ m}$ in $10\text{ s}$ at a constant speed. Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what is the useful output power of the crane in watts?

Show answer & explanation

Answer: 3000

Answer

The useful output power of the crane is

3000\text{ W}

.

The work done in lifting the load vertically is equal to the gravitational potential energy gained,

W = mgh = 250 \times 10 \times 12 = 30,000\text{ J}

. Power is the rate of doing work, so

P = \frac{W}{t} = \frac{30,000}{10} = 3000\text{ W}

.

Step-by-Step Solution

1

Identify the given values

Mass

m = 250\text{ kg}

, height

h = 12\text{ m}

, time

t = 10\text{ s}

, acceleration due to gravity

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

.

Extract values needed for work and power calculations.

2

Calculate the work done in lifting the load

W = mgh = 250 \text{ kg} \times 10 \text{ m s}^{-2} \times 12 \text{ m} = 30,000\text{ J}

.

The work done against gravity equals the gain in gravitational potential energy.

3

Calculate the power output

P = \frac{W}{t} = \frac{30,000\text{ J}}{10\text{ s}} = 3000\text{ W}

.

Power is defined as the rate at which work is done (

P = \frac{W}{t}

).

Key Concept

Power as the rate of doing work against gravity

Question 7988Question →

Given the matrices $A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 4 \\ 2 & -1 \end{pmatrix}$ , what is the product matrix $AB$ ?

\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}

\begin{pmatrix} 2 & 4 \\ 0 & -3 \end{pmatrix}

\begin{pmatrix} 2 & 13 \\ 4 & -1 \end{pmatrix}

\begin{pmatrix} 4 & 9 \\ 6 & -3 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}$

Answer

The matrix product

AB

is

\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}

.

To find

AB

, each element

c_{ij}

is formed by taking the dot product of row

i

of matrix

A

and column

j

of matrix

B

. Performing these steps gives

c_{11} = 2(1) + 1(2) = 4

,

c_{12} = 2(4) + 1(-1) = 7

,

c_{21} = 0(1) + 3(2) = 6

, and

c_{22} = 0(4) + 3(-1) = -3

, yielding

\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}

.

Step-by-Step Solution

1

Multiply the first row of

A

by the first column of

B

to find element

(1,1)

.

2(1) + 1(2) = 2 + 2 = 4

Matrix multiplication rule requires taking the dot product of rows from the first matrix and columns from the second matrix.

2

Multiply the first row of

A

by the second column of

B

to find element

(1,2)

.

2(4) + 1(-1) = 8 - 1 = 7

Evaluates the element in row 1, column 2 of the resulting matrix.

3

Multiply the second row of

A

by the first column of

B

to find element

(2,1)

.

0(1) + 3(2) = 0 + 6 = 6

Evaluates the element in row 2, column 1 of the resulting matrix.

4

Multiply the second row of

A

by the second column of

B

to find element

(2,2)

.

0(4) + 3(-1) = 0 - 3 = -3

Evaluates the element in row 2, column 2 of the resulting matrix.

Key Concept

Matrix Multiplication

Estimated Time:1m 0s

Question 7989Question →

A concave mirror forms a real image that is twice the size of an object. When the object is shifted $10\text{ cm}$ closer to the mirror, a virtual image of the same magnification is produced. What is the focal length of the mirror?

10\text{ cm}

15\text{ cm}

20\text{ cm}

30\text{ cm}

Show answer & explanation

Answer: $10\text{ cm}$

Answer

The focal length of the concave mirror is

10\text{ cm}

.

For a concave mirror forming a real image of magnification

2

,

v_1 = 2u_1

, yielding

u_1 = 1.5f

. When the object is moved

10\text{ cm}

closer, a virtual image of magnification

2

is formed, so

v_2 = -2u_2

, yielding

u_2 = 0.5f

. Subtracting the two object positions (

1.5f - 0.5f = 10\text{ cm}

) directly gives

f = 10\text{ cm}

.

Step-by-Step Solution

1

Set up the mirror equation for the first case (real image).

u_1 = \frac{3}{2}f

For a real inverted image with magnification

m = 2

,

v_1 = +2u_1

. Substituting into

\frac{1}{f} = \frac{1}{u_1} + \frac{1}{v_1}

gives

\frac{1}{f} = \frac{1}{u_1} + \frac{1}{2u_1} = \frac{3}{2u_1}

.

2

Set up the mirror equation for the second case (virtual image).

u_2 = \frac{1}{2}f

For a virtual erect image with magnification

m = 2

, sign convention dictates

v_2 = -2u_2

. Substituting into

\frac{1}{f} = \frac{1}{u_2} + \frac{1}{v_2}

gives

\frac{1}{f} = \frac{1}{u_2} - \frac{1}{2u_2} = \frac{1}{2u_2}

.

3

Use the given displacement between the two object positions to solve for

f

.

f = 10\text{ cm}

The object is moved

10\text{ cm}

closer, so

u_1 - u_2 = 10\text{ cm}

. Substituting the expressions yields

\frac{3}{2}f - \frac{1}{2}f = 10\text{ cm} \implies f = 10\text{ cm}

.

Key Concept

Mirror Formula and Sign Convention for Spherical Mirrors

Estimated Time:2m 0s

Question 7990Question →

The volume flow rate $Q$ of a viscous liquid through a pipe depends on the radius $r$ of the pipe, the coefficient of viscosity $\eta$ , and the pressure gradient $\frac{\Delta P}{L}$ according to the dimensional equation $Q = k r^x \eta^y \left(\frac{\Delta P}{L}\right)^z$ , where $k$ is a dimensionless constant. What is the value of the exponent $x$ ?

Show answer & explanation

Answer: 4

Answer

The value of the exponent

x

is 4.

Applying the principle of dimensional homogeneity, the dimensions of volume flow rate

[Q] = L^3 T^{-1}

are equated to

[r]^x [\eta]^y \left[\frac{\Delta P}{L}\right]^z = L^x (M L^{-1} T^{-1})^y (M L^{-2} T^{-2})^z

. Equating powers yields

y + z = 0

for mass,

-y - 2z = -1

for time, and

x - y - 2z = 3

for length. Solving these simultaneous equations gives

z = 1

,

y = -1

, and

x = 4

.

Step-by-Step Solution

1

Identify the base dimensions of each physical quantity in the given equation.

Flow rate

[Q] = M^0 L^3 T^{-1}

, radius

[r] = L

, viscosity

[\eta] = M L^{-1} T^{-1}

, and pressure gradient

\left[\frac{\Delta P}{L}\right] = M L^{-2} T^{-2}

.

Expressing each quantity in terms of fundamental dimensions (

M

,

L

,

T

) is necessary for dimensional analysis.

2

Substitute dimensions into the power-law equation and collect powers of base dimensions.

M^0 L^3 T^{-1} = M^{y+z} L^{x-y-2z} T^{-y-2z}

.

The principle of dimensional homogeneity requires both sides of a physically valid equation to have identical dimensions.

3

Set up and solve linear equations for the exponents

x

,

y

, and

z

.

Solving

y + z = 0

,

-y - 2z = -1

, and

x - y - 2z = 3

yields

z = 1

,

y = -1

, and

x = 4

.

Equating powers of

M

,

T

, and

L

allows step-by-step determination of each unknown exponent.

Key Concept

Dimensions of Physical Quantities and Dimensional Analysis

Question 7991Question →

Match each vibrating acoustic system setup on the left with the correct mathematical expression for its resonant frequency ( $f$ ) on the right, where $v$ is the speed of sound in air, $L$ is the physical length of the pipe or string, $e$ is the end correction per open end, $T$ is tension, and $\mu$ is linear mass density.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The fundamental mode of a pipe closed at one end with end correction matches

f = \frac{v}{4(L + e)}

; the fundamental mode of a stretched string matches

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

; the fundamental mode of a pipe open at both ends with end correction at both ends matches

f = \frac{v}{2(L + 2e)}

; and the first overtone of a closed pipe without end correction matches

f = \frac{3v}{4L}

.

Each setup corresponds directly to its derived wave equation: closed pipes produce fundamental frequency

f = \frac{v}{4(L+e)}

for one open end, open pipes produce

f = \frac{v}{2(L+2e)}

for two open ends, stretched strings depend on tension and mass per unit length as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

, and the first overtone of a closed pipe is its third harmonic

f = \frac{3v}{4L}

.

Step-by-Step Solution

1

Analyze boundary conditions and effective acoustic length for closed and open pipes.

A closed pipe has one displacement antinode at the open end and one node at the closed end, adding an effective end correction

e

to its physical length

L

(

L_{\text{eff}} = L + e

). An open pipe has two open ends, giving an effective length

L_{\text{eff}} = L + 2e

.

Air displacement antinodes occur slightly outside open pipe boundaries by a distance

e

per open end.

2

Derive the frequency formula for the fundamental mode of a closed pipe with end correction.

For the fundamental mode,

L + e = \frac{\lambda}{4} \implies \lambda = 4(L + e)

. Frequency

f = \frac{v}{\lambda} = \frac{v}{4(L + e)}

.

The distance between a node and an adjacent antinode is one-quarter of a wavelength.

3

Derive the fundamental frequency for a stretched string fixed at both ends.

L = \frac{\lambda}{2} \implies \lambda = 2L

. Using wave velocity

v = \sqrt{\frac{T}{\mu}}

,

f = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

.

Nodes exist at both fixed ends in a vibrating string, making the fundamental wavelength twice the length.

4

Derive the fundamental frequency of an open pipe considering both end corrections.

L + 2e = \frac{\lambda}{2} \implies \lambda = 2(L + 2e)

, so

f = \frac{v}{2(L + 2e)}

.

Antinodes occur at both open ends, placing half a wavelength within the effective acoustic length.

5

Determine the first overtone frequency for a closed pipe without end correction.

The first overtone is the third harmonic (

n = 3

), so

L = \frac{3\lambda}{4} \implies \lambda = \frac{4L}{3}

, which gives

f = \frac{3v}{4L}

.

Closed pipes support only odd integer multiples of the fundamental frequency.

Key Concept

Standing Waves and Resonance in Air Columns and Strings

Question 7992Question →

A ticker-tape timer connected to a $50\text{ Hz}$ alternating current mains supply is used to measure the time interval of a laboratory cart moving down an inclined plane. On the printed tape, $10$ distinct tick spaces are recorded between the initial and final position. A electronic stopwatch running concurrently has a positive zero error of $+0.04\text{ s}$ . What is the true elapsed time interval represented by the ticker tape after properly correcting for the stopwatch zero error?

0.16\text{ s}

0.24\text{ s}

0.20\text{ s}

0.22\text{ s}

Show answer & explanation

Answer: $0.16\text{ s}$

Answer

The true elapsed time interval is

0.16\text{ s}

.

The ticker-tape frequency of

50\text{ Hz}

gives a time interval of

0.02\text{ s}

per space. Ten tick spaces correspond to an uncorrected time of

0.20\text{ s}

. To correct for a positive zero error of

+0.04\text{ s}

, the zero error must be subtracted from the uncorrected reading, yielding a true elapsed time of

0.16\text{ s}

.

Step-by-Step Solution

1

Calculate the period of a single tick interval

T = \frac{1}{f} = \frac{1}{50\text{ Hz}} = 0.02\text{ s}

Frequency

f

is the inverse of period

T

for a ticker-tape timer.

2

Determine the uncorrected total elapsed time from tick spaces

t_{\text{uncorrected}} = 10 \times 0.02\text{ s} = 0.20\text{ s}

Elapsed time equals the number of tick spaces multiplied by the period of one space.

3

Apply the zero error correction

t_{\text{true}} = t_{\text{uncorrected}} - \text{zero error} = 0.20\text{ s} - 0.04\text{ s} = 0.16\text{ s}

True Reading = Observed Reading - (Zero Error).

Key Concept

Measurement of time using ticker-tape timers and instrument zero error correction

Estimated Time:1m 30s

Question 7993Question →

A hiker starts at camp $C$ and walks $5\text{ km}$ due East to checkpoint $A$ . From checkpoint $A$ , the hiker then walks $3\text{ km}$ on a bearing of $150^\circ$ to reach checkpoint $B$ . What is the direct distance from camp $C$ to checkpoint $B$ ?

7\text{ km}

\sqrt{34}\text{ km}

\sqrt{19}\text{ km}

8\text{ km}

Show answer & explanation

Answer: $7\text{ km}$

Answer

7\text{ km}

The interior angle at checkpoint

A

is calculated using three-figure bearings as

270^\circ - 150^\circ = 120^\circ

. Applying the Cosine Rule

CB^2 = 5^2 + 3^2 - 2(5)(3)\cos(120^\circ)

yields

25 + 9 + 15 = 49

. Taking the square root gives the direct distance of

7\text{ km}

.

Step-by-Step Solution

1

Determine the interior angle

\angle CAB

at checkpoint

A

\angle CAB = 120^\circ

Due East corresponds to a bearing of

090^\circ

. Coming into

A

from

C

means line

AC

points West (

270^\circ

). The bearing of

B

from

A

is

150^\circ

. The interior angle between vector

AC

pointing West (

270^\circ

) and vector

AB

on bearing

150^\circ

is

270^\circ - 150^\circ = 120^\circ

.

2

Apply the Cosine Rule to find length

CB

CB^2 = CA^2 + AB^2 - 2(CA)(AB)\cos(\angle CAB)

We have two sides (

CA = 5\text{ km}

,

AB = 3\text{ km}

) and the included angle (

\angle CAB = 120^\circ

).

3

Substitute the known values into the Cosine Rule formula

CB^2 = 5^2 + 3^2 - 2(5)(3)\cos(120^\circ) = 25 + 9 - 30(-0.5) = 34 + 15 = 49

Since

\cos(120^\circ) = -\cos(60^\circ) = -0.5

, the negative sign inside the cosine product cancels with the subtraction sign in the formula.

4

Take the positive square root to find the distance

CB = \sqrt{49} = 7\text{ km}

Distance must be positive.

Key Concept

Cosine Rule for non-right triangles in bearings problems

Question 7994Question →

Determine whether the following statement is true or false: When a bar magnet is placed in the Earth's magnetic field with its north pole pointing towards geographic North, the neutral points formed by the mutual cancellation of the magnet's field and the Earth's horizontal field lie along the magnet's equatorial line (broadside-on position).

True

False

Show answer & explanation

Answer: True

Answer

True. When a bar magnet has its north pole pointing North, the magnetic field along its equatorial (broadside-on) line points South, opposing the Earth's horizontal magnetic field to create neutral points.

The statement is correct because neutral points are formed at positions where the external magnetic field of a magnet is equal in magnitude and opposite in direction to the Earth's horizontal magnetic field component (

B_H

). When the north pole of a bar magnet points North, its field along the broadside-on (equatorial) axis is directed South (opposite to

B_H

), leading to field cancellation and the creation of two neutral points on that axis.

Step-by-Step Solution

1

Identify the direction of the Earth's horizontal magnetic field component (

B_H

).

Earth's horizontal field (

B_H

) acts along the magnetic meridian from magnetic South to magnetic North.

Neutral points can only occur where two magnetic fields act in opposite directions and have equal magnitudes.

2

Determine the direction of the bar magnet's magnetic field on its broadside-on (equatorial) axis.

Outside the magnet, magnetic flux lines travel from North to South. Along the broadside-on line, the magnet's field points southward.

Field lines curve from the north pole back into the south pole externally.

3

Compare the magnetic field directions on the broadside-on axis.

The southward field of the magnet opposes the northward Earth field (

B_H

). At points where their magnitudes match, the net magnetic field becomes zero.

Vector cancellation requires antiparallel vectors of equal magnitude.

Key Concept

Neutral Points of a Bar Magnet in Earth's Magnetic Field

Question 7995Question →

A student records the period of oscillation of a simple pendulum by measuring the time taken for 20 complete swings as $40.0\text{ s}$ . If the absolute error in this time measurement is $\pm 0.8\text{ s}$ , what is the percentage error in the measured time?

0.2\%

2.0\%

5.0\%

8.0\%

Show answer & explanation

Answer: $2.0\%$

Answer

2.0\%

The correct answer is

2.0\%

. Percentage error is defined as the absolute error divided by the measured value, expressed as a percentage:

\frac{0.8\text{ s}}{40.0\text{ s}} \times 100\% = 2.0\%

.

Step-by-Step Solution

1

Identify the given measurement and absolute error.

Measured time

t = 40.0\text{ s}

and absolute uncertainty

\Delta t = 0.8\text{ s}

.

These quantities are needed to determine the relative error of the time measurement.

2

Apply the percentage error formula:

\text{Percentage Error} = \left(\frac{\Delta t}{t}\right) \times 100\%

.

\text{Percentage Error} = \left(\frac{0.8\text{ s}}{40.0\text{ s}}\right) \times 100\% = 0.02 \times 100\% = 2.0\%

.

Multiplying the fractional error by 100 converts it into a percentage representation.

Key Concept

Percentage Error in Physical Measurements

Estimated Time:45s

Question 7996Question →

A body of mass $2.5\text{ kg}$ moving at a speed of $4.0\text{ m s}^{-1}$ along a straight horizontal path is brought to rest in $2.0\text{ s}$ by a constant retarding force. What is the magnitude of this retarding force in newtons?

Show answer & explanation

Answer: 5

Answer

The magnitude of the retarding force is

5.0\text{ N}

.

According to Newton's second law of motion, net force is equal to the rate of change of momentum (

F = \frac{\Delta p}{\Delta t} = \frac{m(v-u)}{t}

). Substituting

m = 2.5\text{ kg}

,

u = 4.0\text{ m s}^{-1}

,

v = 0\text{ m s}^{-1}

, and

t = 2.0\text{ s}

gives

F = \frac{2.5 \times (0 - 4.0)}{2.0} = -5.0\text{ N}

. The magnitude of this force is

5.0\text{ N}

.

Step-by-Step Solution

1

Determine the initial momentum and final momentum of the body.

Initial momentum

p_i = 2.5 \times 4.0 = 10.0\text{ kg m s}^{-1}

, and final momentum

p_f = 0\text{ kg m s}^{-1}

.

Linear momentum is defined as the product of mass and velocity (

p = mv

).

2

Calculate the magnitude of the force applied using the impulse-momentum relationship

F = \frac{\Delta p}{\Delta t}

.

Magnitude of force

F = \frac{|0 - 10.0|}{2.0} = 5.0\text{ N}

.

Newton's second law states that the rate of change of momentum is equal to the net external force applied.

Key Concept

Newton's Second Law and Impulse-Momentum Relationship

Estimated Time:45s

Question 7997Question →

A water wave traveling in deep water has a wavelength of $0.80\text{ m}$ and a speed of $2.4\text{ m/s}$ . Upon entering a shallow region, its speed drops to $1.8\text{ m/s}$ . What are the frequency and wavelength of the wave in the shallow region?

Frequency =

3.0\text{ Hz}

, Wavelength =

0.60\text{ m}

Frequency =

2.25\text{ Hz}

, Wavelength =

0.80\text{ m}

Frequency =

2.25\text{ Hz}

, Wavelength =

0.60\text{ m}

Frequency =

0.33\text{ Hz}

, Wavelength =

0.60\text{ m}

Show answer & explanation

Answer: Frequency = $3.0\text{ Hz}$ , Wavelength = $0.60\text{ m}$

Answer

Frequency =

3.0\text{ Hz}

, Wavelength =

0.60\text{ m}

When a wave passes from deep to shallow water (refraction), its frequency remains unchanged because frequency is fixed by the source. Using

v = f\lambda

, the initial frequency is

f = \frac{2.4}{0.80} = 3.0\text{ Hz}

. In shallow water, the new wavelength is

\lambda' = \frac{1.8}{3.0} = 0.60\text{ m}

.

Step-by-Step Solution

1

Calculate the frequency of the wave in deep water using the wave equation

v = f \lambda

.

f = \frac{v}{\lambda} = \frac{2.4\text{ m/s}}{0.80\text{ m}} = 3.0\text{ Hz}

.

The frequency depends on the wave source and can be determined from the given initial speed and wavelength.

2

Apply the boundary condition for wave refraction.

The frequency in shallow water remains

f = 3.0\text{ Hz}

.

When a wave travels from one medium to another, its frequency remains constant.

3

Calculate the new wavelength in shallow water using

\lambda' = \frac{v'}{f}

.

\lambda' = \frac{1.8\text{ m/s}}{3.0\text{ Hz}} = 0.60\text{ m}

.

The wavelength changes proportionally with speed when frequency is constant.

Key Concept

Constancy of wave frequency during refraction across medium boundaries

Question 7998Question →

Match each of the physical quantities given on the left with its corresponding fundamental SI status or base unit resolution on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Luminous intensity matches with Fundamental quantity measured in candela (cd); Linear momentum matches with Derived quantity expressed in kg·m·s⁻¹; Thermodynamic temperature matches with Fundamental quantity measured in kelvin (K); Specific heat capacity matches with Derived quantity expressed in m²·s⁻²·K⁻¹.

Luminous intensity and thermodynamic temperature are fundamental physical quantities with SI units candela (

\text{cd}

) and kelvin (

\text{K}

) respectively. Linear momentum (

p = mv

) and specific heat capacity (

c = \frac{Q}{m \Delta T}

) are derived physical quantities whose fundamental SI base unit resolutions are

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

and

\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}

respectively.

Step-by-Step Solution

1

Identify fundamental physical quantities and their base SI units

Luminous intensity (measured in

\text{cd}

) and thermodynamic temperature (measured in

\text{K}

) are base physical quantities that cannot be expressed in terms of other quantities.

The standard SI system establishes 7 base independent quantities.

2

Resolve derived quantities into fundamental SI base units

Linear momentum (

p = mv

) has units

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

. Specific heat capacity (

c = \frac{Q}{m\Delta T}

) has units

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

.

Derived physical quantities are formed by combining fundamental quantities algebraically according to physical laws.

3

Match each physical quantity to its correct description

Luminous intensity

\rightarrow

candela (

\text{cd}

); Linear momentum

\rightarrow

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

; Thermodynamic temperature

\rightarrow

kelvin (

\text{K}

); Specific heat capacity

\rightarrow

\text{m}^2\cdot\text{s}^{-2}\cdot\text{K}^{-1}

.

Each pair directly aligns the physical quantity with its base classification and unit derivation.

Key Concept

Classification of fundamental and derived physical quantities and their resolution into SI base units

Question 7999Question →

Monochromatic radiation carrying photons of energy $4.8\text{ eV}$ illuminates a cesium surface inside a photoelectric cell. If the work function of cesium is $2.1\text{ eV}$ , determine the stopping potential, in volts, needed to reduce the photoelectric current to zero.

Show answer & explanation

Answer: 2.7

Answer

The stopping potential required to reduce the photoelectric current to zero is

2.7\text{ V}

.

Einstein's photoelectric equation states that incident photon energy

E

equals the work function

W_0

plus the maximum kinetic energy

K_{\text{max}}

of the photoelectrons (

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

). Rearranging gives

K_{\text{max}} = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}

. Since

K_{\text{max}} = e V_s

, an electron-volt value of kinetic energy numerically equals the stopping potential in volts, giving a stopping potential of

2.7\text{ V}

.

Step-by-Step Solution

1

Calculate the maximum kinetic energy (

K_{\text{max}}

) of the emitted photoelectrons.

K_{\text{max}} = E - W_0 = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}

According to Einstein's photoelectric equation, incident photon energy is divided into overcoming the work function of the metal and providing kinetic energy to the liberated electron.

2

Determine the stopping potential (

V_s

) from the maximum kinetic energy.

V_s = \frac{K_{\text{max}}}{e} = \frac{2.7\text{ eV}}{e} = 2.7\text{ V}

The stopping potential

V_s

is the opposing potential difference needed to stop the fastest moving photoelectrons, defined by

K_{\text{max}} = e V_s

.

Key Concept

Photoelectric Effect and Work Function

Question 8000Question →

A binary operation $\Delta$ defined on the set of real numbers $\mathbb{R}$ is given by $a \Delta b = a + b - \frac{ab}{4}$ . What is the inverse of $2$ under this operation?

-4

4

-2

-\frac{4}{3}

Show answer & explanation

Answer: $-4$

Answer

The inverse of

2

under the binary operation is

-4

.

To find the inverse of

2

, we first determine the identity element

e

using

a \Delta e = a

, which yields

e = 0

. Setting

2 \Delta x = 0

gives

2 + x - \frac{2x}{4} = 0

, which simplifies to

2 + \frac{x}{2} = 0

, resulting in

x = -4

.

Step-by-Step Solution

1

Determine the identity element

e

of the operation.

e = 0

By definition,

a \Delta e = a

. Thus,

a + e - \frac{ae}{4} = a \implies e\left(1 - \frac{a}{4}\right) = 0

, which gives

e = 0

for all real

a

.

2

Set up the inverse equation for the element

2

.

2 + x - \frac{2x}{4} = 0

Let

x

be the inverse of

2

. By definition,

2 \Delta x = e

, where

e = 0

.

3

Simplify and solve for

x

.

x = -4

Simplify

2 + x - \frac{x}{2} = 0 \implies 2 + \frac{x}{2} = 0 \implies \frac{x}{2} = -2 \implies x = -4

.

Key Concept

Inverse Element in Binary Operations

Answer

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