All practice questions

Question 8001Question →

An electric cell of electromotive force $E$ and internal resistance $r$ is connected in series with a fixed resistor of resistance $8.0\text{ }\Omega$ and a galvanometer of internal resistance $40.0\text{ }\Omega$ . When a shunt resistor of resistance $10.0\text{ }\Omega$ is connected in parallel across the galvanometer, the total current supplied by the cell increases by $50\%$ . What is the internal resistance $r$ of the cell in ohms?

Show answer & explanation

Answer: 48

Answer

The internal resistance of the cell is

48.0\text{ }\Omega

.

By determining the initial total circuit resistance

R_{T1} = 48.0 + r

and the post-shunt total circuit resistance

R_{T2} = 16.0 + r

, we apply Ohm's law with

I_2 = 1.5 I_1

. Equating total supply voltage

E

gives

\frac{E}{16.0 + r} = 1.5 \times \frac{E}{48.0 + r}

, which simplifies directly to

r = 48.0\text{ }\Omega

.

Step-by-Step Solution

1

Formulate the initial total resistance of the series circuit

R_{T1} = 8.0\text{ }\Omega + 40.0\text{ }\Omega + r = 48.0 + r

Before the shunt is added, the fixed resistor, galvanometer, and internal resistance of the cell are all connected in series.

2

Calculate the effective resistance of the shunted galvanometer

R_p = \frac{40.0 \times 10.0}{40.0 + 10.0} = 8.0\text{ }\Omega

Connecting the shunt in parallel across the galvanometer forms a parallel network.

3

Formulate the final total circuit resistance after shunting

R_{T2} = 8.0\text{ }\Omega + 8.0\text{ }\Omega + r = 16.0 + r

The new circuit consists of the fixed resistor, the parallel shunted galvanometer combination, and the internal resistance in series.

4

Set up the ratio equation for total circuit currents based on the given

50\%

current increase

\frac{E}{16.0 + r} = 1.5 \left(\frac{E}{48.0 + r}\right)

An increase of

50\%

means

I_2 = 1.5 I_1

. According to Ohm's law, total current is inversely proportional to total circuit resistance for a constant EMF

E

.

5

Solve the algebraic equation for internal resistance

r

r = 48.0\text{ }\Omega

Cross-multiplying yields

48.0 + r = 24.0 + 1.5r

, leading directly to

0.5r = 24.0

, so

r = 48.0\text{ }\Omega

.

Key Concept

Internal Resistance and Circuit Modification via Galvanometer Shunting

Question 8002Question →

A constant horizontal force of $25\text{ N}$ is applied to push a box across a smooth horizontal surface through a displacement of $8\text{ m}$ in the direction of the force. What is the total work done on the box?

200\text{ J}

40\text{ J}

3.125\text{ J}

33\text{ J}

Show answer & explanation

Answer: $200\text{ J}$

Answer

The work done on the box is

200\text{ J}

.

Work done is defined as the product of force and distance moved in the direction of the force (

W = F \times d

). Substituting the given values

F = 25\text{ N}

and

d = 8\text{ m}

gives

W = 200\text{ J}

.

Step-by-Step Solution

1

Identify the given physical quantities

Force

F = 25\text{ N}

, displacement

d = 8\text{ m}

, angle

\theta = 0^\circ

Work depends on force and displacement along the line of action

2

Apply the work formula

W = F \cdot d \cos(\theta)

W = 25\text{ N} \times 8\text{ m} \times \cos(0^\circ) = 200\text{ J}

Since force and displacement are in the same direction,

\cos(0^\circ) = 1

Key Concept

Work Done by a Constant Force

Question 8003Question →

In a manufacturing plant, two independent automated assembly units, $U_1$ and $U_2$ , undergo safety inspection. The probability that unit $U_1$ fails the inspection is $\frac{1}{5}$ , while the probability that unit $U_2$ fails the inspection is $\frac{1}{4}$ . What is the probability that at least one of the two units fails the inspection? Express your answer as a decimal.

Show answer & explanation

Answer: 0.4

Answer

0.4

The probability that at least one unit fails is determined using the addition law for non-mutually exclusive independent events:

P(U_1 \cup U_2) = P(U_1) + P(U_2) - P(U_1 \cap U_2) = 0.20 + 0.25 - (0.20 \times 0.25) = 0.45 - 0.05 = 0.40

. Alternatively, using the complement rule gives

1 - P(\text{neither fails}) = 1 - (1 - 0.20)(1 - 0.25) = 1 - 0.60 = 0.40

.

Step-by-Step Solution

1

Identify the individual failure probabilities

P(U_1) = 0.20

and

P(U_2) = 0.25

These values represent the single event failure probabilities given in the problem statement.

2

Calculate the probability that neither unit fails

P(U_1') \times P(U_2') = (1 - 0.20) \times (1 - 0.25) = 0.80 \times 0.75 = 0.60

Because the units operate independently, their complement events (passing inspection) are also independent.

3

Determine the probability of at least one unit failing

1 - 0.60 = 0.40

The event that at least one unit fails is the complementary event of neither unit failing.

Key Concept

Addition and Multiplication Laws of Probability for Independent Compound Events

Question 8004Question →

Pair each physical quantity in List I with its equivalent SI unit expressed purely in terms of fundamental base units in List II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Electric Charge matches

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

; Linear Momentum matches

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

; Young's Modulus matches

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

; Magnetic Flux Density matches

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

.

Each physical quantity is resolved to fundamental SI base units using defining equations: Electric charge (

Q=It

) yields

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

, linear momentum (

p=mv

) yields

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

, Young's modulus (stress/strain) yields

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

, and magnetic flux density (

B=\frac{F}{IL}

) yields

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

.

Step-by-Step Solution

1

Derive base units for Electric Charge

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

From

Q = I t

, electric current has the fundamental unit ampere (A) and time has the fundamental unit second (s).

2

Derive base units for Linear Momentum

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

From

p = m v

, mass is in kilograms (kg) and velocity is in meters per second (m/s).

3

Derive base units for Young's Modulus

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

Tensile strain is dimensionless. Tensile stress is force per area:

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

.

4

Derive base units for Magnetic Flux Density

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

From magnetic force

F = I L B

, solving for

B = \frac{F}{I L}

gives

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

.

Key Concept

Expressing derived physical quantities in terms of fundamental SI base units

Question 8005Question →

A person standing at a stationary position between two tall parallel vertical walls fires a starter pistol. The person hears the first echo reflected from the nearer wall after $1.2\text{ s}$ and the second echo from the farther wall after $1.8\text{ s}$ . Given that the speed of sound in air is $340\text{ m/s}$ , what is the total distance between the two walls in meters?

Show answer & explanation

Answer: 510

Answer

The total distance between the two walls is

510\text{ m}

.

Because sound travels to each wall and reflects back to the observer, the distance to each wall is given by

d = \frac{v \cdot t}{2}

. The distance to the nearer wall is

d_1 = \frac{340 \times 1.2}{2} = 204\text{ m}

, and the distance to the farther wall is

d_2 = \frac{340 \times 1.8}{2} = 306\text{ m}

. Since the observer is between the two walls, the total separation between the walls is

204\text{ m} + 306\text{ m} = 510\text{ m}

.

Step-by-Step Solution

1

Calculate the distance from the observer to the nearer wall.

d_1 = 204\text{ m}

The sound travels to the nearer wall and back in

1.2\text{ s}

, covering twice the distance to that wall.

2

Calculate the distance from the observer to the farther wall.

d_2 = 306\text{ m}

The sound travels to the farther wall and back in

1.8\text{ s}

, covering twice the distance to that wall.

3

Add the two individual distances to find the total distance between the walls.

D = d_1 + d_2 = 510\text{ m}

The observer is positioned between the two walls, so the separation distance is the sum of both distances.

Key Concept

Echo and Speed of Sound Propagation between Parallel Boundaries

Question 8006Question →

A conducting wire of length $1.5\,\text{m}$ and initial resistance $8.0\,\Omega$ is stretched uniformly until its length is doubled while its total volume remains unchanged. If a steady potential difference of $16\,\text{V}$ is applied across the ends of the stretched wire, what is the electric current flowing through it?

0.50\,\text{A}

1.00\,\text{A}

2.00\,\text{A}

8.00\,\text{A}

Show answer & explanation

Answer: $0.50\,\text{A}$

Answer

The electric current flowing through the stretched wire is

0.50\,\text{A}

.

When a cylindrical wire is stretched to double its length (

L_2 = 2L_1

) while maintaining constant volume, its cross-sectional area halves (

A_2 = A_1/2

). From the resistance formula

R = \rho L / A

, doubling length and halving area causes resistance to increase by a factor of four (

R_2 = 4 R_1 = 32.0\,\Omega

). By Ohm's law (

I = V/R

), applying

16\,\text{V}

across

32.0\,\Omega

yields a current of

0.50\,\text{A}

.

Step-by-Step Solution

1

Relate volume conservation to cross-sectional area.

Since volume

V = A_1 L_1 = A_2 L_2

remains constant when

L_2 = 2L_1

, the new cross-sectional area is

A_2 = A_1 / 2

.

Stretching a wire increases its length while proportionally reducing its area to preserve volume.

2

Calculate the new resistance of the stretched wire.

The new resistance is

R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L_1}{A_1 / 2} = 4 R_1 = 4 \times 8.0\,\Omega = 32.0\,\Omega

.

Resistance is directly proportional to length and inversely proportional to cross-sectional area.

3

Apply Ohm's law to calculate current.

I = \frac{V}{R_2} = \frac{16\,\text{V}}{32.0\,\Omega} = 0.50\,\text{A}

.

Electric current is calculated by dividing potential difference by total resistance.

Key Concept

Effect of wire stretching on electrical resistance under volume conservation

Question 8007Question →

The sum of the first $n$ terms of an arithmetic progression (AP) is given by $S_n = 2n^2 + 3n$ . What is the $8^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 33

Answer

The

8^{\text{th}}

term of the arithmetic progression is

33

.

The

n^{\text{th}}

term of any sequence can be calculated using the identity

T_n = S_n - S_{n-1}

. Substituting

n = 8

gives

S_8 = 2(8)^2 + 3(8) = 152

, and substituting

n = 7

gives

S_7 = 2(7)^2 + 3(7) = 119

. Therefore,

T_8 = 152 - 119 = 33

.

Step-by-Step Solution

1

Calculate the sum of the first 8 terms (

S_8

) by substituting

n = 8

into

S_n = 2n^2 + 3n

.

S_8 = 2(8)^2 + 3(8) = 128 + 24 = 152

.

The sum formula provides the total sum of terms from

T_1

to

T_8

.

2

Calculate the sum of the first 7 terms (

S_7

) by substituting

n = 7

into

S_n = 2n^2 + 3n

.

S_7 = 2(7)^2 + 3(7) = 98 + 21 = 119

.

The sum formula provides the total sum of terms from

T_1

to

T_7

.

3

Subtract

S_7

from

S_8

to determine the value of the

8^{\text{th}}

term (

T_8

).

T_8 = S_8 - S_7 = 152 - 119 = 33

.

The difference between the sum of the first

n

terms and the sum of the first

(n-1)

terms yields the

n^{\text{th}}

term (

T_n = S_n - S_{n-1}

).

Key Concept

Finding the nth term of a sequence from the sum of the first n terms using

T_n = S_n - S_{n-1}

.

Estimated Time:1m 30s

Question 8008Question →

A rectangular glass block of thickness $6.0\text{ cm}$ has a refractive index of $1.50$ . Calculate the apparent depth, in centimeters, of a mark placed at the bottom of the block when viewed normally from above.

Show answer & explanation

Answer: 4

Answer

The apparent depth of the mark is

4.0\text{ cm}

.

The refractive index of a medium relative to air is given by the ratio of real depth to apparent depth (

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

). Substituting the given values gives

\text{Apparent Depth} = \frac{6.0\text{ cm}}{1.50} = 4.0\text{ cm}

.

Step-by-Step Solution

1

Identify the relationship between refractive index, real depth, and apparent depth for normal view

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

By definition of optical refraction when looking normally from an optically less dense medium (air) into a denser medium (glass).

2

Substitute the known values (

n = 1.50

,

\text{Real Depth} = 6.0\text{ cm}

) and solve for the apparent depth

\text{Apparent Depth} = \frac{6.0\text{ cm}}{1.50} = 4.0\text{ cm}

Dividing the real thickness by the refractive index yields the perceived (apparent) depth.

Key Concept

Real and Apparent Depth in Refraction

Question 8009Question →

A hiker standing at a distance from a tall vertical cliff shouts and hears an echo $0.60\text{ s}$ later. If the speed of sound in air is $330\text{ m/s}$ , what is the distance between the hiker and the cliff?

198\text{ m}

99\text{ m}

550\text{ m}

49.5\text{ m}

Show answer & explanation

Answer: $99\text{ m}$

Answer

The distance between the hiker and the cliff is

99\text{ m}

.

An echo is a reflected sound wave. The sound travels from the hiker to the cliff and back, covering a total distance of

2d

in time

t

. Therefore,

2d = v \times t

, which gives

d = \frac{330\text{ m/s} \times 0.60\text{ s}}{2} = 99\text{ m}

.

Step-by-Step Solution

1

Identify given parameters and echo relation

Speed of sound

v = 330\text{ m/s}

, time elapsed

t = 0.60\text{ s}

. An echo involves sound traveling to the wall and back (

2d

).

The total distance traveled by the sound wave during time

t

is twice the distance to the reflecting surface.

2

Calculate the one-way distance

d = \frac{v \times t}{2} = \frac{330 \times 0.60}{2} = 99\text{ m}

.

Dividing the total distance by 2 yields the actual distance from the hiker to the cliff.

Key Concept

Echo distance calculation

Question 8010Question →

A body of mass $3.0\text{ kg}$ moving due east along a smooth horizontal track at $8.0\text{ m s}^{-1}$ collides head-on with a $2.0\text{ kg}$ body moving due west at $4.0\text{ m s}^{-1}$ . Immediately after the impact, the $2.0\text{ kg}$ body rebounds due east with a speed of $5.0\text{ m s}^{-1}$ . If the duration of the impact is $0.02\text{ s}$ , what is the magnitude of the average impact force exerted on the $3.0\text{ kg}$ body?

100\text{ N}

1200\text{ N}

900\text{ N}

1500\text{ N}

Show answer & explanation

Answer: $900\text{ N}$

Answer

The magnitude of the average impact force exerted on the

3.0\text{ kg}

body is

900\text{ N}

.

By designating East as positive, initial velocities are

u_1 = +8.0\text{ m s}^{-1}

and

u_2 = -4.0\text{ m s}^{-1}

. Conservation of linear momentum gives

(3.0 \times 8.0) + (2.0 \times -4.0) = 3.0 v_1 + (2.0 \times 5.0)

, yielding

v_1 = +2.0\text{ m s}^{-1}

. The change in momentum of the

3.0\text{ kg}

body is

\Delta p = 3.0 \times (2.0 - 8.0) = -18.0\text{ N s}

. Dividing the magnitude of this impulse by the impact time (

0.02\text{ s}

) yields an average force of

900\text{ N}

.

Step-by-Step Solution

1

Set up momentum conservation by assigning directional signs to velocities.

Taking East as positive (

+

):

u_1 = +8.0\text{ m s}^{-1}

,

u_2 = -4.0\text{ m s}^{-1}

,

v_2 = +5.0\text{ m s}^{-1}

.

Linear momentum is a vector quantity, so direction must be taken into account.

2

Calculate the initial total momentum and solve for the final velocity of the

3.0\text{ kg}

body (

v_1

).

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \implies (3.0 \times 8.0) + (2.0 \times -4.0) = 3.0 v_1 + (2.0 \times 5.0) \implies 16.0 = 3.0 v_1 + 10.0 \implies v_1 = +2.0\text{ m s}^{-1}

.

Total momentum is conserved in the absence of external forces.

3

Calculate the magnitude of the impulse and average force acting on the

3.0\text{ kg}

body.

\Delta p_1 = m_1(v_1 - u_1) = 3.0 \times (2.0 - 8.0) = -18.0\text{ N s}

. Force magnitude

F = \frac{|\Delta p_1|}{\Delta t} = \frac{18.0}{0.02} = 900\text{ N}

.

The average force equals the rate of change of linear momentum.

Key Concept

Law of Conservation of Linear Momentum and Impulse-Momentum Theorem

Estimated Time:2m 0s

Question 8011Question →

An industrial compression chamber fitted with a frictionless piston contains $0.25\text{ m}^3$ of an ideal gas at an initial pressure of $1.6 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The gas expands while being heated to a final temperature of $87^\circ\text{C}$ until its pressure drops to $8.0 \times 10^4\text{ Pa}$ . What is the final volume occupied by the gas?

0.15\text{ m}^3

0.42\text{ m}^3

0.60\text{ m}^3

1.61\text{ m}^3

Show answer & explanation

Answer: $0.60\text{ m}^3$

Answer

0.60\text{ m}^3

The combined gas law states that

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

. Converting the temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 360\text{ K}

. Rearranging for

V_2

gives

V_2 = 0.25 \times \left(\frac{1.6 \times 10^5}{8.0 \times 10^4}\right) \times \left(\frac{360}{300}\right) = 0.25 \times 2.0 \times 1.2 = 0.60\text{ m}^3

.

Step-by-Step Solution

1

Convert all given temperatures from degrees Celsius to Kelvin.

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

and

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

.

Gas laws require thermodynamic (absolute) temperature in Kelvin.

2

State the combined gas law relating initial and final states.

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

The mass of gas is fixed while pressure, volume, and temperature all change.

3

Rearrange the equation to solve for the final volume

V_2

.

V_2 = V_1 \times \left(\frac{P_1}{P_2}\right) \times \left(\frac{T_2}{T_1}\right)

Isolating the target unknown variable.

4

Substitute the given numerical values into the rearranged equation.

V_2 = 0.25\text{ m}^3 \times \left(\frac{1.6 \times 10^5\text{ Pa}}{8.0 \times 10^4\text{ Pa}}\right) \times \left(\frac{360\text{ K}}{300\text{ K}}\right) = 0.25 \times 2.0 \times 1.2 = 0.60\text{ m}^3

.

Simplifying the ratios yields the correct final volume.

Key Concept

Combined Gas Law

Question 8012Question →

The specific heat capacity of a gas undergoing an isothermal expansion is zero because the temperature of the gas does not change during the process.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. During an isothermal process, the specific heat capacity of a gas is infinitely large (

\infty

), not zero.

The statement is false because the specific heat capacity

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

of a gas during an isothermal process (

\Delta T = 0, Q \neq 0

) is infinitely large (

\infty

). A specific heat capacity of zero occurs during an adiabatic process where no thermal energy enters or leaves the system (

Q = 0

) while temperature changes.

Step-by-Step Solution

1

Apply the fundamental defining equation for specific heat capacity.

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

, where

Q

is the quantity of heat transferred,

m

is the mass, and

\Delta T

is the change in temperature.

Specific heat capacity measures the amount of heat required per unit mass to produce a unit temperature change under a defined thermodynamic process.

2

Identify the thermodynamic boundary conditions for an isothermal expansion.

The temperature remains constant throughout the expansion, so

\Delta T = 0

, but heat energy

Q > 0

must be supplied to offset the work done by the expanding gas.

According to the First Law of Thermodynamics,

\Delta U = Q - W

. For an ideal gas undergoing an isothermal process,

\Delta U = 0

, which mandates

Q = W \neq 0

.

3

Evaluate the limit of

c

as

\Delta T \to 0

for a non-zero heat input

Q

.

c = \frac{Q}{m \cdot 0} \to \infty

.

Dividing a finite non-zero quantity of heat by a zero temperature change yields an infinitely large heat capacity.

4

Contrast this result with the condition required for a specific heat capacity of zero.

For

c = 0

, the heat input must be zero (

Q = 0

) while

\Delta T \neq 0

, which defines an adiabatic process.

Zero heat capacity means temperature changes without any heat transfer.

Key Concept

Thermodynamic process dependence of specific heat capacity

Question 8013Question →

An electric train traveling along a straight track uniformly slows down from a speed of $30\text{ m/s}$ to $10\text{ m/s}$ over a distance of $200\text{ m}$ . What is the magnitude of the deceleration of the train in $\text{m/s}^2$ ?

Show answer & explanation

Answer: 2

Answer

The magnitude of the deceleration is

2\text{ m/s}^2

.

Using the kinematic equation

v^2 = u^2 + 2as

, substituting

v = 10\text{ m/s}

,

u = 30\text{ m/s}

, and

s = 200\text{ m}

gives

100 = 900 + 400a

, which simplifies to

400a = -800

, yielding

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

. The magnitude of deceleration is

2\text{ m/s}^2

.

Step-by-Step Solution

1

Identify the given kinematic variables.

Initial velocity

u = 30\text{ m/s}

, final velocity

v = 10\text{ m/s}

, displacement

s = 200\text{ m}

.

Choosing the appropriate equation of motion requires knowing which variables are given and which is unknown.

2

Apply the third equation of motion relating initial velocity, final velocity, acceleration, and distance.

v^2 = u^2 + 2as

This formula connects

u

,

v

,

a

, and

s

without needing time

t

.

3

Substitute the given values into the equation and solve for acceleration

a

.

(10)^2 = (30)^2 + 2(a)(200) \implies 100 = 900 + 400a \implies 400a = -800 \implies a = -2\text{ m/s}^2

.

Performing algebraic operations to isolate the acceleration parameter.

4

State the magnitude of the deceleration.

The magnitude of deceleration is

2\text{ m/s}^2

.

Deceleration represents the rate of speed reduction, which corresponds to the magnitude of negative acceleration.

Key Concept

Uniformly Accelerated Motion Equations

Question 8014Question →

A body is suspended from a spring balance inside an elevator that is accelerating upwards at $2.0\text{ m s}^{-2}$ on Earth, giving a scale reading of $60\text{ N}$ . The body is then taken to a planet where the acceleration due to gravity is $6.0\text{ m s}^{-2}$ and placed on an ideal beam balance against standard masses calibrated on Earth. Taking $g = 10.0\text{ m s}^{-2}$ on Earth, what is the reading obtained on the beam balance?

5.0\text{ kg}

6.0\text{ kg}

7.5\text{ kg}

3.0\text{ kg}

Show answer & explanation

Answer: $5.0\text{ kg}$

Answer

The reading obtained on the beam balance is

5.0\text{ kg}

.

In an upward accelerating elevator, the spring balance registers an apparent weight of

W_{app} = m(g + a)

. Substituting the given values gives

60\text{ N} = m(10\text{ m s}^{-2} + 2\text{ m s}^{-2})

, which yields a true mass

m = 5.0\text{ kg}

. When measured on another planet using an ideal beam balance, the gravitational force on the unknown mass balances against standard masses (

m \cdot g_p = m_{standard} \cdot g_p

). Since local gravity

g_p

cancels from both sides, the beam balance measures the invariant mass of

5.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the true mass of the body using the scale reading in the accelerating elevator.

m = 5.0\text{ kg}

Inside an upward accelerating elevator, the apparent weight registered by the spring balance is

W_{app} = m(g + a)

. Substituting

W_{app} = 60\text{ N}

,

g = 10.0\text{ m s}^{-2}

, and

a = 2.0\text{ m s}^{-2}

yields

60 = m(10 + 2)

, which gives

m = 5.0\text{ kg}

.

2

Determine the mass reading on a beam balance on the new planet.

The beam balance reads

5.0\text{ kg}

.

A beam balance compares the gravitational force on the object with standard masses:

m \cdot g_p = m_{standard} \cdot g_p \implies m_{standard} = m

. Because local acceleration due to gravity

g_p

cancels out on both sides, a beam balance measures true invariant mass regardless of location or gravity.

Key Concept

Mass vs Weight Measurement: Accelerating Frames and Beam Balance Invariance

Estimated Time:1m 0s

Question 8015Question →

A student investigating the physical properties of a uniform metallic rod records its mass, length, thermodynamic temperature, and mass density. Which of the recorded physical quantities is classified as a derived quantity?

Mass density

Mass

Length

Thermodynamic temperature

Show answer & explanation

Answer: Mass density

Answer

Mass density is the derived physical quantity.

Mass density is defined as mass per unit volume (

\rho = \frac{m}{V}

). Because it is obtained by combining the fundamental quantities of mass and length, it is a derived physical quantity.

Step-by-Step Solution

1

Identify the seven fundamental SI physical quantities

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

Fundamental quantities are independent physical quantities that cannot be defined in terms of other physical quantities.

2

Classify each physical quantity given in the scenario

Mass, length, and thermodynamic temperature are fundamental quantities. Mass density is defined as mass divided by volume (

\rho = \frac{m}{V}

).

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

Key Concept

Classification of Fundamental and Derived Quantities

Question 8016Question →

A proton of mass $1.67 \times 10^{-27}\text{ kg}$ and an electron of mass $9.11 \times 10^{-31}\text{ kg}$ , both carrying charges of equal magnitude, are accelerated from rest through the same electric potential difference. Calculate the ratio of the de Broglie wavelength of the electron to that of the proton.

Show answer & explanation

Answer: 42.8

Answer

The ratio of the de Broglie wavelength of the electron to that of the proton is 42.8.

The de Broglie wavelength of a particle accelerated through potential difference

V

is given by

\lambda = \frac{h}{\sqrt{2mqV}}

. Since both particles carry equal charge

q

and experience the same potential

V

, the ratio of their wavelengths simplifies to

\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \approx 42.8

.

Step-by-Step Solution

1

Relate kinetic energy to accelerating potential difference

E_k = qV

Electric potential energy converted into kinetic energy during acceleration from rest.

2

Express de Broglie wavelength in terms of particle mass, charge, and potential difference

\lambda = \frac{h}{\sqrt{2mqV}}

Combining

p = \sqrt{2mE_k}

with de Broglie's formula

\lambda = \frac{h}{p}

.

3

Formulate the wavelength ratio of electron to proton

\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}

Planck's constant

h

, elementary charge

q

, and potential difference

V

are identical for both particles and cancel out.

4

Substitute numerical values and compute final ratio

\sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \approx 42.8

Square root of the proton-to-electron mass ratio yields the inverse ratio of their wavelengths.

Key Concept

De Broglie wavelength relation to particle mass under constant accelerating potential

Question 8017Question →

An electric ceiling fan operates on a $220\,\text{V}$ mains supply and draws a steady current of $0.5\,\text{A}$ . What is the electrical power rating of the fan?

110\,\text{W}

440\,\text{W}

220.5\,\text{W}

110\,\text{J}

Show answer & explanation

Answer: $110\,\text{W}$

Answer

The electrical power rating of the fan is

110\,\text{W}

.

Electrical power

P

delivered to a circuit element is given by

P = VI

, where

V

is potential difference and

I

is current. Substituting

220\,\text{V}

and

0.5\,\text{A}

gives

110\,\text{W}

.

Step-by-Step Solution

1

Identify the given physical quantities

Voltage

V = 220\,\text{V}

and current

I = 0.5\,\text{A}

.

These are the basic electrical parameters required to calculate power.

2

Apply the electrical power formula

P = V \times I

Electrical power is defined as the product of potential difference across a device and current flowing through it.

3

Substitute values and compute the power

P = 220\,\text{V} \times 0.5\,\text{A} = 110\,\text{W}

Multiplying potential difference by current yields the rate of energy conversion in Watts.

Key Concept

Electrical Power (

P = VI

)

Estimated Time:45s

Question 8018Question →

A sample of an ideal gas has an average translational kinetic energy of $E_k$ per molecule at an initial temperature of $27^\circ\text{C}$ . If the gas is heated at constant volume until the average kinetic energy per molecule doubles to $2E_k$ , what is the final temperature of the gas in degrees Celsius?

54^\circ\text{C}

327^\circ\text{C}

600^\circ\text{C}

927^\circ\text{C}

Show answer & explanation

Answer: $327^\circ\text{C}$

Answer

The final temperature of the gas is

327^\circ\text{C}

.

In the kinetic theory of matter, average translational kinetic energy per molecule is directly proportional to absolute temperature (

E_k \propto T

). Initial temperature

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

. Doubling the kinetic energy doubles the Kelvin temperature to

600\text{ K}

. Subtracting

273

yields

327^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin.

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

Kinetic theory calculations require thermodynamic temperature measured on the absolute Kelvin scale.

2

Apply the proportional relationship between average translational kinetic energy and temperature.

Since

E_k \propto T

, doubling

E_k

means

T_2 = 2 \times T_1 = 2 \times 300\text{ K} = 600\text{ K}

.

The average kinetic energy of gas molecules is directly proportional to the absolute temperature.

3

Convert the final temperature from Kelvin back to degrees Celsius.

t_2 = 600\text{ K} - 273 = 327^\circ\text{C}

The question asks for the final temperature specifically in degrees Celsius.

Key Concept

Average translational kinetic energy of an ideal gas molecule is directly proportional to its absolute temperature (

E_k = \frac{3}{2} k_B T

).

Estimated Time:1m 30s

Question 8019Question →

Two fishing boats leave a harbor $H$ at the same time. Boat $A$ travels on a bearing of $020^\circ$ for a distance of $8\text{ km}$ , while Boat $B$ travels on a bearing of $140^\circ$ for a distance of $7\text{ km}$ . What is the distance between Boat $A$ and Boat $B$ in kilometers?

Show answer & explanation

Answer: 13

Answer

13 km

The distance between the two boats forms the third side of triangle

HAB

, where

HA = 8\text{ km}

,

HB = 7\text{ km}

, and the included angle at the harbor

H

is

\angle AHB = 140^\circ - 020^\circ = 120^\circ

. By the Cosine Rule,

AB^2 = 8^2 + 7^2 - 2(8)(7)\cos(120^\circ) = 64 + 49 - 112(-0.5) = 169

. Taking the square root gives

AB = 13\text{ km}

.

Step-by-Step Solution

1

Calculate the included angle between the direction vectors of the two boats from the harbor.

Included angle

\angle AHB = 140^\circ - 020^\circ = 120^\circ

The angle between two bearings originating from the same point is the difference between their bearing angles.

2

State the Cosine Rule for side

AB

in triangle

HAB

.

AB^2 = HA^2 + HB^2 - 2 \cdot HA \cdot HB \cdot \cos(\angle AHB)

The Cosine Rule calculates an unknown side when two sides and their included angle (SAS) are given.

3

Substitute given side lengths

HA = 8\text{ km}

,

HB = 7\text{ km}

, and angle

\angle AHB = 120^\circ

.

AB^2 = 8^2 + 7^2 - 2(8)(7)\cos(120^\circ) = 64 + 49 - 112(-0.5) = 169

Since

120^\circ

is in the second quadrant,

\cos(120^\circ) = -0.5

, which changes the subtracted term to addition.

4

Compute the principal square root of 169.

AB = \sqrt{169} = 13\text{ km}

Distance is a non-negative scalar quantity.

Key Concept

Applying the Cosine Rule to solve bearing problems

Question 8020Question →

A simple pendulum completes $20$ full oscillations in a time interval of $40\text{ s}$ . What is the period of oscillation of the pendulum in seconds?

Show answer & explanation

Answer: 2

Answer

The period of oscillation of the pendulum is

2.0\text{ s}

.

The period

T

of a repeating motion is the time taken to complete one full cycle. Dividing the total measured time (

40\text{ s}

) by the number of oscillations (

20

) gives

2.0\text{ s}

.

Step-by-Step Solution

1

Identify the given total time and total number of oscillations

Total time

t = 40\text{ s}

, number of oscillations

N = 20

The period is defined as the time taken for one single complete oscillation.

2

Apply the period formula

T = \frac{t}{N}

T = \frac{40\text{ s}}{20} = 2.0\text{ s}

Dividing the total measured time by the total count of oscillations yields the time per oscillation.

Key Concept

Period of a Simple Pendulum

Answer

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