All practice questions

Question 12821Question →

A spring balance is attached to the ceiling of an elevator accelerating downwards at $2.0\text{ m s}^{-2}$ . Suspended from the hook of the spring balance is a light, frictionless pulley carrying two masses of $3.0\text{ kg}$ and $1.0\text{ kg}$ connected by a light inextensible string. Taking the acceleration due to gravity $g = 10.0\text{ m s}^{-2}$ , what is the reading registered by the spring balance in newtons?

Show answer & explanation

Answer: 24

Answer

The reading registered by the spring balance is 24 N.

In a frame accelerating downwards at

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

, the effective acceleration due to gravity is reduced to

g' = g - a = 8.0\text{ m s}^{-2}

. Within this frame, the tension in the Atwood machine string is

T = \frac{2(3.0)(1.0)}{3.0 + 1.0} \times 8.0 = 12.0\text{ N}

. Since two string segments act downward on the light pulley suspended from the spring balance, the total tension force registered by the balance scale is

2T = 24.0\text{ N}

.

Step-by-Step Solution

1

Determine the effective local acceleration due to gravity inside the accelerating elevator

g' = 8.0\text{ m s}^{-2}

Because the elevator accelerates downward at

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

, objects inside experience an apparent gravitational acceleration of

g' = g - a

.

2

Compute the tension in the string supporting the two masses in the modified gravitational field

T = 12.0\text{ N}

For an Atwood machine system in effective gravity

g'

, string tension is

T = \frac{2 m_1 m_2}{m_1 + m_2} g' = \frac{2(3.0)(1.0)}{4.0} \times 8.0 = 12.0\text{ N}

.

3

Calculate the downward pull on the spring balance

F = 24.0\text{ N}

The spring balance supports the frictionless pulley, which experiences a downward force from two upward string segments, making the total measured weight force equal to

2T = 24.0\text{ N}

.

Key Concept

Apparent weight measurement and tension forces in accelerating frames

Question 12822Question →

A student uses a stopwatch with a positive zero error of $+0.30\text{ s}$ to measure the time taken for a trolley to travel down an inclined plane. If the stopwatch display reads $15.70\text{ s}$ at the end of the trial, what is the actual time taken by the trolley?

15.40\text{ s}

16.00\text{ s}

15.70\text{ s}

15.4\text{ s}

Show answer & explanation

Answer: $15.40\text{ s}$

Answer

The actual time taken by the trolley is

15.40\text{ s}

.

The correct answer is

15.40\text{ s}

. When a measuring instrument has a positive zero error, it means the scale reads a value greater than zero before measurement begins. Therefore, the true value is found by subtracting the zero error from the observed reading:

15.70\text{ s} - 0.30\text{ s} = 15.40\text{ s}

.

Step-by-Step Solution

1

Identify the observed reading and the zero error of the instrument.

Observed reading =

15.70\text{ s}

, Zero error =

+0.30\text{ s}

.

Instrument readings must be corrected for systematic errors before recording actual values.

2

Apply the zero error correction formula:

\text{Actual Value} = \text{Observed Reading} - \text{Zero Error}

.

\text{Actual Time} = 15.70\text{ s} - (+0.30\text{ s}) = 15.40\text{ s}

.

A positive zero error indicates the timer started above zero, so the initial offset must be subtracted.

Key Concept

Zero Error Correction in Time Measurement Instruments

Question 12823Question →

An alternating current (AC) circuit operating at a frequency of $50\ \text{Hz}$ contains a resistor of resistance $R = 30\ \Omega$ , an inductor of inductance $L = \frac{0.9}{\pi}\ \text{H}$ , and a capacitor of capacitance $C = \frac{200}{\pi}\ \mu\text{F}$ connected in series. What is the total impedance of the circuit?

50\ \Omega

70\ \Omega

170\ \Omega

40\ \Omega

Show answer & explanation

Answer: $50\ \Omega$

Answer

The total impedance of the AC circuit is

50\ \Omega

.

The inductive reactance is

X_L = 2\pi (50)\left(\frac{0.9}{\pi}\right) = 90\ \Omega

and the capacitive reactance is

X_C = \frac{1}{2\pi (50)\left(\frac{200 \times 10^{-6}}{\pi}\right)} = 50\ \Omega

. Since resistance

R

and net reactance

(X_L - X_C = 40\ \Omega)

are perpendicular vectors in a phasor diagram, the total impedance is calculated using the Pythagorean relation:

Z = \sqrt{30^2 + 40^2} = 50\ \Omega

.

Step-by-Step Solution

1

Calculate the inductive reactance (

X_L

)

X_L = 2\pi f L = 2\pi \times 50 \times \frac{0.9}{\pi} = 90\ \Omega

Inductive reactance depends on supply frequency and inductance.

2

Calculate the capacitive reactance (

X_C

)

X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times \frac{200 \times 10^{-6}}{\pi}} = \frac{1}{0.02} = 50\ \Omega

Capacitive reactance is inversely proportional to supply frequency and capacitance.

3

Determine the net reactance (

X

)

X = X_L - X_C = 90\ \Omega - 50\ \Omega = 40\ \Omega

Inductive and capacitive reactances are

180^\circ

out of phase.

4

Calculate total impedance (

Z

) using phasor addition

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \Omega

Resistance and net reactance are

90^\circ

out of phase, requiring right-triangle vector summation.

Key Concept

Total Impedance in a Series RLC AC Circuit

Estimated Time:2m 0s

Question 12824Question →

What is the total number of sigma ( $\sigma$ ) bonds in a single molecule of prop-2-enal (acrolein, $\text{CH}_2=\text{CH}-\text{CHO}$ )?

Show answer & explanation

Answer: 7; seven; 7 sigma bonds; 7 sigma

Answer

7

Expanding prop-2-enal (

\text{CH}_2=\text{CH}-\text{CHO}

) reveals four

\text{C}-\text{H}

single bonds, one

\text{C}-\text{C}

single bond, one

\text{C}=\text{C}

double bond (composed of one

\sigma

and one

\pi

bond), and one

\text{C}=\text{O}

double bond (composed of one

\sigma

and one

\pi

bond). Summing all head-on orbital overlaps yields a total of 7 sigma (

\sigma

) bonds.

Step-by-Step Solution

1

Draw the expanded structural formula of prop-2-enal.

The expanded formula showing all individual atoms and bonds is

\text{H}_2\text{C}=\text{CH}-\text{C}(=\text{O})\text{H}

.

Expanding the structural formula ensures that all implicit single bonds, double bonds, and hydrogen attachments are explicitly visible for counting.

2

Count all carbon-hydrogen (

\text{C}-\text{H}

) single sigma bonds.

There are 2

\text{C}-\text{H}

bonds on the terminal alkene carbon, 1

\text{C}-\text{H}

bond on the central alkene carbon, and 1

\text{C}-\text{H}

bond on the aldehyde carbon, giving a total of 4

\text{C}-\text{H}

\sigma

bonds.

Every single covalent bond formed with hydrogen involves head-on

sp^2-s

orbital overlap and constitutes one

\sigma

bond.

3

Count the sigma bonds among the carbon-carbon and carbon-oxygen links.

The

\text{C}=\text{C}

double bond contains 1

\sigma

bond, the

\text{C}-\text{C}

single bond contains 1

\sigma

bond, and the

\text{C}=\text{O}

double bond contains 1

\sigma

bond, giving 3 heavy-atom

\sigma

bonds.

Every covalent bond—whether single, double, or triple—contains exactly one

\sigma

bond resulting from axial head-on orbital overlap.

4

Sum the total number of sigma bonds.

4\text{ (C-H \sigma\ bonds)} + 3\text{ (C-C and C-O \sigma\ bonds)} = 7\text{ \sigma\ bonds}

.

Adding all localized head-on orbital overlaps gives the total count of sigma bonds in the molecule.

Key Concept

Determination of sigma (

\sigma

) and pi (

\pi

) bond counts in organic structures

Question 12825Question →

Two long, straight, parallel horizontal conductors are separated vertically by a distance of $2.0\text{ cm}$ . The upper conductor has a mass per unit length of $0.04\text{ kg/m}$ and carries a steady current of $50\text{ A}$ . Assuming the currents in the two conductors flow in opposite directions so that the resulting magnetic force is repulsive, what current (in amperes) must flow through the lower conductor to magnetically levitate and balance the weight of the upper conductor? (Take $g = 9.8\text{ m/s}^2$ and $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ )

Show answer & explanation

Answer: 784

Answer

The required current in the lower conductor is

784\text{ A}

.

Equating magnetic repulsion per unit length

\frac{\mu_0 I_1 I_2}{2\pi d}

to weight per unit length

\lambda g

gives

\frac{(2 \times 10^{-7}) \times 50 \times I_2}{0.02} = 0.04 \times 9.8

, which simplifies to

5 \times 10^{-4} I_2 = 0.392

, yielding

I_2 = 784\text{ A}

.

Step-by-Step Solution

1

Equate the upward repulsive magnetic force per unit length to the downward gravitational weight per unit length.

\frac{\mu_0 I_1 I_2}{2\pi d} = \lambda g

For the upper conductor to levitate in vertical static equilibrium, the upward magnetic force per meter must exactly balance its weight per meter.

2

Substitute all given physical values in SI units into the force balance equation.

\frac{(4\pi \times 10^{-7}) \times 50 \times I_2}{2\pi \times 0.02} = 0.04 \times 9.8

Converting distance

d = 2.0\text{ cm} = 0.02\text{ m}

and using mass density

\lambda = 0.04\text{ kg/m}

sets up a single equation with unknown

I_2

.

3

Simplify both sides of the equation.

5 \times 10^{-4} I_2 = 0.392

Calculating

\frac{2 \times 10^{-7} \times 50}{0.02} = 5 \times 10^{-4}\text{ N/(A}\cdot\text{m)}

and

0.04 \times 9.8 = 0.392\text{ N/m}

.

4

Solve for the unknown current

I_2

.

I_2 = \frac{0.392}{5 \times 10^{-4}} = 784\text{ A}

Dividing the weight per meter by the magnetic force coefficient yields the exact required current magnitude.

Key Concept

Interaction force between parallel current-carrying conductors and mechanical equilibrium

Question 12826Question →

A nitrogen nucleus $^{14}_{7}\text{N}$ has a nuclear mass of $13.9992\text{ u}$ . Given that the mass of a proton is $1.0073\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , what is the total binding energy of the nucleus in $\text{MeV}$ ? (Take $1\text{ u} = 931\text{ MeV}$ )

Show answer & explanation

Answer: 105.02

Answer

The total binding energy of the

^{14}_{7}\text{N}

nucleus is

105.02\text{ MeV}

.

The total binding energy is computed by finding the total rest mass of 7 protons and 7 neutrons (14.1120 u), subtracting the actual nuclear mass of Nitrogen-14 (13.9992 u) to obtain a mass defect of 0.1128 u, and converting this mass defect to energy by multiplying by 931 MeV/u, yielding 105.02 MeV.

Step-by-Step Solution

1

Determine the number of constituent protons and neutrons

Z = 7

protons and

N = 14 - 7 = 7

neutrons

The atomic number is 7 and the mass number is 14.

2

Calculate the total mass of the constituent nucleons

M_{\text{nucleons}} = (7 \times 1.0073\text{ u}) + (7 \times 1.0087\text{ u}) = 7.0511\text{ u} + 7.0609\text{ u} = 14.1120\text{ u}

The sum of the individual rest masses of all isolated protons and neutrons.

3

Compute the mass defect

\Delta m = 14.1120\text{ u} - 13.9992\text{ u} = 0.1128\text{ u}

Mass defect is the difference between total mass of isolated nucleons and the bound nuclear mass.

4

Convert the mass defect into energy in MeV

E_b = 0.1128\text{ u} \times 931\text{ MeV/u} = 105.0168\text{ MeV} \approx 105.02\text{ MeV}

Applying the mass-energy equivalence factor

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

.

Key Concept

Mass Defect and Binding Energy

Estimated Time:2m 0s

Question 12827Question →

A porter carries a suitcase of mass $5\text{ kg}$ along a horizontal platform for a distance of $10\text{ m}$ at a constant speed, and then lifts it vertically upward through a height of $2\text{ m}$ onto a shelf. Taking $g = 10\text{ m s}^{-2}$ , what is the total work done by the porter on the suitcase?

100\text{ J}

600\text{ J}

500\text{ J}

10\text{ J}

Show answer & explanation

Answer: $100\text{ J}$

Answer

100\text{ J}

During horizontal motion at constant speed, the upward force exerted by the porter is perpendicular to the horizontal displacement, so no work is performed horizontally (

W_{\text{horizontal}} = F s \cos 90^\circ = 0\text{ J}

). During vertical lifting, the force acts in the direction of displacement, yielding

W_{\text{vertical}} = mgh = 5\text{ kg} \times 10\text{ m s}^{-2} \times 2\text{ m} = 100\text{ J}

. The total work done is therefore

100\text{ J}

.

Step-by-Step Solution

1

Calculate the work done during horizontal motion

W_{\text{horizontal}} = 0\text{ J}

The supporting force acts vertically upward at an angle of

90^\circ

to the horizontal displacement, giving

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

.

2

Calculate the work done in lifting the suitcase vertically

W_{\text{vertical}} = mgh = 5 \times 10 \times 2 = 100\text{ J}

Work done against gravity equals the increase in gravitational potential energy.

3

Sum the work done in both stages

W_{\text{total}} = 0 + 100 = 100\text{ J}

Total work done is the scalar sum of work performed along each segment of motion.

Key Concept

Work done by a constant force depends on the direction of displacement (

W = F s \cos \theta

). Perpendicular forces do no work.

Question 12828Question →

Which category of industrial chemicals is characterized by production in small batch quantities, high purity levels, and high unit cost for specialized uses such as pharmaceuticals?

Fine chemicals

Heavy chemicals

Basic chemicals

Fertilizer chemicals

Show answer & explanation

Answer: Fine chemicals

Answer

Fine chemicals

Fine chemicals are defined by small-scale batch manufacturing, extremely high purity standards, high cost per unit mass, and targeted uses such as pharmaceuticals, perfumes, and analytical reagents.

Step-by-Step Solution

1

Analyze the features given in the stem: small batch production, high degree of purity, and high unit cost for specialized applications.

These properties uniquely describe fine chemicals.

Unlike heavy chemicals which are synthesized continuously in large tonnage, fine chemicals are produced in limited batches under strict purity control.

Key Concept

Classification of Heavy and Fine Chemicals

Question 12829Question →

Complete the following statement on acid-base conjugate pairs under the Brønsted-Lowry theory.

Fill in the blanks below

When the hydrogen carbonate ion (

HCO_3^-

) acts as a Brønsted-Lowry base by accepting a proton, it forms as its conjugate acid. Conversely, when it acts as a Brønsted-Lowry acid by donating a proton, it forms the ion as its conjugate base.

Show answer & explanation

Answer

The conjugate acid formed when

HCO_3^-

accepts a proton is carbonic acid (

H_2CO_3

), and the conjugate base formed when it donates a proton is the carbonate ion (

CO_3^{2-}

).

According to the Brønsted-Lowry theory, an acid is a proton (

H^+

) donor and a base is a proton acceptor. The hydrogen carbonate ion (

HCO_3^-

) is amphiprotic. When acting as a base by accepting

H^+

, it forms its conjugate acid, carbonic acid (

H_2CO_3

). When acting as an acid by donating

H^+

, it leaves behind its conjugate base, the carbonate ion (

CO_3^{2-}

).

Step-by-Step Solution

1

Identify the species formed when

HCO_3^-

acts as a proton acceptor (base).

According to the Brønsted-Lowry definition, a base accepts a proton (

H^+

). Adding

H^+

to

HCO_3^-

yields

H_2CO_3

(carbonic acid).

Accepting a proton increases the number of hydrogen atoms by 1 and raises the net electric charge by +1 (from -1 to 0).

2

Identify the species formed when

HCO_3^-

acts as a proton donor (acid).

According to the Brønsted-Lowry definition, an acid donates a proton (

H^+

). Removing

H^+

from

HCO_3^-

yields

CO_3^{2-}

(carbonate ion).

Donating a proton decreases the number of hydrogen atoms by 1 and reduces the net electric charge by 1 (from -1 to -2).

Key Concept

Brønsted-Lowry Acid-Base Theory and Amphiprotic Conjugate Pairs

Estimated Time:1m 30s

Question 12830Question →

A pipe closed at one end vibrates in its first overtone. An open pipe vibrating in its fundamental mode has a frequency equal to that of the closed pipe. Neglecting end corrections, what is the ratio of the length of the open pipe to the length of the closed pipe?

2 : 3

3 : 2

4 : 3

1 : 3

Show answer & explanation

Answer: $2 : 3$

Answer

The ratio of the length of the open pipe to the length of the closed pipe is

2 : 3

.

The first overtone of a closed pipe corresponds to its 3rd harmonic, giving a frequency of

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

. Equating this to the fundamental frequency of an open pipe (

f = \frac{v}{2L_o}

) yields

\frac{v}{2L_o} = \frac{3v}{4L_c}

, which simplifies to

\frac{L_o}{L_c} = \frac{2}{3}

or

2 : 3

.

Step-by-Step Solution

1

Write the frequency formula for the first overtone of the closed pipe.

For a closed pipe of length

L_c

, odd harmonics are produced (

n = 1, 3, 5, \dots

). The first overtone is the third harmonic (

n = 3

):

f_{\text{closed}} = \frac{3v}{4L_c}

Closed air columns produce only odd harmonics, where the fundamental is

n=1

and the first overtone is

n=3

.

2

Write the fundamental frequency formula for the open pipe.

For an open pipe of length

L_o

, all harmonics are produced (

m = 1, 2, 3, \dots

). The fundamental frequency (

m = 1

) is:

f_{\text{open}} = \frac{v}{2L_o}

Open air columns have antinodes at both ends, yielding a fundamental wavelength of

\lambda = 2L_o

.

3

Equate the two frequencies and solve for the ratio

\frac{L_o}{L_c}

.

\frac{v}{2L_o} = \frac{3v}{4L_c}

Cancel the speed of sound

v

from both sides:

\frac{1}{2L_o} = \frac{3}{4L_c}

Cross-multiply:

6L_o = 4L_c \implies \frac{L_o}{L_c} = \frac{4}{6} = \frac{2}{3}

The question states that the frequencies of the two pipe configurations are equal.

Key Concept

Harmonics in Open and Closed Air Columns

Question 12831Question →

A simple machine lifts a load of $200\text{ N}$ when an effort of $50\text{ N}$ is applied to it. What is the mechanical advantage of the machine?

Show answer & explanation

Answer: 4; 4.0

Answer

The mechanical advantage of the machine is

4

.

Mechanical Advantage (MA) is the ratio of the force exerted by the machine (load) to the force applied to the machine (effort). Dividing

200\text{ N}

by

50\text{ N}

yields a dimensionless mechanical advantage of

4

.

Step-by-Step Solution

1

Identify the given values from the problem statement

Load (

L

) =

200\text{ N}

, Effort (

E

) =

50\text{ N}

Mechanical advantage is defined as the ratio of load to effort.

2

Apply the formula for Mechanical Advantage (MA)

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{200\text{ N}}{50\text{ N}} = 4

Dividing the output force by the input force gives the force amplification factor of the machine.

Key Concept

Mechanical Advantage of Simple Machines

Estimated Time:45s

Question 12832Question →

If a fixed mass of unsaturated air in a closed container is cooled at constant total pressure, its relative humidity increases until it reaches the dew point; upon cooling further below the dew point, condensation occurs such that the relative humidity remains at 100% while the saturated vapour pressure continues to decrease.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

Cooling unsaturated air reduces its maximum moisture-holding capacity (SVP). Before saturation, actual vapour pressure is fixed, so relative humidity rises to 100% at the dew point. Continuous cooling past this point forces condensation, decreasing the actual vapour pressure alongside the SVP to keep the air continuously saturated at 100% relative humidity.

Step-by-Step Solution

1

Analyze the relationship between cooling and relative humidity for unsaturated air.

As temperature decreases, the saturated vapour pressure (SVP) decreases while the actual partial vapour pressure remains constant, causing relative humidity (actual VP / SVP × 100%) to increase.

Relative humidity is inversely proportional to the saturated vapour pressure at a fixed moisture content.

2

Identify the state reached when relative humidity reaches 100%.

The air reaches saturation at the dew point temperature where actual vapour pressure equals SVP.

By definition, the dew point is the temperature at which water vapour in air begins to condense.

3

Determine the thermodynamic behavior during continuous cooling below the dew point.

Excess vapour condenses into liquid water, decreasing actual vapour pressure to continuously match the lower SVP at each reduced temperature.

Air cannot maintain an unsaturated or supersaturated equilibrium state in the presence of condensate, keeping relative humidity locked at 100%.

Key Concept

Vapour saturation, dew point determination, and condensation mechanics during air cooling.

Estimated Time:1m 30s

Question 12833Question →

A converging lens has a focal length of $25\text{ cm}$ . What is the optical power of the lens in dioptres?

+4.0\text{ D}

-4.0\text{ D}

+0.04\text{ D}

+25.0\text{ D}

Show answer & explanation

Answer: $+4.0\text{ D}$

Answer

The optical power of the lens is

+4.0\text{ D}

.

The optical power

P

of a lens in dioptres (

\text{D}

) is calculated using

P = \frac{1}{f}

, where

f

is the focal length in metres. Converting

25\text{ cm}

to metres gives

0.25\text{ m}

. Since the lens is converging, its focal length is positive (

+0.25\text{ m}

). Thus,

P = \frac{1}{+0.25} = +4.0\text{ D}

.

Step-by-Step Solution

1

Convert the focal length from centimetres to metres

f = 25\text{ cm} = 0.25\text{ m}

The unit of optical power (dioptre,

\text{D}

) requires the focal length to be expressed in metres.

2

Apply the sign convention for a converging lens

f = +0.25\text{ m}

A converging (convex) lens has a real principal focus, so its focal length is positive by sign convention.

3

Calculate the optical power using the formula

P = \frac{1}{f}

P = \frac{1}{+0.25} = +4.0\text{ D}

Optical power is defined as the inverse of the focal length in metres.

Key Concept

Power of a Thin Lens

Estimated Time:45s

Question 12834Question →

Match each physical component or thermal phenomenon listed on the left with its corresponding primary heat transfer mechanism and operational principle on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: the evacuated space matches the prevention of conduction and convection due to lack of medium; silvered inner glass matches the minimization of radiation via reflective low-emissivity surfaces; the thick copper base matches enhanced conduction via free electron diffusion; and the offshore land breeze matches natural convection driven by air density differences.

Each phenomenon is matched correctly to its fundamental physical requirement: vacuum interspace eliminates conduction and convection by removing matter; silvered surfaces prevent radiation loss via reflection; copper base accelerates conduction through electron movement; land breeze is a fluid density-driven convection current.

Step-by-Step Solution

1

Analyze the vacuum interspace mechanism

Since conduction requires direct particle collisions and convection requires fluid flow, removing air creates a vacuum that eliminates both conduction and convection.

Both conduction and convection depend on a physical medium.

2

Analyze the silvered glass walls function

Radiation does not require a medium and is governed by surface emissivity. Shiny, silvered surfaces reflect thermal radiation back into the vessel.

Polished metallic coatings decrease thermal radiation emissivity and increase reflectivity.

3

Analyze the copper cooking base conduction property

Metals like copper transfer heat rapidly across solid structures using free electrons alongside atomic lattice vibrations.

Free electron diffusion makes copper an exceptionally good conductor of heat.

4

Analyze the coastal land breeze phenomenon

At night, land loses thermal energy faster than water, causing cooler dense air above land to slide under warmer rising air over the sea.

This fluid circulation driven by thermal expansion and buoyancy differences is natural convection.

Key Concept

Distinct physical requirements and microscopic mechanisms of conduction, convection, and thermal radiation.

Estimated Time:2m 0s

Question 12835Question →

A simple machine with a velocity ratio of $5$ requires an effort of $200\text{ N}$ to raise a load of $800\text{ N}$ . What is the efficiency of the machine in percentage?

Show answer & explanation

Answer: 80%; 80; 80 percent; 80 %

Answer

The efficiency of the machine is

80\%

.

The mechanical advantage is calculated by dividing the load (

800\text{ N}

) by the effort (

200\text{ N}

), yielding

\text{MA} = 4

. Dividing the mechanical advantage by the velocity ratio (

5

) and multiplying by

100\%

gives an efficiency of

80\%

.

Step-by-Step Solution

1

Calculate the Mechanical Advantage (MA) of the machine

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{800\text{ N}}{200\text{ N}} = 4

Mechanical Advantage is defined as the ratio of the force overcome (load) to the force applied (effort).

2

Calculate the Efficiency using Mechanical Advantage and Velocity Ratio

\text{Efficiency} = \left(\frac{\text{MA}}{\text{VR}}\right) \times 100\% = \left(\frac{4}{5}\right) \times 100\% = 80\%

Efficiency is the ratio of Mechanical Advantage to Velocity Ratio expressed as a percentage.

Key Concept

Efficiency of a Simple Machine

Estimated Time:1m 30s

Question 12836Question →

A simple pendulum suspended in a physics laboratory has a length of $0.64\text{ m}$ . Given that acceleration due to gravity $g = 10\text{ m/s}^2$ and taking $\pi^2 = 10$ , what is the period of oscillation of the pendulum in seconds?

Show answer & explanation

Answer: 1.6

Answer

The period of oscillation of the simple pendulum is

1.6\text{ s}

.

Using the pendulum period relationship

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

, substituting

L = 0.64\text{ m}

,

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

, and

\pi = \sqrt{10}

yields

T = 2\sqrt{10}\sqrt{\frac{0.64}{10}} = 2\sqrt{0.64} = 1.6\text{ s}

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum.

The period formula is

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

.

The period depends on the length of the pendulum

L

and acceleration due to gravity

g

.

2

Substitute the given values into the formula.

T = 2\pi \sqrt{\frac{0.64\text{ m}}{10\text{ m/s}^2}}

.

Given parameters are

L = 0.64\text{ m}

and

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

.

3

Simplify the equation using

\pi = \sqrt{10}

.

T = 2\sqrt{10} \times \sqrt{0.064} = 2 \sqrt{10 \times 0.064} = 2 \sqrt{0.64} = 2 \times 0.8 = 1.6\text{ s}

.

Using

\pi^2 = 10

simplifies the calculation cleanly without requiring a calculator.

Key Concept

Simple Pendulum Period of Oscillation

Question 12837Question →

A dip circle is set up in a vertical plane that is inclined at an angle of $60^\circ$ to the magnetic meridian. The needle comes to rest at an apparent angle of dip of $45^\circ$ . If the actual horizontal component of the Earth's magnetic field in the magnetic meridian is $4.0 \times 10^{-5}\text{ T}$ , what is the vertical component of the Earth's magnetic field at that location?

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

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

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

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

Show answer & explanation

Answer: $2.0 \times 10^{-5}\text{ T}$

Answer

The vertical component of the Earth's magnetic field is

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

.

In a vertical plane inclined at an angle

\alpha

to the magnetic meridian, the vertical component

B_V

remains constant while the effective horizontal component becomes

B_H' = B_H \cos \alpha

. Substituting

B_H = 4.0 \times 10^{-5}\text{ T}

and

\alpha = 60^\circ

yields

B_H' = 2.0 \times 10^{-5}\text{ T}

. Using the formula for apparent dip

\tan \theta' = B_V / B_H'

with

\theta' = 45^\circ

gives

\tan 45^\circ = 1

, which confirms

B_V = B_H' = 2.0 \times 10^{-5}\text{ T}

.

Step-by-Step Solution

1

Determine the effective horizontal component of the magnetic field in the plane of inclination

B_H' = B_H \cos \alpha = (4.0 \times 10^{-5}\text{ T}) \times \cos 60^\circ = 2.0 \times 10^{-5}\text{ T}

When a dip circle is rotated by an angle

\alpha

away from the magnetic meridian, the horizontal component acting along the plane of the dip circle is reduced to

B_H \cos \alpha

.

2

Relate the apparent angle of dip to the vertical component and effective horizontal component

\tan \theta' = \frac{B_V}{B_H'}

The vertical component

B_V

remains unchanged regardless of the vertical plane's orientation.

3

Substitute the given values to solve for

B_V

B_V = B_H' \tan 45^\circ = (2.0 \times 10^{-5}\text{ T}) \times 1 = 2.0 \times 10^{-5}\text{ T}

Since

\tan 45^\circ = 1

, the vertical component is equal to the resolved horizontal component.

Key Concept

Apparent Dip Angle and Resolution of Earth's Magnetic Field Components

Estimated Time:2m 0s

Question 12838Question →

A straight horizontal wire of length $0.40\text{ m}$ and mass $0.12\text{ kg}$ carries a steady electric current directed towards the East. The wire is situated in a uniform horizontal magnetic field of $0.60\text{ T}$ directed at an angle of $30^\circ$ North of East. Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what minimum current must flow through the wire for the magnetic force to act vertically upward and balance the weight of the wire?

5.0\text{ A}

10.0\text{ A}

1.0\text{ A}

5.8\text{ A}

Show answer & explanation

Answer: $10.0\text{ A}$

Answer

10.0\text{ A}

The magnetic force on a current-carrying conductor is given by

F = I L B \sin\theta

. With current flowing East and the magnetic field directed

30^\circ

North of East, the angle

\theta = 30^\circ

. The right-hand rule confirms that

\vec{L} \times \vec{B}

points vertically upward. Setting the upward force equal to the weight

mg

, we get

I \times 0.40 \times 0.60 \times \sin 30^\circ = 0.12 \times 10

, which yields

I = 10.0\text{ A}

.

Step-by-Step Solution

1

Calculate the weight of the horizontal wire

W = mg = 0.12\text{ kg} \times 10\text{ m s}^{-2} = 1.2\text{ N}

The magnetic force must balance the downward gravitational force acting on the wire.

2

Determine the angle

\theta

between the current direction and magnetic field

\theta = 30^\circ

The current flows East and the magnetic field points

30^\circ

North of East, so the angle between the vector length element and magnetic field is

30^\circ

.

3

Express the magnetic force using

F = I L B \sin\theta

F = I \times 0.40\text{ m} \times 0.60\text{ T} \times \sin 30^\circ = 0.12 I

The cross product

\vec{I L} \times \vec{B}

gives the magnitude

I L B \sin\theta

and a vertical upward direction according to the right-hand rule.

4

Equate the magnetic force to the weight and solve for

I

0.12 I = 1.2 \implies I = 10.0\text{ A}

For complete vertical equilibrium, the upward magnetic force must equal the downward weight.

Key Concept

Magnetic force on a current-carrying conductor in a uniform magnetic field

Estimated Time:2m 0s

Question 12839Question →

If two solid spheres, $A$ and $B$ , constructed from the same uniform metallic material, have radii in the ratio $2:1$ respectively, then supplying equal quantities of heat energy to both spheres will cause sphere $B$ to experience a temperature rise eight times that of sphere $A$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because the mass and heat capacity of a uniform solid sphere scale with the cube of its radius (

r^3

), giving sphere

A

eight times the heat capacity of sphere

B

. For an equal input of thermal energy, sphere

B

undergoes eight times the temperature rise of sphere

A

.

The statement is correct because volume scales as

r^3

, giving sphere

A

eight times the mass and heat capacity of sphere

B

. As a result, sphere

B

undergoes eight times the temperature increase of sphere

A

when absorbing equal heat energy.

Step-by-Step Solution

1

Relate the masses of the two spheres using their radii ratio.

m_A = \rho V_A = \rho \cdot \frac{4}{3}\pi (2r_B)^3 = 8 \left(\rho \cdot \frac{4}{3}\pi r_B^3\right) = 8 m_B

.

Mass is proportional to volume for uniform density, and volume scales as

r^3

.

2

Express the heat capacities of both spheres.

C_A = m_A c = 8 m_B c = 8 C_B

.

Heat capacity

C = mc

is an extensive property proportional to mass, while specific heat capacity

c

is constant for a given material.

3

Compare the temperature rises for equal heat energy

Q

.

\Delta T_B = \frac{Q}{C_B} = \frac{Q}{\frac{1}{8}C_A} = 8 \left(\frac{Q}{C_A}\right) = 8 \Delta T_A

.

Temperature rise

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

is inversely proportional to heat capacity when heat input

Q

is identical.

Key Concept

Extensive nature of heat capacity and its scaling with volume (

r^3

) versus intensive specific heat capacity.

Estimated Time:1m 30s

Question 12840Question →

When

16.8\text{ g}

of sodium hydrogentrioxocarbonate(IV) (

\text{NaHCO}_3

) is strongly heated in a closed system until decomposition is complete according to the equation:

2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)

What is the total volume of gaseous products liberated at standard temperature and pressure (STP)?
[

M_r\text{ of NaHCO}_3 = 84\text{ g mol}^{-1}

; Molar volume of gas at STP

= 22.4\text{ dm}^3\text{ mol}^{-1}

]

2.24\text{ dm}^3

4.48\text{ dm}^3

4.80\text{ dm}^3

8.96\text{ dm}^3

Show answer & explanation

Answer: $4.48\text{ dm}^3$

Answer

The total volume of gaseous products liberated at STP is

4.48\text{ dm}^3

.

Decomposing

16.8\text{ g}

(

0.20\text{ mol}

) of

\text{NaHCO}_3

yields

0.10\text{ mol}

of

\text{H}_2\text{O}(g)

and

0.10\text{ mol}

of

\text{CO}_2(g)

, totaling

0.20\text{ mol}

of gaseous products. At STP,

0.20\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 4.48\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the number of moles of

\text{NaHCO}_3

decomposed.

\text{Moles of NaHCO}_3 = \frac{16.8\text{ g}}{84\text{ g mol}^{-1}} = 0.20\text{ mol}

.

Converting the given mass into moles using molar mass.

2

Determine the mole ratio between

\text{NaHCO}_3

and total gaseous products.

From

2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)

,

2\text{ moles}

of

\text{NaHCO}_3

produce

1\text{ mole}

of

\text{H}_2\text{O}(g)

and

1\text{ mole}

of

\text{CO}_2(g)

, giving

2\text{ moles}

of total gaseous products.

Both water vapor (at high decomposition temperature) and carbon(IV) oxide exist in the gaseous state.

3

Calculate total moles and total volume of gas at STP.

\text{Total moles of gas} = 0.20\text{ mol}

.

\text{Total volume} = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 4.48\text{ dm}^3

.

Multiplying total gaseous moles by the molar volume at STP.

Key Concept

Thermal decomposition stoichiometry of sodium hydrogentrioxocarbonate(IV)

Estimated Time:1m 30s

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept