Introduction to Physics
EXERCISES (Q-1 to Q-10)
EXERCISES (Q-1 to Q-10)
1. Find the dimensions of
(a) linear momentum,
(b) frequency and
(c) pressure.
ANSWER: (a) Linear momentum = mass x velocity
= mass x distance/time
→p = m*l/t
The dimension of mass = [M]
The dimension of velocity = [L/T] = [LT⁻¹]
Hence the dimensions of linear momentum,
[p] = [M]*[LT⁻¹] = [MLT⁻¹]
(b) Frequency = 1/Time period
→f = 1/T
The dimension of frequency [f] = [1/T] = [T⁻¹]
(c) Pressure = Force per unit area
→ P = Force/Area = Mass*acceleration/Area
The dimension of mass = [M]
The dimension of acceleration = [LT⁻²]
The dimension of area = [L²]
Hence the dimension of pressure,
[P] = [M]*[LT⁻²]/[L²] =[ML⁻¹T⁻²]
2. Find the dimensions of
(a) angular speed ⍵,
(b) angular acceleration α
(c) Torque Г and
(d) moment of inertia I.
Some of the equations involving these quantities are
⍵ = (θ₂-θ₁)/(t₂-t₁), α = (⍵₂-⍵₁)/(t₂-t₁),
Г = F*r and I = mr².
The symbols have standard meanings.
ANSWER: (a) Angular speed ⍵ = (θ₂ - θ₁)/(t₂ - t₁)
The dimension of [θ₂ - θ₁] =Dimension of angle which is dimensionless
The dimension of [t₂ - t₁] = Dimension of time =[T]
Hence [⍵] = 1/[T] = [T⁻¹]
(b) Angular acceleration α = (⍵₂ - ⍵₁)/(t₂ - t₁) The dimension of (⍵₂ - ⍵₁) = The dimension of angular speed = [T⁻¹]
The dimension of (t₂ - t₁) = The dimension of time =[T]
Hence the dimension of angular acceleration,
[α] = [T⁻¹]/[T] = [T⁻²]
(c) Torque = Force *Arm (distance),
i.e. Г = F*r
The dimension of force [F] = mass*acceleration =[M]*[LT⁻²] =[MLT⁻²]
The dimension of arm, [r] = [L]
Hence the dimension of torque,
[Г] = [MLT⁻²]*[L] =[ML²T⁻²]
(d) The moment of inertia, I =mr²
The dimension of mass, [m] = [M]
The dimension of radius/distance, [r] = [L]
Hence the dimension of the moment of inertia,
[I] = [M]*[L]² =[ML²]
3. Find the dimensions of
(a) electric field E,
(b) magnetic field B and
(c) magnetic permeability µ₀.
The relevant equations are
F =qE, F = qvB, and B = µ₀I/2πa
where F is a force, q is a charge, v is speed, I is current, and a is distance.
ANSWER: (a) F = qE, hence
Electric field E = F/q
The dimension of force [F] = [MLT⁻²]
The dimension of charge, [q] = [TI]
{Since charge = current*time}
Hence the dimension of electric field,
[E] = [MLT⁻²]/[TI] = [MLT⁻³I⁻¹]
(b) Since F = qvB,
Magnetic field, B = F/qv
Now, [F] = [MLT⁻²]
[q] = [TI]
Speed, [v] = [LT⁻¹]
Hence the dimension of magnetic field,
[B] = [MLT⁻²]/[TI]*[LT⁻¹] = [MT⁻²I⁻¹]
(c) Given, B = µ₀I/2πa
→µ₀ =2πaB/I
The dimension of distance, [a] =[L]
The dimension of the magnetic field, [B] =[MT⁻²I⁻¹]
The dimension of current, [I] = [I] and π is dimensionless. Hence the dimension of magnetic permeability,
[µ₀] = [L][MT⁻²I⁻¹]/[I] =[MLT⁻²I⁻²]
4. Find the dimensions of
(a) electric dipole moment p and
(b) magnetic dipole moment M.
The defining equations are p = q.d, and M = IA
where d is distance, A is area, q is a charge and I is current.
ANSWER: (a) The electric dipole moment, p = q*d
The dimensions of charge, [q] =[TI]
The dimensions of distance, [d] =[L]
Hence the dimensions of electric dipole moment,
[p] = [TI][L] = [LTI]
(b) The magnetic dipole moment, M = IA
The dimension of the area, [A] =[L²]
Hence the dimensions of the magnetic dipole moment, [M] = [I][L²] =[L²I]
5. Find the dimensions of Planck's constant h from the equation E = h𝞶 where E is the energy and 𝞶 is the frequency.
ANSWER: Since, E = h𝞶
→h = E/𝞶
Energy = Force*distance {same as work done}
→[E] = [MLT⁻²][L] =[ML²T⁻²]
The dimensions of frequency, [𝞶] = [T⁻¹]
Hence the dimensions of Planck's constant,
[h] = [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]
6. Find the dimensions of
(a) specific heat capacity c,
(b) the coefficient of linear expansion α and
(c) the gas constant R.
Some of the equations involving these quantities are
Q =mc(T₂-T₁), Lₜ = L₀[1+α(T₂-T₁)] and PV = nRT.
ANSWER: (a) Since Q = mc(T₂ - T₁)
→c = Q/m(T₂ - T₁)
The dimension of (T₂-T₁) = The dimension of temperature = [K]
The dimension of mass, [m] =[M]
The dimensions of heat, Q = the dimension of energy =[ML²T⁻²]
Hence the dimensions of specific heat capacity,
[c] = [ML²T⁻²]/[M][K] =[L²T⁻²K⁻¹]
(b) Since Lₜ = L₀ {1+α(T₂-T₁)}
→Lₜ - L₀ = L₀α(T₂-T₁)
→α = (Lₜ - L₀)/{L₀(T₂-T₁)}
The dimensions of (Lₜ-L₀) and L₀ = the dimensions of length =[L]
The dimensions of (T₂-T₁) = dimension of temperature =[K]
Hence the dimensions of the coefficient of linear expansion, [α] =[L]/[L][K] =[K⁻¹]
(c) Since PV = nRT
→R = PV/nT
The dimensions of pressure, [P] =Force/area =[MLT⁻²]/[L²] =[ML⁻¹T⁻²]
The dimensions of volume, [V] =[L³]
The dimension of temperature, T = [K]
The dimensions of amount of gas, [n] = [(mol)]
Hence the dimensions of the gas constant,
[R] = [ML⁻¹T⁻²][L³]/[(mol)][K]
=[ML²T⁻²K⁻¹(mol)⁻¹]
7. Taking force, length and time to be the fundamental quantities find the dimensions of
(a) density,
(b) pressure
(c) momentum, and
(d) energy.
ANSWER: Note that instead of mass force is taken as a fundamental quantity. Hence we shall express the mass in terms of force where ever it comes. For example, Force= mass*acceleration
→mass = Force/acceleration
(a) Density = mass per unit volume = mass/volume
= Force/(acceleration*volume)
Dimension of force = [F]
Dimensions of acceleration = [LT⁻²]
Dimensions of volume = [L³]
Hence the dimensions of density = [F]/{[LT⁻²][L³]}
=[FL⁻⁴T²]
Hence the dimensions of density = [F]/{[LT⁻²][L³]}
=[FL⁻⁴T²]
(b) Pressure = Force/Area = [F]/[L²] =[FL⁻²]
(c) Momentum = Mass*velocity
={Force/(Acceleration)}*{Length/Time}
={[F]/[LT⁻²]}*{[L]/[T]}
=[FL⁻¹T²]*[LT⁻¹]
=[FT]
(d) Energy = Force*Distance =[F]*[L] =[FL]
8. Suppose the acceleration due to gravity at a place is 10 m/s². Find its value in cm/(minute)².
ANSWER: 1 m = 100 cm, and 1 s = 1/60 minute
Putting these in the units,
10 m/s² = 10*(100 cm)/(1/60 minute)²
= 1000*60² cm/(minute)²
= 1000*3600 cm/(minute)²
= 3.6x10⁶ cm/(minute)²
9. The average speed of a snail is 0.020 miles/hour and that of a leopard is 70 miles/hour. Convert these speeds in SI units.
ANSWER: The SI unit of length is m (meter), and of time is s (seconds).
1 mile = 1.6 km = 1600 m, 1 hour = 3600 s
Hence 1 mile/hour = (1600 m)/(3600 s)
= 4/9 m/s
So, average speed of a snail = 0.020 miles/hour
= 0.020*4/9 m/s =0.08/9 m/s =0.00889 m/s
= 8.89x10⁻³ m/s
And the average speed of a leopard =70 miles/hour
=70*4/9 m/s =280/9 m/s =31.11 m/s
10. The height of the mercury column in a barometer in a Calcutta laboratory was recorded to be 75 cm. Calculate this pressure in SI and CGS units using the following data: Specific gravity of mercury = 13.6, Density of water = 10³ kg/m³, g = 9.8 m/s² at Calcutta. Pressure = hρg in usual symbols.
ANSWER: Here h = 75 cm =0.75 m
Density of mercury, ρ = specific gravity*Density of water =13.6*10³ kg/m³
g = 9.8 m/s²
Hence the pressure at the base of mercury column in SI units =hρg
=0.75*13.6x10³*9.8 N/m²
=99960 N/m²
≈100000 N/m²
=10x10⁴ N/m²
In CGS units, units of mass = g, length = cm and time = s. We know that 1 kg = 1000 g, 1 m = 100 cm.
Hence, 1 N =1 kg-m/s² = (1000 g)(100 cm)/s²
=10⁵ g-cm/s² =10⁵ dyne
So, 1 N/m² =(10⁵ dyne)/(100 cm)²
=10 dyne/cm²
So in CGS unit pressure = 10x10⁴ N/m²
=10x10⁴*(10 dyne/cm²)
=10x10⁵ dyne/cm²
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Links to the Chapters
Links to the Chapters
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
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Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
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Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
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Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
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Click here for → OBJECTIVE-II
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CHAPTER- 6 - Friction
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Click here for → Friction - OBJECTIVE-II
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CHAPTER- 5 - Newton's Laws of Motion
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Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
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Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
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Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
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Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
