01Concept in plain language
A physical quantity does not change when you change the unit. Only the label changes.
The rod that is 1 metre long is also 100 centimetres long — same rod, two descriptions.
Write the measurement as a product: quantity = n × u, where n is the
numerical value and u is the unit. Because the quantity itself is fixed, the product must stay fixed:
n₁ u₁ = n₂ u₂
Everything in this topic is a consequence of that one line. It says that n and u are inversely
related: make the unit smaller and the number gets bigger. A metre becomes 100
centimetres because the centimetre is a hundred times smaller.
Why dimensions enter
For a simple length the ratio of units is a single number. For a derived quantity the unit is built
out of several base units, each with its own exponent. Energy, for instance, is M L²
T⁻², so its unit contains one power of the mass unit, two of the length unit and minus two
of the time unit. Each of those ratios must be applied with its own exponent:
n₂ = n₁ (M₁/M₂)a (L₁/L₂)b (T₁/T₂)c
Here a, b and c are simply read off the dimensional formula of the quantity. That is the whole
method — which is why this topic sits directly on top of the dimensional-formulae file.
The one thing to get right: which unit goes on top
The old unit goes in the numerator, the new unit in the denominator. Every wrong answer in this topic comes from inverting one of those ratios.
Sanity check that costs two seconds: is the new unit smaller than the old one? Then the new number must be bigger. If your answer disagrees, you inverted something.
03The four-step method
- Write the dimensional formula of the quantity. This gives you the exponents a, b, c.
Nothing else about the quantity matters.
- Express both sets of units in a common form. If one system uses grams and the other
kilograms, convert so that both ratios are computed in the same unit. This is where most arithmetic
errors originate, not in the physics.
- Form each ratio as old ÷ new and raise it to its exponent, keeping the sign.
- Multiply everything by n₁, then sanity-check the direction: smaller new unit
→ bigger new number.
The α, β, γ questions are the easiest of all
When a question says 'the unit of mass is α kg, the unit of length is β m, the unit of time is γ s', you do not need to compute anything. Take the dimensional formula, flip the sign of every exponent, and write α, β, γ in place of M, L, T.
energy = M¹ L² T⁻² → flip → α⁻¹ β⁻² γ⁺²
That is exactly NCERT's answer for the calorie: 4.2 α⁻¹ β⁻² γ². The 4.2 is just n₁ carried along unchanged.
A different question shape: constructing a base unit
Sometimes the question fixes the units of force, length and time and asks for the unit of mass. Nothing is being converted here, so n₁u₁ = n₂u₂ does not apply. Instead rearrange the dimensional formula for the base unit you want — M = F T² / L — and substitute the given unit sizes directly.
06Exceptions and traps
The inverted ratio
By far the most common error. Old unit on top, new unit underneath. Verify with the direction test: smaller new unit means bigger new number.
Negative exponents flip the direction of that one factor
In G, the mass exponent is −1, so the mass factor is 10⁻³ rather than 10³. In density, the length exponent is −3, so the length factor is 10⁶ rather than 10⁻⁶. Copy the signs from the dimensional formula before substituting.
Prefixes take the same power as the unit
1 cm² is 10⁻⁴ m², not 10⁻² m². 1 cm³ is 10⁻⁶ m³. Anywhere an area or a volume appears, the conversion factor is squared or cubed.
Mixed units inside one ratio
If the new unit of mass is quoted in grams while the SI unit is the kilogram, convert one of them before dividing. 1 kg / 100 g is not 1/100; it is 1000 g / 100 g = 10.
Do not force the conversion formula onto every question
'The unit of force is 100 N, of length 10 m, of time 100 s — find the unit of mass' is not a conversion. Rearrange M = FT²/L and substitute. Using n₁u₁ = n₂u₂ here produces nonsense.
Exact versus rounded values
NCERT uses 1 y = 365.25 d = 3.156 × 10⁷ s and c = 3 × 10⁸ m s⁻¹. Options are usually spaced far enough apart that rounding does not matter, but when two options differ only in the third digit, use the more precise values.
09Twenty worked questions
About the labels
Items tagged PYQ here are the NCERT chapter-end exercises themselves, which NTA has drawn on repeatedly. Every conversion on this page was recomputed by an independent script before printing.
Q01SI to CGS
One joule expressed in ergs is:
- 10⁷ erg
- 10⁵ erg
- 10⁶ erg
- 10⁴ erg
Show full solution
Given
1 J = 1 kg m² s⁻²; 1 erg = 1 g cm² s⁻².
Asked
The numerical value of 1 J in the CGS system.
Concept
Energy has dimensions M L² T⁻². Substitute the ratio of the old unit to the new unit, each raised to its own exponent.
Formula
n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)² (T₁/T₂)⁻²
Baby steps
- M₁/M₂ = kg/g = 10³.
- L₁/L₂ = m/cm = 10², so (L₁/L₂)² = 10⁴.
- T₁/T₂ = s/s = 1, so the time factor is 1.
- n₂ = 1 × 10³ × 10⁴ × 1 = 10⁷.
- So 1 J = 10⁷ erg.
Why not others
| (a) | Correct. |
| (b) | This is the newton-to-dyne factor, not the joule-to-erg factor. |
| (c) | This comes from squaring 10² as 10³. |
| (d) | This uses only the length factor. |
Shortcut
Learn the two anchors and derive the rest: 1 N = 10⁵ dyne, 1 J = 10⁷ erg. Power follows immediately, since 1 W = 1 J s⁻¹ = 10⁷ erg s⁻¹.
Where it goes wrong
The erg is a far smaller unit than the joule, so the number must get bigger. If your answer is less than 1, the ratio was inverted.
Q02SI to CGS
One newton expressed in dynes is:
- 10⁷ dyne
- 10⁶ dyne
- 10³ dyne
- 10⁵ dyne
Show full solution
Given
1 N = 1 kg m s⁻²; 1 dyne = 1 g cm s⁻².
Asked
The numerical value of 1 N in CGS.
Concept
Force has dimensions M L T⁻², so only one power of length appears — unlike energy.
Formula
n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)¹ (T₁/T₂)⁻²
Baby steps
- Mass factor: kg/g = 10³.
- Length factor: m/cm = 10² (first power only).
- Time factor: 1.
- n₂ = 10³ × 10² = 10⁵. So 1 N = 10⁵ dyne.
Why not others
| (a) | That is the joule-to-erg factor; energy carries L². |
| (b) | Comes from a slip in one of the two powers. |
| (c) | Uses only the mass factor. |
| (d) | Correct. |
Shortcut
Newton and joule differ by exactly one power of length, i.e. by a factor of 10² when moving to CGS: 10⁵ and 10⁷.
Where it goes wrong
Do not memorise these as isolated facts. Rebuild them from the dimensional formula and you will never confuse the two.
Q03PYQ · NCERT Ex. 1.2(a)NCERT exercise
Fill in the blank: 1 kg m² s⁻² = … g cm² s⁻².
- 10⁵
- 10⁶
- 10⁷
- 10⁴
Show full solution
Given
The same quantity written in SI base units and in CGS base units.
Asked
The CGS numerical value.
Concept
This is the joule-to-erg conversion written out in base units, so the exponents are handed to you directly — no dimensional analysis is needed.
Formula
n₂ = n₁ (kg/g)¹ (m/cm)² (s/s)⁻²
Baby steps
- kg → g: multiply by 10³.
- m² → cm²: multiply by (10²)² = 10⁴.
- s⁻² is unchanged.
- Total: 10³ × 10⁴ = 10⁷.
Why not others
| (a) | Force-level factor. |
| (b) | One power of ten short. |
| (c) | Correct — and this is why 1 J = 10⁷ erg. |
| (d) | Comes from converting m² as if it were m. |
Shortcut
When the units are already written out with their exponents, just convert each factor separately and multiply. The dimensional formula is visible on the page.
Where it goes wrong
Converting m² using 10² instead of 10⁴. The prefix is squared along with the unit.
Q04PYQ · NCERT Ex. 1.2(d)NCERT exercise
G = 6.67 × 10⁻¹¹ N m² kg⁻². Its value in cm³ s⁻² g⁻¹ is:
- 6.67 × 10⁻¹¹
- 6.67 × 10⁻⁸
- 6.67 × 10⁻⁵
- 6.67 × 10⁻¹⁴
Show full solution
Given
G = 6.67 × 10⁻¹¹ in SI; [G] = M⁻¹ L³ T⁻².
Asked
The CGS numerical value of G.
Concept
Note the exponents in [G]: mass carries −1, so the mass ratio appears to the power −1 and makes the number smaller.
Formula
n₂ = n₁ (M₁/M₂)⁻¹ (L₁/L₂)³ (T₁/T₂)⁻²
Baby steps
- Mass: (10³)⁻¹ = 10⁻³.
- Length: (10²)³ = 10⁶.
- Time: 1.
- n₂ = 6.67 × 10⁻¹¹ × 10⁻³ × 10⁶
- = 6.67 × 10⁻¹¹⁺³ = 6.67 × 10⁻⁸.
Why not others
| (a) | No conversion applied at all. |
| (b) | Correct. |
| (c) | Comes from using +3 for the mass factor instead of −3. |
| (d) | Comes from subtracting 3 instead of adding 6 − 3. |
Shortcut
Check the sign of each exponent in the dimensional formula before you start. A negative exponent flips the direction of that factor.
Where it goes wrong
G is the classic case where one exponent is negative. Students apply 10³ for mass out of habit and land three orders of magnitude away.
Q05PYQ · NCERT Ex. 1.2(c)NCERT exercise
3.0 m s⁻² expressed in km h⁻² is:
- 3.888 × 10⁴
- 1.08 × 10⁴
- 3.888 × 10⁵
- 1.296 × 10⁴
Show full solution
Given
a = 3.0 m s⁻²; convert to km h⁻².
Asked
The numerical value in the new units.
Concept
Acceleration is L T⁻². The time unit is getting much larger, and it enters to the power −2, so the number grows sharply.
Formula
n₂ = n₁ (L₁/L₂)¹ (T₁/T₂)⁻²
Baby steps
- Length: m/km = 10⁻³.
- Time: s/h = 1/3600, and (1/3600)⁻² = 3600² = 1.296 × 10⁷.
- n₂ = 3.0 × 10⁻³ × 1.296 × 10⁷.
- = 3.0 × 1.296 × 10⁴ = 3.888 × 10⁴ km h⁻².
Why not others
| (a) | Correct. |
| (b) | Comes from using 3600 once instead of squared. |
| (c) | One power of ten too large. |
| (d) | This is 3600² × 10⁻³ without the factor of 3.0. |
Shortcut
Two separate factors, applied one at a time: divide by 1000 for km, then multiply by 3600² for h⁻². Doing them together in one line is where sign errors creep in.
Where it goes wrong
Squaring 3600 gives 1.296 × 10⁷, not 1.296 × 10⁶. Work it as 36² × 10⁴ = 1296 × 10⁴.
Q06PYQ · NCERT Ex. 1.3New system
A calorie equals 4.2 J, where 1 J = 1 kg m² s⁻². In a new system the unit of mass is α kg, the unit of length is β m and the unit of time is γ s. The magnitude of a calorie in the new units is:
- 4.2 αβ²γ⁻²
- 4.2 α⁻¹β²γ⁻²
- 4.2 α⁻¹β⁻²γ⁻²
- 4.2 α⁻¹β⁻²γ²
Show full solution
Given
1 cal = 4.2 kg m² s⁻². New units: mass α kg, length β m, time γ s.
Asked
The magnitude of a calorie in the new system.
Concept
Every exponent in the dimensional formula reappears with its sign flipped, because the number and the unit always move in opposite directions.
Formula
n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)² (T₁/T₂)⁻²
Baby steps
- M₁/M₂ = 1 kg / α kg = α⁻¹.
- L₁/L₂ = 1 m / β m = β⁻¹, so squared it is β⁻².
- T₁/T₂ = 1 s / γ s = γ⁻¹, raised to −2 it is γ².
- n₂ = 4.2 α⁻¹ β⁻² γ², exactly as NCERT states.
Why not others
| (a) | Every sign is wrong — this is the reciprocal. |
| (b) | The length exponent kept the wrong sign. |
| (c) | The time exponent kept the wrong sign; −2 applied to γ⁻¹ gives γ⁺². |
| (d) | Correct. |
Shortcut
Take the dimensional formula M¹L²T⁻², flip every sign to get M⁻¹L⁻²T⁺², then write α, β, γ in place of M, L, T. Two seconds, no algebra.
Where it goes wrong
The double negative on time. The dimensional exponent is −2 and the ratio is already γ⁻¹, so the two negatives multiply to give γ+2.
Q07PYQ · NCERT Ex. 1.1(a)NCERT exercise
The volume of a cube of side 1 cm, expressed in m³, is:
- 10⁻²
- 10⁻⁴
- 10⁻⁶
- 10⁻³
Show full solution
Asked
Its volume in cubic metres.
Concept
A prefix is raised to the same power as the unit it attaches to. For volume, cube the conversion factor.
Formula
V = a³ ; 1 cm = 10⁻² m
Baby steps
- 1 cm = 10⁻² m.
- V = (10⁻² m)³.
- = 10⁻⁶ m³.
- So a cubic centimetre is a millionth of a cubic metre, and 1 L = 10³ cm³ = 10⁻³ m³.
Why not others
| (a) | This squares nothing and treats volume as a length. |
| (b) | This is the area conversion, 1 cm² = 10⁻⁴ m². |
| (c) | Correct. |
| (d) | This is the litre-to-cubic-metre factor, a different quantity. |
Shortcut
Three lines that solve most prefix errors: 1 cm = 10⁻² m, 1 cm² = 10⁻⁴ m², 1 cm³ = 10⁻⁶ m³.
Where it goes wrong
Writing 1 cm³ = 10⁻² m³. The exponent on the prefix is multiplied by the power, not left alone.
Q08PYQ · NCERT Ex. 1.1(b)NCERT exercise
The total surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm, expressed in mm², is closest to:
- 1.5 × 10³
- 1.5 × 10⁴
- 1.5 × 10²
- 1.5 × 10⁵
Show full solution
Given
Solid cylinder, r = 2.0 cm, h = 10.0 cm.
Asked
Total surface area in square millimetres.
Concept
'Solid' means both circular ends are included. Compute in cm² first, then convert, squaring the length factor.
Formula
A = 2πr(r + h) ; 1 cm² = 10² mm²
Baby steps
- A = 2πr(r + h) = 2π(2.0)(2.0 + 10.0).
- = 2π × 2.0 × 12.0 = 48π cm².
- = 150.8 cm².
- 1 cm² = 10² mm², so A = 1.508 × 10⁴ mm² ≈ 1.5 × 10⁴ mm².
Why not others
| (a) | Comes from converting cm² using 10¹ instead of 10². |
| (b) | Correct. |
| (c) | This is the answer left in cm². |
| (d) | One power of ten too many. |
Shortcut
Total surface of a cylinder = 2πr(r + h). The curved part alone is 2πrh; the word solid is what adds the two 2πr² ends.
Where it goes wrong
Reading 'solid cylinder' as curved surface only would give 2π(2)(10) = 40π = 125.7 cm² and the wrong option. The adjective carries the mark.
Q09PYQ · NCERT Ex. 1.1(c)NCERT exercise
A vehicle moving with a speed of 18 km h⁻¹ covers, in 1 s, a distance of:
- 5 m
- 10 m
- 18 m
- 2.5 m
Show full solution
Given
Speed = 18 km h⁻¹; time = 1 s.
Asked
Distance covered in metres.
Concept
Convert the speed to m s⁻¹ using the standard factor 5/18, then multiply by the time.
Formula
1 km h⁻¹ = 1000/3600 m s⁻¹ = 5/18 m s⁻¹
Baby steps
- 18 km h⁻¹ = 18 × (5/18) m s⁻¹.
- = 5 m s⁻¹.
- Distance in 1 s = 5 × 1.
- = 5 m.
Why not others
| (a) | Correct. |
| (b) | This uses 36 km h⁻¹. |
| (c) | This ignores the conversion completely. |
| (d) | This applies the factor 5/18 twice. |
Shortcut
Multiply by 5/18 to go from km h⁻¹ to m s⁻¹; multiply by 18/5 to go back. Useful anchors: 18 km h⁻¹ = 5 m s⁻¹, 36 = 10, 54 = 15, 72 = 20.
Where it goes wrong
Using 5/18 in the wrong direction. Sanity check: a metre per second is a brisk walk, and 18 km h⁻¹ is a slow cycle, so 5 is plausible while 18 is not.
Q10PYQ · NCERT Ex. 1.1(d)NCERT exercise
The relative density of lead is 11.3. Its density in kg m⁻³ is:
- 11.3
- 1.13 × 10³
- 1.13 × 10⁵
- 1.13 × 10⁴
Show full solution
Given
Relative density of lead = 11.3; density of water = 10³ kg m⁻³.
Asked
Density of lead in kg m⁻³.
Concept
Relative density is a dimensionless ratio to water. Multiply by the density of water in whatever unit is asked for.
Formula
ρ = (relative density) × ρwater
Baby steps
- ρwater = 1 g cm⁻³ = 10³ kg m⁻³.
- ρlead = 11.3 × 1 g cm⁻³ = 11.3 g cm⁻³.
- In SI: 11.3 × 10³ kg m⁻³.
- = 1.13 × 10⁴ kg m⁻³.
Why not others
| (a) | This is the dimensionless ratio, not a density. |
| (b) | This is the density of water itself. |
| (c) | One order of magnitude too large. |
| (d) | Correct. |
Shortcut
1 g cm⁻³ = 10³ kg m⁻³. Check it from the dimensions: (10⁻³ kg)/(10⁻² m)³ = 10⁻³/10⁻⁶ = 10³.
Where it goes wrong
Relative density has no unit, so answering '11.3 kg m⁻³' confuses the ratio with the density. The question asks you to convert, not to copy.
Q11SI to CGS
One pascal expressed in dyne cm⁻² is:
- 10³
- 10²
- 10
- 10⁻¹
Show full solution
Given
1 Pa = 1 N m⁻²; 1 N = 10⁵ dyne; 1 m² = 10⁴ cm².
Asked
The CGS value of one pascal.
Concept
Pressure is M L⁻¹ T⁻². The negative length exponent partly cancels the mass factor, leaving only a factor of ten.
Formula
n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)⁻¹ (T₁/T₂)⁻²
Baby steps
- Mass: 10³. Length: (10²)⁻¹ = 10⁻². Time: 1.
- n₂ = 10³ × 10⁻² = 10.
- Cross-check: 10⁵ dyne / 10⁴ cm² = 10 dyne cm⁻². ✓
- So 1 Pa = 10 dyne cm⁻², and 1 bar = 10⁶ dyne cm⁻².
Why not others
| (a) | Comes from ignoring the negative sign on the length exponent. |
| (b) | One power of ten too many. |
| (c) | Correct. |
| (d) | Comes from inverting the whole result. |
Shortcut
Two routes must agree: dimensional substitution, and the direct N/m² → dyne/cm² division. Running both is a free self-check on any conversion.
Where it goes wrong
Pressure is the classic case where the mass and length factors nearly cancel. Expect a small answer, not a huge one.
Q12PYQ · NCERT Ex. 1.5New system
A new unit of length is chosen such that the speed of light in vacuum is unity. If light takes 8 min 20 s to travel from the Sun to the Earth, the Sun–Earth distance in the new unit is:
- 300
- 500
- 8.33
- 1000
Show full solution
Given
c = 1 in the new units; time taken = 8 min 20 s.
Asked
The Sun–Earth distance in the new unit of length.
Concept
If speed is numerically 1, then distance = speed × time is numerically equal to the time in seconds. The new unit of length is simply the distance light covers in one second.
Formula
d = c t, with c = 1
Baby steps
- t = 8 min 20 s = 8 × 60 + 20 = 500 s.
- d = c × t = 1 × 500.
- = 500 new units of length.
- Sanity check in SI: 3 × 10⁸ × 500 = 1.5 × 10¹¹ m, the astronomical unit. ✓
Why not others
| (a) | This uses only the 300 from 5 minutes. |
| (b) | Correct. |
| (c) | This is the time in minutes. |
| (d) | This doubles the time. |
Shortcut
Whenever a constant is set to unity, that constant's units disappear and two quantities become numerically identical. Setting c = 1 makes distance and time interchangeable — the same idea behind the light year.
Where it goes wrong
Converting 8 min 20 s as 8.20 minutes. It is 8 min + 20 s = 500 s, not 8.2 × 60.
Q13SI to CGS
One watt expressed in erg s⁻¹ is:
- 10⁷
- 10⁵
- 10³
- 10⁶
Show full solution
Given
1 W = 1 J s⁻¹; 1 J = 10⁷ erg.
Asked
The CGS value of one watt.
Concept
Power is M L² T⁻³. The extra time factor contributes nothing, because the second is the same unit in both systems.
Formula
n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)² (T₁/T₂)⁻³
Baby steps
- Mass: 10³. Length: (10²)² = 10⁴. Time: 1⁻³ = 1.
- n₂ = 10³ × 10⁴ = 10⁷.
- So 1 W = 10⁷ erg s⁻¹.
- Same factor as the joule, because the second is shared by SI and CGS.
Why not others
| (a) | Correct. |
| (b) | Force-level factor. |
| (c) | Mass factor only. |
| (d) | One power of ten short. |
Shortcut
Any quantity whose only difference from energy is a power of time converts by the same factor 10⁷ in CGS, because both systems use the second.
Where it goes wrong
The time exponent looks like extra work but the ratio is 1, so it can never change the answer. Recognising that early saves time in the exam.
Q14New system
In a certain system the unit of mass is 10 kg, the unit of length is 10 m and the unit of time is 1 s. The magnitude of 1 joule in this system is:
- 10³
- 10
- 10⁻¹
- 10⁻³
Show full solution
Given
New units: mass 10 kg, length 10 m, time 1 s. Energy has dimensions M L² T⁻².
Asked
The numerical value of 1 J in the new system.
Concept
The new units are all larger than the SI units, so the number must come out smaller.
Formula
n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)² (T₁/T₂)⁻²
Baby steps
- M₁/M₂ = 1 kg / 10 kg = 10⁻¹.
- L₁/L₂ = 1 m / 10 m = 10⁻¹, squared gives 10⁻².
- T₁/T₂ = 1, so the time factor is 1.
- n₂ = 1 × 10⁻¹ × 10⁻² = 10⁻³.
Why not others
| (a) | Every sign inverted. |
| (b) | Only the mass factor applied, and inverted. |
| (c) | The length was not squared. |
| (d) | Correct. |
Shortcut
Sanity check before computing: are the new units bigger or smaller than SI? Bigger units → smaller number. Here all are bigger or equal, so the answer must be below 1.
Where it goes wrong
Applying the ratio the wrong way up. Write the old unit on top and the new unit underneath, every time.
Q15New system
In a new system the unit of force is 100 N, the unit of length is 10 m and the unit of time is 100 s. The unit of mass in this system is:
- 10³ kg
- 10⁴ kg
- 10⁵ kg
- 10⁶ kg
Show full solution
Given
New units: force 100 N, length 10 m, time 100 s.
Asked
The corresponding unit of mass.
Concept
Work with the units themselves rather than with a numerical value. Rearrange F = ma so that mass is the subject, then substitute the given unit sizes.
Formula
F = M L T⁻² → M = F T² / L
Baby steps
- M = F T² / L.
- Substitute the new unit sizes: (100 N)(100 s)² / (10 m).
- = 100 × 10⁴ / 10 kg.
- = 10⁵ kg.
Why not others
| (a) | Comes from using T instead of T². |
| (b) | Comes from multiplying by 10 instead of dividing. |
| (c) | Correct. |
| (d) | One power of ten too many. |
Shortcut
Whenever the question fixes force, length and time instead of mass, length and time, rearrange the dimensional formula for the missing base unit and substitute directly.
Where it goes wrong
Do not use the n₁u₁ = n₂u₂ formula here. Nothing is being converted — you are being asked to construct one of the base units of the new system.
Q16Energy units
Given 1 eV = 1.6 × 10⁻¹⁹ J, one joule expressed in electron volts is:
- 1.6 × 10¹⁹
- 6.25 × 10¹⁸
- 6.02 × 10²³
- 1.6 × 10⁻¹⁹
Show full solution
Given
1 eV = 1.6 × 10⁻¹⁹ J.
Asked
1 J expressed in electron volts.
Concept
Simple reciprocal. The electron volt is a tiny unit, so one joule contains a very large number of them.
Formula
1 J = 1 / (1.6 × 10⁻¹⁹) eV
Baby steps
- 1 J = 1 ÷ (1.6 × 10⁻¹⁹) eV.
- 1/1.6 = 0.625.
- = 0.625 × 10¹⁹.
- = 6.25 × 10¹⁸ eV.
Why not others
| (a) | Right order of magnitude, wrong mantissa — this is 1/1.6 mishandled. |
| (b) | Correct. |
| (c) | This is Avogadro's number, which has nothing to do with the conversion. |
| (d) | This is the eV-to-joule factor, i.e. the question read backwards. |
Shortcut
1/1.6 = 0.625 is worth knowing outright; it recurs in photoelectric and nuclear calculations throughout the syllabus.
Where it goes wrong
Answering 1.6 × 10⁻¹⁹ because it is the number printed in the question. Ask which unit is smaller: the eV, so the count must be enormous.
Q17SI to CGS
Young's modulus of steel is 2 × 10¹¹ N m⁻². In dyne cm⁻² this is:
- 2 × 10¹²
- 2 × 10¹⁰
- 2 × 10¹³
- 2 × 10⁹
Show full solution
Given
Y = 2 × 10¹¹ N m⁻²; [Y] = M L⁻¹ T⁻².
Concept
Young's modulus has the dimensions of pressure, so the conversion factor is the same one: a single factor of ten.
Formula
1 N m⁻² = 10 dyne cm⁻²
Baby steps
- From the pressure conversion, 1 Pa = 10 dyne cm⁻².
- Y = 2 × 10¹¹ × 10 dyne cm⁻².
- = 2 × 10¹² dyne cm⁻².
- Verify by exponents: 10³ (mass) × 10⁻² (length) = 10. ✓
Why not others
| (a) | Correct. |
| (b) | Comes from dividing by ten instead of multiplying. |
| (c) | Two powers of ten too many. |
| (d) | Comes from applying the length factor twice. |
Shortcut
All pressure-dimensioned quantities — pressure, stress, every modulus, energy density — share the factor 10 between SI and CGS.
Where it goes wrong
Do not derive the factor again from scratch under exam pressure. Recognise the dimensional family and reuse the factor.
Q18Density
The density of water is 1 g cm⁻³. In kg m⁻³ this is:
- 1
- 10
- 100
- 1000
Show full solution
Given
ρ = 1 g cm⁻³; [ρ] = M L⁻³.
Asked
The SI value of the density of water.
Concept
Mass goes down by 10³ when converting g to kg, but volume goes down by 10⁶, so the ratio goes up by 10³.
Formula
n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)⁻³
Baby steps
- M₁/M₂ = g/kg = 10⁻³.
- L₁/L₂ = cm/m = 10⁻², and (10⁻²)⁻³ = 10⁶.
- n₂ = 1 × 10⁻³ × 10⁶ = 10³.
- So water has a density of 1000 kg m⁻³.
Why not others
| (a) | No conversion applied. |
| (b) | Only one factor of ten applied. |
| (c) | Two factors applied. |
| (d) | Correct. |
Shortcut
Anchor value: water is 1 g cm⁻³ = 10³ kg m⁻³. Mercury is 13.6 g cm⁻³ = 1.36 × 10⁴ kg m⁻³, and both appear constantly in fluid and pressure questions.
Where it goes wrong
Note that this conversion runs the opposite way to most: the number gets larger even though we are moving to larger units, because the length exponent is negative.
Q19Power units
Given that 1 horsepower = 746 W, its value in erg s⁻¹ is:
- 7.46 × 10⁷
- 7.46 × 10⁸
- 7.46 × 10⁹
- 7.46 × 10¹⁰
Show full solution
Given
1 hp = 746 W; 1 W = 10⁷ erg s⁻¹.
Asked
1 hp in CGS power units.
Concept
Chain the two conversions. Handle the mantissa and the power of ten separately.
Formula
1 hp = 746 × 10⁷ erg s⁻¹
Baby steps
- 1 W = 10⁷ erg s⁻¹.
- 1 hp = 746 × 10⁷ erg s⁻¹.
- 746 = 7.46 × 10².
- So 1 hp = 7.46 × 10² × 10⁷ = 7.46 × 10⁹ erg s⁻¹.
Why not others
| (a) | The factor of 746 was dropped. |
| (b) | One power of ten short — 746 contributes 10², not 10¹. |
| (c) | Correct. |
| (d) | One power of ten too many. |
Shortcut
Convert the unit first, then attach the mantissa. Mixing the two in one step is where powers of ten go astray.
Where it goes wrong
Writing 746 as 7.46 × 10¹. Count the digits before the decimal point: 746 has three, so the exponent is 2.
Q20New system
A physical quantity has dimensions M L⁻¹ T⁻² and a value of 100 in SI units. In a system where the unit of mass is 100 g, the unit of length is 10 cm and the unit of time is 1 minute, its value is:
- 3.6 × 10³
- 3.6 × 10⁵
- 3.6 × 10⁴
- 3.6 × 10⁶
Show full solution
Given
n₁ = 100, [X] = M L⁻¹ T⁻². New units: 100 g, 10 cm, 1 min.
Asked
The numerical value in the new system.
Concept
Convert every new unit into the corresponding SI unit first, so that both sides of each ratio are in the same unit. Then apply the exponents with their signs.
Formula
n₂ = n₁ (M₁/M₂)¹ (L₁/L₂)⁻¹ (T₁/T₂)⁻²
Baby steps
- Mass: 1 kg / 100 g = 1000 g / 100 g = 10, exponent +1 → factor 10.
- Length: 1 m / 10 cm = 100 cm / 10 cm = 10, exponent −1 → factor 10⁻¹.
- Time: 1 s / 60 s = 1/60, exponent −2 → factor 60² = 3600.
- n₂ = 100 × 10 × 0.1 × 3600.
- = 100 × 3600 = 3.6 × 10⁵.
Why not others
| (a) | Comes from dropping the factor of 100 in n₁. |
| (b) | Correct. |
| (c) | Comes from using 60 instead of 60². |
| (d) | Comes from an extra factor of ten in the mass or length term. |
Shortcut
Put every unit into a common form (all grams, all centimetres, all seconds) before forming any ratio. Mixed units inside a ratio are the main source of error in this question type.
Where it goes wrong
Notice that the mass factor of 10 and the length factor of 1/10 cancel exactly, leaving only the time term. Spotting that cancellation early turns a four-step calculation into one step.