NEET 2027 · Physics · Class 11 · Chapter 1
Sections 1.4 to 1.6, the summary, and a walk through all seventeen exercises.
Same format as Part 1. The grey strip is what the book says, written from scratch in plain words. Below it, what that means explains it. Anything with a dark heading bar can be changed and tried.
A dimension is a recipe. It says which of the seven base ingredients went into a quantity, and how much of each.
Speed is made from one length and one time — the time divided rather than multiplied. So the recipe for speed is: one length, minus one time. Nothing else goes in.
This is not the same as the unit. Metres per second, feet per second and kilometres per hour are three different units, but they all have the same recipe. Dimensions are about what kind of quantity it is, not about which measuring stick you happened to use.
The square brackets are a signal. Writing [F] does not mean "force". It means "the dimensions of force". The brackets say: I am talking about the recipe now, not the amount.
In mechanics you will almost always use only the first three — M, L and T. The other four appear when electricity, heat, light or chemistry come in.
The word to notice is powers. The recipe is written as a set of exponents.
Volume is length × length × length, so length is raised to the power 3. Mass and time do not appear at all, so their powers are 0. The full recipe is M⁰ L³ T⁰.
Those zeros are not decoration. M⁰ is a statement: mass has nothing to do with volume. Writing the zeros makes the recipe complete.
Follow the force one carefully, because it is the pattern for everything else.
Start with the equation that defines the quantity: F = ma. Then swap each symbol for its recipe. Mass is M. Acceleration is a length divided by a time twice over, so L T⁻². Multiply them and you get M L T⁻².
Three steps, always the same: write the defining equation, make your quantity the subject, swap symbols for recipes.
Set the three powers and see which quantity you have made.
This sentence is telling you what dimensions deliberately throw away.
They throw away the size. They throw away direction. They throw away what the quantity is for. All that survives is the kind of thing it is.
So a starting speed, a finishing speed and a change in speed all look identical to this method. That is not a flaw — it is what makes the method simple. But it does mean the method can never tell two same-recipe quantities apart, and that limitation comes back in §1.6.
It is also why numbers like ½ and 2π simply vanish. They are pure numbers with no ingredients in them at all.
These two terms are often muddled, and the difference is small but real.
The dimensional formula is the recipe on its own: [M L T⁻²].
The dimensional equation is the whole statement, with the quantity named on the left: [F] = [M L T⁻²].
Formula is the answer. Equation is the sentence containing the answer.
Look at density, [M L⁻³ T⁰]. Density is mass ÷ volume, and volume carries L³. Dividing means the exponent turns negative, so L³ becomes L⁻³.
That is the general habit worth building: multiply and the exponents add; divide and they subtract. Everything else in this chapter is bookkeeping on top of that.
The whole of mechanics can be built as one chain rather than twenty facts to memorise:
| Quantity | Built from | Recipe |
|---|---|---|
| velocity | length ÷ time | L T⁻¹ |
| acceleration | velocity ÷ time | L T⁻² |
| force | mass × acceleration | M L T⁻² |
| work | force × length | M L² T⁻² |
| power | work ÷ time | M L² T⁻³ |
| pressure | force ÷ area | M L⁻¹ T⁻² |
Four small moves — divide by time, multiply by mass, multiply by length, divide by time — cover most of the subject.
You can add a length to a length. You cannot add a length to a time — the result would not mean anything, so no correct equation ever asks you to.
The second half says: treat units like symbols in algebra. If metres appear on the top and on the bottom, cross them out. This is exactly what you already do with x's in a fraction.
"Homogeneity" just means "all the same kind". Every term in a correct equation must be the same kind of thing.
This gives you a free check on any formula. Reduce each term to its recipe. If one term comes out different from the others, the equation is wrong and you have caught the mistake without knowing anything about the physics.
It takes about fifteen seconds and it catches dropped squares, upside-down fractions and terms multiplied where they should have been divided. Worth doing on anything you rearranged yourself.
Go through it term by term. x is a length. x₀ is a length. v₀t is (L T⁻¹) × T, and the T's cancel, leaving L. And ½at² is (L T⁻²) × T², where T⁻² and T² cancel, again leaving L.
Four terms, all L. Consistent.
Notice what the ½ did: nothing at all. It is a pure number and this test cannot see it. That fact becomes important on the very next line.
This is the single most examined sentence in the section, so it is worth getting exactly right.
The test works in one direction only.
Fails the test → definitely wrong. No number you put in front can rescue it.
Passes the test → might be right, might not. Because pure numbers are invisible, T = 2π√(l/g) and T = 5√(l/g) both pass, and only one of them describes a real pendulum.
One sentence to carry into the exam: this method can prove an equation wrong, but it can never prove one right.
This looks like a footnote but it is one of the most useful rules in the chapter.
Whatever sits inside a sine, cosine, tangent, logarithm or exponential must be a pure number. And whatever comes out is a pure number too.
So in a wave written as y = A sin(ωt − kx), the whole bracket must be dimensionless. That forces ωt to be dimensionless, so ω must be T⁻¹. And kx must be dimensionless, so k must be L⁻¹. You have worked out the dimensions of both constants without being told anything about them.
Whenever a question hands you an unfamiliar formula with unknown constants in it, look first for an addition sign or a special function. That is where the free information is hiding.
Anything added to or subtracted from something else must have the same dimensions as it.
So in an expression like (1 − v²), the 1 is a pure number, which forces v² to be a pure number too — and that is how you know a c² has gone missing underneath. In (V − b), the b must be a volume. In (√x + B), the B must be L½.
Attack the additions before anything else.
This is the second, more ambitious use. Instead of checking a formula, you build one.
Two conditions are attached. You must already know which quantities matter, and the answer must be a product of them raised to powers — not a sum of separate terms.
"Up to three" is not arbitrary. In mechanics you only get three equations, one each from M, L and T, so you can only solve for three unknown powers.
Walk through it slowly, because this exact structure appears in many questions.
Step one. Write the guess: the period is some pure number k, times length to some power, times g to some power, times mass to some power. You do not know the powers yet.
Step two. Replace everything with recipes. The left side is just T¹. The right side works out to Lx+y T−2y Mz.
Step three. Match one ingredient at a time.
Put them back and you get T = k √(l/g).
The interesting result is z = 0. Mass dropped out. The method has just told you that a heavy bob and a light bob swing at the same rate — a real physical fact, found without any experiment.
And the honest limitation: k = 2π cannot be found this way. Pure numbers are invisible to the method, exactly as in §1.6.1.
You are trying to make T = k lx gy mz come out with the recipe of a time, which is M⁰ L⁰ T¹. Move the sliders until all three ingredients match.
A good demonstration of both what the method can and cannot do, in one example.
K = ma is a force, not an energy. Out. K = m²v³ has the wrong recipe entirely. Out. And K = ½mv² + ma adds an energy to a force, which is illegal no matter what the total comes to. Out.
But ½mv² and (3/16)mv² are identical as far as this test is concerned, because ½ and 3/16 are both invisible. The method has narrowed four wrong answers down to two candidates and then stopped.
Exam habit worth building: scan the options for an illegal addition first. You can eliminate those at a glance, before doing any work at all.
Against all that, it has one real strength: when it says an equation is wrong, it is right.
The chapter closes with eleven numbered points. Most restate what you have already read. These are the ones that carry new emphasis:
Seven is the number to remember: metre, kilogram, second, ampere, kelvin, mole, candela. Everything else in physics is built from these.
Carry the units through your working instead of dropping them and adding them back at the end. They cancel like ordinary symbols, and if the units at the end are not what you expected, you have found a mistake.
The chapter ends by repeating the asymmetry, which tells you how much weight it carries. If you remember one sentence from §1.6, make it this one.
Method and answer for each. Open one at a time and try it yourself first.
(a) 1 cm = 10⁻² m, so a cube of side 1 cm has volume (10⁻²)³ = 10⁻⁶ m³. Cube the conversion factor, do not just apply it once.
(b) A solid cylinder means both flat ends count: A = 2πr(r + h) = 2π(2.0)(12.0) = 48π = 150.8 cm². Then 1 cm² = 10² mm², so A ≈ 1.5 × 10⁴ mm².
(c) 18 km h⁻¹ × 5/18 = 5 m s⁻¹, so in one second it covers 5 m.
(d) Relative density is a ratio with no unit. 11.3 × 1 g cm⁻³ = 11.3 g cm⁻³ = 1.13 × 10⁴ kg m⁻³.
Use n₂ = n₁ (M₁/M₂)a (L₁/L₂)b (T₁/T₂)c, reading a, b, c off the dimensional formula.
(a) Energy: a = 1, b = 2. Factor = 10³ × 10⁴ = 10⁷, so 1 kg m² s⁻² = 10⁷ g cm² s⁻².
(b) 1 light year = 9.46 × 10¹⁵ m, so 1 m = 1.057 × 10⁻¹⁶ ly.
(c) 3.0 m s⁻² → km h⁻²: divide by 10³ for km, multiply by 3600² for h⁻². Result 3.888 × 10⁴ km h⁻².
(d) G has a = −1, b = 3, so the mass factor is 10⁻³ and the length factor 10⁶. Answer 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹. Watch that negative mass exponent.
1 cal = 4.2 kg m² s⁻². The new units are α kg, β m and γ s.
Energy has the recipe M¹ L² T⁻². Every ratio is old ÷ new, so M₁/M₂ = 1/α = α⁻¹, and similarly for the others. Applying the exponents gives α⁻¹, β⁻², and γ⁻¹ raised to −2, which is γ⁺².
Answer: 4.2 α⁻¹ β⁻² γ².
The quick way: take the dimensional formula M¹L²T⁻² and flip every sign, then write α, β, γ in place of M, L, T. That gives α⁻¹β⁻²γ² immediately.
The point is that big and small are meaningless on their own. Each statement needs a standard to compare against.
So instead of "atoms are very small", say atoms are much smaller than the tip of a pin. Instead of "a jet plane moves with great speed", say a jet plane moves faster than a car. Instead of "the mass of Jupiter is very large", say Jupiter's mass is very large compared with Earth's.
Parts (d), (e) and (f) already contain a comparison, so they are fine as they stand.
If the speed of light is numerically 1, then distance = speed × time becomes distance = time, in numbers.
t = 8 min 20 s = 8 × 60 + 20 = 500 s. So the distance is 500 new units.
Check it in SI: 3 × 10⁸ × 500 = 1.5 × 10¹¹ m, which is the Sun–Earth distance. Correct.
Careful: 8 min 20 s is 500 s, not 8.2 minutes.
Work out all three least counts in the same unit and pick the smallest.
Vernier, 20 divisions: 1 mm ÷ 20 = 0.05 mm.
Screw gauge, pitch 1 mm, 100 divisions: 1 ÷ 100 = 0.01 mm.
Optical instrument, one wavelength of light: about 6 × 10⁻⁴ mm.
The optical instrument wins by a wide margin.
Magnification = image size ÷ object size, so object = image ÷ magnification.
3.5 mm ÷ 100 = 0.035 mm, which is 35 µm. A sensible answer for a hair.
The microscope makes things look bigger, so the true value must come out smaller. If your answer is larger than 3.5 mm you divided the wrong way. The twenty observations are only there to justify taking an average — they do not enter the arithmetic.
(a) A single thread is thinner than the 1 mm least count of a metre scale. So wind n turns tightly around a pencil with no gaps, measure the total length L of the coil, and the diameter is L/n. The error is shared among n turns as well.
(b) No. More divisions does lower the least count arithmetically, but the circular scale has a fixed circumference, so the lines eventually become too close to tell apart. Random errors also exceed the least count, so the extra resolution buys nothing real.
(c) Random errors are equally likely either way, so they partly cancel when averaged. With 100 readings there is more cancelling than with 5, giving a more reliable mean. Note that this does not help with systematic errors such as a zero error.
Areal magnification = 1.55 m² ÷ 1.75 cm². Convert first: 1.55 m² = 1.55 × 10⁴ cm².
Ratio = 1.55 × 10⁴ ÷ 1.75 = 8857.
Linear magnification is the square root: √8857 ≈ 94.
Areas scale as the square of lengths, so you must take the square root at the end.
(a) 0.007 m² → 1 (leading zeros never count)
(b) 2.64 × 10²⁴ kg → 3 (only the front number counts)
(c) 0.2370 g cm⁻³ → 4 (trailing zero after a decimal counts)
(d) 6.320 J → 4
(e) 6.032 N m⁻² → 4 (trapped zero counts)
(f) 0.0006032 m² → 4 (four leading zeros discarded, trapped zero kept)
First convert the thickness: 2.01 cm = 0.0201 m. That has 3 significant figures, fewer than the other two, so it decides the answer.
Surface area = 2(lb + bt + tl) = 2(4.25517 + 0.0202005 + 0.0851034) = 8.7209 m² → 8.72 m².
Volume = 4.234 × 1.005 × 0.0201 = 0.08552892 m³ → 0.0855 m³.
In 0.0855 the significant digits are 8, 5 and 5. The zeros in front are placeholders.
(a) Convert to kilograms: 2.30 + 0.02015 + 0.02017 = 2.34032 kg. This is an addition, so count decimal places. The box mass is known to 2, so the answer is 2.34 kg.
(b) 20.17 − 20.15 = 0.02 g. Both inputs are known to 2 decimal places, so the answer keeps 2 decimal places — which here leaves only one significant figure.
Two four-figure masses have produced a one-figure difference. Subtraction is the only operation that can do this.
The relation is written m = m₀ / (1 − v²)1/2, with c missing.
The 1 in the bracket is a pure number, so whatever is subtracted from it must also be a pure number. But v² has the recipe L² T⁻². To cancel that you must divide by something with the same recipe, and c² = (L T⁻¹)² = L² T⁻² fits exactly.
Answer: m = m₀ / (1 − v²/c²)1/2.
Also check the top: m and m₀ are both masses, so no stray c is allowed there.
A hydrogen atom is about 0.5 Å across in radius, so r = 0.5 × 10⁻¹⁰ m.
Volume of one atom = (4/3)πr³ ≈ 5.24 × 10⁻³¹ m³.
For a mole, multiply by 6.022 × 10²³: ≈ 3.2 × 10⁻⁷ m³.
Molar volume at STP = 22.4 L = 22.4 × 10⁻³ m³. Atomic volume of a mole, from 1.14, ≈ 3.2 × 10⁻⁷ m³.
Ratio = 22.4 × 10⁻³ ÷ 3.2 × 10⁻⁷ ≈ 7 × 10⁴.
The ratio is huge because in a gas the molecules themselves take up only about one part in 10⁵ of the container. Almost all of a gas is empty space — which is exactly why gases can be squeezed and liquids cannot.
What you notice is the angle a thing moves through, not the distance.
A tree 10 m from the train sweeps through a large angle in a second, so it appears to rush past. A hill 10 km away moves through a tiny angle in that same second, so it appears almost still.
The Moon is so far off that its angle barely changes at all, which makes it look as though it is travelling along with you.
Treat the Sun as a sphere. R = 7.0 × 10⁸ m, so R³ = 3.43 × 10²⁶ m³.
V = (4/3)πR³ ≈ 1.44 × 10²⁷ m³.
ρ = 2.0 × 10³⁰ ÷ 1.44 × 10²⁷ ≈ 1.4 × 10³ kg m⁻³.
That is slightly denser than water, and firmly in the range of solids and liquids rather than gases — even though the Sun is a plasma. The reason is gravity: the Sun's own weight squeezes it to that density despite the enormous temperature.
Cubing 7.0 × 10⁸ gives 3.43 × 10²⁶. Multiply the exponent by three; that is where this one usually goes wrong.