Applications of Dimensions — NEET Physics MCQs with Solutions
Free NEET Physics Applications of Dimensions MCQs with step-by-step solutions (36 questions). Part of Units and Dimensions. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Applications of Dimensions · medium · theory
The principle of homogeneity of dimensions states that
A. All terms must be positive
B. Each term in a valid physical equation has the same dimensions ✓ Correct
C. Both sides must have the same units only
D. Dimensions cancel out
Solution: Only quantities with the same dimensions can be equated or added, which is the basis of dimensional analysis.
Q2 — Applications of Dimensions · easy · theory
Dimensional analysis can be used to
A. Check the dimensional correctness of an equation ✓ Correct
B. Find dimensionless constants
C. Derive equations with trigonometric functions
D. Prove an equation is numerically exact
Solution: A key use is verifying that an equation is dimensionally consistent.
Q3 — Applications of Dimensions · easy · theory
In the equation $v = u + at$, checking dimensions shows each term has the dimensions of
A. Time
B. Velocity $[LT^{-1}]$ ✓ Correct
C. Acceleration
D. Distance
Solution: u is a velocity, and $at = [LT^{-2}][T] = [LT^{-1}]$, so every term is a velocity.
Q4 — Applications of Dimensions · easy · theory
The equation $s = ut + \frac{1}{2}at^2$ is dimensionally
A. Correct (each term has dimensions of length) ✓ Correct
B. Dimensionless
C. Incorrect
D. Correct only for u = 0
Solution: $ut = [LT^{-1}][T] = [L]$ and $at^2 = [LT^{-2}][T^2] = [L]$, matching s.
Q5 — Applications of Dimensions · easy · theory
A major LIMITATION of dimensional analysis is that it
A. Cannot find dimensionless constants like ½ or 2π ✓ Correct
B. Cannot check units
C. Cannot handle length
D. Always gives wrong results
Solution: Dimensionless numerical factors are invisible to dimensional analysis.
Q6 — Applications of Dimensions · medium · theory
Dimensional analysis cannot derive relations that involve
A. Products of quantities
B. Ratios
C. Trigonometric, exponential or logarithmic functions ✓ Correct
D. Powers of quantities
Solution: Such functions require dimensionless arguments and cannot be built by dimensional analysis alone.
Q7 — Applications of Dimensions · medium · theory
The time period $T$ of a simple pendulum depends on length $l$ and $g$. By dimensional analysis, $T \propto$
A. $\sqrt{lg}$
B. $\sqrt{l/g}$ ✓ Correct
C. $l/g$
D. $g/l$
Solution: Matching dimensions gives $T = k\sqrt{l/g}$ (the constant $k = 2\pi$ cannot be found dimensionally).
Q8 — Applications of Dimensions · medium · theory
If a physical equation is dimensionally correct, then it is
A. Not necessarily numerically correct ✓ Correct
B. Always wrong
C. Dimensionless
D. Always exactly correct
Solution: Dimensional correctness is necessary but not sufficient — numerical constants may still be wrong.
Q9 — Applications of Dimensions · easy · theory
If an equation is dimensionally INCORRECT, then it is
A. Numerically correct
B. Possibly correct
C. Dimensionless
D. Definitely wrong ✓ Correct
Solution: A dimensionally inconsistent equation cannot be physically valid.
Q10 — Applications of Dimensions · easy · theory
The argument of a trigonometric function (like $\sin\theta$) must be
A. A velocity
B. Dimensionless ✓ Correct
C. A time
D. A length
Solution: Only dimensionless quantities can appear inside sin, cos, log or exponential functions.
Q11 — Applications of Dimensions · medium · theory
In $y = A\sin(\omega t)$, the product $\omega t$ must be
A. A time
B. A frequency
C. A length
D. Dimensionless ✓ Correct
Solution: It is the argument of a sine function, so $\omega t$ is dimensionless (hence $\omega$ has dimensions $[T^{-1}]$).
Q12 — Applications of Dimensions · easy · theory
Dimensional analysis is used to convert a physical quantity from
A. Scalar to vector
B. Energy to force
C. One system of units to another ✓ Correct
D. One dimension to another
Solution: The relation $n_1u_1 = n_2u_2$ lets us change unit systems, e.g. SI to CGS.
Q13 — Applications of Dimensions · easy · theory
To convert 1 newton into dynes, we use $1\,\text{N} = ?$
A. $10^{-5}$ dyne
B. $10^5$ dyne ✓ Correct
C. $10^3$ dyne
D. $10^7$ dyne
Solution: 1 N = 1 kg·m/s² = (10³ g)(10² cm)/s² = 10⁵ dyne.
Q14 — Applications of Dimensions · easy · theory
To convert 1 joule into ergs, $1\,\text{J} = ?$
A. $10^5$ erg
B. $10^{-7}$ erg
C. $10^7$ erg ✓ Correct
D. $10^3$ erg
Solution: 1 J = 1 kg·m²/s² = (10³ g)(10⁴ cm²)/s² = 10⁷ erg.
Q15 — Applications of Dimensions · easy · theory
The number of dimensionless quantities that dimensional analysis can determine is
A. All of them
B. One
C. Zero (they are invisible to it) ✓ Correct
D. Two
Solution: Pure numbers have no dimensions, so dimensional analysis cannot pin them down.
Q16 — Applications of Dimensions · easy · theory
A student writes $F = mv^2$. Dimensionally this is
A. Correct only for large v
B. Dimensionless
C. Correct
D. Wrong (dimensions do not match force) ✓ Correct
Solution: $mv^2 = [ML^2T^{-2}]$ (energy), not $[MLT^{-2}]$ of force — so it is dimensionally wrong.
Q17 — Applications of Dimensions · easy · theory
The equation $\frac{1}{2}mv^2 = mgh$ is dimensionally
A. Incorrect
B. Dimensionless
C. Correct only if v = 0
D. Correct (both are energy) ✓ Correct
Solution: Both sides have dimensions $[ML^2T^{-2}]$.
Q18 — Applications of Dimensions · easy · theory
Dimensional analysis works best when a quantity depends on
A. Trigonometric functions
B. A sum of many terms
C. Only a few other quantities (a product of powers) ✓ Correct
D. Exponentials
Solution: It is powerful for relations of the form $Q = k\,a^x b^y c^z$.
Q19 — Applications of Dimensions · medium · theory
If force depends on mass $m$, velocity $v$ and radius $r$ as $F = k m^a v^b r^c$, dimensional analysis gives (for centripetal force)
A. $a=2, b=1, c=-1$
B. $a=1, b=2, c=-1$ ✓ Correct
C. $a=1, b=1, c=1$
D. $a=1, b=-2, c=1$
Solution: Matching $[MLT^{-2}]$ with $[M]^a[LT^{-1}]^b[L]^c$ gives $a=1, b=2, c=-1$, i.e. $F = kmv^2/r$.
Q20 — Applications of Dimensions · easy · theory
The dimensional method can check an equation but cannot
A. Detect wrong dimensions
B. Guarantee it is completely correct ✓ Correct
C. Convert units
D. Find dimensions of a quantity
Solution: Right dimensions do not guarantee right numerical factors or added dimensionless terms.
Q21 — Applications of Dimensions · medium · theory
The value of a physical quantity remains the same when units change, so
A. $u_1 = u_2$
B. $n_1 = n_2$
C. $n_1u_1 = 0$
D. $n_1u_1 = n_2u_2$ ✓ Correct
Solution: A smaller unit needs a larger numerical value, keeping the product (the physical quantity) constant.
Q22 — Applications of Dimensions · easy · theory
In the expression $e^{-kt}$, the quantity $kt$ must be
A. An energy
B. Dimensionless ✓ Correct
C. A time
D. A length
Solution: Exponents must be dimensionless, so $k$ has dimensions $[T^{-1}]$.
Q23 — Applications of Dimensions · easy · theory
Which of the following CAN be found by dimensional analysis?
A. The value of π
B. The value of ½
C. The dependence of a quantity on others (as powers) ✓ Correct
D. A logarithmic term
Solution: It reveals the power-law dependence but not pure numbers.
Q24 — Applications of Dimensions · medium · theory
The equation $T = 2\pi\sqrt{m/k}$ (spring period). Dimensionally, $\sqrt{m/k}$ has dimensions of
A. Frequency
B. Time ✓ Correct
C. Length
D. Mass
Solution: $m/k = [M]/[MT^{-2}] = [T^2]$, so its square root is a time.
Q25 — Applications of Dimensions · easy · theory
The kinetic energy of a body is $10$ J in SI units. In CGS units (erg) it equals
A. $10^6$ erg
B. $10$ erg
C. $10^8$ erg ✓ Correct
D. $10^7$ erg
Solution: Since 1 J = 10⁷ erg, 10 J = 10 × 10⁷ = 10⁸ erg.
Q26 — Applications of Dimensions · easy · theory
Dimensional homogeneity implies that in $A = B + C$
A. A is the largest
B. A is dimensionless
C. B and C are dimensionless
D. A, B and C all have the same dimensions ✓ Correct
Solution: Terms being added and their sum must share identical dimensions.
Q27 — Applications of Dimensions · medium · theory
The gravitational constant $G$, speed of light $c$ and Planck constant $h$ can be combined to give a quantity with dimensions of
A. Charge
B. Length (the Planck length) ✓ Correct
C. Temperature
D. Current
Solution: A famous application: $\sqrt{Gh/c^3}$ has dimensions of length (the Planck length).
Q28 — Applications of Dimensions · easy · theory
A dimensionally correct equation may still be wrong because
A. It can miss numerical constants or extra dimensionless terms ✓ Correct
B. Units cannot be checked
C. It has no variables
D. Dimensions are always wrong
Solution: Dimensional analysis is blind to pure numbers and dimensionless additive terms.
Q29 — Applications of Dimensions · easy · theory
To use $n_1u_1 = n_2u_2$ for converting units, we need the quantity's
A. Sign
B. Dimensional formula ✓ Correct
C. Numerical value only
D. Direction
Solution: The dimensional formula tells how the unit scales when base units change.
Q30 — Applications of Dimensions · medium · theory
The main advantage of dimensional analysis is that it
A. Replaces experiments
B. Gives exact numerical answers
C. Gives a quick check and can suggest the form of a relation ✓ Correct
D. Works for all equations
Solution: It is a fast consistency check and can indicate power dependencies, though not exact constants.