Applications of Dimensions — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Applications of Dimensions MCQs with step-by-step solutions (36 questions). Part of Units and Dimensions. Practise online on Prepizo — no login needed.
▶ Practise Applications of Dimensions online (free)
Questions with solutions
Q1 — Applications of Dimensions · medium · theory
The principle of homogeneity of dimensions states that
A. Dimensions cancel out
B. Both sides must have the same units only
C. Each term in a valid physical equation has the same dimensions ✓ Correct
D. All terms must be positive
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. Find dimensionless constants
B. Prove an equation is numerically exact
C. Check the dimensional correctness of an equation ✓ Correct
D. Derive equations with trigonometric functions
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. Acceleration
B. Distance
C. Velocity $[LT^{-1}]$ ✓ Correct
D. Time
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. Correct only for u = 0
C. Dimensionless
D. Incorrect
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. Always gives wrong results
D. Cannot handle length
Solution: Dimensionless numerical factors are invisible to dimensional analysis.
Q6 — Applications of Dimensions · medium · theory
Dimensional analysis cannot derive relations that involve
A. Powers of quantities
B. Ratios
C. Products of quantities
D. Trigonometric, exponential or logarithmic functions ✓ Correct
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. $\sqrt{l/g}$ ✓ Correct
B. $\sqrt{lg}$
C. $g/l$
D. $l/g$
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. Always exactly correct
B. Not necessarily numerically correct ✓ Correct
C. Dimensionless
D. Always wrong
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. Dimensionless
B. Possibly correct
C. Definitely wrong ✓ Correct
D. Numerically 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. A time
C. A length
D. Dimensionless ✓ Correct
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. Dimensionless ✓ Correct
B. A length
C. A time
D. A frequency
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. One system of units to another ✓ Correct
C. One dimension to another
D. Energy to force
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. $10^7$ dyne
C. $10^3$ dyne
D. $10^5$ dyne ✓ Correct
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. One
B. Two
C. Zero (they are invisible to it) ✓ Correct
D. All of them
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. Dimensionless
B. Wrong (dimensions do not match force) ✓ Correct
C. Correct
D. Correct only for large v
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. Correct only if v = 0
C. Correct (both are energy) ✓ Correct
D. Dimensionless
Solution: Both sides have dimensions $[ML^2T^{-2}]$.
Q18 — Applications of Dimensions · easy · theory
Dimensional analysis works best when a quantity depends on
A. A sum of many terms
B. Only a few other quantities (a product of powers) ✓ Correct
C. Trigonometric functions
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. $a=1, b=-2, c=1$
B. $a=1, b=1, c=1$
C. $a=2, b=1, c=-1$
D. $a=1, b=2, c=-1$ ✓ Correct
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. Convert units
B. Find dimensions of a quantity
C. Detect wrong dimensions
D. Guarantee it is completely correct ✓ Correct
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. $n_1u_1 = 0$
B. $n_1 = n_2$
C. $u_1 = u_2$
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. Dimensionless ✓ Correct
B. An energy
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 dependence of a quantity on others (as powers) ✓ Correct
B. The value of π
C. The value of ½
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. Length
B. Time ✓ Correct
C. Frequency
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. $10$ erg
B. $10^6$ erg
C. $10^7$ erg
D. $10^8$ erg ✓ Correct
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. B and C are dimensionless
B. A is the largest
C. A, B and C all have the same dimensions ✓ Correct
D. A is dimensionless
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. Length (the Planck length) ✓ Correct
B. Charge
C. Current
D. Temperature
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. Units cannot be checked
B. It has no variables
C. Dimensions are always wrong
D. It can miss numerical constants or extra dimensionless terms ✓ Correct
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. Direction
B. Numerical value only
C. Dimensional formula ✓ Correct
D. Sign
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. Works for all equations
B. Replaces experiments
C. Gives a quick check and can suggest the form of a relation ✓ Correct
D. Gives exact numerical answers
Solution: It is a fast consistency check and can indicate power dependencies, though not exact constants.