Dimensional Analysis & Applications — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Dimensional Analysis & Applications MCQs with step-by-step solutions (21 questions). Part of Units and Measurements (Std 11). Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Dimensional Analysis & Applications · easy · theory
Dimensional analysis can be used to:
A. Find the value of dimensionless constants
B. Determine whether a quantity is a vector
C. Prove that an equation is physically correct
D. Check the correctness of an equation, derive relations and convert units ✓ Correct
Solution: These three uses are its standard applications; the other options are its limitations.
Q2 — Dimensional Analysis & Applications · easy · theory
The principle of homogeneity of dimensions states that:
A. Every term on both sides of a correct equation has the same dimensions ✓ Correct
B. Only fundamental quantities have dimensions
C. Dimensions can be added like numbers
D. All physical quantities have the same dimensions
Solution: Quantities of unlike dimensions can never be added or equated.
Q3 — Dimensional Analysis & Applications · medium · theory
A limitation of dimensional analysis is that it cannot:
A. Convert a quantity from one system of units to another
B. Determine dimensionless constants such as $2\pi$ ✓ Correct
C. Check the dimensional correctness of an equation
D. Give the dimensions of a derived quantity
Solution: It gives the form of a relation but leaves any pure numerical factor undetermined.
Q4 — Dimensional Analysis & Applications · medium · theory
Dimensional analysis fails to derive a relation when the quantity depends on:
A. Only one other quantity
B. More than three other quantities ✓ Correct
C. Exactly two other quantities
D. No other quantity
Solution: With only three base dimensions in mechanics, at most three unknown exponents can be found.
Q5 — Dimensional Analysis & Applications · hard · theory
Dimensional analysis cannot be applied to equations involving:
A. Ratios of physical quantities
B. Powers of physical quantities
C. Trigonometric, logarithmic or exponential functions ✓ Correct
D. Products of physical quantities
Solution: The argument of such a function must itself be dimensionless, which restricts the method.
Q6 — Dimensional Analysis & Applications · medium · theory
An equation that is dimensionally correct:
A. Must contain no constants
B. Must always be physically correct
C. May still be physically wrong ✓ Correct
D. Must be an equation of motion
Solution: For example $s = ut + at^2$ is dimensionally sound but physically wrong — the factor $\dfrac{1}{2}$ is missing.
Q7 — Dimensional Analysis & Applications · easy · theory
In a dimensionally correct equation, quantities that are added together must have:
A. No dimensions at all
B. Different dimensions
C. The same numerical value
D. The same dimensions ✓ Correct
Solution: This follows directly from the principle of homogeneity.
Q8 — Dimensional Analysis & Applications · medium · theory
Dimensional analysis cannot distinguish between:
A. Energy and power
B. Length and time
C. Mass and force
D. A scalar and a vector of the same dimensions ✓ Correct
Solution: Work and torque, for instance, share a dimensional formula despite one being a scalar and the other a vector.
Q9 — Dimensional Analysis & Applications · medium · numerical
The formula $T = 2\pi\sqrt{\dfrac{l}{g}}$ for the period of a simple pendulum is:
A. Correct only in SI units
B. Correct only for small lengths
C. Dimensionally correct ✓ Correct
D. Dimensionally incorrect
Solution: $\sqrt{\dfrac{[L]}{[LT^{-2}]}} = \sqrt{[T^2]} = [T]$, matching the left-hand side.
Q10 — Dimensional Analysis & Applications · easy · numerical
The equation $v = u + at$ is:
A. Dimensionally incorrect
B. Dimensionally correct ✓ Correct
C. Valid only in CGS units
D. Valid only for zero acceleration
Solution: Each term has the dimension $[L^1T^{-1}]$: $[LT^{-2}][T] = [LT^{-1}]$ for the last term.
Q11 — Dimensional Analysis & Applications · medium · numerical
One newton expressed in the CGS system is:
A. $10^3\text{ dyne}$
B. $10^7\text{ dyne}$
C. $10^{-5}\text{ dyne}$
D. $10^5\text{ dyne}$ ✓ Correct
Solution: $1\text{ N} = (10^3\text{ g})(10^2\text{ cm})(\text{s})^{-2} = 10^5\text{ dyne}$.
Q12 — Dimensional Analysis & Applications · medium · numerical
One joule expressed in erg is:
A. $10^3\text{ erg}$
B. $10^{-7}\text{ erg}$
C. $10^5\text{ erg}$
D. $10^7\text{ erg}$ ✓ Correct
Solution: $1\text{ J} = 10^3\text{ g} \times (10^2\text{ cm})^2 \times \text{s}^{-2} = 10^7\text{ erg}$.
Q13 — Dimensional Analysis & Applications · hard · numerical
To convert a quantity of dimensions $[M^aL^bT^c]$ between two systems, the numerical values are related by:
A. $n_2 = \dfrac{n_1}{M_1L_1T_1}$
B. $n_2 = n_1\left(\dfrac{M_2}{M_1}\right)^a\left(\dfrac{L_2}{L_1}\right)^b\left(\dfrac{T_2}{T_1}\right)^c$
C. $n_2 = n_1\left(\dfrac{M_1}{M_2}\right)^a\left(\dfrac{L_1}{L_2}\right)^b\left(\dfrac{T_1}{T_2}\right)^c$ ✓ Correct
D. $n_2 = n_1(M_1L_1T_1)$
Solution: The product of numerical value and unit size is invariant, which gives this conversion rule.
Q14 — Dimensional Analysis & Applications · medium · numerical
The equation $s = ut + \dfrac{1}{2}at^2$ is:
A. Correct only when $u = 0$
B. Dimensionally incorrect
C. Dimensionally correct, each term having the dimension of length ✓ Correct
D. Correct only when $a = 0$
Solution: $[LT^{-1}][T] = [L]$ and $[LT^{-2}][T^2] = [L]$, matching the left side.
Q15 — Dimensional Analysis & Applications · hard · numerical
Which of the following equations is dimensionally incorrect?
A. $s = ut + \dfrac{1}{2}at^2$
B. $v = u + at^2$ ✓ Correct
C. $F = ma$
D. $v^2 = u^2 + 2as$
Solution: $[LT^{-2}][T^2] = [L]$, which cannot be added to a velocity $[LT^{-1}]$.
Q16 — Dimensional Analysis & Applications · hard · numerical
Assuming the period of a simple pendulum depends on its length $l$ and on $g$, dimensional analysis gives:
A. $T \propto lg$
B. $T \propto \dfrac{l}{g}$
C. $T \propto \sqrt{\dfrac{g}{l}}$
D. $T \propto \sqrt{\dfrac{l}{g}}$ ✓ Correct
Solution: Writing $T = kl^ag^b$ and matching dimensions gives $a = \dfrac{1}{2}$ and $b = -\dfrac{1}{2}$; the constant $2\pi$ cannot be found this way.
Q17 — Dimensional Analysis & Applications · hard · numerical
One pascal expressed in $\text{dyne}/\text{cm}^2$ is:
A. $10^5$
B. $10$ ✓ Correct
C. $10^4$
D. $10^{-1}$
Solution: $1\text{ Pa} = \dfrac{10^5\text{ dyne}}{10^4\text{ cm}^2} = 10\,\text{dyne}/\text{cm}^2$.
Q18 — Dimensional Analysis & Applications · medium · numerical
The relation $E = mc^2$ is:
A. Correct only in CGS units
B. Dimensionally incorrect
C. Dimensionally correct, both sides having $[M^1L^2T^{-2}]$ ✓ Correct
D. Correct only for photons
Solution: $[M][LT^{-1}]^2 = [M^1L^2T^{-2}]$, which is the dimension of energy.
Q19 — Dimensional Analysis & Applications · medium · numerical
One watt expressed in $\text{erg}/\text{s}$ is:
A. $10^7$ ✓ Correct
B. $10^3$
C. $10^{-7}$
D. $10^5$
Solution: Since $1\text{ J} = 10^7\text{ erg}$, one joule per second is $10^7$ erg per second.
Q20 — Dimensional Analysis & Applications · medium · numerical
The numerical factor $2\pi$ in the pendulum formula:
A. Depends on the system of units used
B. Cannot be obtained by dimensional analysis ✓ Correct
C. Has the dimension of time
D. Follows directly from dimensional analysis
Solution: Dimensional methods give the form of a relation but never its dimensionless constants.
Q21 — Dimensional Analysis & Applications · easy · numerical
The dimensional formula of the quantity $\dfrac{1}{2}mv^2$ is the same as that of:
A. Momentum
B. Force
C. Work ✓ Correct
D. Power
Solution: Kinetic energy and work share the dimension $[M^1L^2T^{-2}]$.