Blog Log in Get Started Free

Measurement MCQs for NEET — Physics Questions with Answers

Practice free Measurement (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

Dimensions of potential energy are

MLT^–1

$M L^{2} T^{- 2}$

$M L^{- 1} T^{- 2}$

$M L^{- 1} T^{- 1}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$M L^{2} T^{- 2}$

● ● ● ● ●

The dimensions of shear modulus are

MLT^–1

$M L^{2} T^{- 2}$

$M L^{- 1} T^{- 2}$

$M L T^{- 2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Shear modulus = $\frac{Shearing stress}{Shearing strain}$ $= \frac{F}{A θ} = [M L^{- 1} T^{- 2}]$

● ● ● ● ●

Pressure gradient has the same dimension as that of

Velocity gradient

Potential gradient

Energy gradient

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Velocity gradient $= \frac{v}{x} = \frac{[L T^{- 1}]}{[L]} = [T^{- 1}]$

Potential gradient $= \frac{V}{x} = \frac{[M L^{2} T^{- 3} A^{- 1}]}{[L]}$ $= [M L T^{- 3} A^{- 1}]$

Energy gradient $= \frac{E}{x} = \frac{[M L^{2} T^{2}]}{[L]} = [M L T^{- 2}]$

and pressure gradient $= \frac{P}{x} = \frac{[M L^{- 1} T^{- 2}]}{[L]} = [M L^{- 2} T^{- 2}]$

● ● ● ● ●

If force (F), length (L) and time (T) are assumed to be fundamental units, then the dimensional formula of the mass will be

$F L^{- 1} T^{2}$

$F L^{- 1} T^{- 2}$

$F L^{- 1} T^{- 1}$

$F L^{2} T^{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Let $m = K F^{a} L^{b} T^{c}$

Substituting the dimension of

$[F] = [M L T^{- 2}],$ $[C] = [L] and [T] = [T]$

and comparing both sides, we get $m = F L^{- 1} T^{- 2}$

● ● ● ● ●

The dimensions of universal gas constant is

$[M L^{2} T^{- 2} θ^{- 1}]$

$[M^{2} L T^{- 2} θ]$

$[M L^{3} T^{- 1} θ^{- 1}]$

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$∵ R = \frac{P V}{T}$ $= [\frac{M L^{- 1} T^{- 2} \times L^{3}}{θ}] = [M L^{2} T^{- 2} θ^{- 1}]$

● ● ● ● ●

In the relation $y = a \cos (ω t - k x)$ , the dimensional formula for k is

$[M^{0} L^{- 1} T^{- 1}]$

$[M^{0} L T^{- 1}]$

$[M^{0} L^{- 1} T^{0}]$

$[M^{0} L T]$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$[K x]$ = Dimension of $ω t =$ (dimensionless) hence $K = \frac{1}{X} = \frac{1}{L} = [L^{- 1}]$

$∴ [K] = [L^{- 1}]$

● ● ● ● ●

Position of a body with acceleration 'a' is given by $x = K a^{m} t^{n},$ here t is time. Find dimension of m and n.

m = 1, n = 1

m = 1, n = 2

m = 2, n = 1

m = 2, n = 2

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

As $x = K a^{m} \times t^{n}$

$[M^{0} L T^{0}] = {[L T^{- 2}]}^{m} {[T]}^{n} = [L^{m} T^{- 2 m + n}]$

$∴$ m = 1 and $- 2 m + n = 0$ $\Rightarrow n = 2$ .

● ● ● ● ●

"Pascal-Second" has dimension of

Force

Energy

Pressure

Coefficient of viscosity

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$N S m^{- 2} = N m^{- 2} \times S =$ Pascal-second.

● ● ● ● ●

In a system of units if force (F), acceleration (A) and time (T) are taken as fundamental units then the dimensional formula of energy is

FA²T

FAT²

F²AT

FAT

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$E = K F^{a} A^{b} T^{c}$

$[M L^{2} T^{- 2}] = {[M L T^{- 2}]}^{a} {[L T^{- 2}]}^{b} {[T]}^{c}$

$[M L^{2} T^{- 2}] = [M^{a} L^{a + b} T^{- 2 a - 2 b + c}]$

$∴ a = 1, a + b = 2 \Rightarrow b = 1$

and $- 2 a - 2 b + c = - 2 \Rightarrow c = 2$

$∴ E = K F A T^{2}$ .

● ● ● ● ●

The ratio of the dimension of Planck's constant and that of moment of inertia is the dimension of

Frequency

Velocity

Angular momentum

Time

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\frac{h}{I} = [\frac{M L^{2} T^{- 1}}{M L^{2}}] = [T^{- 1}]$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Measurement question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.