NEET Medical Units and Measurements PYQs previous year questions and solutions help students build a strong foundation in basic measurement concepts essential for physics. This section covers important topics such as physical quantities, SI units, significant figures, dimensional analysis, and errors in measurements through exam-oriented questions. By practicing these PYQs with detailed solutions, students can improve accuracy, understand question patterns, and strengthen problem-solving skills required for NEET.
Access NEET Units and Measurements PYQs Previous Year Questions and Solutions
TOPIC 1 : Units
The angle of 1′ (minute of arc) in radian is nearly equal to
(a) $2.91 \times 10^{-4}$ rad
(b) $4.85 \times 10^{-4}$ rad
(c) $4.80 \times 10^{-6}$ rad
(d) $1.75 \times 10^{-2}$ rad
[NEET (Oct.) 2020]
Ans. (a)
$1 \text{ minute} = \dfrac{1}{60} \text{ degree} = \dfrac{1}{60} \times \dfrac{\pi}{180} \text{ rad}$
$= 2.91 \times 10^{-4} \text{ rad}$
The unit of thermal conductivity is :
(a) $J \text{ m}^{-1} K^{-1}$
(b) $W m K^{-1}$
(c) $W m^{-1} K^{-1}$
(d) $J m K^{-1}$
[NEET (National) 2019]
Ans. (c)
The rate of heat flow through a conductor of length $L$ and area of cross-section $A$ is given by
$\dfrac{dQ}{dt} = KA \dfrac{\Delta T}{L}$ J/s or watt
where, $K = \text{coefficient of thermal conductivity and}$
$\Delta T = \text{change in temperature}$
$K = \dfrac{L}{A \Delta T} \dfrac{dQ}{dt}$
$\text{Unit of K } = \dfrac{\text{metre}}{\text{(metre)}^2 \times \text{kelvin}} \times \text{watt}$
$= W m^{-1} K^{-1}$
The unit of permittivity of free space, $\epsilon_0$ is
(a) coulomb/newton-metre
(b) newton-metre$^2$ / coulomb$^2$
(c) coulomb$^2$/newton-metre$^2$
(d) coulomb/(newton-metre)
[CBSE AIPMT 2004]
Ans. (c)
According to Coulomb’s law, the electrostatic force
$F = \dfrac{1}{4\pi\epsilon_0} \times \dfrac{q_1 q_2}{r^2}$
$q_1 \text{ and } q_2 = \text{charges, } r = \text{distance between charges}$
and $\epsilon_0 = \text{permittivity of free space}$
$\Rightarrow \epsilon_0 = \dfrac{1}{4\pi} \times \dfrac{q_1 q_2}{r^2 F}$
Substituting the units for $q, r \text{ and } F$, we obtain unit of $\epsilon_0$
$= \dfrac{\text{coulomb} \times \text{coulomb}}{\text{newton-(metre)}^2} = \dfrac{\text{(coulomb)}^2}{\text{newton-(metre)}^2}$
The value of Planck’s constant in SI unit is
(a) $6.63 \times 10^{-34} \text{ J-s}$
(b) $6.63 \times 10^{-34} \text{ kg-m/s}$
(c) $6.63 \times 10^{-34} \text{ kg-m}^2$
(d) $6.63 \times 10^{-34} \text{ J-s}$
[CBSE AIPMT 2002]
Ans. (d)
The value of Planck’s constant is $6.63 \times 10^{-34}$ and J-s is unit of the Planck’s constant.
In a particular system, the unit of length, mass and time are chosen to be 10 cm, 10 g and 0.1 s respectively. The unit of force in this system will be equivalent to
(a) 0.1 N
(b) 1 N
(c) 10 N
(d) 100 N
[CBSE AIPMT 1994]
Ans. (a)
Force $F = [MLT^{-2}]$
$= (10 \text{ g})(10 \text{ cm})(0.1 \text{ s})^{-2}$
Changing these units into MKS system
$F = (10^{-2} \text{ kg})(10^{-2} \text{ m})(10^{-1} \text{ s})^{-2}$
$= 10^{-1} \text{ N} = 0.1 \text{ N}$
TOPIC 2 : Errors in Measurement and Significant Figure
A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading: 0 mm
Circular scale reading: 52 divisions
Given that, 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is
(a) 0.52 cm
(b) 0.026 cm
(c) 0.26 cm
(d) 0.052 cm
[NEET 2021]
Ans. (d)
Given, the main scale reading, $MSR = 0$
The circular scale reading, $CSR = 52 \text{ divisions}$
Now, we shall determine the least count of the screw gauge.
$LC = \dfrac{p}{n}$
Here, $p$ is the pitch of the screw, $n$ is the number of circular divisions in one complete revolution.
$LC = \dfrac{1}{100} \text{ mm}$
$\Rightarrow LC = 0.01 \text{ mm} \Rightarrow LC = 0.001 \text{ cm}$
Thus, the least count of the screw gauge is 0.001 cm.
Therefore, diameter of the wire of screw gauge,
$D = MSR + (CSR \times LC)$
$D = 0 + (52 \times 0.001) = 0.052 \text{ cm}$
Time intervals measured by a clock give the following readings: 1.25 s, 1.24 s, 1.27 s, 1.21 s and 1.28 s. What is the percentage relative error of the observations?
(a) 2%
(b) 4%
(c) 16%
(d) 1.6%
[NEET (Oct.) 2020]
Ans. (d)
Mean time interval $\bar{T} = \dfrac{1.25+1.24+1.27+1.21+1.28}{5} = \dfrac{6.25}{5} = 1.25 \text{ s}$
Mean absolute error, $\Delta \bar{T} = \dfrac{|\Delta T_1| + |\Delta T_2| + |\Delta T_3| + |\Delta T_4| + |\Delta T_5|}{5}$
$\Rightarrow = \dfrac{|1.25-1.25| + |1.25-1.24| + |1.25-1.27| + |1.25-1.21| + |1.25-1.28|}{5}$
$\Rightarrow = \dfrac{0 + 0.01 + 0.02 + 0.04 + 0.03}{5} = \dfrac{0.1}{5} = 0.02 \text{ s}$
Percentage relative error $= \dfrac{\Delta \bar{T}}{\bar{T}} \times 100 = \dfrac{0.02}{1.25} \times 100 = 1.6\%$
A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is
(a) 0.25 mm
(b) 0.5 mm
(c) 1.0 mm
(d) 0.01 mm
[NEET (Sep.) 2020]
Ans. (b)
Given, least count = 0.01 mm
Number of divisions on circular scale = 50
Pitch of the screw gauge = least count $\times$ number of divisions on circular scale
$= 0.01 \times 50 = 0.5 \text{ mm}$
Taking into account of the significant figures, what is the value of 9.99 m – 0.0099 m?
(a) 9.98 m
(b) 9.980 m
(c) 9.9 m
(d) 9.9801 m
[NEET (Sep.) 2020]
Ans. (a)
The difference between 9.99 m and 0.0099 m is $9.99 – 0.0099 = 9.9801 \text{ m}$
Taking significant figures into account, as both values have two significant figures after decimal (in 9.99), their difference will also have two significant figures after decimal, i.e., 9.98 m.
The main scale of a vernier calliper has $n$ divisions/cm. $n$ divisions of the vernier scale coincide with $(n-1)$ divisions of main scale. The least count of the vernier callipers is
(a) $\dfrac{1}{(n+1)(n-1)} \text{ cm}$
(b) $\dfrac{1}{n} \text{ cm}$
(c) $\dfrac{1}{n^2} \text{ cm}$
(d) $\dfrac{1}{n(n+1)} \text{ cm}$
[NEET (Odisha) 2019]
Ans. (c)
$n(VSD) = (n-1)MSD \Rightarrow 1 VSD = \dfrac{(n-1)}{n} MSD$
Least Count (LC) = 1 MSD – 1 VSD
$= 1 MSD – \dfrac{(n-1)}{n} MSD = \dfrac{1}{n} MSD$
Here, $1 MSD = \dfrac{1}{n} \text{ cm} \Rightarrow LC = \dfrac{1}{n} \times \dfrac{1}{n} \text{ cm} = \dfrac{1}{n^2} \text{ cm}$
A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of -0.004 cm, the correct diameter of the ball is
(a) 0.053 cm
(b) 0.525 cm
(c) 0.521 cm
(d) 0.529 cm
[NEET 2013]
Ans. (d)
Given, $LC = 0.001 \text{ cm}$, $MSR = 5 \text{ mm} = 0.5 \text{ cm}$, $VSR = 25$
Zero error $= -0.004 \text{ cm}$
Final reading = $MSR + (VSR \times LC) – \text{zero error}$
$= 0.5 + (25 \times 0.001) – (-0.004) = 0.5 + 0.025 + 0.004 = 0.529 \text{ cm}$
In an experiment, four quantities $a, b, c$ and $d$ are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity $P$ is calculated $P = \dfrac{a^3 b^2}{cd}$. Error in $P$ is
(a) 14%
(b) 10%
(c) 7%
(d) 4%
[NEET 2013]
Ans. (a)
$\dfrac{\Delta P}{P} \times 100 = \left( 3\dfrac{\Delta a}{a} + 2\dfrac{\Delta b}{b} + \dfrac{\Delta c}{c} + \dfrac{\Delta d}{d} \right) \times 100$
$= 3(1\%) + 2(2\%) + 3\% + 4\% = 3 + 4 + 3 + 4 = 14\%$
If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be
(a) 4%
(b) 6%
(c) 8%
(d) 2%
[CBSE AIPMT 2008]
Ans. (b)
$V = \dfrac{4}{3}\pi r^3 \Rightarrow \dfrac{\Delta V}{V} \times 100 = 3 \times \dfrac{\Delta r}{r} \times 100$
$= 3 \times 2\% = 6\%$
The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement of mass and length are 4% and 3% respectively, the maximum error in the measurement of density will be
(a) 7%
(b) 9%
(c) 12%
(d) 13%
[CBSE AIPMT 1996]
Ans. (d)
$\rho = \dfrac{m}{V} = \dfrac{m}{l^3} \Rightarrow \dfrac{\Delta \rho}{\rho} \times 100 = \pm \left( \dfrac{\Delta m}{m} + 3\dfrac{\Delta l}{l} \right) \times 100\%$
$= \pm (4 + 3 \times 3) = \pm 13\%$
The percentage errors in the measurement of mass and speed are 2% and 3% respectively. The error in kinetic energy obtained by measuring mass and speed, will be
(a) 12%
(b) 10%
(c) 8%
(d) 2%
[CBSE AIPMT 1995]
Ans. (c)
$K = \dfrac{1}{2}mv^2 \Rightarrow \dfrac{\Delta K}{K} \times 100 = \dfrac{\Delta m}{m} \times 100 + 2 \times \dfrac{\Delta v}{v} \times 100$
$= 2\% + 2 \times 3\% = 8\%$
In a vernier callipers $N$ divisions of vernier scale coincide with $N-1$ divisions of main scale (in which length of one division is 1 mm). The least count of the instrument should be
(a) $N$
(b) $N-1$
(c) $\dfrac{1}{10N}$
(d) $\dfrac{1}{(N-1)}$
[CBSE AIPMT 1994]
Ans. (c)
$N VSD = (N-1) MSD \Rightarrow 1 VSD = \left( \dfrac{N-1}{N} \right) MSD$
$LC = 1 MSD – 1 VSD = \left( 1 – \dfrac{N-1}{N} \right) MSD = \dfrac{1}{N} MSD$
$= \dfrac{1}{N} \times 0.1 \text{ cm} = \dfrac{1}{10N} \text{ cm}$
A certain body weighs 22.42 g and has a measured volume of 4.7 cc. The possible error in the measurement of mass and volume are 0.01 g and 0.1 cc. Then, maximum error in the density will be
(a) 22%
(b) 2%
(c) 0.2%
(d) 0.02%
[CBSE AIPMT 1991]
Ans. (b)
$\dfrac{\Delta \rho}{\rho} = \left( \dfrac{\Delta m}{m} + \dfrac{\Delta V}{V} \right) \times 100 = \left( \dfrac{0.01}{22.42} + \dfrac{0.1}{4.7} \right) \times 100 \approx 2\%$
TOPIC 3 : Dimensions
If force [F], acceleration [a] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.
(a) $[F][a][T]$
(b) $[F][a][T^2]$
(c) $[F][a][T^{-1}]$
(d) $[F][a^{-1}][T]$
[NEET 2021]
Ans. (b)
$[E] = [F]^a [a]^b [T]^c$
$[ML^2T^{-2}] = [MLT^{-2}]^a [LT^{-2}]^b [T]^c$
Equating powers: $a=1, a+b=2 \Rightarrow b=1, -2a-2b+c=-2 \Rightarrow -2-2+c=-2 \Rightarrow c=2$
$[E] = [F^1][a^1][T^2]$
If E and G respectively denote energy and gravitational constant, then $|\dfrac{E}{G}|$ has the dimensions of
(a) $[M^2][L^{-1}][T^0]$
(b) $[M][L^{-1}][T^{-1}]$
(c) $[M][L^0][T^0]$
(d) $[M^2][L^{-2}][T^{-1}]$
[NEET 2021]
Ans. (a)
$[E] = [ML^2T^{-2}]$
$[G] = [M^{-1}L^3T^{-2}]$
$[\dfrac{E}{G}] = \dfrac{[ML^2T^{-2}]}{[M^{-1}L^3T^{-2}]} = [M^2L^{-1}T^0]$
Dimensions of stress are
(a) $[ML^2T^{-2}]$
(b) $[ML^0T^{-2}]$
(c) $[ML^{-1}T^{-2}]$
(d) $[MLT^{-2}]$
[NEET (Sep.) 2020]
Ans. (c)
$\text{Stress} = \dfrac{\text{Force}}{\text{Area}} = \dfrac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$
A physical quantity of the dimensions of length that can be formed out of $c, G$ and $\dfrac{e^2}{4\pi\epsilon_0}$ is
(a) $\dfrac{1}{c^2} \left[ G\dfrac{e^2}{4\pi\epsilon_0} \right]^{1/2}$
(b) $c^2 \left[ G\dfrac{e^2}{4\pi\epsilon_0} \right]^{1/2}$
(c) $\dfrac{1}{c^2} \left[ \dfrac{e^2}{G 4\pi\epsilon_0} \right]^{1/2}$
(d) $\dfrac{1}{c} G \dfrac{e^2}{4\pi\epsilon_0}$
[NEET 2017]
Ans. (a)
$\left[ \dfrac{e^2}{4\pi\epsilon_0} \right] = [ML^3T^{-2}]$, $[G] = [M^{-1}L^3T^{-2}]$, $[c] = [LT^{-1}]$
$\dfrac{1}{c^2} \sqrt{G \dfrac{e^2}{4\pi\epsilon_0}} = [L^{-2}T^2] \sqrt{[M^{-1}L^3T^{-2}][ML^3T^{-2}]} = [L^{-2}T^2][L^3T^{-2}] = [L]$
If energy (E), velocity (v) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be
(a) $[Ev^{-2}T^{-1}]$
(b) $[Ev^{-1}T^{-2}]$
(c) $[Ev^{-2}T^{-2}]$
(d) $[E^{-2}v^{-1}T^{-3}]$
[CBSE AIPMT 2015]
Ans. (c)
$[S] = [MT^{-2}]$
$[S] = [E]^a [v]^b [T]^c \Rightarrow [MT^{-2}] = [ML^2T^{-2}]^a [LT^{-1}]^b [T]^c$
$a=1, 2a+b=0 \Rightarrow b=-2, -2a-b+c=-2 \Rightarrow -2+2+c=-2 \Rightarrow c=-2$
$[S] = [Ev^{-2}T^{-2}]$
If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as $[\eta^x \rho^y r^z]$, then the values of $x, y$ and $z$ are
(a) $1, -1, -1$
(b) $-1, -1, 1$
(c) $-1, -1, -1$
(d) $1, 1, 1$
[CBSE AIPMT 2015]
Ans. (a)
$[LT^{-1}] = [ML^{-1}T^{-1}]^x [ML^{-3}]^y [L]^z$
$x+y=0, -x-3y+z=1, -x=-1 \Rightarrow x=1, y=-1, z=-1$
If force (F), velocity (v) and time (T) are taken as fundamental units, then the dimensions of mass are
(a) $[FvT^{-1}]$
(b) $[FvT^{-2}]$
(c) $[Fv^{-1}T^{-1}]$
(d) $[Fv^{-1}T]$
[CBSE AIPMT 2014]
Ans. (d)
$F = \dfrac{mv}{t} \Rightarrow m = \dfrac{Ft}{v} = [Fv^{-1}T]$
The dimensions of $(\mu_0\epsilon_0)^{-1/2}$ are
(a) $[L^{1/2}T^{-1/2}]$
(b) $[L^{-1}T]$
(c) $[LT^{-1}]$
(d) $[L^{1/2}T^{1/2}]$
[CBSE AIPMT 2012]
Ans. (c)
$c = \dfrac{1}{\sqrt{\mu_0\epsilon_0}} \Rightarrow [c] = [LT^{-1}]$
The dimensions of $\dfrac{1}{2}\epsilon_0 E^2$, where $\epsilon_0$ is permittivity of free space and E is electric field, are
(a) $[ML^2T^{-2}]$
(b) $[ML^{-1}T^{-2}]$
(c) $[ML^2T^{-1}]$
(d) $[MLT^{-1}]$
[CBSE AIPMT 2010]
Ans. (b)
This is energy density $= \dfrac{\text{Energy}}{\text{Volume}} = \dfrac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$
If the dimensions of a physical quantity are given by $[M^a L^b T^c]$, then the physical quantity will be
(a) pressure if $a=1, b=-1, c=-2$
(b) velocity if $a=1, b=0, c=-1$
(c) acceleration if $a=1, b=1, c=-2$
(d) force if $a=0, b=-1, c=-2$
[CBSE AIPMT 2009]
Ans. (a)
$\text{Pressure} = \text{Force/Area} = [ML^{-1}T^{-2}]$
Which two of the following five physical parameters have the same dimensions?
(i) Energy density (ii) Refractive index (iii) Dielectric constant (iv) Young’s modulus (v) Magnetic field
(a) (ii) and (iv)
(b) (iii) and (v)
(c) (i) and (iv)
(d) (i) and (v)
[CBSE AIPMT 2008]
Ans. (c)
Energy density = $[ML^{-1}T^{-2}]$; Young’s modulus = $[ML^{-1}T^{-2}]$
Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and of current I, would be
(a) $[ML^2T^{-3}I^{-1}]$
(b) $[ML^2T^{-2}]$
(c) $[ML^2T^{-1}I^{-1}]$
(d) $[ML^2T^{-3}I^{-2}]$
[CBSE AIPMT 2007]
Ans. (d)
$R = \dfrac{V}{I} = \dfrac{W}{qI} = \dfrac{[ML^2T^{-2}]}{[IT][I]} = [ML^2T^{-3}I^{-2}]$
The velocity v of a particle at time t is given by $v = at + \dfrac{b}{t+c}$. The dimensions of a, b and c are respectively
(a) $[LT^{-2}], [L] \text{ and } [T]$
(b) $[L^2], [T] \text{ and } [LT^2]$
(c) $[LT^2], [LT] \text{ and } [L]$
(d) $[L], [LT] \text{ and } [T^2]$
[CBSE AIPMT 2006]
Ans. (a)
$[at] = [v] \Rightarrow [a] = [LT^{-2}]$; $[c] = [t] = [T]$; $[\dfrac{b}{t}] = [v] \Rightarrow [b] = [L]$
The ratio of the dimensions of Planck’s constant and that of the moment of inertia is the dimension of
(a) frequency
(b) velocity
(c) angular momentum
(d) time
[CBSE AIPMT 2005]
Ans. (a)
$[h] = [ML^2T^{-1}]$; $[I] = [ML^2]$ $\Rightarrow [\dfrac{h}{I}] = [T^{-1}]$ (frequency)
The dimensions of universal gravitational constant are
(a) $[M^{-1}L^3T^{-2}]$
(b) $[ML^2T^{-1}]$
(c) $[M^{-2}L^3T^{-2}]$
(d) $[M^{-2}L^2T^{-1}]$
[CBSE AIPMT 2004]
Ans. (a)
$G = \dfrac{Fr^2}{m_1 m_2} = \dfrac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$
Planck’s constant has the dimensions of
(a) linear momentum
(b) angular momentum
(c) energy
(d) power
[CBSE AIPMT 2001]
Ans. (b)
$[h] = [ML^2T^{-1}]$; Angular momentum $= [I\omega] = [ML^2T^{-1}]$
A pair of physical quantities having same dimensional formula is
(a) force and torque
(b) work and energy
(c) force and impulse
(d) linear momentum and angular momentum
[CBSE AIPMT 2000]
Ans. (b)
Work = $[ML^2T^{-2}]$; Energy = $[ML^2T^{-2}]$
The dimensional formula for magnetic flux is
(a) $[ML^2T^{-2}A^{-1}]$
(b) $[ML^3T^{-2}A^{-2}]$
(c) $[M^0L^{-2}T^2 A^{-2}]$
(d) $[ML^2T^{-1}A^2]$
[CBSE AIPMT 1999]
Ans. (a)
$\phi = BA = \left(\dfrac{F}{il}\right)A = \dfrac{[MLT^{-2}]}{[A][L]} [L^2] = [ML^2T^{-2}A^{-1}]$
The dimensions of coefficient of viscosity $\eta$ are
(a) $[ML^{-3}]$
(b) $[MLT^{-2}]$
(c) $[MT^{-1}]$
(d) $[ML^{-1}T^{-1}]$
[CBSE AIPMT 1997]
Ans. (d)
$F = 6\pi\eta r v \Rightarrow [\eta] = \dfrac{[MLT^{-2}]}{[L][LT^{-1}]} = [ML^{-1}T^{-1}]$
Which of the following will have the dimensions of time?
(a) $LC$
(b) $\dfrac{R}{L}$
(c) $\dfrac{L}{R}$
(d) $\dfrac{C}{L}$
[CBSE AIPMT 1996]
Ans. (c)
$\dfrac{L}{R}$ is the time constant.
If $(p + \dfrac{a}{V^2}) = b\dfrac{\theta}{V}$, the dimensions of $a$ will be
(a) $[ML^5T^{-2}]$
(b) $[M^{-1}L^3T^2]$
(c) $[ML^{-3}T^{-1}]$
(d) $[MLT]$
[CBSE AIPMT 1996]
Ans. (a)
$[p] = [\dfrac{a}{V^2}] \Rightarrow [a] = [p][V^2] = [ML^{-1}T^{-2}][L^6] = [ML^5T^{-2}]$
Which of the following is a dimensional constant?
(a) Refractive index
(b) Poisson’s ratio
(c) Relative density
(d) Gravitational constant
[CBSE AIPMT 1995]
Ans. (d)
$G$ has both dimensions and a constant value.
The dimensions of $\eta$ in $\eta = \dfrac{p(r^2-x^2)}{4vl}$ are
(a) $[M^0L^0T^0]$
(b) $[MLT^{-1}]$
(c) $[ML^2T^{-2}]$
(d) $[ML^{-1}T^{-1}]$
[CBSE AIPMT 1993]
Ans. (d)
$[\eta] = \dfrac{[ML^{-1}T^{-2}][L^2]}{[LT^{-1}][L]} = [ML^{-1}T^{-1}]$
The constant $\alpha$ in $p = p_0 \exp(-\alpha t^2)$
(a) is dimensionless
(b) has dimensions $[T^{-2}]$
(c) has dimensions $[T^2]$
(d) has dimensions of $p$
[CBSE AIPMT 1992]
Ans. (b)
$[\alpha t^2] = [M^0L^0T^0] \Rightarrow [\alpha] = [T^{-2}]$
If $p$ is pressure, $c$ is speed and $S$ is intensity, $p^x S^y c^z$ is dimensionless if
(a) $x=1, y=1, z=1$
(b) $x=-1, y=1, z=1$
(c) $x=1, y=-1, z=1$
(d) $x=1, y=1, z=-1$
[CBSE AIPMT 1992]
Ans. (c)
$[ML^{-1}T^{-2}]^x [MT^{-3}]^y [LT^{-1}]^z = [M^0L^0T^0]$
$x+y=0, -x+z=0 \Rightarrow x=1, y=-1, z=1$
The dimensional formula for permeability of free space $\mu_0$ is
(a) $[MLT^{-2}A^{-2}]$
(b) $[ML^{-1}T^2A^{-2}]$
(c) $[ML^{-1}T^{-2}A^2]$
(d) $[MLT^{-2}A^{-1}]$
[CBSE AIPMT 1991]
Ans. (a)
Using Biot-Savart Law: $[\mu_0] = [MLT^{-2}A^{-2}]$
Frequency $f = C m^x k^y$. The values of $x$ and $y$ are
(a) $x=1/2, y=1/2$
(b) $x=-1/2, y=-1/2$
(c) $x=1/2, y=-1/2$
(d) $x=-1/2, y=1/2$
[CBSE AIPMT 1990]
Ans. (d)
$[T^{-1}] = [M]^x [MT^{-2}]^y \Rightarrow x+y=0, -2y=-1 \Rightarrow y=1/2, x=-1/2$
The dimensions of coefficient of viscosity $\eta$ in $F = -\eta A \dfrac{dv}{dz}$ are
(a) $[ML^{-2}T^{-2}]$
(b) $[M^0L^0T^0]$
(c) $[ML^2T^{-2}]$
(d) $[ML^{-1}T^{-1}]$
[CBSE AIPMT 1990]
Ans. (d)
$[\eta] = \dfrac{[MLT^{-2}]}{[L^2][T^{-1}]} = [ML^{-1}T^{-1}]$
The dimensional formula of pressure is
(a) $[MLT^{-2}]$
(b) $[ML^{-1}T^2]$
(c) $[ML^{-1}T^{-2}]$
(d) $[MLT^2]$
[CBSE AIPMT 1990]
Ans. (c)
$[P] = [ML^{-1}T^{-2}]$
The dimensional formula of torque is
(a) $[ML^2T^{-2}]$
(b) $[MLT^{-2}]$
(c) $[ML^{-1}T^{-2}]$
(d) $[ML^{-2}T^{-2}]$
[CBSE AIPMT 1989]
Ans. (a)
$[\tau] = [ML^2T^{-2}]$
If $x = at + bt^2$, the unit of $b$ is
(a) $km/s$
(b) $km \cdot s$
(c) $km/s^2$
(d) $km \cdot s^2$
[CBSE AIPMT 1989]
Ans. (c)
$[bt^2] = [x] \Rightarrow \text{unit of } b = km/s^2$
Dimensional formula of self-inductance is
(a) $[MLT^{-2}A^{-2}]$
(b) $[ML^2T^{-1}A^{-2}]$
(c) $[ML^2T^{-2}A^{-2}]$
(d) $[ML^2T^{-2}A^{-1}]$
[CBSE AIPMT 1989]
Ans. (c)
$e = L \dfrac{di}{dt} \Rightarrow [L] = \dfrac{[ML^2T^{-2}]}{[A^2 T^{-1}] \cdot [T^{-1}]} = [ML^2T^{-2}A^{-2}]$ (Correction: $V = L di/dt \Rightarrow [L] = [V][T][I]^{-1} = [ML^2T^{-3}A^{-1}][T][A]^{-1} = [ML^2T^{-2}A^{-2}]$)
Which one has dimensions different from the remaining three?
(a) Energy per unit volume
(b) Force per unit area
(c) Voltage $\times$ charge per unit volume
(d) Angular momentum
[CBSE AIPMT 1989]
Ans. (d)
(a), (b), (c) are all $[ML^{-1}T^{-2}]$; (d) is $[ML^2T^{-1}]$
The dimensional formula for angular momentum is
(a) $[M^0L^2T^{-2}]$
(b) $[ML^2T^{-1}]$
(c) $[MLT^{-1}]$
(d) $[ML^2T^{-2}]$
[CBSE AIPMT 1988]
Ans. (b)
$[L] = [ML^2T^{-1}]$
The dimensional formula of CR is
(a) $[M^0L^0T]$
(b) $[M^0L^0T^0]$
(c) $[M^0L^0T^{-1}]$
(d) Not expressible
[CBSE AIPMT 1988]
Ans. (a)
$CR$ is the time constant, so its dimension is $[T]$.
Important Chapter Links
Strengthen your preparation with NEET Units and Measurements PYQs Previous Year Questions and Solutions along with full Class 11 Physics coverage, including important topics such as physical quantities, systems of units, dimensional analysis, significant figures, and errors in measurements. This section is closely connected with other Class 11 chapters like motion in one two and three dimensions, laws of motion, work energy and power, and gravitation, helping you build a strong conceptual foundation. Practicing these previous year questions not only improves accuracy and problem solving skills but also helps you understand exam patterns, making it an essential resource for complete NEET preparation and effective revision of the entire Class 11 syllabus.
FAQs – NEET Units and Measurements PYQs Previous Year Questions and Solutions
Why are NEET Units and Measurements PYQs important?
NEET Units and Measurements PYQs help students understand exam patterns, frequently asked concepts, and improve accuracy in solving numerical problems based on SI units, dimensions, and measurement errors.
What topics are covered in Units and Measurements for NEET?
This chapter includes physical quantities, systems of units, SI units, significant figures, dimensional analysis, errors in measurements, accuracy, precision, and propagation of errors.
How do PYQs help in scoring better in NEET Physics?
Practicing PYQs strengthens conceptual clarity, improves speed and accuracy, and helps students become familiar with the types of questions asked in the NEET exam.
Are numerical problems from Units and Measurements easy in NEET?
Yes, most questions are concept-based and scoring if fundamentals like unit conversion, dimensional analysis, and significant figures are clear.
How should I prepare Units and Measurements for NEET?
Focus on understanding concepts, practice PYQs regularly, revise formulas and units, and pay special attention to significant figures and error calculations.
What is the role of dimensional analysis in NEET Physics?
Dimensional analysis helps verify equations, derive relationships between physical quantities, and check the correctness of formulas in numerical problems.
Is Units and Measurements important for other chapters in Physics?
Yes, it is the foundation for all chapters like motion, laws of motion, work energy, and gravitation, as correct units and measurements are required in every calculation.
Who has prepared these NEET Units and Measurements PYQs and solutions?
This content is prepared by Physics Anand Classes, authored by Er Neeraj Anand, and published by Anand Technical Publishers under the brand Anand Classes.
Do these PYQs include step-by-step solutions?
Yes, all questions are provided with clear, step-by-step solutions to help students understand the concepts and improve problem-solving skills.
Can I rely only on PYQs for Units and Measurements preparation?
PYQs are extremely important, but they should be combined with theory revision and practice questions to ensure complete understanding and better performance in NEET.