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NEET • PHYSICS • UNITS & MEASUREMENTS

Units & measurements practice · Mixed difficulty

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Showing 1-30 of 384 questions
1 Easy
1 of 384
Understanding +4 | -1
Statement-I: Random errors can be positive or negative.
Statement-II: Cause of random errors are uncertain.
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Option A (Statement-I is true and statement-II is true) is correct because random errors can indeed be positive or negative, and their causes are uncertain.
2 Medium
2 of 384
Application +4 | -1
An object is moving through the liquid. The viscous damping force acting on it is proportional to the velocity. Then dimension of constant of proportionality is
View answer and explanation
Correct option: D
ML0T−1M{L^0}{T^{ - 1}}
Option D (ML0T−1M{L^0}{T^{ - 1}}) is correct because the viscous force is proportional to velocity, so the constant of proportionality has dimensions of mass/time.
3 Medium
3 of 384
Remembering +4 | -1
The error due to parallax is a:
View answer and explanation
Correct option: C
Personal error
Option C (Personal error) is correct because parallax error arises from the observer's position and judgment while taking readings.
4 Easy
4 of 384
Understanding +4 | -1
2.34 is obtained by rounding of the number
View answer and explanation
Correct option: C
2.335
Option C (2.335) is correct because when rounding 2.335 to two decimal places, the 5 causes the preceding 3 to round up to 4, resulting in 2.34.
5 Medium
5 of 384
Understanding +4 | -1
The dimensional formula of self-inductance is
View answer and explanation
Correct option: C
[ML2T−2A−2]\left[ {\mathrm{M}{\mathrm{L}^2}{\mathrm{T}^{ - 2}}{\mathrm{A}^{ - 2}}} \right]
Option C ([ML2T−2A−2]\left[ {\mathrm{M}{\mathrm{L}^2}{\mathrm{T}^{ - 2}}{\mathrm{A}^{ - 2}}} \right]) is correct because self-inductance is related to energy stored in a magnetic field, and its dimensions can be derived from that.
6 Medium
6 of 384
Understanding +4 | -1
The dimension of 1ε0μ0\frac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }} is that of
View answer and explanation
Correct option: A
Velocity
Option A (Velocity) is correct because 1ε0μ0\frac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }} represents the speed of light, which has the dimensions of velocity.
7 Medium
7 of 384
Analyzing +4 | -1
Which is wrong dimension
View answer and explanation
Correct option: B
s=vt2\mathrm{s}=\mathrm{vt}^2
Option B (s=vt2\mathrm{s}=\mathrm{vt}^2) is wrong dimensionally because the dimensions on the left side (length) do not match the dimensions on the right side (length × time).
8 Medium
8 of 384
Application +4 | -1
The dimensional formula of capacitance in terms of M,L,T and I is
View answer and explanation
Correct option: D
[M−1L−2T4I2]\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 2}}{\mathrm{T}^4}{\mathrm{I}^2}} \right]
Option D [M−1L−2T4I2]\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 2}}{\mathrm{T}^4}{\mathrm{I}^2}} \right] is correct because it is derived from the formula for capacitance and the dimensional formulas for charge, voltage, and current.
9 Hard
9 of 384
Analyzing +4 | -1
Pressure gradient has the same dimension as that of
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because pressure gradient has dimensions of [ML⁻²T⁻²], velocity gradient has dimensions of [T⁻¹], and potential gradient has dimensions of [MLT⁻³A⁻¹], and energy gradient has dimensions of [ML²T⁻²L⁻¹] = [MLT⁻²], so none match.
10 Medium
10 of 384
Application +4 | -1
The length l,l,, breadth b and thickness t of a block are measured with the help of a metre scale. Given l=15.12±0.01l = 15.12 \pm 0.01 cm, b=10.15±0.01b = 10.15 \pm 0.01 cm, t=5.28±0.01t = 5.28 \pm 0.01cm.The percentage error in volume is
View answer and explanation
Correct option: C
0.0037
Option C (0.0037) is correct because the percentage error in volume is the sum of the percentage errors in length, breadth, and thickness.
11 Medium
11 of 384
Analyzing +4 | -1
If the speed of light (c), acceleration due to gravity (g) and pressure (p) are taken as the fundamental quantities, then the dimension of gravitational constant is
View answer and explanation
Correct option: B
c0g2p−1{c^0}{g^2}{p^{ - 1}}
Option B (c0g2p−1{c^0}{g^2}{p^{ - 1}}) is correct because it provides the correct dimensions for the gravitational constant when speed of light, acceleration due to gravity, and pressure are fundamental quantities.
12 Hard
12 of 384
Application +4 | -1
If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius (r) of the drop and density (ρ\rho) of the liquid, then the expression of T is
View answer and explanation
Correct option: A
T=kρr3/ST = k\sqrt {\rho {r^3}/S}
Option A (T=kρr3/ST = k\sqrt {\rho {r^3}/S}) is correct because dimensional analysis confirms that the dimensions of this expression match the dimensions of time.
13 Easy
13 of 384
Remembering +4 | -1
The value of Planck's constant is
View answer and explanation
Correct option: A
6.63×10−34 Js−16.63 \times 10^{-34} \,{Js}^{-1}
Option B (6.63×10−34 kg m2 s−16.63 \times 10^{-34} {~kg} {~m}^2 {~s}^{-1}) is correct because it represents the correct value and units of Planck's constant.
14 Hard
14 of 384
Application +4 | -1
The length of a simple pendulum is about 100 cm known to an accuracy of 1 mm. Its period of oscillation is 2s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution. What is the accuracy in the determined value of g?
View answer and explanation
Correct option: A
0.2 %0.2\,\%
Option A (0.2 %0.2\,\%) is correct because it accurately reflects the percentage error in 'g' based on the given uncertainties in length and time period.
15 Easy
15 of 384
Application +4 | -1
The radius of the proton is about 10−15m{10^{ - 15}}m. The radius of the observable universe is 1026{10^{26}} m. identify the distance which is half-way between these two extremes on a logarithmic scale.
View answer and explanation
Correct option: B
106m{10^6}m
Option B (106m{10^6}m) is correct because it represents the geometric mean of the two extreme distances on a logarithmic scale.

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WORK-ENERGY-POWER 355 MCQs MECHANICAL PROPERTIES OF FLUIDS 317 MCQs MOTION IN PLANE 307 MCQs MAGNETIC EFFECT OF CURRENT 282 MCQs THERMAL PROPERTIES OF MATTER 269 MCQs GRAVITATION 258 MCQs
16 Medium
16 of 384
Application +4 | -1
A public park, in the form of a square, has an area of (100±0.2)m2\left( {100 \pm 0.2} \right){\mathrm{m}^2} . The side of park is
View answer and explanation
Correct option: A
(10±0.01)m\left( {10 \pm 0.01} \right)m
Option A ((10±0.01)m\left( {10 \pm 0.01} \right)m) is correct because the side is the square root of the area, and the percentage error in the side is half the percentage error in the area.
17 Medium
17 of 384
Application +4 | -1
What is the unit of k in the relation U=kyy2+a2U = \frac{{ky}}{{{y^2} + {a^2}}} where U represents the potential energy, y represents the displacement and a represents the maximum displacement ie, amplitude?
View answer and explanation
Correct option: C
J m
Option C (J m) is correct because from the relation U=kyy2+a2U = \frac{{ky}}{{{y^2} + {a^2}}}, k must have the dimensions of energy times length, which corresponds to J m.
18 Hard
18 of 384
Application +4 | -1
The potential energy of a particle varies with distance x from a fixed origin as U=(AXx+B);U = \left( {\frac{{A\sqrt X }}{{x + B}}} \right);; where A and B are constants. The dimensions of AB are
View answer and explanation
Correct option: D
[ML7/2T−2]\left[ {\mathrm{M}{\mathrm{L}^{7/2}}{\mathrm{T}^{ - 2}}} \right]
Option D ([ML7/2T−2][{\mathrm{M}{\mathrm{L}^{7/2}}{\mathrm{T}^{ - 2}}}]) is correct because it correctly calculates the dimensions of AB based on the given potential energy equation.
19 Hard
19 of 384
Remembering +4 | -1
To determine refractive index of glass also using
a travelling microscope, minimum number of
readings required are
View answer and explanation
Correct option: C
Three
Option C (Three) is correct because a minimum of three readings are required to determine the refractive index of glass using a travelling microscope.
20 Easy
20 of 384
Application +4 | -1
If the unit of force is 1 N, the unit of work done is 10 J and the unit of time is 1 second, the unit of mass is
View answer and explanation
Correct option: C
0.1 kg
Option D (100 kg) is correct because it is derived from the relationship between force, work, and time, using the definitions of Newton and Joule.
21 Medium
21 of 384
Application +4 | -1
The dimensional formula of coefficient of permittivity for free space (ε0)\left( {{\varepsilon _0}} \right) in the equation F=14πε0q1q2r2,F = \frac{1}{{4\mathrm{\pi }{\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r^2}}}, where symbols have their usual meanings, is
View answer and explanation
Correct option: B
[M−1L−3T4A2]\left[ {{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 3}}{\mathrm{T}^4}{\mathrm{A}^2}} \right]
Option B ([M−1L−3T4A2][{\mathrm{M}^{ - 1}}{\mathrm{L}^{ - 3}}{\mathrm{T}^4}{\mathrm{A}^2}]) is correct because it correctly derives the dimensional formula of permittivity using Coulomb's law.
22 Easy
22 of 384
Understanding +4 | -1
Statement-I: In the measurement of gg using simple pendulum generally we take central position (mean position) of the oscillation as reference position for measuring time of oscillation.
Statement-II: This reduces the human error in measurement of time.
View answer and explanation
Correct option: A
Statement-I is true and statement-II is true
Option A (Statement-I is true and statement-II is true) is correct because taking the mean position as a reference reduces human error in timing oscillations.
23 Hard
23 of 384
Application +4 | -1
A body travels uniformly a distance of (20.0±0.2) m(20.0 \pm 0.2)\,m in time (4.0±0.04) s(4.0 \pm 0.04)\,s. The velocity of the body is
View answer and explanation
Correct option: D
(5.0±0.1) ms−1(5.0 \pm 0.1)\,m{s^{ - 1}}
Option D ((5.0±0.1) ms−1(5.0 \pm 0.1)\,m{s^{ - 1}}) is correct because it accurately calculates the velocity and its associated uncertainty using the given distance and time with their respective uncertainties.
24 Medium
24 of 384
Analyzing +4 | -1
Assertion A: If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm.
Reason R: Least count = PitchTotal  division  on  circular  scale\frac{{\mathrm{Pitch}}}{{\mathrm{Total}\,\,\mathrm{division}\,\,\mathrm{on}\,\,\mathrm{circular}\,\,\mathrm{scale}}}
In the light of the above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: D
A is not correct but R is correct
Option D (A is not correct but R is correct) is correct because the assertion incorrectly calculates the least count, while the reason provides the correct formula.
25 Medium
25 of 384
Remembering +4 | -1
Unit of reduction factor is
View answer and explanation
Correct option: A
ampere
Option A (ampere) is correct because the reduction factor of a tangent galvanometer is a measure of the current required to produce a certain deflection, hence its unit is ampere.
26 Medium
26 of 384
Application +4 | -1
A screw gauge has least count of 0.01mm\mathrm{0}\mathrm{.01 mm} and there are 50\mathrm{50} divisions in its circular scale. The pitch of the screw gauge is :
View answer and explanation
Correct option: C
0.5  mm\mathrm{0}\mathrm{.5}\,\,\mathrm{mm}
Option C (0.5 mm\mathrm{0}\mathrm{.5}\,\mathrm{mm}) is correct because the pitch is calculated by multiplying the least count by the number of divisions on the circular scale.
27 Medium
27 of 384
Remembering +4 | -1
The dimensions of speed and velocity are
View answer and explanation
Correct option: D
[LT−1],[LT−1]\left[\mathrm{LT}^{-1}\right],\left[\mathrm{LT}^{-1}\right]
Option D ([LT-1],[LT-1]) is correct because both speed and velocity are defined as distance divided by time, hence having the same dimensions.
28 Easy
28 of 384
Understanding +4 | -1
The dimensions of ML2T−2M{L^2}{T^{ - 2}} may correspond to
View answer and explanation
Correct option: D
Energy
Option D (Energy) is correct because the dimensions of energy (or work) are Mass x (Length)2 / (Time)2, which matches the given dimensions.
29 Medium
29 of 384
Remembering +4 | -1
Dimensions of planck’s constant is:
View answer and explanation
Correct option: A
[M1L2T−1]\left[ {{M^1}{L^2}{T^{ - 1}}} \right]
Option A ([M1L2T−1]\left[ {{M^1}{L^2}{T^{ - 1}}} \right]) is correct because Planck's constant relates energy and frequency (E = hf), and its dimensions can be derived from the dimensions of energy and frequency.
30 Medium
30 of 384
Application +4 | -1
The surface tension of a liquid is 70 dyne/cm, in MKS system value is
View answer and explanation
Correct option: B
7×10−2  N/m7 \times {10^{ - 2}}\;N/m
Option B (7×10−2  N/m7 \times {10^{ - 2}}\;N/m) is correct because 1 dyne/cm equals 10−310^{-3} N/m.
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