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NEET • PHYSICS • UNITS & MEASUREMENTS

Units & measurements practice · Mixed difficulty

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Showing 91-120 of 384 questions

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91 Easy
91 of 384
Understanding +4 | -1
Dimensions of the following three quantities are the same
View answer and explanation
Correct option: D
Pressure, stress, coefficient of elasticity
Option D (Pressure, stress, coefficient of elasticity) is correct because all three quantities are defined as force per unit area, hence they share the same dimensions [ML⁻¹T⁻²].
92 Medium
92 of 384
Remembering +4 | -1
A sextant is used to measure
View answer and explanation
Correct option: B
Height of an object
Option B (Height of an object) is correct because a sextant is an instrument used to measure the angle between an astronomical object and the horizon, which is then used to calculate the height of the object.
93 Medium
93 of 384
Remembering +4 | -1
Light Year is a unit of
View answer and explanation
Correct option: C
Distance
Option C (Distance) is correct because a light-year is defined as the distance light travels in one year.
94 Medium
94 of 384
Understanding +4 | -1
The dimensions [MLT−2A−2]\mathrm{[ML}{\mathrm{T}^{ - 2}}{\mathrm{A}^{ - 2}}\mathrm{]} belong to the:
View answer and explanation
Correct option: D
Magnetic permeability
Option D (Magnetic permeability) is correct because the dimensions of magnetic permeability are [MLT−2A−2][MLT^{-2}A^{-2}].
95 Medium
95 of 384
Understanding +4 | -1
Which one of the following is not a unit of Young’s modulus?
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Correct option: A
Nm−1\mathrm{N}{\mathrm{m}^{ - 1}}
Option A (Nm−1\mathrm{N}{\mathrm{m}^{ - 1}}) is the correct answer because it represents force per unit length, not stress per unit strain, which is Young's modulus.
96 Medium
96 of 384
Application +4 | -1
If L,C and R denote inductance, capacitance and resistance respectively, then which of the following combination has the dimension of time?
View answer and explanation
Correct option: C
LR\frac{L}{R}
Option C (L/R) is correct because the ratio of inductance to resistance has the dimensions of time.
97 Easy
97 of 384
Application +4 | -1
Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be :
View answer and explanation
Correct option: A
[M]=[T−1C−2 h][\mathrm{M}]=\left[\mathrm{T}^{-1} \mathrm{C}^{-2} \mathrm{~h}\right]
Option A ([M]=[T−1C−2 h][\mathrm{M}]=\left[\mathrm{T}^{-1} \mathrm{C}^{-2} \mathrm{~h}\right]) is correct because by expressing mass in terms of time, velocity, and angular momentum, we can derive the correct dimensional formula using the definitions of these quantities.
98 Medium
98 of 384
Application +4 | -1
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 1mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is
View answer and explanation
Correct option: B
0.052 cm0.052\,cm
Option B (0.052 cm0.052\,cm) is correct because it is the diameter calculated from the screw gauge readings, considering the least count and circular scale division.
99 Medium
99 of 384
Understanding +4 | -1
Dimensions of charge are
View answer and explanation
Correct option: D
TATA
Option D (TATA) is correct because charge is current multiplied by time, and current is represented by AA and time by TT.
100 Easy
100 of 384
Application +4 | -1
The relative error in the determination of the surface area of a sphere is α\alpha. Then the relative error in the determination of it’s volume is
View answer and explanation
Correct option: A
32α\frac{3}{2}\alpha
Option A (32α\frac{3}{2}\alpha) is correct because the volume is proportional to the radius cubed, and the surface area is proportional to the radius squared, so the relative error in volume is 3/2 times the relative error in surface area.
101 Easy
101 of 384
Understanding +4 | -1
The physical quantity angular momentum has the same dimensions as that of
View answer and explanation
Correct option: D
Planck’s constant
Option D (Planck’s constant) is correct because both angular momentum and Planck's constant have the dimensions of [ML²T⁻¹].
102 Medium
102 of 384
Understanding +4 | -1
The fundamental physical quantities that have same dimensions in the dimensional formulae of torque and angular momentum are
View answer and explanation
Correct option: C
Mass, length
Option C (Mass, length) is correct because both torque and angular momentum have mass and length in their dimensional formulas.
103 Medium
103 of 384
Application +4 | -1
The dimensional formula of 1ε0e2hc\frac{1}{{{\varepsilon _0}}}\frac{{{e^2}}}{{hc}} is
View answer and explanation
Correct option: A
[M0L0T0A0]\left[ {{\mathrm{M}^0}{\mathrm{L}^0}{\mathrm{T}^0}{\mathrm{A}^0}} \right]
Option A ([M⁰L⁰T⁰A⁰]) is correct because the expression is dimensionless, representing a pure number (fine-structure constant).
104 Medium
104 of 384
Remembering +4 | -1
“Pascal-Second” has dimension of
View answer and explanation
Correct option: D
Coefficient of viscosity
The correct answer is Coefficient of viscosity because Pascal-Second is the SI unit for viscosity.
105 Medium
105 of 384
Application +4 | -1
What will be the unit of time in that system in which the unit of length is metre, unit of mass is kg and unit of force is kg wt?
View answer and explanation
Correct option: D
19.8 sec⁡\frac{1}{{\sqrt {9.8} }}\,\sec
Option D (19.8 sec⁡\frac{1}{{\sqrt {9.8} }}\,\sec) is correct because it is derived from equating the units of force in the new system to the standard SI units.
106 Easy
106 of 384
Understanding +4 | -1
Statement-I: Pressure can be subtracted from pressure gradient
Statement-II: Only like quantities can be added or subtracted from each other
View answer and explanation
Correct option: C
Statement-I is false and statement-II is true
Option C (Statement-I is false and statement-II is true) is correct because pressure and pressure gradient have different dimensions, and only quantities with the same dimensions can be added or subtracted.
107 Medium
107 of 384
Analyzing +4 | -1
A physical quantity x depends on quantities y and z as follows: x=Ay+Btan⁡C zx = Ay + B\tan C\,z where A, B and C are constants. Which of the following do not have the same dimensions
View answer and explanation
Correct option: D
x and A
Option D (x and A) is correct because the dimensions of x are not necessarily the same as A, since A is multiplied by y, and y may have different dimensions than x.
108 Easy
108 of 384
Analyzing +4 | -1
If A,BA, B and CC are physical quantities, having different dimensions, which of the following combination can never be meaningful quantity?
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Correct option: A
(A−B)/C(A-B) /C
Option A ((A-B)/C) is correct because for (A-B) to be meaningful, A and B must have the same dimensions, but the question states that A, B, and C have different dimensions.
109 Easy
109 of 384
Analyzing +4 | -1
Which of the following pairs does not have similar dimensions
View answer and explanation
Correct option: C
Tension and surface tension
Option C (Tension and surface tension) is correct because tension is a force and surface tension is force per unit length, hence they have different dimensions.
110 Medium
110 of 384
Application +4 | -1
In a vernier callipers, one main scale division is x  cmx\;cm and nn division of the vernier scale coincide with (n−1)\left( {n - 1} \right) divisions of the main scale. The least count (in cm) of the callipers is
View answer and explanation
Correct option: C
xn\frac{x}{n}
Option C (xn\frac{x}{n}) is correct because the least count is calculated by finding the difference between one main scale division and one vernier scale division.
111 Medium
111 of 384
Remembering +4 | -1
The unit of the coefficient of viscosity in S.I. system is
View answer and explanation
Correct option: D
kg/m−skg/m - s
Option D (kg/m−skg/m - s) is correct because it represents the correct SI unit for the coefficient of viscosity.
112 Medium
112 of 384
Remembering +4 | -1
The dimensions of surface tension are
View answer and explanation
Correct option: D
MT−2M{T^{ - 2}}
Option D (MT−2MT^{-2}) is correct because surface tension is force per unit length, and its dimensional formula reflects this relationship.
113 Hard
113 of 384
Application +4 | -1
If L1=(2.02±0.01) m{L_1} = (2.02 \pm 0.01)\,m and L2=(1.02±0.01) m{L_2} = (1.02 \pm 0.01)\,m then L1+2L2{L_1} + 2{L_2} is (in m)
View answer and explanation
Correct option: B
4.06±0.034.06 \pm 0.03
Option B (4.06±0.034.06 \pm 0.03) is correct because it accurately calculates the sum and its associated error.
114 Easy
114 of 384
Application +4 | -1
If the measurement of volume of solid sphere using the formula V=43πr3V = \frac{4}{3}\pi {r^3} if the error committed in the measurement of the radius r is 1.5%, the percentage error in the volume measurement is
View answer and explanation
Correct option: C
0.045
Option C (0.045) is correct because the percentage error in volume is three times the percentage error in the radius.
115 Medium
115 of 384
Application +4 | -1
The dimension of quantity (L/RCV) is
View answer and explanation
Correct option: C
[A−1]\left[ {{A^{ - 1}}} \right]
Option C ([A−1{A^{ - 1}}]) is correct because it involves finding the dimension of (L/RCV) where L is inductance, R is resistance, C is capacitance, and V is voltage. Since L/R has the dimension of time [T], CV has the dimension of charge [AT], and L/RCV becomes [T/AT] = [A-1].
116 Medium
116 of 384
Application +4 | -1
The position of a particle at time t is given by the relation x(t)=(v0α  )(1−eαt),  x\left( t \right) = \left( {\begin{array}{cccccccccccccccccccc} {\frac{{{v_0}}}{\alpha }}\\ \; \end{array}} \right)\left( {1 - {e^{\alpha t}}} \right),\; where v0{v_0} is constant and α>0\alpha > 0 . The dimensions of v0{v_0} and α\alpha are respectively
View answer and explanation
Correct option: A
M0L1T−1 and T−1{M^0}{L^1}{T^{ - 1}}\,and\,{T^{ - 1}}
Option A (M0L1T−1 and T−1{M^0}{L^1}{T^{ - 1}}\,and\,{T^{ - 1}}) is correct because the dimensions of x(t) must be length (L), which implies v0{v_0} has dimensions of LT−1LT^{-1} and α\alpha has dimensions of T−1T^{-1} to make the exponent dimensionless.
117 Hard
117 of 384
Application +4 | -1
If pressure P, velocity V and time T are taken as fundamental physical quantities, the dimensional formulae of the force is
View answer and explanation
Correct option: A
PV2T2P{V^2}{T^2}
Option A (PV2T2P{V^2}{T^2}) is correct because it yields the dimensions of force when pressure, velocity, and time are taken as fundamental quantities.
118 Medium
118 of 384
Analyzing +4 | -1
The physical quantity which is not a unit of energy is
View answer and explanation
Correct option: B
MeV-sec
Option B (MeV-sec) is not a unit of energy because MeV is a unit of energy, and multiplying it by seconds gives a unit of energy multiplied by time, not just energy.
119 Easy
119 of 384
Understanding +4 | -1
Dimensions of impulse are same as that of
View answer and explanation
Correct option: B
Momentum
The correct answer is Momentum because impulse is defined as the change in momentum.
120 Easy
120 of 384
Application +4 | -1
The vernier constant of Vernier callipers is 0.1 mm0.1 \mathrm{~mm} and it has zero error of (−0.05)cm(-0.05) \mathrm{cm}. While measuring diameter of a sphere, the main scale reading is 1.7 cm1.7 \mathrm{~cm} and coinciding vernier division is 5 . The corrected diameter will be ×10−2 cm\times 10^{-2} \mathrm{~cm}.
View answer and explanation
Correct option: A
180
Option A (180) is correct because the corrected diameter is calculated by adding the main scale reading and vernier scale reading, then correcting for the zero error.
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