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Quiz 6
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Q.1
Equation for the state of a gas is where a, b, c and R are constants. the isothermal can be represented by P = AVm - BVn where A and B depends only on temperature and .. [CBSE1995]
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a) m = - and n=-1
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b) m =c and n= 1
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c) m =-c and n=1
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d) m = c and n = -1
Explanation
Dividing equation by by Vc and adjusting terms we get Now P = AVm - BVn --(2) comparing above two equations we get m --c and n = -1 Answer: (a)
Q.2
If Force(F), Area(A) and density(D) are taken as the fundamental units, the representation of Young's Modulus will be
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a) F-1A-1D-1
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b) FA-2D2
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c) FA-1D0
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d) FA-1D
Explanation
Let Y = [ Fa Ab Dc] Dimensional formula of Young's modulus = [ML-1T-2 ] Thus [ML-1T-2 ]= [ MLT-2]a[L2]b[ML-3]c [ML-1T-2 ] = [ Ma+c La+2b-3c T-2a-3c By comparing a+c = 1 a+2b-3c = -1 -2a-2c = -2 on solving c= 0, a = 1, b= -1 Answer: (c)
Q.3
The dimensions of physical quantity X in the equation Force= X/(density) is given by
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a) ML4T-2
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b) M2L-2T-1
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c) M2L-2T-2
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d) ML-2T-1
Explanation
Form given equation X = F˜(Density) Dimension of X = [MLT-2][ML-3] Dimension of X = [M2L-2T-2 Answer: (c)
Q.4
If dimensional formula of physical quantity id Ma LbTc. If a = 1, b = -1, c = -2 then the physical quantity is ... ...
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a) Pressure
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b) Force
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c) Angular momentum
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d) Energy
Explanation
Pressure = Force /Area Answer : (a)
Q.5
Dimensional formula of root means square of velocity is ...
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a) M0L1T-1
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b) M0L-1T-1
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c) M0L0T-1
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d) M0L2T-1
Explanation
Answer : (a)
Q.6
SI unit of pressure is ... ...
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a) pascal
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b) dyne/cm2
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c) cm of Hg
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d) atmosphere
Explanation
Answer : (a)
Q.7
Dimensions of coefficient of thermal conductivity is
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a) ML2T-2K-1
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b) MLT-3K-1
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c) MLT-2K-1
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d) MLT-3K
Explanation
Heat conducted by rod is given by formula here k is Coefficient of conductivity T1 and T2 are temperature A: Area norml to the heat flow d= length of rod Since equation is dimensionality valid Answer:(b)
Q.8
The damping force on a body is proportional to the velocity of the body. In this position the proportionality constant has dimensions of ..
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a) M2T-1
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b) MLT-2
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c) kg/ sec
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d) It is the dimensionless constant
Explanation
F = k V k = F/V Thus dimension of k = [MLT-2] / [LT-1] Dimension of k = MT-1= kg/sec Answer: (c)
Q.9
Curie is a unit of ..
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a) Energy of γ rays
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b) Half life
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c) Radioactivity
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d) Intensity of γ rays
Explanation
Answer: (c)
Q.10
Oersted is the unit of
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a) Dip
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b) Magnetic intensity
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c) Magnetic moment
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d) pole strength
Explanation
Answer: (b)
Q.11
Ampere-hour is unit of
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a) Quantity of electricity
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b) Strength of electric current
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c) power
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d) energy
Explanation
Qty of charge = Current × time Answer:(a)
Q.12
Dimension of coefficient of viscosity re
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a) ML2T-2
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b) ML2T-1
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c) ML-1T-1
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d) MLT
Explanation
Viscus Force F = η A (dV/dx) [MLT-2 = η[L2] [ T-1] η = ML-1T-1 Answer: (c)
Q.13
Unit of Stefan's constant is
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a) Js-1
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b) Jm-2s-1K-4
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c) Jm-2
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d) Js
Explanation
Stefan's law: Radiant energy per emitted by unit area of surface in unit second is proportional to the fourth power of absolute temperature of the body E=σT4 --(1) Here T is temperature Now E = Energy / (Area-time) Unit of E = Jm-2s-1 Thus σ = E/T4 unit of σ = Jm-2s-1 / K4 Unit of σ = Jm-2s-1 K-4 Answer: (b)
Q.14
Dimensional formula for Boltzmann constant K ..
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a) [ML2T-2θ-1 mole-1]
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b) [ML2T-2θ-1 ]
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c) [M2LT-2θ]
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d) [ML0T-2θ-1 mole-1]
Explanation
According to Ideal gas equation PV = nRT R = PV/nT PV have dimension of work thus dimension of R is ML2T-2θ-1mole-1 Now Boltzman constant = R/NA Here NA is Avogadro's number Thus dimension of Boltzmann constant = [ML2T-2θ-1 mole-1] Answer: (a)
Q.15
the dimensional formula for bulk modulus of elasticity is .. [MNR1986]
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a) ML-2T-2
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b) ML-3T-2
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c) ML2T-2
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d) ML-1T-2
Explanation
Bulk modulus = (-ΔP V) / ΔV As V/ΔV is dimension less Dimension of Bulk modulus = Dimension of pressure pressure = F/A Dimension of pressure = [ML-1T-2] Answer:(d)
Q.16
Which of the following is different from others
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a) Plank's constant
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b) Coefficient of viscosity
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c) Force constant
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d) Poission's ratio
Explanation
Poission's ratio don't have dimensions Answer: (d)
Q.17
In Ohm's experiment, the value of an unknown resistance were found to be 4.12Ω, 4.08Ω, 4.22Ω and 4.14Ω. Calculate absolute error, relative error and percentage error. in these measurement. ... .. [ Guj. T.B. ]
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a) 4.14, 0.04, 0.096%
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b) 0.04, 0.0096, 0.96%
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c) 0.04, 0.96, 96%
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d) 4.14, 0.4, 0.0096%
Explanation
Average resistance = (4.12+4.08+4.22+4.14)/4 = 4.14Ω Absolute error of each observation ΔR1 = 4.14 - 4.12 = 0.02 Ω ΔR2 = 4.14 - 4.08 = 0.06 Ω ΔR3 = 4.14 - 4.22 = -0.08 Ω ΔR4 = 4.14 - 4.14 = 0.00 Ω Average absolute error, considering only magnitude Average absolute error = (0.2 + 0.6 + 0.08)/4 = 0.04Ω Resistance R = (4.14 ±0.04)Ω Relative error = Absolute error/ average value = 0.04/4.14 = 0.0096 Percentage error = Relative error ×100 Percentage error = 0.0096 ×100 = 0.96% Answer : (b)
Q.18
If the length of a cylinder is l = (4.00 ± 0.01) cm, radius r = ( 0.250 ±0.001) cm and mass m = ( 6.25 ± 0.01) g. Calculate the percentage error in determination of density... .. [ Guj T.B.]
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a) 12%
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b) 11.2%
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c) 1.21%
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d) 12.1%
Explanation
Formula for density is Relative error Percentage error = Relative error × 100 Percentage error = 12.1 ×10-3 × 100 = 1.21% Answer : (c)
Q.19
The acceleration due to gravity (g) is determined by using simple pendulum of length l = (100 ± 0.1 ) cm. if the time period is T = (2 ± 0.01 )s, find the maximum percentage error in the measurement of g
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a) 21%
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b) 0.0021%
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c) 0.011 %
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d) 1.1 %
Explanation
formula for periodic time of simple pendulum Relative error Percentage of error = relative error × 100 Percentage error = 11 × 10-3 × 100 Percentage error = 1.1% Answer : (d)
Q.20
The length breadth and thickness of metal sheet are 4.234 m., 1.005 m and 2.01 cm respectively. Calculate the total area and volume of the sheet to the correct number of significant figures.
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a) 8.7216 m2, 0.855 m3
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b) 8.72 m2, 0.086 m3
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c) 8.721 m2, 0.0855 m3
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d) 8.7205 m2, 0.8553 m3
Explanation
length = 4.234 m breadth b = 1.005 m height = h = 2.01 cm = 0.0201 m Total area = 2 [ (l × b) + (b × h) + (h × l) ] Total area = 8.7209478 m3 Since minimum significant figure are three in data thus answer should have only three significant figure Correct answer = 8.72 m2 Volume = l × b × h Volume = 4.234 ×1.005 × 0.0201 = 0.0855289 Since minimum significant figure are three in data thus answer should have only three significant figure V = 0.086 m3 Answer : (b)
Q.21
The electric force between two electric charges is given by by formula Here r is the distance between changes q1 and qGive the unit and dimensional formula of ε0
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a) N-1C2m-2, M-1 L-3T4 A2
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b) N2C-1m2, M-1 L3T4 A-2
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c) N1C-2m3, M1 L3T-4 A-2
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d) N1C-2m3, M1 L-3T-4 A2
Explanation
Answer : (a)
Q.22
Check dimensional validity of the following equations A) Pressure P = h ρ g All the terms have usual meaning
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a) A and B both are dimensionally valid
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b) A and B are dimensionally invalid
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c) A is dimensionally invalid and B is dimensionally valid
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d) A is dimensionally valid and B is dimensionally invalid
Explanation
A) Pressure P= h ρ g Dimensions of LHS = [P] = M1 L-1 T-2 Dimension of RHS = [h ρ g ] = [ M1L3T0] [M0 L1 T-2] [ M0 L1 T0] Dimensions of RHS = M1 L-1 T-2 Since dimensions of LHS = RHS A) is dimensionally valid For B) Dimension of LHS = [s] = [M0 L1 T0] Now RHS have two terms first v0 t Dimension of first RHS term = [ M0 L1 T-1] [ M0L0 T1] Dimension of first term = [M0 L1 T0] Dimensions of second terms = [at]2 = [a]2 [t]2 = [M0L1 T-2]2 [M0 L0 T1]2 = [M0L2T-2] Thus second terms of RHS is not equal to LHS thus B is dimensionally in-valid Answer : (d)
Q.23
Check dimensional validity of the following equations
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a) A and B are dimesionally valid
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b) A and B are dimensionally invalid
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c) A is dimensionally valid nad B is dimensionally invalid
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d) A is dimensionally invalid, B is dimensionally valid
Explanation
First equation A) Dimensions of LHS = [Fs] = [ M1L1 T-2] [ M0L1 T1] = [M1 L2 T-2 ] Dimensions of RHS [mv02] = [mv2 ] = [M1 L0 T0] [ M0L1T-2]2 =M1L2T-2 Thus LHS = RHS A) dimensionally valid Second equation B) Dimensions of LHS = [F] = [M1L1T-2] Dimensions of RHS = Dimensions of LHS ≠ RHS B) is dimensionally invalid Answer : (c)
Q.24
Above equation is dimensionally valid equation. Obtain dimensional formula of a, b c. Here v is velocity v0 is initial velocity, t is time
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a) LT-2, L, T
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b) L1, T L2 T1
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c) L2T1, L1T2, L
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d) l, LT, T2
Explanation
Addition and subtraction can be carried out with physical quantities having same dimensions only there is term t + c , c will have same dimensions of t or [c] = [M0 L0 T1] Now terms b/(t+c) should have dimensions of velocity [M0L1 T-1 ] = [b] / [M0 L0 T1] On simplification we get [b] = M1L1T0 Dimension of at should havae dimension of v thus [at] = [v] [a] [ M0 L0 T1 ] = [ M0 L1 T-1] on simplification we get [a] = M0 L1 T-2 Answer : (a)
Q.25
An object is falling freely under the gravitational force. Its velocity after travelling a distance h is v. if speed depends upon gravitational acceleration g and distance. hnece v ∝ .... [ Guj. T.B. ]
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a) √(gh)
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b) gh
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c) g√h
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d) h√g
Explanation
Suppose velocity of object v ∝ ga hb v = kga hb let k be dimension less constant, equation is dimensionally valid [v] = [g]a [ h]b [M0L1T-1 ] = [M0L1T-2]b [ M0L1T0]b0L1T-1 ] = [M0La+b T-2b Comparing the base we get a+b = 1 and -2a = -1 Thus a = 1/2 and b = 1/2 Thus g ∝ √(gh) Answer : (a)
Q.26
A gas bubble formed due to an explosion under water oscillates with a period T proportional to Pa ρbEc. Where P is the static pressure, ρ is density of water and E is the total energy of the explosion. Find value of a , b and c ... ... [ Guj T.B]
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a) 1, 1, 1
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b) 1/4, 1/2, 1/3
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c) -5/6, 1/2, 1/3
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d) 1/3, 1/2, -5/6
Explanation
T ∝ k Pa ρbEc Let proportionality constant be dimension less [M0 L0T1] = [ M1 L-1 T-2]a [M1 L-3T0]b [M1L2 T-2]c [M0 L0T1] = Ma+b+c L-a-3b+2c T-2a-2c comparing indices of M a+b+c = 0 Comparing indices of L -a-3b+2c = 0 Comparing indices of T -2a-2c = 1 On simplification of above equation we get a= -5/6, b = 1/2 and c = 1/3 Answer : (c )
Q.27
If the velocity of light, acceleration due to gravity nad normal pressure are chosen as fundamental units, find the unit of mass, length and time. Given velocity of light c = 3×108, g = 10ms-2 and normal atmospheric pressure P = 105 N/m2
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a) 8.1 ×1035 kg, 9 ×1015 m, 3 ×107s
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b) 9.12 ×1015 kg, 3 ×107 m, 8 ×1035s
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c) 3.02 ×107 kg, 8.1 ×1035 m, 9 ×1015s
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d) 1035 kg, 9 ×1015 m, 3 ×107s
Explanation
Given [c] = L1T-1 = 3 × 108 ms-1 ... .... eq(1) [g] = L1T-2 = 10 ms-2 ... .... eq(2) [P] = M1 L-1T-2 = 105 Nm-2 ... ... eq(3) Dividing eq(1) by eq(2) Substituting value of T in eq(1) L1T-1 = 3 × 108 L = 3 ×108 ×3×107 = 9 ×1015 m Substituting value L and T in eq(3) M1L-1T-2 = 105 M1 = L1T2×105 M1 = 9×1015×(3×107)2 ×105=81×1035 kg Answer : (a)
Q.28
In M.K.S. system, if the value of gravitational constant is 6.67 × 10-11Nm2 kg-2 then what is its value in .... ... dyne cm2 g-2, i c.g.s system 1kg = 103 gm Answer : (b)
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a) 6.67 ×10-7
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b) 6.67 ×10-8
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c) 6.67 ×10-9
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d) 6.67 ×10-10
Explanation
Use following conversion to get answer 1N = 105 dyne 1 m = 102 cm
Q.29
A string stretched with tension T has frequency 'f' which depends on length l and mass per unit length m then f ∝ ... .... [ Guj T.B.]
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a)
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b)
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c)
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d)
Explanation
Let f = k Talbmc Here let k be dimension less constant thus Dimension of LHS = Dimensions of RHS M0L0T-1 = [M1L1 T-2]a [L1]b[M1L-1]c On simplification a = 1/2, b = -1, c = -1/2 Answer : (b)
Q.30
A stationary object is accelerated. the distance d travelled by it in time t depends upon the acceleration a and time t. From above description, obtain the relation between d,a and t with the help of dimensional analysis , k= 1/2 .... ... [ Guj T.B. ]
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a) d = at
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b) d = (1/2)at2
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c) d = a/2t
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d) d = a2t2
Explanation
let d =k ap tq dimension of LHS = RHS L1 = [L1T-2]p [M0L0T1]q On simplification we get p= 1 and q = 2 also given k =1/2 thus option b is correct Answer : (b)
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