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Units, Dimension And Measurement Mcq
Quiz 9
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Q.1
If the frequency (F), velocity (V) and density (D) are taken as fundamental quantities then the dimensional formula for momentum in this new system will be ....
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a) D-1V-4F3
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b) D1V4F-3
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c) D1V1F-2
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d) D2V2F2
Explanation
F = T-1 V = L1T-1 ∴ L = VF-1 D = M1L-3 ∴ M = D1 L3 Momentum = M1L1T-1 Momentum = [D1 (VF-1)3][ VF-1] [F] Momentum = D1V4F-3 Answer : (b)
Q.2
In a new system of units if the units of distance, a mass and time are taken as 10cm, 100g and 0.1s respectively then the new unit of force in this system of unit will be ... newton
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a) 10
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b) 1.0
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c) 0.1
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d) 102
Explanation
Dimensions of Force = M1L1T-2 (100 g)1(10cm)1 (0.1 s)-2 = 1o5 dyne = 1N Answer : (b)
Q.3
Solid angle made by hemisphere at its centre is ... ...
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a) 2π sr
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b) π sr
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c) 4π sr
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d) 1 sr
Explanation
Solid angle Ω = Area/r2 = 2π sr Answer : (a)
Q.4
103 g/cm3 = ... .. kg/m3
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a) 10-6
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b) 101
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c) 103
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d) 106
Explanation
Answer : (d)
Q.5
Dimensional formula of Stefan-Boltzmann's constant is ... . Take dimension of temperature as J
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a) M1L2T-1K4
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b) M1L0T-3K-4
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c) M1L2T-2K4
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d) M1L2T-3K4
Explanation
Stefan-Boltzmann constant = Power/ Area ×(temperature)4 Answer : (b)
Q.6
Error in measurement of length of a cube is 30% then error in measurement of area of one face is ...
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a) 12%
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b) 6%
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c) 9%
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d) 3%
Explanation
Answer : (b)
Q.7
Density of liquid in CGS unit system is 0.625 g/cm3, find its value in SI unit system
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a) 0.625
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b) 0.0625
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c) 0.00625
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d) 625
Explanation
Answer : (d)
Q.8
Which of the following measurement is more precise
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a) 5.00 mm
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b) 5.00 cm
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c) 5.00 m
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d) 5.00 km
Explanation
Answer:(a)
Q.9
If the units of length and force are increased four times, then the unit of energy will
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a) increase 8 times
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b) increase 16 times
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c) decreases 16 times
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d) increase 4 times
Explanation
Answer:(a)
Q.10
The main scale of an instrument has 20 divisions in 1 cm and 24 divisions of main scale coincides with 25 divisions of vernier scale. The least count of the instrument is ..
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a) 0.001 cm
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b) 0.02 cm
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c) 0.002 cm
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d) 0.01 cm
Explanation
Least count of main scale = 1cm/20 = 0.05 cm = 0.5 mm 24 division main scale = 25 division of vernier scale 24×0.5 = 25S S = 0.48 Least count of vernier = least count of main – S Least count = 0.5mm -0.48mm = 0.02 mm = 0.002cm Answer:(c)
Q.11
While measuring the length of a cylinder by vernier callipers, the MSR is found to be 4.6 cm and 6th vernier division on the main scale. If the least count of the vernier is 0.01 cm and zero error is –0.02 cm then the length of the cylinder is
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a) 4.56 cm
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b) 4.68 cm
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c) 4.64 cm
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d) 4.66 cm
Explanation
Answer:(b)
Q.12
If force (F), velocity(V) and time (T) are takenas fundamental units, then the dimensions of mass are
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a) [F V–1 T–1]
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b) [F V–1 T]
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c) [F V T–1]
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d) [F V T–2]
Explanation
F = ma M = [FV-1T] Answer:(b)
Q.13
If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensionalformula of surface tension will be :
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a) [E-2 V-1 T-3]
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b) [E V-2 T-1]
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c) [E V-1 T-2]
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d) [E V-2 T-2]
Explanation
Surface tension = [MT-2] [MT-2]= k [E]a [V]c [T]d [MT-2]= k [ML2T-2]a [LT-1]c [T]d [MT-2]=k [Ma L2a-c T-2a-c+d] a=1 2a-c = 0 ∴ c= -2 -2 = -2a-c+d ∴ d= -2 Surface tension = [EV-2T-2] Answer:(b)
Q.14
In dimension of critical velocity νc, of liquid flowing through a tube are expressed as (ηxρyrz), whereη, ρ and r are the coefficient of viscosity of liquid,density of liquid and radius of the tube respectively,then the values of x, y and z are given by :
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a) 1, 1, 1
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b) 1, -1, -1
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c) -1, -1, 1
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d) -1, -1, -1
Explanation
νc ∝ [ηx ρy rz ] [LT-1 ] ∝[(M1 L-1 T-1 )x (M1 L-3 )y (L1 )z ] [LT-1 ] ∝[M(x+y) L(-x-3y+z) T(-x) ] ∴ x= 1, y =-1, -x-3y+z=1 -1+3+y=1 ∴ y=1 Answer:(b)
Q.15
Planck's constant (h), speed of light in vacuum (c)and Newton's gravitational constant (G) are threefundamental constants. Which of the followingcombinations of these has the dimension of length?
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a)
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b)
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c)
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d)
Explanation
Solution L=ha cb Gc L=(M1 L2 T-1 )a (L1 T-1 )b (M-1 L3 T-2)c L=M(a-c) L(2a+b+3c) T(-a-b-2c) a-c=0 -a-b-2c = 0 2a+b+3c =1 On solving above equation a =1/2 , c= 1/2, b= -3/2 Answer:(c)
Q.16
A physical quantity of the dimensions of length thatcan be formed out of c, G and is [c is velocityof light, G is universal constant of gravitation and e is charge] …[ NEET 2017]
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a)
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b)
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c)
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d)
Explanation
Dimension of c = LT-1 Dimension of G = M-1L3T-2 Dimensions of Using formula Now -b+c = 0 a+3b+3c = 1 -a-2b-2c= 0 On solving equations we get a = -2 b = 1/2, c = 1/2 Answer:(a)
Q.17
The energy of a system as a function of time t is given as E(t) = A2 exp(-αt), where α = 0.2 s-The measurement of A has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t) at t = 5 s is … [ IIT Advance 2015]
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a) 2
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b) 4
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c) 6
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d) 8
Explanation
ΔE =2AΔA exp(-αt)+ A2 exp(-αt) (-Δt) Answer:(b)
Q.18
Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = Time and A = electric current, then …. [ IIT Mains -2016]
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a) [ε0] = [M−1 L−3 T2 A]
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b) [ε0] = [M−1 L−3 T4 A2]
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c) [ε0] = [M−1 L2 T−1 A−2]
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d) [ε0] = [M−1 L2 T−1 A]
Explanation
Answer:(b)
Q.19
The diameter of a cylinder is measured using a Vernier caliper with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is
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a) 5.112 cm
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b) 5.124 cm
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c) 5.136 cm
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d) 5.148 cm
Explanation
Main scale reading = 5.1 cm Main scale least count = 0.050 cm Vernier scale least count = 2.45/50=0.049 Least count . = 0.050 − 0.049 = = 0.001 cm V.S.R. = 24 ∴ d = (5.1) cm + 24 × 0.001 = 5.124 cm Answer:(b)
Q.20
)Using the expression 2d sin θ = λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90º. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0°.
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a) the absolute error in d remains constant.
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b) the absolute error in d increases.
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c) the fractional error in d remains constant.
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d) the fractional error in d decreases.
Explanation
2d sinθ = λ 2(Δdsinθ+dcosθΔθ)= Δλ As Δλ =0 Δdsinθ+dcosθ Δθ=0 Given Δθ = constant Thus fractional error decreases with increase in θ Option “d” correct Answer:(d)
Q.21
)Match List I with List II and select the correct answer using the codes given below the lists:
List I
List II
P. Boltzmann constant
[ML
2
T
−1
]
Q. Coefficient of viscosity
[ML
−1
T
−1
]
R. Planck constant
[MLT
−3
T
−1
]
S. Thermal conductivity
[ML
2
T
−2
K
−1
]
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b) P → 3 ; Q → 2 ; R → 1 ; S → 4
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a) P → 3 ; Q → 1 ; R → 2 ; S → 4
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c) P → 4 ; Q → 2 ; R → 1 ; S → 3
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d) P → 4 ; Q → 1 ; R → 2 ; S → 3
Explanation
Energy of gas molecule P → 4 Q. Coefficient of viscosity By Strokes law F = 6πrη v Q → 2 R. Planck constant : E = hν R → 1 S. Thermal conductivity K=[M1 L 1 T-3 K-1 ] S → 3 (P) → 4; (Q) → (2); (R) → (1); (S) → (3) Answer:(c)
Q.22
During Searle’s experiment, zero of the Vernier scale lies between 3.20 × 10−2 m and3.25 × 10-2 m of the main scale. The 20th division of the Vernier scale exactly coincideswith one of the main scale divisions. When an additional load of 2 kg is applied to thewire, the zero of the Vernier scale still lies between 3.20 × 10−2m and 3.25 × 10−2 m of themain scale but now the 45th division of Vernier scale coincides with one of the main scaledivisions. The length of the thin metallic wire is 2 m and its cross−sectional area is 8 × 10−7 mThe least count of the Vernier scale is 1.0 × 10−5m. The maximum percentageerror in the Young’s modulus of the wire is ..[ IIT Advance 2014]
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a) 2
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b) 6
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c) 5
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d) 8
Explanation
Main scale division = 0.05×10-2 m= 0.5 mm Given :The least count of the Vernier scale is 1.0 × 10−5m First Reading Li = (3.2 × 10−2 + 20 × 1.0 × 10−5) m = (3.2 + 0.02) × 10−2 = 3.22 × 10−2 m Second reading Lf = (3.2 + 45 × 10−3) × 10−2 m = 3.245 × 10−2 m change in length x = Lf − Li = 0.025 × 10−2 m List count of vernier is the error in measurement We have taken reading for two time to find change in length Thus error Δx= 2×1.0 ˜ 10−5m =2×10−5m Error in length Young’s modulus is given as We are assuming l , F, A to be known with proper accuracy. Answer:(d)
Q.23
To find the distance d over which a signal can be seen clearly in foggyconditions, arailways engineer uses dimensional analysis and assumes that the distance depends on themass density ρ of the fog, intensity (power/area) S of the light from the signal and itsfrequency f. The engineer finds that d is proportional to S1/n. The value of n is … [ IIT Advance 2014]
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a) 1
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b) 2
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c) 3
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d) 4
Explanation
d ∝ ρa sb fc a+b =0 -3a = 1 3b+c = 0 b = 1/3 S1/n = S1/3 Thus n = 3 Answer:(c)
Q.24
Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equaldivisions and a screw gauge with 100 divisions on its circular scale. In the Verniercallipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale andin the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then …[ IIT Advance 2015]
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a) If the pitch of the screw gauge is twice the least count of the Vernier callipers, theleast count of the screw gauge is 0.01 mm
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b) If the pitch of the screw gauge is twice the least count of the Vernier callipers, theleast count of the screw gauge is 0.005 mm.
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c) If the least count of the linear scale of the screw gauge is twice the least count of theVernier callipers, the least count of the screw gauge is 0.01 mm.
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d) If the least count of the linear scale of the screw gauge is twice the least count of theVernier callipers, the least count of the screw gauge is 0.005 mm.
Explanation
Smallest division of Main scale = 1/8 = 0.125 cm 5 divisions of the vernier scale = 4 divisions of the main scale = 4 × 0.125 = 0.5 cm 1 divisions of 1 vernier = 0.5/5 = 0.1cm Least count of vernier = 1 main scale division - 1 vernier scale division = 0.125 - 0.1 = 0.025 cm Least count of screw gauge =pitch /100 If pitch of the screw gauge is twice the least count of vernier calipers then the least countof screw gauge =0.05/100=0.0005cm=0.005mm Hence (b) option is correct Also, least count of linear scale of screw gauge = 2 × 0.025 cm = 0.05 cm ⇒ pitch= 2 × 0.05 cm = 0.1 cm = 1 mm ∴ Least Count = 1mm/100=0.01mm Hence (c) is correct. Answer:(b,c)
Q.25
than one correct answer Q269) Planck's constant h, speed of light c and gravitational constant G are used to form a unit oflength L and a unit of mass M. Then the correct option(s) is(are) ... [IIT Advance 2016]
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a) M ∝ √ c
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b) M ∝ √G
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c) L ∝ √h
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d) L ∝ √G
Explanation
Let L ∝ hxcyGz ∴ L ∝ (ML2T-1)x (LT-1)y (M-1L3T-2)z L ∝ Mx-z L2x+y+3z T-x-y-2z - x - z = 0 …………… (1) 2x + y + 3z = 1 …………… (2) x + y + 1z = 0 …………… (3) On solving x =1/2 ,y= -3/2,z=1/2 ∴ Option (C) and (D) are correct. Similarly Let M ∝ hxcyGz ∴ M ∝ Mx-z L2x+y+3z T-x-y-2z ∴ x - z = 1 …………… (1) 2x + y + 3z …………… (2) x + y + 2z = 0 …………… (3) Equation (2) - Equation (3) gives x + z = 0 ∴ 2x = 1 x=1/2 ,z= -1/2 ,y= 1/2 ∴ M ∝ √h , m ∝ √c Option (A) is correct. Answer: a, c, d Answer:(a, c, d)
Q.26
In terms of potential difference V, electric current I, permittivity &epslon0 , permeability µ0 and speed of light c, the dimensionally correct equation(s) is(are) [ IIT Advance 2015]
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a) µ0 I2 = ε0 V2
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b) ε0I = µ0V
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c) I = ε0 cV
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d) µ0cI = ε0 V
Explanation
Answer:(a, c)
Q.27
A length-scale (l) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for is(are) dimensionally correct? [ IIT Advance 2016]
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a)
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b)
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c)
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d)
Explanation
[ε]=[M-1 L-3 A2 T4 ] [kB]=[ML2 T-2 θ-1] [T]=[θ] and [n]=[L-3] [q]=[AT] Answer:(b, d)
Q.28
There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2, respectively, are .. [IIT Advance 2016]
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a) 2.87 and 2.86
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b) 2.85 and 2.82
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c) 2.87 and 2.87
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d) 2.87 and 2.83
Explanation
Least count of main C1 = 0.1cm 10 of sliding scale = 9×0.1 1 of sliding scale = 0.9/10 = 0.09 cm Least count of vernier =Least count of main scale – least count of sliding scale Least count of vernier =0.1 -0.09 = 0.01 Reading of C1 = Main scale reading + sliding scale × least count C1 = 2.8+ 7×0.01 = 2.8 +0.07 = 2.87 cm Second C2 Least count of main scale = 1/10 =0.1cm 10 of sliding scale = 11 × 0.1 1 of sliding scale = 1.1/10 =0.11 cm Least count of main scale – least count of sliding scale =0.1 -0.11 = -0.01 ( since least count is negative add least count of main scale reading) Reading of C2 = Main scale reading +0.1+ sliding scale × least count C2 = 2.8+ 7×(-0.01) = 2.8+0.1 -0.07 = 2.83 cm C1 = 2.87 cm C2 = 2.83 Answer:(d)
Q.29
than one correct answer Q274) In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is = 2π . The values of R and r are measured to be (60 ± 1) mm and (10 ± 1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true? [ IIT Advance 2016]
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a) The error in the measurement of r is 10%
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b) The error in the measurement of T is 3.57%
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c) The error in the measurement of T is 2%
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d) The error in the determined value of g is 11%
Explanation
R = 10±1 mm thus error in r = 10% (option a correct) Sum of observation of T = 2.78 Average reading = 2.19 /5=0.556 =0.56 as least count is 0.01 s Absolute error = (0.04 +0+0.01+0.03)/5= 0.016 s = 0.02 s Relative error = 0.02/0.56 =3.57% Error in g Taking derivative of g Option d correct Answer:(a, d)
Q.30
A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is δT = 0.01 second and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 ms–2 and the velocity of sound is 300 ms–Then the fractional error in the measurement, &deltaL/L, is closest to
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a) 0.2 %
0%
b) 5 %
0%
c) 3 %
0%
d) 1 %
Explanation
Total time taken Time taken by stone to reach water Time taken by sound to travel distance t2 =L/Vs Answer:(d)
0 h : 0 m : 1 s
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