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Quiz 4
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Q.1
The least count of a stop watch is 1/5 second. The time of oscillations of a pendulum is measured to be 25 second. Minimum % error in the measurement of time will be ..
0%
a) 0.1%
0%
b) 0.8%
0%
c) 1.8%
0%
d) 8%
Explanation
Given ΔT = 0.2 sec relative error in time = (ΔT / T ) % error = relative error ×100 ∴ (0.2 / 25)×100 = 0.8% Answer:(b)
Q.2
The dimensions of e.m.f in MKSQ system are..
0%
a) MT-2Q-2
0%
b) ML2T-2Q-2
0%
c) ML2T-2Q-1
0%
d) ML2T-2Q2
Explanation
emf = work/charge = [ML2T-2] / [Q] Answer: (c)
Q.3
Volume of one drop of mercury is 1.76cc volume of 25 such drop will be
0%
a) 44cc
0%
b) 0.44cc
0%
c) 4.4cc
0%
d) 44.00cc
Explanation
Answer: (a)
Q.4
The dimension of R in the expression Q=Qo(1-e-t/RC) is given by
0%
a) [ML2T-3A-2]
0%
b) [ML2T-2A-3]
0%
c) [T]
0%
d) [M-1L-2T-3A-1]
Explanation
t/RC must be dimension less t/RC = [M0L0T0A0] ∴ Dimension of t = Dimension of RC since C = Q/V Dimensional formula of C = [ AT]/ ML2T-3A-1 ] C= A2T4M-1L-2 Now Dimension of R = dimension of t / dimension of C [R] = [T] / A2T4M-1L-2 [R] = ML2 T-3A-2 Answer: (a)
Q.5
If K represents kinetic energy, V velocity and T time, and these are chosen as the fundamental unit then, the units of surface tension will be..
0%
a) KV-2T-2
0%
b) KV-1T-2
0%
c) K2V-1T-3
0%
d) KV-2T-1
Explanation
Surface tension S = Energy / Area Now V×T have the dimension of length L (VT)2 have the dimension of Area L2 Thus Dimension of Surface tension S = K/ V2T2 Dimension of S = KV-2T-2 Answer:(a)
Q.6
The velocity of water wave is found to be proportional to λaρbgc, where λ is the wave length, ρ is density of water and g is gravitational acceleration. Then the values of a,b,c are
0%
a) 1,0, 1
0%
b) 1/2, 1, 0
0%
c) 1/2, 0, 1/2
0%
d) 1/2, 0, 1
Explanation
From given V = kλaρbgc by substituting dimensions in above formula we get [LT-1] = [L]a [ML-3]b[LT-2]c LT-1 = LaMbL-3bLCT-2c LT-1= MbLa-3b+cT-2c from above b= 0 a-3b+c=1 -2c=-1 thus c= 1/2, and a = 1/2 Answer: (c)
Q.7
E, m, L and G denote energy, mass, angular momentum and gravitational constant respectively. Then the dimensions of El2/m5G2 are
0%
a) angle
0%
b) length
0%
c) mass
0%
d) time
Explanation
Dimensions of energy E = [ ML2 Dimension of angular monetum L = [ ML2T-1] Dimension of mass = [M] Dimension of gravitational constant G = [M-1 L3T-2] Substituting dimension in given formula we get From given option only angle is dimensional hence option a Answer: (a)
Q.8
When 2.0734 is added to 35.2 the sum is
0%
a) 37.2734
0%
b) 37.273
0%
c) 37.3
0%
d) 37.27
Explanation
Since the least number after decimal is 1, the final result will have same. On wrounding 37.2734 to one place after decimal it becomes 37.3 Answer: (c)
Q.9
A body travels uniformly a distance of (12.6±0.2)m in a time (3.0±0.3) s. the velocity of body within error limits is
0%
a) (4.2±0.5) m/s
0%
b) (4.2±0.3) m/s
0%
c) (4.2±0.4)m/s
0%
d) (4.2±0.6)m/s
Explanation
V= s/t V = 12.6/3.0 = 4.2 m/s note ΔV value is rounded to one place of decimal V = (4.2 ± 0.5) m/s Answer:(a)
Q.10
The number of particles crossing per unit area perpendicular to X-axis in unit time is where n1 and n2 are number of particles per unit volume for the value of x1 and x2 respectively. The dimensions of diffusion constant D are
0%
a) M0LT2
0%
b) M0L2T-4
0%
c) M0LT-3
0%
d) M0L2T-1
Explanation
as N = number of particles/ Area×times Dimension of N = [ L-2T-1] n1 and n2 are number of particles per unit volume Dimensions of n = [ L-3] x1 and x2 are positions Dimension of x = [L] By substituting dimensions in given equation we get [ L-2T-1] = -D {[ L-3] / [L] On simplification we get Dimension of D = M0L2T-1 Answer: (d)
Q.11
When the number 6.03587 is rounded up to the second place of decimals, it becomes
0%
a) 6.035
0%
b) 6.04
0%
c) 6.03
0%
d) none of these
Explanation
Answer:(b)
Q.12
if velocity(V), acceleration (A) and Force(F) are taken as fundamental quantities instead of mass(M) Length(L) and Time(T) the dimensions of Young'modulus of elasticity would be ( With MLT young modulus = [ML-1T-2]
0%
a) FA2V-2
0%
b) FA2V-3
0%
c) FA2V-4
0%
d) FA2V-5
Explanation
Let Y = FaAbVc ∴ [ML-1T-2] = [MLT -2]a [LT-2]b[LT-1]c [ML-1T-2] = [MaLaT -2a] [LbT-2b][LT-1c] From above a=1 a+b+c = -1 -2a -2b -c = -2 -a-b = -3 b = 2 and c=-4 Thus dimensional formula for Y in F,A,V is FA2V-4 Answer: (c)
Q.13
Suppose the acceleration due to gravity at a place is 10m/sIts value in cm/(minute)2 is ..
0%
a) 36×105
0%
b)3.6×105
0%
c) 6×104
0%
d) 36×104
Explanation
Answer: (a)
Q.14
the volume of sphere is 1.76cmthe volume of 25 such spheres taking into accountthe significant figure is
0%
a) 0.44×102cm3
0%
b) 44.0 cm3
0%
c) 44cm3
0%
d) 44.00cm3
Explanation
Total volume = 1.76×25 = 44.0cm3 since three significant numbers are in the volume one sphere Answer: (b)
Q.15
The dimensional formula for permeability 'µ' is given by..
0%
a) MLT-2A-2
0%
b) M0L-1T
0%
c) M0L2T-1A2
0%
d) None of the above
Explanation
Answer:(a)
Q.16
The dimensions of self inductance are [IIT 1983]
0%
a) MLT-2
0%
b) ML2T-1A-2
0%
c) ML2T-2A-2
0%
d) ML2T-2A-1
Explanation
Answer: (c)
Q.17
A resistance of 6Ω with tolerance of 10% and another of 4Ω with tolerance of 10% are connected in series. The tolerance of combination is about
0%
a) 5%
0%
b) 10%
0%
c) 12%
0%
d) 15%
Explanation
We know that Rs = R1 + R2 Rs = 6+4 = 10Ω Here ΔR1 = 0.6 and R2 = 0.4 on substituting and solving the equation we get Answer: (b)
Q.18
In Q105 question , the tolerance when resistance connected in parallel is
0%
a) 10%
0%
b) 20%
0%
c) 30%
0%
d) 40%
Explanation
For parallel connection resistance is Error In Rp Answer: (c)
Q.19
In vernier callipers N divisions of vernier scale coincide with (N-1) divisions of main scale in which length of division is 1 mm. The least count of the instrument in cm is
0%
a) N
0%
b) N-1
0%
c) 1/10N
0%
d) (1/N) -1
Explanation
Least count = (1/N) × 1M.S. M.S. = 1 mm = 1/10 cm Least count = 1/10N Answer:(c)
Q.20
The dimensions of a rectangular block measured with callipers having least count of 0.01cm are 5mm×10mm×5mm. The maximum percentage error in the measurement of the volume of the block is [ AMU 1995]
0%
a) 20%
0%
b) 15%
0%
c) 10%
0%
d) 5%
Explanation
Least count = error = 0.01cm = 0.1mm Volume = l×b×h Answer: (d)
Q.21
The equation of a wave is given by where ω is an angular velocity and V is the linear velocity. The dimension of k is : (MPPMT1993)
0%
a) LT
0%
b) T
0%
c) T-1
0%
d) T2
Explanation
Dimension of x/v = k K = L/ LT-1 k = T Answer: (b)
Q.22
One 'lux' is equal to [CPMT 1993]
0%
a) lumen/m2
0%
b) lumen /cm2
0%
c) candela/m2
0%
d) candela/cm2
Explanation
Answer: (a)
Q.23
Erg m-1 can be the unit of of measurement of [DPET 1993]
0%
a) Force
0%
b) Momentum
0%
c) Power
0%
d) Acceleration
Explanation
Erg is unit of energy dimensional formula = ML2T-2 Thus Erg / m = MLT-2 Is a dimension of force Answer:(b)
Q.24
The dimensional formula for Entropy is
0%
a) MLT-2θ-1
0%
b) ML2
0%
c) ML2T-2θ-1
0%
d) ML-2T-2θ-1
Explanation
Entropy = enrgy/ temperature Dimension of entropy = ML2T-2 / θ Dimension of entropy = ML2T-2θ-1 Answer: (c)
Q.25
The physical quantity that has no dimensions .. [ EAMCET 1995]
0%
a) Angular velocity
0%
b) Linear momentum
0%
c) Angular Momentum
0%
d) Strain
Explanation
Starain = Δl / l thus no dimensions Answer: (d)
Q.26
Electron volt is the unit of .. [MPPMT 1993]
0%
a) Charge
0%
b) Potential difference
0%
c) Momentum
0%
d) Energy
Explanation
Answer: (d)
Q.27
SI unit of gas constant are .. [ MPPMT 1987]
0%
a) Watt K-1mol-1
0%
b) Newton K-1mol-1
0%
c) Joule K-1mol-1
0%
d) Erg K-1mol-1
Explanation
Answer:(c)
Q.28
Universal time is based on [SCRA 1989]
0%
a) Rotation of earth on its axis
0%
b) Earth's orbital motion around the earth
0%
c) Vibration of caesium atom
0%
d) Oscillation of quartz crystal
Explanation
Answer: (c)
Q.29
Which of the following is most accurate measurement?
0%
a) 3×10-2 cm
0%
b) 0.030cm
0%
c) 300×10-4 cm
0%
d) 30×10-3
Explanation
Error in (a) is 1 in 3 Error in (b) is 1 in 30 error in(c) is 1 in 300 Error in (d) is 1 in 30 Answer: (c)
Q.30
Which of the following is not expressed in proper units [MNR 1995]
0%
a) Stress / Strain = N/m2
0%
b) surface tension = N/m
0%
c) Energy = Kg×m /sec
0%
d) Pressure = N/m2
Explanation
Energy = mass×acceleration ×displacement Dimension formula of energy is ML2T-2 Answer: (c)
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