Units & Measurements

NEET Physics · 152 questions · Page 4 of 16 · Click an option or "Show Solution" to reveal answer

Q31
The diameter of a spherical bob, when measured with vernier callipers yielded the following values : 3.33mathrm cm,3.32mathrm cm,3.34mathrm cm,3.33mathrm cm3.33 \\mathrm{~cm}, 3.32 \\mathrm{~cm}, 3.34 \\mathrm{~cm}, 3.33 \\mathrm{~cm} and 3.32mathrm cm3.32 \\mathrm{~cm}. The mean diameter to appropriate significant figures is :
A 3.328 cm3.328 \mathrm{~cm}
B 3.3 cm3.3 \mathrm{~cm}
C 3.33 cm3.33 \mathrm{~cm}
D 3.32 cm3.32 \mathrm{~cm}
Correct Answer
Option C
Solution
\\text{ } \\begin{aligned} \\text{ Mean diameter } & =\\frac{\\mathrm{d}_1+\\mathrm{d}_2+\\mathrm{d}_3+\\mathrm{d}_4+\\mathrm{d}_5}{5} \\ & =\\frac{3.33+3.32+3.34+3.33+3.32}{5} \\ & =3.328 \\approx 3.33 \\end{aligned}
Q32
The mechanical quantity, which has dimensions of reciprocal of mass (mathrmM1)(\\mathrm{M}^{-1}) is :
A angular momentum
B coefficient of thermal conductivity
C torque
D gravitational constant
Correct Answer
Option D
Solution

Angular momentum

=left[mathrmML2mathrm T1right]=\\left[\\mathrm{ML}^2 \\mathrm{~T}^{-1}\\right]

Coeff of thermal conductivity

=left[mathrmMLT3mathrm K1right]=\\left[\\mathrm{MLT}^{-3} \\mathrm{~K}^{-1}\\right]

Torque

=left[mathrmML2mathrm T2right]=\\left[\\mathrm{ML}^2 \\mathrm{~T}^{-2}\\right]

Gravitational constant

=left[mathrmM1mathrm L3mathrm T2right]=\\left[\\mathrm{M}^{-1} \\mathrm{~L}^3 \\mathrm{~T}^{-2}\\right]

So, gravitational constant has power of

(1)(-1)

of

mathrmM\\mathrm{M}

.

Q33
The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are :
A Personal errors
B Least count errors
C Random errors
D Instrumental errors
Correct Answer
Option C
Solution

Error arise due to unpredictable fluctuation in temperature and voltage supply are rightarrow\\rightarrow random errors.

Q34
A metal wire has mass (0.4pm0.002) mathrmg(0.4 \\pm 0.002) ~\\mathrm{g}, radius (0.3pm0.001) mathrmmm(0.3 \\pm 0.001) ~\\mathrm{mm} and length (5pm0.02) mathrmcm(5 \\pm 0.02) ~\\mathrm{cm}. The maximum possible percentage error in the measurement of density will nearly be :
A 1.3%
B 1.6%
C 1.4%
D 1.2%
Correct Answer
Option B
Solution
rho=fracMV\\rho=\\frac{M}{V}
rho=fracMpir2ell\\rho=\\frac{M}{\\pi r^{2} \\ell}
fracDeltarhorho=fracDeltamathrmMmathrmM+frac2Deltamathrmrmathrmr+fracDeltaellell\\frac{\\Delta \\rho}{\\rho}=\\frac{\\Delta \\mathrm{M}}{\\mathrm{M}}+\\frac{2 \\Delta \\mathrm{r}}{\\mathrm{r}}+\\frac{\\Delta \\ell}{\\ell}
fracDeltarhorho\\frac{\\Delta \\rho}{\\rho} \\%=\\left[\\frac{0.002}{0.4}+\\frac{2(0.001)}{(0.3)}+\\frac{0.02}{5}\\right] \\times 100 \\%
=frac12=\\frac{1}{2} \\%+\\frac{2}{3} \\%+\\frac{2}{5} \\%
=1.6=1.6 \\%
Q35
The physical quantity that has the same dimensional formula as pressure is
A Coefficient of viscosity
B Force
C Momentum
D Young's modulus of elasticity
Correct Answer
Option D
Solution

Pressure

[P]=left[FoverAright]=left[MLT2overL2right]=[ML1T2][P] = \\left[ {{F \\over A}} \\right] = \\left[ {{{ML{T^{ - 2}}} \\over {{L^2}}}} \\right] = [M{L^{ - 1}}{T^{ - 2}}]

And, dimensions of Young's modulus

[Y]=left[StressoverStrainright]=[ML1T2][Y] = \\left[ {{{Stress} \\over {Strain}}} \\right] = [M{L^{ - 1}}{T^{ - 2}}]

[Force]

=[MLT2]= [ML{T^{ - 2}}]

[Momentum]

=[MLT1]= [ML{T^{ - 1}}]

[Coefficient of viscosity]

=[ML1T1]= [M{L^{ - 1}}{T^{ - 1}}]
Q36
The percentage error in the measurement of g is : (Given that g=4pi2LoverT2g = {{4{\\pi ^2}L} \\over {{T^2}}}, L=(10,pm,0.1)L = (10\\, \\pm \\,0.1) cm, T=(100,pm,1)T = (100\\, \\pm \\,1) s)
A 7%
B 2%
C 5%
D 3%
Correct Answer
Option D
Solution

Given

g=4p2LoverT2g = {{4{p^2}L} \\over {{T^2}}}

Fractional error in value of g

Deltagoverg=DeltaLoverL+2DeltaToverT{{\\Delta g} \\over g} = {{\\Delta L} \\over L} + 2{{\\Delta T} \\over T}
RightarrowDeltagovergtimes100=DeltaLoverLtimes100+2DeltaToverTtimes100\\Rightarrow {{\\Delta g} \\over g} \\times 100 = {{\\Delta L} \\over L} \\times 100 + 2{{\\Delta T} \\over T} \\times 100
RightarrowDeltagovergtimes100=0.1over10times100+2times1over100times100=3\\Rightarrow {{\\Delta g} \\over g} \\times 100 = {{0.1} \\over {10}} \\times 100 + 2 \\times {1 \\over {100}} \\times 100 = 3\\%

Hence, percentage error in measurement of g is 3%

Q37
The dimensions [MLT -2 A -2 ] belong to the
A Magnetic flux
B Self inductance
C Magnetic permeability
D Electric permittivity
Correct Answer
Option C
Solution

Dimensional formula of magnetic permeability is [MLT -2 A -2 ]

Q38
Plane angle and solid angle have
A Units but no dimensions
B Dimensions but no units
C No units and no dimensions
D Both units and dimensions
Correct Answer
Option A
Solution

Plane angle =

{{Arc} \\over {Radius}} = {{[L]} \\over {[L]}}\\overset{{} \}\\longrightarrow

Unit = Radian

=[M0L0T0]= [{M^0}{L^0}{T^0}]

Solid angle =

{{Area} \\over {{{(Radius)}^2}}}\\overset{{} \}\\longrightarrow

Unit = Steradian

=L2overL2=[M0L0T0]= {{{L^2}} \\over {{L^2}}} = [{M^0}{L^0}{T^0}]

therefore\\therefore Both have units but no dimensions.

Q39
The area of a rectangular field (in m 2 ) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is
A 138 ×\times 10 1
B 1382
C 1382.5
D 14 ×\times 10 2
Correct Answer
Option D
Solution

Area = Length times\\times Breadth = 55.3 times\\times 25 m 2 = 1382.5 m 2 = 14 times\\times 10 2 m 2 (Rounding off of two significant figures)

Q40
The E and G respectively denote energy and gravitational constant, then EoverG{E \\over G} has the dimensions of :
A [M 2 ] [L -2 ] [T -1 ]
B [M 2 ] [L -1 ] [T 0 ]
C [M] [L -1 ] [T -1 ]
D [M] [L 0 ] [T 0 ]
Correct Answer
Option B
Solution

E = energy = [ML 2 T -2 ] G = Gravitational constant = [M -1 L 3 T -2 ] So,

EoverG=[E]over[G]=ML2T2overM1L3T2=[M2L1T0]{E \\over G} = {{[E]} \\over {[G]}} = {{M{L^2}{T^{ - 2}}} \\over {{M^{ - 1}}{L^3}{T^{ - 2}}}} = [{M^2}{L^{ - 1}}{T^0}]
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