Units & Measurements

NEET Physics · 152 questions · Page 12 of 16 · Click an option or "Show Solution" to reveal answer

Q111
A force defined by F=αt2+βtF=\alpha t^2+\beta t acts on a particle at a given time tt. The factor which is dimensionless, if α\alpha and β\beta are constants, is:
A βt/α\beta t / \alpha
B αt/β\alpha t / \beta
C αβt\alpha \beta t
D αβ/t\alpha \beta / t
Correct Answer
Option B
Solution

The force acting on a particle at a given time t t is defined as F=αt2+βt F = \alpha t^2 + \beta t , where α \alpha and β \beta are constants.

We need to determine which factor is dimensionless if α \alpha and β \beta are constants.

Using the principle of homogeneity of dimensions: [F]=[αt2]=[βt][α]=[F][t2]and[β]=[F][t][α][t]=[β]αtβ=dimensionless \begin{aligned} & [F] = [\alpha t^2] = [\beta t] \\ & [\alpha] = \dfrac{[F]}{[t^2]} \quad \text{and} \quad [\beta] = \dfrac{[F]}{[t]} \\ & \therefore \quad [\alpha][t] = [\beta] \\ & \therefore \quad \dfrac{\alpha t}{\beta} = \text{dimensionless} \end{aligned}

Q112
The diameter of a spherical bob, when measured with vernier callipers yielded the following values : 3.33 cm,3.32 cm,3.34 cm,3.33 cm3.33 \mathrm{~cm}, 3.32 \mathrm{~cm}, 3.34 \mathrm{~cm}, 3.33 \mathrm{~cm} and 3.32 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
  Mean diameter =d1+d2+d3+d4+d55=3.33+3.32+3.34+3.33+3.325=3.3283.33\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}
Q113
The mechanical quantity, which has dimensions of reciprocal of mass (M1)(\mathrm{M}^{-1}) is :
A angular momentum
B coefficient of thermal conductivity
C torque
D gravitational constant
Correct Answer
Option D
Solution

Angular momentum

=[ML2 T1]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]

Coeff of thermal conductivity

=[MLT3 K1]=\left[\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\right]

Torque

=[ML2 T2]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]

Gravitational constant

=[M1 L3 T2]=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]

So, gravitational constant has power of

(1)(-1)

of

M\mathrm{M}

.

Q114
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 random errors.

Q115
A metal wire has mass (0.4±0.002) g(0.4 \pm 0.002) ~\mathrm{g}, radius (0.3±0.001) mm(0.3 \pm 0.001) ~\mathrm{mm} and length (5±0.02) cm(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
ρ=MV\rho=\frac{M}{V}
ρ=Mπr2\rho=\frac{M}{\pi r^{2} \ell}
Δρρ=ΔMM+2Δrr+Δ\frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{M}}{\mathrm{M}}+\frac{2 \Delta \mathrm{r}}{\mathrm{r}}+\frac{\Delta \ell}{\ell}
Δρρ%=[0.0020.4+2(0.001)(0.3)+0.025]×100%\frac{\Delta \rho}{\rho} \%=\left[\frac{0.002}{0.4}+\frac{2(0.001)}{(0.3)}+\frac{0.02}{5}\right] \times 100 \%
=12%+23%+25%=\frac{1}{2} \%+\frac{2}{3} \%+\frac{2}{5} \%
=1.6%=1.6 \%
Q116
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]=[FA]=[MLT2L2]=[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]=[StressStrain]=[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}}]
Q117
The percentage error in the measurement of g is : (Given that g=4π2LT2g = {{4{\pi ^2}L} \over {{T^2}}}, L=(10±0.1)L = (10\, \pm \,0.1) cm, T=(100±1)T = (100\, \pm \,1) s)
A 7%
B 2%
C 5%
D 3%
Correct Answer
Option D
Solution

Given

g=4p2LT2g = {{4{p^2}L} \over {{T^2}}}

Fractional error in value of g

Δgg=ΔLL+2ΔTT{{\Delta g} \over g} = {{\Delta L} \over L} + 2{{\Delta T} \over T}
Δgg×100=ΔLL×100+2ΔTT×100\Rightarrow {{\Delta g} \over g} \times 100 = {{\Delta L} \over L} \times 100 + 2{{\Delta T} \over T} \times 100
Δgg×100=0.110×100+2×1100×100=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%

Q118
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 ]

Q119
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 =

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

Unit = Radian

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

Solid angle =

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

Unit = Steradian

=L2L2=[M0L0T0]= {{{L^2}} \over {{L^2}}} = [{M^0}{L^0}{T^0}]

\therefore Both have units but no dimensions.

Q120
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 Breadth = 55.3 ×\times 25 m 2 = 1382.5 m 2 = 14 ×\times 10 2 m 2 (Rounding off of two significant figures)

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