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Units & Measurements

JEE Physics · 184 questions · Page 1 of 19 · Click an option or "Show Solution" to reveal answer

Q1
A silver wire has a mass (0.6 ±\pm 0.006) g, radius (0.5 ±\pm 0.005) mm and length (4 ±\pm 0.04) cm. The maximum percentage error in the measurement of its density will be :
A 4%
B 3%
C 6%
D 7%
Correct Answer
Option A
Solution
ρ=mV=mπr2×l\rho = {m \over V} = {m \over {\pi {r^2} \times l}}

∴\therefore % error in

ρ=(0.0060.6+2×0.0050.5+0.044)×100\rho = \left( {{{0.006} \over {0.6}} + 2 \times {{0.005} \over {0.5}} + {{0.04} \over 4}} \right) \times 100
=4%= 4\%
Q2
The dimension of mutual inductance is :
A [ML2T−2A−1][M{L^2}{T^{ - 2}}{A^{ - 1}}]
B [ML2T−3A−1][M{L^2}{T^{ - 3}}{A^{ - 1}}]
C [ML2T−2A−2][M{L^2}{T^{ - 2}}{A^{ - 2}}]
D [ML2T−3A−2][M{L^2}{T^{ - 3}}{A^{ - 2}}]
Correct Answer
Option C
Solution

∵\because

U=12Mi2U = {1 \over 2}M{i^2}
⇒[M]=[U][i2]=ML2T−2A2\Rightarrow [M] = {{[U]} \over {[{i^2}]}} = {{M{L^2}{T^{ - 2}}} \over {{A^2}}}
=[ML2T−2A−2]= [M{L^2}{T^{ - 2}}{A^{ - 2}}]
Q3
The equation of a circle is given by x2+y2=a2x^2+y^2=a^2, where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation : (x−At)2+(y−tB)2=a2{(x - At)^2} + {\left( {y - {t \over B}} \right)^2} = {a^2}. The dimensions of t is given as [T−1][\mathrm{T^{-1}]}.
A A=[L−1T−1],B=[LT−1]\mathrm{A=[L^{-1}T^{-1}],B=[LT^{-1}]}
B A=[L−1T−1],B=[LT]\mathrm{A=[L^{-1}T^{-1}],B=[LT]}
C A=[LT],B=[L−1T−1]\mathrm{A=[LT],B=[L^{-1}T^{-1}]}
D A=[L−1T],B=[LT−1]\mathrm{A=[L^{-1}T],B=[LT^{-1}]}
Correct Answer
Option C
Solution

Here, At is distance, so dimensions of

[At]=[x]=[L][At] = [x] = [L]

Given. The dimensions of t is

[T−1][\mathrm{T^{-1}]}

[A×T−1]=[L]⇒[A]=[LT]{\left[A \times \mathrm{T}^{-1}\right]=[\mathrm{L}] \Rightarrow[A]=[\mathrm{LT}]}

[tB]=[y]=[L]\left[ {{t \over B}} \right] = [y] = [L]

⇒T−1[B]=[L]⇒B=[L−1 T−1]\Rightarrow \dfrac{\mathrm{T}^{-1}}{[B]}=[L] \Rightarrow B=\left[\mathrm{L}^{-1} \mathrm{~T}^{-1}\right]

Q4
The measured value of the length of a simple pendulum is 20 cm20 \mathrm{~cm} with 2 mm2 \mathrm{~mm} accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N%\mathrm{N} \%. The value of N\mathrm{N} is:
A 6
B 5
C 4
D 8
Correct Answer
Option A
Solution
T=2πℓgg=4π2ℓT2Δgg=Δℓℓ+2ΔTT=0.220+2(140)=1.220\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\ & \mathrm{g}=\frac{4 \pi^2 \ell}{\mathrm{T}^2} \\ & \frac{\Delta \mathrm{g}}{\mathrm{g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{T}}{\mathrm{T}} \\ & =\frac{0.2}{20}+2\left(\frac{1}{40}\right) \\ & =\frac{1.2}{20} \end{aligned}

Percentage change

=1.220×100=6%=\frac{1.2}{20} \times 100=6 \%
Q5
Consider two physical quantities AA and BB related to each other as E=B−x2AtE=\dfrac{B-x^2}{A t} where E,xE, x and tt have dimensions of energy, length and time respectively. The dimension of ABA B is
A L−2M1T0L^{-2} M^1 T^0
B L2M−1T1L^2 M^{-1} T^1
C L0M−1T1L^0 M^{-1} T^1
D L−2M−1T1L^{-2} M^{-1} T^1
Correct Answer
Option B
Solution
[B]=L2 A=x2tE=L2TML2 T−2=1MT−1[ A]=M−1 T[AB]=[L2M−1 T1]\begin{aligned} & {[\mathrm{B}]=\mathrm{L}^2} \\ & \mathrm{~A}=\frac{\mathrm{x}^2}{\mathrm{tE}}=\frac{\mathrm{L}^2}{\mathrm{TML}^2 \mathrm{~T}^{-2}}=\frac{1}{\mathrm{MT}^{-1}} \\ & {[\mathrm{~A}]=\mathrm{M}^{-1 \mathrm{~T}}} \\ & {[\mathrm{AB}]=\left[\mathrm{L}^2 \mathrm{M}^{-1} \mathrm{~T}^1\right]} \end{aligned}
Q6
Applying the principle of homogeneity of dimensions, determine which one is correct, where TT is time period, GG is gravitational constant, MM is mass, rr is radius of orbit.
A T2=4π2rGM2T^2=\dfrac{4 \pi^2 r}{G M^2}
B T2=4π2r3T^2=4 \pi^2 r^3
C T2=4π2r3GMT^2=\dfrac{4 \pi^2 r^3}{G M}
D T2=4π2r2GMT^2=\dfrac{4 \pi^2 r^2}{G M}
Correct Answer
Option C
Solution

To determine which option is correct based on the principle of homogeneity of dimensions, we need to ensure that both sides of the equation have the same dimensions.

The time period (

TT

) is measured in units of time (

TT

), the gravitational constant (

GG

) has units of

m3kg−1s−2\text{m}^3\text{kg}^{-1}\text{s}^{-2}

, mass (

MM

) has units of mass (

MM

), and the radius of orbit (

rr

) has units of length (

LL

). Let's analyze each option: Option A:

T2=4π2rGM2T^2=\frac{4 \pi^2 r}{G M^2}

The dimensions of the right-hand side of the equation are

L(L3MT2)M2=LL3M−1T−2M2=LL3T−2=L−2T2\frac{L}{\left(\frac{L^3}{MT^2}\right)M^2}=\frac{L}{L^3M^{-1}T^{-2}M^2}=\frac{L}{L^3T^{-2}}=L^{-2}T^2

, which do not match with

T2T^2

(time squared). Thus, Option A is incorrect. Option B:

T2=4π2r3T^2=4 \pi^2 r^3

The dimensions of the right-hand side are

L3L^3

, which clearly do not match the dimensions

T2T^2

of the squared time period. So, Option B is incorrect. Option C:

T2=4π2r3GMT^2=\frac{4 \pi^2 r^3}{G M}

The dimensions of the right-hand side of the equation are

L3(L3MT2)M=L3L3T−2=T2\frac{L^3}{\left(\frac{L^3}{MT^2}\right)M}=\frac{L^3}{L^3T^{-2}}=T^2

, which match the dimensions of the squared time period

T2T^2

. Therefore, Option C is correct based on the principle of homogeneity of dimensions. Option D:

T2=4π2r2GMT^2=\frac{4 \pi^2 r^2}{G M}

The dimensions of the right-hand side are

L2(L3MT2)M=L2L3M−1T−2M=L−1T2\frac{L^2}{\left(\frac{L^3}{MT^2}\right)M}=\frac{L^2}{L^3M^{-1}T^{-2}M}=L^{-1}T^2

, which do not match with

T2T^2

(time squared).

Hence, Option D is incorrect.

Thus, based on the principle of homogeneity of dimensions, Option C is the correct one:

T2=4π2r3GMT^2=\frac{4 \pi^2 r^3}{G M}

.

This equation also corresponds to Kepler's third law of planetary motion, which relates the orbital period of a planet to its orbital radius, considering the mass of the central body.

Q7
Match List I with List II: .tg .tg List I List II A. Torque I. kgm−1 s−2\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-2} B. Energy density II. kg ms−1\mathrm{kg} \,\mathrm{ms}^{-1} C. Pressure gradient III. kg m−2 s−2\mathrm{kg}\, \mathrm{m}^{-2} \mathrm{~s}^{-2} D. Impulse IV. kg m2 s−2\mathrm{kg} \,\mathrm{m}^{2} \mathrm{~s}^{-2} Choose the correct answer from the options given below:
A A-IV, B-I, C-III, D-II
B A-IV, B-I, C-II, D-III
C A-I, B-IV, C-III, D-II
D A-IV, B-III, C-I, D-II
Correct Answer
Option A
Solution

Torque →\to kg m

2^2

s

−2^{-2}

(IV) Energy density →\to kg m

−1^{-1}

s

−2^{-2}

(I) Pressure gradient →\to kg m

−2^{-2}

s

−2^{-2}

(III) Impulse →\to kg m s

−1^{-1}

(II)

Q8
Which of the following is not a dimensionless quantity?
A Relative magnetic permeability (μ\mur)
B Power factor
C Permeability of free space (μ\mu0)
D Quality factor
Correct Answer
Option C
Solution

[μ\mur] = 1 as μ\mur =

μμm{\mu \over {{\mu _m}}}

[power factor (cos ϕ\phi)] = 1

μ0=B0H{\mu _0} = {{{B_0}} \over H}

(unit = NA−-2) : Not dimensionless [μ\mu0] = [MLT−-2A−-2] quality factor

(Q)=Energy storedEnergy dissipated per cycle(Q) = {{Energy\,stored} \over {Energy\,dissipated\,per\,cycle}}

So Q is unitless & dimensionless.

Q9
If E and H represents the intensity of electric field and magnetising field respectively, then the unit of E/H will be :
A ohm
B mho
C joule
D newton
Correct Answer
Option A
Solution

Unit of

EH{E \over H}

is

volt/metreAmpere/metre=voltAmpere=ohm{{volt/metre} \over {Ampere/metre}} = {{volt} \over {Ampere}} = ohm
Q10
If Z=A2B3C4Z = {{{A^2}{B^3}} \over {{C^4}}}, then the relative error in Z will be :
A ΔAA+ΔBB+ΔCC{{\Delta A} \over A} + {{\Delta B} \over B} + {{\Delta C} \over C}
B 2ΔAA+3ΔBB−4ΔCC{{2\Delta A} \over A} + {{3\Delta B} \over B} - {{4\Delta C} \over C}
C 2ΔAA+3ΔBB+4ΔCC{{2\Delta A} \over A} + {{3\Delta B} \over B} + {{4\Delta C} \over C}
D ΔAA+ΔBB−ΔCC{{\Delta A} \over A} + {{\Delta B} \over B} - {{\Delta C} \over C}
Correct Answer
Option C
Solution
Z=A2B3C4Z = {{{A^2}{B^3}} \over {{C^4}}}

∴\therefore

ΔZZ=2ΔAA+3×ΔBB+4ΔCC{{\Delta Z} \over Z} = {{2\Delta A} \over A} + 3 \times {{\Delta B} \over B} + {{4\Delta C} \over C}
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