Units & Measurements — Page 2 | JEE Physics

Q11

Identify the pair of physical quantities that have same dimensions:

A velocity gradient and decay constant

B Wien's constant and Stefan constant

C angular frequency and angular momentum

D wave number and Avogadro number

Correct Answer

Option A

Solution

Velocity gradient

= {{dv} \over {dx}}

$\Rightarrow$ Dimensions are

{{[L{T^{ - 1}}]} \over {[L]}} = [{T^{ - 1}}]

Decay constant $\lambda$ has dimensions of

[{T^{ - 1}}]

because of the relation

{{dN} \over {dt}} = - \lambda

N $\Rightarrow$ Velocity gradient and decay constant have same dimensions.

Q12

Identify the pair of physical quantities which have different dimensions:

A Wave number and Rydberg's constant

B Stress and Coefficient of elasticity

C Coercivity and Magnetisation

D Specific heat capacity and Latent heat

Correct Answer

Option D

Solution

[S] = {{[C]} \over {[m] \times [\Delta T]}}

and,

[L] = {{[Q]} \over {[m]}}

$\Rightarrow$ They have different dimensions.

Q13

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error in determining the density is:

A 6%

B 2.5%

C 3.5%

D 4.5%

Correct Answer

Option D

Solution

Density of a material (d) =

{M \over {{L^3}}}

$\therefore$ Error in density,

{{\Delta d} \over d} = {{\Delta M} \over M} + 3{{\Delta L} \over L}

{{\Delta d} \over d} \times 100 = {{\Delta M} \over M} \times 100 + 3{{\Delta L} \over L} \times 100

\Rightarrow {{\Delta d} \over d} \times 100 = 1.5\% + 3\left( 1 \right)\%

= 4.5 %

Q14

In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1 g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young's Modulus determined by this experiment?

A 0.14%

B 9%

C 1.4%

D 0.9%

Correct Answer

Option C

Solution

{{\Delta Y} \over Y} = \left( {{{\Delta m} \over m}} \right) + \left( {{{\Delta g} \over g}} \right) + \left( {{{\Delta A} \over A}} \right) + \left( {{{\Delta l} \over l}} \right) + \left( {{{\Delta L} \over L}} \right)

= \left( {{{1g} \over {1kg}}} \right) + 0 + 2\left( {{{\Delta r} \over r}} \right) + \left( {{{\Delta l} \over l}} \right) + \left( {{{\Delta L} \over L}} \right)

= \left( {{{1g} \over {1kg}}} \right) + 2\left( {{{0.001cm} \over {0.2cm}}} \right) + \left( {{{0.001cm} \over {0.5cm}}} \right) + \left( {{{0.001m} \over {1m}}} \right)

= \left( {{1 \over {1000}}} \right) + 2\left( {{{1 \times 10} \over {2 \times {{10}^3}}}} \right) + \left( {{1 \over 5} \times {{{{10}^2}} \over {{{10}^3}}}} \right) + \left( {{1 \over {{{10}^3}}}} \right)

= {1 \over {1000}} + {1 \over {100}} + {2 \over {{{10}^3}}} + {1 \over {{{10}^3}}}

= {{1 + 10 + 2 + 1} \over {1000}} = {{14} \over {1000}} \times 100\%

= 1.4\%

Q15

If 50 Vernier divisions are equal to 49 main scale divisions of a traveling microscope and one smallest reading of main scale is

0.5 \mathrm{~mm}

, the Vernier constant of traveling microscope is

A 0.01 mm

B 0.01 cm

C 0.1 mm

D 0.1 cm

Correct Answer

Option A

Solution

\begin{aligned} & 50 \mathrm{~V}+\mathrm{S}=49 \mathrm{~S}+\mathrm{S} \\ & \mathrm{S}=50(\mathrm{~S}-\mathrm{V}) \\ & 5=50(\mathrm{~S}-\mathrm{V}) \\ & \mathrm{S}-\mathrm{V}=\frac{0.5}{50}=\frac{1}{100}=0.01 \mathrm{~mm} \end{aligned}

Q16

The dimension of magnetic field in M, L, T and C (coulomb) is given as

A MLT-1C-1

B MT2C-2

C MT-1C-1

D MT-2C-1

Correct Answer

Option C

Solution

We know that, Lorentz force

\left| {\overrightarrow F } \right| = \left| {q\overrightarrow v \times \overrightarrow B } \right|

\therefore [B]

=

{[F] \over {[q][v]}}

=

{[{ML{T^{ - 2}}]} \over {[C] \times [L{T^{ - 1}}]}}

= [

M{T^{ - 1}}{C^{ - 1}}

]

Q17

The physical quantities not having same dimensions are

A torque and work

B momentum and Planck's constant

C stress and Young's modulus

D speed and

{\left( {{\mu _0}{\varepsilon _0}} \right)^{ - 1/2}}

Correct Answer

Option B

Solution

Momentum = mv = [

{M{L}{T^{ - 1}}}

] Planck's constant, h =

{E \over v}

=

{[{M{L^2}{T^{ - 2}]}} \over {[{T^{ - 1}}]}}

=

[{M{L^2}{T^{ - 1}}}]

So Momentum and Planck's constant do not have same dimensions.

Q18

Which one of the following represents the correct dimensions of the coefficient of viscosity?

A ML-1T-1

B MLT-1

C ML-1T-2

D ML-2T-2

Correct Answer

Option A

Solution

From stokes law, Viscous force F =

6\pi \eta rv

\Rightarrow \eta = {F \over {6\pi rv}}

$\therefore$

[\eta] = {[{ML{T^{ - 2}}]} \over {[L][L{T^{ - 1}}]}}

\Rightarrow [\eta] = [M{L^{ - 1}}{T^{ - 1}}]

Q19

Out of the following pair, which one does NOT have identical dimensions is

A angular momentum and Planck's constant

B impulse and momentum

C moment of inertia and moment of a force

D work and torque

Correct Answer

Option C

Solution

Moment of inertia, I = Mr2 $\therefore$ [I] = [ML2] Moment of force,

\overrightarrow \tau = \overrightarrow r \times \overrightarrow F

$\therefore$

[\overrightarrow \tau ]

=

[L][ML{T^{ - 2}}]

=

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

Q20

Identify the pair whose dimensions are equal

A torque and work

B stress and energy

C force and stress

D force and work

Correct Answer

Option A

Solution

To determine which pairs have equivalent dimensions, we examine the respective dimensional formulas.

For work (W) specified as $\overrightarrow{F} \cdot \overrightarrow{s}$ : $W = F s \cos \theta$ The dimensions are: $W = [MLT^{-2}][L] = [ML^2T^{-2}]$ For torque ( $\overrightarrow{\tau}$ ) given as $\overrightarrow{r} \times \overrightarrow{F}$ : $\tau = rF \sin \theta$ The dimensions are: $\tau = [L] [MLT^{-2}] = [ML^2T^{-2}]$ Therefore, the dimensions of torque and work are identical.