Units & Measurements — Page 11 | JEE Physics

Q101

A torque meter is calibrated to reference standards of mass, length and time each with

5 \%

accuracy. After calibration, the measured torque with this torque meter will have net accuracy of :

A 15%

B 25%

C 75%

D 5%

Correct Answer

Option B

Solution

We know that, torque applied on a rotating body,

\begin{aligned} \tau &=\text{ Force } \times \text{ Perpendicular distance } \\\\ \Rightarrow \quad[\tau] &=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}] \Rightarrow[\tau]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] \end{aligned}

\Rightarrow {{\Delta \tau } \over \tau } = {{\Delta M} \over M} + 2{{\Delta L} \over L} + 2{{\Delta T} \over T}

= 5 \times 5\% = 25\%

Q102

An expression of energy density is given by

u=\dfrac{\alpha}{\beta} \sin \left(\dfrac{\alpha x}{k t}\right)

, where

\alpha, \beta

are constants,

x

is displacement,

k

is Boltzmann constant and t is the temperature. The dimensions of

\beta

will be :

A

\left[\mathrm{ML}^{2} \mathrm{~T}^{-2} \theta^{-1}\right]

B

\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]

C

\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]

D

\left[\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{0}\right]

Correct Answer

Option D

Solution

u = {\alpha \over \beta }\sin \left( {{{\alpha x} \over {kt}}} \right)

[\alpha] = \left[ {{{kt} \over x}} \right] = {{[Energy]} \over {[Dis\tan ce]}}

[\beta ] = {{[\alpha ]} \over {[u]}}

= {{[Energy]/[Dis\tan ce]} \over {[Energy]/[Volume]}}

= [{L^2}]

Q103

The dimensions of

\left(\dfrac{\mathrm{B}^{2}}{\mu_{0}}\right)

will be : (if

\mu_{0}

: permeability of free space and

B

: magnetic field)

A

\left[\mathrm{M}\, \mathrm{L}^2 \,\mathrm{T}^{-2}\right]

B

\left[\mathrm{M} \,\mathrm{L} \,\mathrm{T}^{-2}\right]

C

\left[\mathrm{M} \,\mathrm{L}^{-1} \,\mathrm{~T}^{-2}\right]

D

\left[\mathrm{M} \,\mathrm{L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]

Correct Answer

Option C

Solution

\left[ {{{{B^2}} \over {{\mu _0}}}} \right]=

[Energy density]

= {{M{L^2}{T^{ - 2}}} \over {{L^3}}} = M{L^{ - 1}}{T^{ - 2}}

Q104

Consider the efficiency of carnot's engine is given by

\eta=\dfrac{\alpha \beta}{\sin \theta} \log_e \dfrac{\beta x}{k T}

, where

\alpha

and

\beta

are constants. If T is temperature, k is Boltzmann constant,

\theta

is angular displacement and x has the dimensions of length. Then, choose the incorrect option :

A Dimensions of

\beta

is same as that of force.

B Dimensions of

\alpha^{-1} x

is same as that of energy.

C Dimensions of

\eta^{-1} \sin \theta

is same as that of

\alpha \beta

.

D Dimensions of

\alpha

is same as that of

\beta

.

Correct Answer

Option D

Solution

Since, dimensions trigonometric function and logarithmic function are dimensionless quantities.

\therefore[\eta]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right]

Also, dimensions of temperature, $[T]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}{ }^0 \mathrm{~K}\right]$ Dimensions of Boltzmann constant, $[k]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\right]$ Dimension of $x=\left[\mathrm{M}^0 \mathrm{LT}^0\right]$ (A)

[\beta ] = \left[ {{{kT} \over x}} \right] = \left[ {{E \over x}} \right] = [ML{T^{ - 2}}] = [F]

(B)

[\alpha \beta ] = [{M^0}{L^0}{T^0}]

{[\alpha ]^{ - 1}} = [\beta ] = \left[ {{{kT} \over x}} \right]

So

{[\alpha ]^{ - 1}}[x] = [kT] = [M{L^2}{T^{ - 2}}]

(C)

\eta \sin \theta = \alpha \beta

So

[\eta \sin \theta ] = [\alpha \beta ]

[\eta ] = [{M^0}{L^0}{T^0}]

it is dimensionless quantity (D)

[\alpha ] \ne [\beta ]

Q105

Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is

\rho

and radius of the drop is r, then

\mathrm{T}=\mathrm{K} \sqrt{\rho \mathrm{r}^{3} / \mathrm{S}^{3 / 2}}

is dimensionally correct, where K is dimensionless. Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period. In the light of above statements, choose the correct answer from the options given below.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Correct Answer

Option D

Solution

We know,

\begin{gathered} {[S]=\left[\mathrm{MT}^{-2}\right]} \\\\ {[\rho]=\left[\mathrm{ML}^{-3}\right]} \\\\ {[r]=\left[\mathrm{L}]\right.} \end{gathered}

\begin{aligned} \therefore \text{ Dimension of } \mathrm{RHS} &=\frac{\left[\mathrm{M}^{\frac{1}{2}} \mathrm{~L}^{-\frac{3}{2}}\right]\left[\mathrm{L}^{\frac{3}{2}}\right]}{\left[\mathrm{MT}^{-2}\right]^{\frac{3}{4}}} \\\\ &=\left[\mathrm{M}^{\left(\frac{1}{2}-\frac{3}{4}\right)} \mathrm{L}^{\left(-\frac{3}{2}+\frac{3}{2}\right)} \mathrm{T}^{\frac{6}{4}}\right]=\mathrm{M}^{-\frac{1}{4}} \mathrm{~L}^{0} \mathrm{~T}^{\frac{3}{2}} \end{aligned}

dimension of L.H.S. = $[\mathrm{T}]$ $\therefore$ Dimension of LHS $\neq$ Dimension of RHS.

Q106

A travelling microscope has 20 divisions per

\mathrm{cm}

on the main scale while its vernier scale has total 50 divisions and 25 vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?

A 0.001 cm

B 0.002 mm

C 0.002 cm

D 0.005 cm

Correct Answer

Option C

Solution

1 MSD =

{1 \over {20}}

cm 1 VSD =

{{24} \over {25}} \times {1 \over {20}}

cm $\therefore$ Least count = 1 MSD $-$ 1 VSD

= {1 \over {20}}\left( {1 - {{24} \over {25}}} \right)

cm

= {1 \over {20}} \times {1 \over {25}}

cm

= 0.002

cm

Q107

If the velocity of light

\mathrm{c}

, universal gravitational constant

\mathrm{G}

and Planck's constant

\mathrm{h}

are chosen as fundamental quantities. The dimensions of mass in the new system is :

A

\left[\mathrm{h}^{1} \mathrm{c}^{1} \mathrm{G}^{-1}\right]

B

\left[\mathrm{h}^{-1 / 2} \mathrm{c}^{1 / 2} \mathrm{G}^{1 / 2}\right]

C

\left[\mathrm{h}^{1 / 2} \mathrm{c}^{1 / 2} \mathrm{G}^{-1 / 2}\right]

D

\left[\mathrm{h}^{1 / 2} \mathrm{c}^{-1 / 2} \mathrm{G}^{1}\right]

Correct Answer

Option C

Solution

\begin{aligned} & {[\mathrm{M}]=[\mathrm{G}]^{\mathrm{x}}[\mathrm{h}]^{\mathrm{y}}[\mathrm{c}]^{\mathrm{z}}} \\\\ & {[\mathrm{M}]=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^x\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y\left[\mathrm{LT}^{-1}\right]^z} \\\\ & {\left[\mathrm{M}^1 \mathrm{~L}^0 \mathrm{~T}^0\right]=\left[\mathrm{M}^{-x+y}\right]\left[\mathrm{L}^{3 x+2 y+z}\right]\left[\mathrm{T}^{-2 x-y-z}\right]} \\\\ & \mathrm{y}-\mathrm{x}=1 ......(1) \\\\ & 3 \mathrm{x}+2 \mathrm{y}+\mathrm{z}=0 .......(2) \\\\ & -2 \mathrm{x}- \mathrm{y}-\mathrm{z}=0 ........(3) \end{aligned}

On solving, $\mathrm{x}=-\dfrac{1}{2}, \mathrm{y}=\dfrac{1}{2}, \mathrm{z}=\dfrac{1}{2}$ So $\mathrm{m}=\sqrt{\dfrac{\mathrm{hc}}{\mathrm{G}}}$

Q108

Dimension of

\dfrac{1}{\mu_{0} \in_{0}}

should be equal to

A

\mathrm{T}^{2} / \mathrm{L}^{2}

B

\mathrm{T} / \mathrm{L}

C

\mathrm{L}^{2} / \mathrm{T}^{2}

D

\mathrm{L} / \mathrm{T}

Correct Answer

Option C

Solution

The term $\dfrac{1}{\mu_{0} \epsilon_{0}}$ appears in the formula for the speed of light $c$ , which is: $c = \sqrt{\dfrac{1}{\mu_{0} \epsilon_{0}}}$ where $\mu_{0}$ is the permeability of free space and $\epsilon_{0}$ is the permittivity of free space.

The speed of light $c$ has dimensions of length over time ( $\mathrm{L} / \mathrm{T}$ ).

Therefore, the term $\dfrac{1}{\mu_{0} \epsilon_{0}}$ has dimensions equal to the square of the speed of light, which is $\mathrm{L}^{2} / \mathrm{T}^{2}$ .

Q109

A cylindrical wire of mass

(0.4 \pm 0.01) \mathrm{g}

has length

(8 \pm 0.04) \mathrm{cm}

and radius

(6 \pm 0.03) \mathrm{mm}

. The maximum error in its density will be:

A 1%

B 5%

C 4%

D 3.5%

Correct Answer

Option C

Solution

The density of a cylindrical wire is given by the formula:

\rho = \frac{m}{V} = \frac{m}{\pi r^2 l}

where $m$ is the mass, $r$ is the radius, and $l$ is the length.

The relative error in a calculated quantity is the sum of the relative errors in the quantities it depends on.

For the density, this is given by:

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}

Given that $\Delta m = 0.01$ g, $m = 0.4$ g, $\Delta r = 0.03$ mm, $r = 6$ mm, $\Delta l = 0.04$ cm, and $l = 8$ cm, we can substitute these values into the formula to find the relative error in the density:

\frac{\Delta \rho}{\rho} = \frac{0.01}{0.4} + 2\frac{0.03}{6} + \frac{0.04}{8} = 0.025 + 0.01 + 0.005 = 0.04

So the relative error in the density is 0.04, or 4%.

Q110

The speed of a wave produced in water is given by

v=\lambda^{a} g^{b} \rho^{c}

. Where

\lambda, g

and

\rho

are wavelength of wave, acceleration due to gravity and density of water respectively. The values of

a, b

and

c

respectively, are :

A

\dfrac{1}{2}, 0, \dfrac{1}{2}

B

1,1,0

C

1,-1,0

D

\dfrac{1}{2}, \dfrac{1}{2}, 0

Correct Answer

Option D

Solution

\begin{aligned} & v=\lambda^{\mathrm{a}} \mathrm{g}^{\mathrm{b}} \rho^{\mathrm{c}} \\\\ & \text{using dimension formula} \\\\ & \Rightarrow\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-1}\right]=\left[\mathrm{L}^1\right]^{\mathrm{a}}\left[\mathrm{L}^1 \mathrm{~T}^{-2}\right]^{\mathrm{b}}\left[\mathrm{M}^1 \mathrm{~L}^{-3}\right]^{\mathrm{c}} \\\\ & \Rightarrow\left[\mathrm{M}^0 \mathrm{~L}^1 \mathrm{~T}^{-1}\right]=\left[\mathrm{M}^{\mathrm{c}} \mathrm{L}^{\mathrm{a}+\mathrm{b}-\mathrm{c}} \mathrm{T}^{-2 \mathrm{~b}}\right] \\\\ & \therefore \mathrm{c}=0, \mathrm{a}+\mathrm{b}-3 \mathrm{c}=1,-2 \mathrm{~b}=-1 \Rightarrow \mathrm{b}=\frac{1}{2} \\\\ & \text{ Now } \mathrm{a}+\mathrm{b}-3 \mathrm{c}=1 \\\\ & \Rightarrow \mathrm{a}+\frac{1}{2}-0=1 \\\\ & \Rightarrow \mathrm{a}=\frac{1}{2} \\\\ & \therefore \mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{1}{2}, \mathrm{c}=0 \end{aligned}