Units & Measurements — Page 10 | JEE Physics

Q91

If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be :

A [FT

-

1 V

-

1]

B [FTV

-

1]

C [FT2 V]

D [FVT

-

1]

Correct Answer

Option B

Solution

[M] = K[F]a [T]b [V]c [M1] = [M1L1T $-$ 2]a [T1]b [L1T $-$ 1]c a = 1, b = 1, c = $-$ 1 $\therefore$ [M] = [FTV $-$ 1]

Q92

Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as

{v_2} = {n \over {{m^2}}}{v_1}

and

{a_2} = {{{a_1}} \over {mn}}

respectively. Here m and n are constants. The relations for distance and time in two systems respectively are :

A

{{{n^3}} \over {{m^3}}}{L_1} = {L_2}

and

{{{n^2}} \over m}{T_1} = {T_2}

B

{L_1} = {{{n^4}} \over {{m^2}}}{L_2}

and

{T_1} = {{{n^2}} \over m}{T_2}

C

{L_1} = {{{n^2}} \over m}{L_2}

and

{T_1} = {{{n^4}} \over {{m^2}}}{T_2}

D

{{{n^2}} \over m}{L_1} = {L_2}

and

{{{n^4}} \over {{m^2}}}{T_1} = {T_2}

Correct Answer

Option A

Solution

[L] = {{[{v^2}]} \over {[a]}}

so

{{{{[{v_2}]}^2}} \over {[{a_2}]}} = {{{{\left[ {{n \over {{m^2}}}{v_1}} \right]}^2}} \over {\left[ {{{{a_1}} \over {mn}}} \right]}}

{{{{[{v_2}]}^2}} \over {[{a_2}]}} = {{{n^3}} \over {{m^3}}}{{{{[{v_1}]}^2}} \over {[{a_1}]}}

or

[{L_2}] = {{{n^3}} \over {{m^3}}}[{L_1}]

Similarly,

[T] = {{[v]} \over {[a]}}

So,

[{T_2}] = {{{n^2}} \over m}[{T_1}]

Q93

The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be :

A [ML

-

1T

-

1]

B [ML

-

1T

-

2]

C [ML2T

-

1]

D [M

-

1L3T0]

Correct Answer

Option A

Solution

[pascal-second] =

{{ML{T^{ - 2}}} \over {{L^2}}} \times T

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

Q94

An expression for a dimensionless quantity P is given by

P = {\alpha \over \beta }{\log _e}\left( {{{kt} \over {\beta x}}} \right)

; where

\alpha

and

\beta

are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of

\alpha

will be :

A [M0 L

-

1 T0]

B [M L0 T

-

2]

C [M L T

-

2]

D [M L2 T

-

2]

Correct Answer

Option C

Solution

Given, $P=\dfrac{\alpha}{\beta} \log _{e}\left[\dfrac{k t}{\beta x}\right]$ The logarithmic term is dimensionless.

Thus, $[k t / \beta x]$ is also dimensionless. i.e.

$\dfrac{[k][t]}{[\beta][x]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]$ .......(i) We have, $E=k t$ Thus, Eq. (i) becomes,

\begin{array}{r} \frac{\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\right]}{[\mathrm{\beta}]\left[\mathrm{L}^{1}\right]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right] \\\\ {[\beta]=\left[\mathrm{MLT}^{-2}\right]} \end{array}

Since, $P$ is also a dimensionless quantity.

\begin{aligned} &\frac{[\alpha]}{[\beta]}=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right] \\\\ &{[\alpha]=[\beta]\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]} \\\\ &{[\alpha]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-2}\right]} \end{aligned}

Q95

The expression given below shows the variation of velocity (v) with time (t),

v=\mathrm{At}^2+\dfrac{\mathrm{Bt}}{\mathrm{C}+\mathrm{t}}

. The dimension of ABC is :

A [M0L1T−2]

B [M0L2T−3]

C [M0L2T−2]

D [M0L1T−3]

Correct Answer

Option B

Solution

\begin{aligned} & {\left[\mathrm{LT}^{-1}\right]=[\mathrm{A}]\left[\mathrm{T}^2\right]=\frac{[\mathrm{B}][\mathrm{T}]}{[\mathrm{C}]+[\mathrm{T}]}} \\ & {[\mathrm{C}]=[\mathrm{T}]} \\ & {[\mathrm{A}]=\left[\mathrm{LT}^{-3}\right]} \\ & {[\mathrm{B}]=\left[\mathrm{LT}^{-1}\right]} \\ & {[\mathrm{ABC}]=\left[\mathrm{L}^2 \mathrm{~T}^{-3}\right]} \end{aligned}

Q96

If n main scale divisions coincide with (n + 1) vernier scale divisions. The least count of vernier callipers, when each centimetre on the main scale is divided into five equal parts, will be :

A

{2 \over {n + 1}}

mm

B

{5 \over {n + 1}}

mm

C

{1 \over {2n}}

mm

D

{1 \over {5n}}

mm

Correct Answer

Option A

Solution

5 parts of main scale division = 1 cm $\therefore$ 1 part of main scale division =

{1 \over 5}

cm $\therefore$ 1 M.S.D. =

{1 \over 5}

cm (n + 1) vernier scale division = n main scale division. $\therefore$ 1 V.S.D. =

{n \over n+1}

M.S.D. =

{n \over n+1}

$\times$ 1 M.S.D. =

{n \over n + 1}

$\times$

{1 \over 5}

cm We know, L.C. = 1 M.S.D. $-$ 1 V,S.D. =

{1 \over 5}

cm $-$

{n \over {5(n + 1)}}

cm =

{{n + 1 - n} \over {5(n + 1)}}

cm =

{1 \over {5(n + 1)}}

cm

Q97

If momentum [P], area

[\mathrm{A}]

and time

[\mathrm{T}]

are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :

A

\left[\mathrm{P} \,\mathrm{A}^{-1} \mathrm{~T}^{0}\right]

B

\left[\mathrm{P} \,\mathrm{A}\mathrm{~T}^{-1}\right]

C

\left[\mathrm{P}\,\mathrm{A}^{-1} \mathrm{~T}\right]

D

\left[\mathrm{P} \,\mathrm{A}^{-1} \mathrm{~T}^{-1}\right]

Correct Answer

Option A

Solution

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

Now if

[\eta ] = {[P]^a}{[A]^b}{[T]^c}

\Rightarrow [M{L^{ - 1}}{T^{ - 1}}] = {[M{L^1}{T^{ - 1}}]^a}{[{L^2}]^b}{[T]^c}

\Rightarrow a = 1,\,a + 2b = - 1,\, - a + c = - 1

\Rightarrow a = 1,\,b = - 1,\,c = 0

\Rightarrow [\eta ] = [P]{[A]^{ - 1}}{[T]^0}

= [P{A^{ - 1}}{T^0}]

Q98

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are

1 \%, 2 \%

and

3 \%

respectively. The maximum percentage error in the detection of the dissipated heat will be :

A 2

B 4

C 6

D 8

Correct Answer

Option D

Solution

Given, $\dfrac{\Delta R}{R} \times 100=1 \%$

\frac{\Delta F}{F} \times 100=2 \% \text{ and } \frac{\Delta t}{t} \times 100=3 \%

We know that, heat produced due to current flowing through a resistor $R$ .

$H =I^{2} R t$ $\Rightarrow \dfrac{\Delta H}{H} =\dfrac{2 \Delta I}{I}+\dfrac{\Delta R}{R}+\dfrac{\Delta t}{t}$ $\Rightarrow \dfrac{\Delta H}{H} \times 100 =2\left(\dfrac{\Delta I}{I} \times 100\right)+\dfrac{\Delta R}{R} \times 100+\dfrac{\Delta t}{t} \times 100$ $=2 \times 2 \%+1 \%+3 \%=8 \%$

Q99

A screw gauge of pitch

0.5 \mathrm{~mm}

is used to measure the diameter of uniform wire of length

6.8 \mathrm{~cm}

, the main scale reading is

1.5 \mathrm{~mm}

and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is : [Screw gauge has 50 divisions on its circular scale]

A 6.8 cm2

B 3.4 cm2

C 3.9 cm2

D 2.4 cm2

Correct Answer

Option B

Solution

Least count

= {{0.5} \over {50}}

mm = 0.01 mm $\therefore$ Diameter, d = 1.5 mm + 7 $\times$ 0.01 = 1.57 mm $\therefore$ Surface area

= (2\pi r) \times l

= \pi dl

= 3.142 \times {{1.57} \over {10}} \times 6.8

cm2 = 3.354 cm2 = 3.4 cm2

Q100

In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and

4^{\text{th }}

Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to

1 \mathrm{~mm}

. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and

6^{\text{th }}

Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be :

A 3.02 cm

B 3.06 cm

C 3.10 cm

D 3.20 cm

Correct Answer

Option C

Solution

Given, In Vernier calipers, 10 VSD $=9$ MSD $\Rightarrow \quad 1$ VSD $=\dfrac{9}{10}$ MSD $\therefore$ Least count of vernier scale,

\begin{aligned} \text{ LC } &=1 \mathrm{MSD}-1 \mathrm{VSD} \\\\ &=1 \mathrm{MSD}-\frac{9}{10} \mathrm{MSD}=\operatorname{MSD}\left(1-\frac{9}{10}\right) \\\\ &=\frac{\text{ MSD }}{10}=\frac{1 \mathrm{~mm}}{10} \quad[\because 1 \mathrm{MSD}=1 \mathrm{~mm}] \\\\ &=0.1 \mathrm{~mm}=0.01 \mathrm{~cm} \end{aligned}

According to given situation, Negative error $=$ Main scale reading - Least count $\times$ Number of coinciding main scale division

=0.1-0.01 \times 4=0.1-0.04=0.06 \mathrm{~cm}

$\therefore$ Diameter of spherical body

\begin{aligned} &=30 \times 0.1+6 \times 0.01+0.06 \\\\ &=3.0+0.06+0.06=3.12 \mathrm{~cm} \end{aligned}

Which is closest to $3.10 \mathrm{~cm}$ .