Units & Measurements — Page 8 | JEE Physics

Q71

The least count of the main scale of a vernier callipers is 1 mm. Its vernier scale is divided into 10 divisions and coincide with 9 divisions of the main scale. When jaws are touching each other, the 7th division of vernier scale coincides with a division of main scale and the zero of vernier scale is lying right side of the zero of main scale. When this vernier is used to measure length of a cylinder the zero of the vernier scale between 3.1 cm and 3.2 cm and 4th VSD coincides with a main scale division. The length of the cylinder is : (VSD is vernier scale division)

A 3.21 cm

B 2.99 cm

C 3.07 cm

D 3.2 cm

Correct Answer

Option C

Solution

Least count = 1 mm or 0.01 cm Zero error = 0 + 0.01 × 7 = 0.07 cm Reading = 3.1 + (0.01 × 4) – 0.07 = 3.1 + 0.04 – 0.07 = 3.1 – 0.03 = 3.07 cm

Q72

If speed V, area A and force F are chosen as fundamental units, then the dimension of Young’s modulus will be

A FA–1V0

B FA2V–1

C FA2V–2

D FA2V–3

Correct Answer

Option A

Solution

Y = k [F]x [A]y [V]z [M1L1T –2] = [MLT–2]x [L2]y [LT–1]z [M1L1T –2] = [M]x [L]x+2y+z[T]–2x–z Comparing power of M, L and T x = 1 ……(1) x + 2y + z = –1 ……(2) –2x – z = –2 ……(3) After solving x = 1 y = –1 z = 0 $\therefore$ Y = FA–1V0

Q73

For the four sets of three measured physical quantities as given below. Which of the following options is correct ? (i) A1 = 24.36, B1 = 0.0724, C1 = 256.2 (ii) A2 = 24.44, B2 = 16.082, C2 = 240.2 (iii) A3 = 25.2, B3 = 19.2812, C3 = 236.183 (iv) A4 = 25, B4 = 236.191, C4 = 19.5

A A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3 = A4 + B4 + C4

B A4 + B4 + C4

<

A1 + B1 + C1

<

A3 + B3 + C3

<

A2 + B2 + C2

C A4 + B4 + C4

<

A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3

D A4 + B4 + C4

>

A3 + B3 + C3 = A2 + B2 + C2

>

A1 + B1 + C1

Correct Answer

Option D

Solution

A1 + B1 + C1 = 24.36 + 0.0724 + 256.2 = 280.6324 = 280.6 (After rounding off) A2 + B2 + C2 = 24.44 + 16.082 + 240.2 = 280.722 = 280.7 (After rounding off) A3 + B3 + C3 = 25.2 + 19.2812 + 236.183 = 280.6642 = 280.7 (After rounding off) A4 + B4 + C4 = 25 + 236.191 + 19.5 = 280.691 = 281 (After rounding off)

Q74

The work done by a gas molecule in an isolated system is given by,

W = \alpha {\beta ^2}{e^{ - {{{x^2}} \over {\alpha kT}}}}

, where x is the displacement, k is the Boltzmann constant and T is the temperature.

\alpha

and

\beta

are constants. Then the dimensions of

\beta

will be :

A

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

B

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

C

[ML{T^{ - 2}}]

D

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

Correct Answer

Option C

Solution

where, k is Boltzmann constant, T is temperature and x is displacement. We know that,

{{{x^2}} \over {\alpha kT}}

is a dimensionless quantity. $\therefore$

\left[ {{{{x^2}} \over {\alpha kT}}} \right] = [{M^0}{L^0}{T^0}] \Rightarrow [\alpha ] = {{[{x^2}]} \over {[k][T]}}

\Rightarrow [\alpha ] = {{[{L^2}]} \over {[k][T]}}

..... (i) Since, dimensions of k are

[k] = [{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}]

...... (ii) Dimensions of temperature are

[T] = [K]

..... (iii) Substituting Eqs. (ii) and (iii) in Eq. (i), we get

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

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

According to dimensional analysis,

[W] = [\alpha {\beta ^2}]

\Rightarrow [{\beta ^2}] = {{[W]} \over {[\alpha ]}}

\Rightarrow [{\beta ^2}] = {{{M^1}{L^2}{T^{ - 2}}]} \over {[{M^{ - 1}}{T^2}]}} = [{M^2}{L^2}{T^{ - 4}}]

\Rightarrow [\beta ] = [ML{T^{ - 2}}]

Q75

Match with :

List - I		List - II
(a)	h (Planck's constant)	(i)	$[ML{T^{ - 1}}]$
(b)	E (kinetic energy)	(ii)	$[M{L^2}{T^{ - 1}}]$
(c)	V (electric potential)	(iii)	$[M{L^2}{T^{ - 2}}]$
(d)	P (linear momentum)	(iv)	$[M{L^2}{I^{ - 1}}{T^{ - 3}}]$

A (a)

\to

(ii), (b)

\to

(iii), (c)

\to

(iv), (d)

\to

(i)

B (a)

\to

(i), (b)

\to

(ii), (c)

\to

(iv), (d)

\to

(iii)

C (a)

\to

(iii), (b)

\to

(ii), (c)

\to

(iv), (d)

\to

(i)

D (a)

\to

(iii), (b)

\to

(iv), (c)

\to

(ii), (d)

\to

(i)

Correct Answer

Option A

Solution

Kinetic Energy,

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

Momentum,

mv = [ML{T^{ - 1}}]

Plank constant :

E = h\gamma

\Rightarrow M{L^2}{T^{ - 2}} = h \times {1 \over T}

\Rightarrow h = [M{L^2}{T^{ - 1}}]

Also,

E = qV

\Rightarrow V = {{[M{L^2}{T^{ - 2}}]} \over {[C]}} = [M{L^2}{T^{ - 2}}{C^{ - 1}}]

Q76

The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on circular scale coincides with the reference line. The radius of the wire is :

A 1.80 mm

B 0.90 mm

C 0.82 mm

D 1.64 mm

Correct Answer

Option C

Solution

Least count =

{{1mm} \over {100}} = 0.01mm

zero error = + 8 $\times$ LC = + 0.08 mm True reading (Diameter) = (1 mm + 72 $\times$ LC) $-$ (Zero error) = (1 mm + 72 $\times$ 0.01 mm) $-$ 0.08 mm = 1.72 mm $-$ 0.08 mm = 1.64 mm Therefore, radius =

{{1.64} \over 2}

= 0.82 mm.

Q77

If e is the electronic charge, c is the speed of light in free space and h is Planck's constant, the quantity

{1 \over {4\pi {\varepsilon _0}}}{{|e{|^2}} \over {hc}}

has dimensions of :

A

[ML{T^{ - 1}}]

B

[ML{T^0}]

C

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

D

[L{C^{ - 1}}]

Correct Answer

Option C

Solution

Given e = electronic charge c = speed of light in free space h = Planck's constant We know, E =

{{hc} \over \lambda }

and

F = {1 \over {4\pi {\varepsilon _0}}}{{{q^2}} \over {{d^2}}}

$\Rightarrow$

{{{q^2}} \over {4\pi {\varepsilon _0}}} = F{d^2}

{1 \over {4\pi {\varepsilon _0}}}{{{e^2}} \over {hc}}

=

{{F{d^2}} \over {E\lambda }}

=

{{Fd.d} \over {E\lambda }}

=

{d \over \lambda }

$=$ dimensionless

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

Q78

In a typical combustion engine the workdone by a gas molecule is given by

W = {\alpha ^2}\beta {e^{{{ - \beta {x^2}} \over {kT}}}}

, where x is the displacement, k is the Boltzmann constant and T is the temperature. If

\alpha

and

\beta

are constants, dimensions of

\alpha

will be :

A

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

B

[ML{T^{ - 1}}]

C

[ML{T^{ - 2}}]

D

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

Correct Answer

Option A

Solution

kT has dimension of energy

{{\beta {x^2}} \over {kT}}

is dimensionless

[\beta ][{L^2}] = [M{L^2}{T^{ - 2}}]

[\beta ] = [M{T^{ - 2}}]

{\alpha ^2}\beta

has dimensions of work

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

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

Q79

If 'C' and 'V' represent capacity and voltage respectively then what are the dimensions of

\lambda

where C/V =

\lambda

?

A

[{M^{ - 3}}{L^{ - 4}}{I^3}{T^7}]

B

[{M^{ - 2}}{L^{ - 3}}{I^2}{T^6}]

C

[{M^{ - 2}}{L^{ - 4}}{I^3}{T^7}]

D

[{M^{ - 1}}{L^{ - 3}}{I^{ - 2}}{T^{ - 7}}]

Correct Answer

Option C

Solution

\lambda = {C \over V} = {{Q/V} \over V} = {Q \over {{V^2}}}

V = {{work} \over Q}

\lambda = {{{Q^3}} \over {{{(work)}^2}}} = {{{{(It)}^3}} \over {{{(F.s)}^2}}}

= {{\left[ {{I^3}{T^3}} \right]} \over {{{\left[ {M{L^2}{T^{ - 2}}} \right]}^2}}} = [{M^{ - 2}}{L^{ - 4}}{I^3}{T^7}]

Q80

One main scale division of a vernier callipers is 'a' cm and nth division of the vernier scale coincide with (n

-

1)th division of the main scale. The least count of the callipers in mm is :

A

{{10a} \over n}

B

{{10na} \over {(n - 1)}}

C

\left( {{{n - 1} \over {10n}}} \right)a

D

{{10a} \over {(n - 1)}}

Correct Answer

Option A

Solution

n VSD = (n $-$ 1) MSD 1 VSD =

\left( {{{n - 1} \over n}} \right)

MSD L.C. = 1 MSD $-$ 1 VSD = 1 MSD $-$

\left( {{{n - 1} \over n}} \right)

MSD = 1 MSD $-$ 1 MSD +

{{MSD} \over n}

=

{{MSD} \over n}

=

{a \over n}

cm =

{{10a} \over n}

mm