Units & Measurements — Page 5 | JEE Physics

Q41

The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants G, h and c. Which of the following correctly gives the Planck length ?

A G

\hbar

2 c3

B G2

\hbar

c

C

{G^{{{1}/{2}}}}{\hbar ^2}c

D

{\left( {{{G\hbar } \over {{c^3}}}} \right)^{{{1}/{2}}}}

Correct Answer

Option D

Solution

Plank length,

\ell

= k Gp

\hbar

q Cr [ Mo L To] = [ M $-$ 1 L3 T $-$ 2 ]p [ M L2 T $-$ 1] q [ L T $-$ 1]r [Mo L To ] = [M $-$ p + q L(3p + 2q + r) T $-$ (2p + q + r)] Comparing both sides, $-$ p + q = 0 3p + 2q + r = 1 $-$ (2p + q + r) = 0 Solving those equation we get, p =

{1 \over 2},

q =

{1 \over 2},

r = - {3 \over 2}

\therefore\,\,\,

\ell

= k G

{^{{1 \over 2}}}

{\hbar ^{{1 \over 2}}}

{C^{ - {3 \over 2}}}

=

{\left( {{{G\hbar } \over {{C^3}}}} \right)^{{1 \over 2}}}

(assume k = 1)

Q42

In a screw gauge,

5

complete rotations of the screw cause it to move a linear distance of

0.25

cm.

There are

100

circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of

4

main scale divisions and

30

circular scale divisions. Assuming negligible zero error, the thickness of the wire is :

A

0.4300

cm

B

0.2150

cm

C

0.3150

cm

D

0.0430

cm

Correct Answer

Option B

Solution

5 complete rotations = 0.25 cms So, 1 complete rotation of screw = 0.05 cm

\therefore\,\,\,\,

1 main scale division = 0.05 cm 1 circular scale =

{{0.05} \over {100}}

= 5 $\times$ 10 $-$ 4 cm Thickness of a wire = 4 main scale and 30 circular scale divisions = 4 $\times$ 0.05 + 30 $\times$ 5 $\times$ 10 $-$ 4 = 0.2150 cm.

Q43

The relative error in the determination of the surface area of sphere is

\alpha

. Then the relative error in the determination of its volume is :

A

{3 \over 2}\alpha

B

{2 \over 3}\alpha

C

{5 \over 2}\alpha

D

\alpha

Correct Answer

Option A

Solution

Relative error in the surface are of the sphere,

{{\Delta S} \over S}

= 2 $\times$

{{\Delta r} \over r}

= $\propto$ (given) Relative error in volume,

{{\Delta V} \over V}

= 3 $\times$

{{\Delta r} \over r}

= 3 $\times$

{1 \over 2}

$\times$

{{\Delta S} \over S}

=

{3 \over 2}

$\times$ $\propto$ =

{3 \over 2}

$\propto$

Q44

The relative uncertainly in the period of a satellite orbiting around the earth is 10-2. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :

A 10

-

2

B 2

\times

10

-

2

C 3

\times

10

-

2

D 6

\times

10

-

2

Correct Answer

Option B

Solution

From kepler's law, T = 2 $\pi$

\sqrt {{{{r^3}} \over {GM}}}

$\Rightarrow$

\,\,\,

T2 =

{{4{\pi ^2}} \over {GM}}{r^3}

$\Rightarrow$

\,\,\,

M =

{{4{\pi ^2}} \over G} \times {{{r^3}} \over {{T^2}}}

\therefore\,\,\,

{{\Delta M} \over M}

= 2

{{\Delta T} \over T}

+ 3

{{\Delta r} \over r}

as

{{\Delta r} \over r} \simeq 0

\therefore\,\,\,

\left| {{{\Delta M} \over M}} \right|

= 2

{{\Delta T} \over T}

= 2 $\times$ 10 $-$ 2

Q45

The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in the measurement of a physical quantity A =

{{{P^3}{Q^2}} \over {\sqrt R S}}.

The maximum percentage error in the value of A will be :

A 6.0%

B 7.5%

C 8.5%

D 6.5%

Correct Answer

Option D

Solution

Given, A =

{{{P^3}{Q^2}} \over {\sqrt R S}}

\therefore\,\,\,\,

{{\Delta A} \over A}

= 3

{{\Delta P} \over P}

+ 2

{{\Delta Q} \over Q}

+

{1 \over 2}

{{\Delta R} \over R}

+

{{\Delta S} \over S}

Maximum percentage error in the value of A is

{{\Delta A} \over A}

$\times$ 100 = 3 $\times$ 0.5 + 2 $\times$ 1 +

{1 \over 2}

$\times$ 3 + 1 $\times$ 1.5 = 6.5 %

Q46

Which of the following combinations has the dimension of electrical resistance (

\in

0 is the permittivity of vacuum and

\mu

0 is the permeability of vacuum)?

A

\sqrt {{{{ \in _0}} \over {{\mu _0}}}}

B

{{{{ \in _0}} \over {{\mu _0}}}}

C

\sqrt {{{{\mu _0}} \over {{ \in _0}}}}

D

{{{{\mu _0}} \over {{ \in _0}}}}

Correct Answer

Option C

Solution

According to Coulomb's law F =

{1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}

$\therefore$

{ \in _0} = {1 \over {4\pi }}{{{q^2}} \over {F{r^2}}}

\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}

=

\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]

Force between two parallel current carrying wires,

{F \over L} = {{{\mu _0}} \over {2\pi }}{{{i^2}} \over r}

$\therefore$

{{\mu _0}}

=

{{2\pi rF} \over {{i^2}L}}

\left[ {{\mu _0}} \right] = {{\left[ L \right]\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{A^2}} \right]\left[ L \right]}}

=

\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]

From Ohm's law, V = IR $\therefore$

R = {V \over I}

= {U \over {It}} \times {1 \over I}

[R] =

{{\left[ U \right]} \over {{{\left[ I \right]}^2}\left[ t \right]}}

=

{{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {{A^2}} \right]\left[ T \right]}}

=

\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]

Let, R $\propto$

{\left[ {{ \in _0}} \right]^a}{\left[ {{\mu _0}} \right]^b}

$\Rightarrow$

\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]

=

\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]

a

\left[ {ML{T^{ - 2}}{A^{ - 2}}} \right]

b $\Rightarrow$

\left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} \right]

= [ M-

a

+ b L-3

a

+ b T4

a

- 2b A2

a

- 2b ] By comparing both sides we get, -

a

+ b = 1 - 3

a

+ b = 2 By solving we get,

a

=

- {1 \over 2}

and b =

{1 \over 2}

$\therefore$ R =

{\left[ {{ \in _0}} \right]^{ - {1 \over 2}}}{\left[ {{\mu _0}} \right]^{{1 \over 2}}}

=

\sqrt {{{{\mu _0}} \over {{ \in _0}}}}

Q47

In the formula X = 5YZ2 , X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units?

A [M–3L–2T8A4]

B [M–2L–2T6A3]

C [M–1L–2T4A2]

D [M–2L0 T–4A–2]

Correct Answer

Option A

Solution

Capacitance (C) =

{{{Q^2}} \over {2E}}

=

{{\left[ {{A^2}{T^2}} \right]} \over {\left[ {M{L^2}{T^{ - 2}}} \right]}}

=

\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]

X Magnetic field (B) =

{F \over {IL}}

=

{{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ A \right]\left[ L \right]}}

=

{\left[ {M{T^{ - 2}}{A^{ - 1}}} \right]}

= Z Given, X = 5YZ2 $\therefore$ Y =

{X \over {5{Z^2}}}

[Y] =

{{\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]} \over {{{\left[ {M{T^{ - 2}}{A^{ - 1}}} \right]}^2}}}

$\therefore$ [Y] = [M–3L–2T8A4]

Q48

The area of a square is 5.29 cm2. The area of 7 such squares taking into account the significant figures is :-

A 37.0 cm2

B 37 cm2

C 37.030 cm2

D 37.03 cm2

Correct Answer

Option A

Solution

The area of one square is 5.29 cm².

To find the area of 7 such squares, we can simply multiply the area of one square by 7: 7 x 5.29 cm² = 37.03 cm² Since the given area has three significant figures, we need to round our answer to three significant figures as well.

Therefore, the area of 7 such squares, taking into account the significant figures, is 37.0 cm².

Q49

In the density measurement of a cube, the mass and edge length are measured as (10.00 ± 0.10) kg and (0.10 ± 0.01) m, respectively. The error in the measurement of density is :

A 0.01 kg/m3

B 0.10 kg/m3

C 0.31 kg/m3

D 0.07 kg/m3

Correct Answer

Option C

Solution

Mass (m) = (10.00 ± 0.10) kg Edge length (

l

) = (0.10 ± 0.01) m Volume of the cube (V) =

{l^3}

Density, $\rho$ =

{m \over V}

{{d\rho } \over \rho } = {{dm} \over m} + {{dV} \over V}

$\Rightarrow$

{{d\rho } \over \rho } = {{dm} \over m} + 3{{dl} \over l}

$\Rightarrow$

{{d\rho } \over \rho } = {{0.10} \over {10.00}} + 3{{0.01} \over {0.10}}

= 0.31

Q50

If surface tension (S), Moment of inertia (I) and Planck's constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be :-

A S1/2I1/2h0

B S3/2I1/2h0

C S1/2I1/2h-1

D S1/2I3/2h-1

Correct Answer

Option A

Solution

We know, surface tension (S) =

{F \over L}

=

{{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ L \right]}}

$\therefore$ [S] =

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

Moment of inertia (I) = mr2 $\therefore$ [I] =

{\left[ {M{L^2}} \right]}

Planck's constant (h) =

{E \over f}

= Et $\therefore$ [h] =

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

Also linear momentum (p) = mv =

{\left[ {ML{T^{ - 1}}} \right]}

Now we have to express p in terms of s, I and h. $\therefore$ Let, [P] = [Sa Ib hc] $\Rightarrow$

{\left[ {ML{T^{ - 1}}} \right]}

=

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

a

{\left[ {M{L^2}} \right]}

b

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

c $\Rightarrow$

{\left[ {ML{T^{ - 1}}} \right]}

= [ Ma + b + c L2b +2c T- 2a - c ] By comparing the dimensions of both sides, we get a + b + c = 1 .........(1) 2b +2c = 1 ..............(2) - 2a - c = -1 ...................(

3) By solving those three equations we get, a =

{1 \over 2}

b =

{1 \over 2}

c = 0 $\therefore$ linear momentum [p] = [S1/2I1/2h0]