Units & Measurements

Given,

AD = c ln (BD)

As log is dimensionless.

So, [BD] = 1   $ \Rightarrow $  [B] = ${1 \over {\left[ D \right]}}$

And  [AD] = [C]

Now checking options one by one

(a)   [B² C²] = [B²] [A² D²] = [A²] [B² D²] = [A²]

$ \therefore $   This is meaningful.

(b)   $\left( {{{A - C} \over D}} \right)$ is not meaningful.

As dimension of A $ \ne $ dimension of C

Hence (A $-$ C)   is not possible.

(c)   $\left[ {{A \over B}} \right]$ = [AD] = [C]

$ \therefore $   ${{A \over B}}$ $-$ C is meaningful.

(d)   $\left[ {{C \over {BD}}} \right]$ = ${{\left[ C \right]} \over {\left[ {BD} \right]}}$ = ${{\left[ C \right]} \over 1}$ = [C]

$\left[ {{{A{D^2}} \over C}} \right]$ = ${{\left[ {AD} \right]\left[ D \right]} \over {\left[ C \right]}}$ = ${{\left[ C \right]\left[ D \right]} \over {\left[ C \right]}}$ = [D]

As dimension of C and D are not same so it is not meaning ful.

Dimension of electric field $[E] = [M{A^{ - 1}}L{T^{ - 3}}]$

Dimensions of magnetic field [B]

$[B] = [M{A^{ - 1}}L{T^{ - 2}}]$

Dimensions of magnetic permebility

$[{\mu _0}] = [M{A^{ - 2}}{T^{ - 2}}L]$

$[S] = {{EB} \over {{\mu _0}}} = {{[M{A^{ - 1}}L{T^{ - 3}}][M{A^{ - 1}}{T^{ - 2}}]} \over {[M{A^{ - 2}}{T^{ - 2}}L]}} = [M{T^{ - 3}}]$

$[x] = {x^\alpha } \Rightarrow {p \over v} = time$

$[v] = {x^\beta }$

$[a] = {x^p} \Rightarrow {v \over a} = time$

$[p] = {x^q}$

$[F] = {x^r}$

$ \Rightarrow \left( {{{position} \over {speed}}} \right) = \left( {{{speed} \over {acceleration}}} \right) = \left( {{p \over F}} \right)$

$ \Rightarrow {{{x^\alpha }} \over {{x^\beta }}} = {{{x^\beta }} \over {{x^\alpha }}} = {{{x^q}} \over {{x^r}}}$

$ \Rightarrow {x^{\alpha - \beta }} = {x^{\beta - p}} = {x^{q - r}}$

$ \therefore $ $\alpha - \beta = \beta - p = q - r$

$\alpha + p = 2\beta $

$q + p = \beta + r$

$p + q - r = \beta $

$[M] = [Mass] = [{M^0}{L^0}{T^0}]$

[J] = [Angular momentum] = [ML²T^-1]

[L] = [Length]

Now, $[M{L^2}{T^{ - 1}}] = [{M^0}{L^0}{T^0}]$

$ \therefore $ [L²] = [T]

Power [P] = [MLT².LT^-1] = [ML²T^-3]

= [L²L^-6]

[P] = [L^-4\

Energy/Work [W] = [MLT^-2.L]

= [L²L^-4] = [L^-2]

Force [F] = [MLT^-2] = [L.L^-4] = [L^-3]

Linear momentum [p] = [MLT^-1] = [L.L^-2]

[p] = [L^-1]

The time period is measured for five successive measurements as follows :

T(s)	0.52	0.56	0.57	0.54	0.59
$\Delta $T	0.04	0	0.01	0.02	0.03

Therefore,

${T_{mean}} = {{0.52 + 0.56 + 0.57 + 0.54 + 0.59} \over 5}$

$ = {{2.78} \over 5} = 0.556 \approx 0.56$

and we know $\Delta$T(absolute error) = $\left| {{T_{mean}} - T} \right|$

Error in reading = $|{T_{mean}} - {T_1}| = 0.04$
$|{T_{mean}} - {T_2}| = 0.00$
$|{T_{mean}} - {T_3}| = 0.01$
$|{T_{mean}} - {T_4}| = 0.02$
$|{T_{mean}} - {T_5}| = 0.03$

$ \Rightarrow \Delta {T_{mean}} = {{0.04 + 0 + 0.01 + 0.02 + 0.03} \over 5} = 0.02$

Thus, the error in the measurement of T is

${{\Delta T} \over T} \times 100 = {{0.02} \over {0.56}} \times 100 = 3.57\% $ .... (1)

Hence, option (B) is correct.

The error in the measurement of r is

${{\Delta r} \over r} \times 100 = {1 \over {10}} \times 100 = 10\% $ ..... (2)

Hence, option (A) is correct.

Now, it is given that

$T = 2\pi \sqrt {{{7(R - r)} \over {5g}}} $

$ \Rightarrow {T^2} = 4{\pi ^2}\left( {{{7(R - r)} \over {5g}}} \right)$

$ \Rightarrow g = {{28{\pi ^2}} \over 5}\left( {{{R - r} \over {{T^2}}}} \right)$

Taking log and differentiating on both the sides of this equation, we get

$\ln g = \ln \left( {{{28{\pi ^2}} \over 5}} \right) + \ln (R - r) - 2\ln T$

$ \Rightarrow {{\Delta g} \over g} = \left( {{{\Delta R + \Delta r} \over {R - r}}} \right) + {{2\Delta T} \over T}$

$ = \left( {{{1 + 1} \over {50}}} \right) + 2 \times 0.0357$

$ \Rightarrow {{\Delta g} \over g} \times 100 = \left( {{1 \over {25}} + 2 \times 0.0357} \right) \times 100$

$ = (4 + 7.14)\% = 11.14\% = 11\% $

Therefore, the error in the determined value g is 11%.

Answer (A), (B), (D)

$[n] = [{L^{ - 3}}];[q] = [AT]$

$[\varepsilon ] = [{M^{ - 1}}{L^{ - 3}}{A^2}{T^4}]$

$[T] = [L]$

$[l] = [L]$

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

(a) RHS

$ = \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}} $

$ = \sqrt {{{[{L^{ - 3}}{A^2}{T^2}]} \over {[{L^{ - 1}}{A^2}{T^2}]}}} = \sqrt {[{L^{ - 2}}]} = [{L^{ - 1}}]$ Wrong

(b) RHS

$ = \sqrt {{{[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]} \over {[{L^{ - 3}}{A^2}{T^2}]}}} $

$ = \sqrt {{{[{L^{ - 1}}{A^2}{T^2}]} \over {[{L^{ - 3}}{A^2}{T^2}]}}} = L$ Correct

(c) RHS

$ = \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 2}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}} $

$ = \sqrt {[{L^3}]} $ Wrong

(d) RHS

$ = \sqrt {{{[{A^2}{T^2}]} \over {[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}][{L^{ - 1}}][{M^1}{L^2}{T^{ - 2}}{K^{ - 1}}][K]}}} $

$ = \sqrt {{{[{A^2}{T^2}]} \over {[{L^{ - 2}}{T^2}{A^2}]}}} = [L]$ Correct

The dimensions of Planck's constant is $h = [{M^1}{L^3}{T^{ - 2}}]$

Speed of light is $c = [{L^1}{T^{ - 1}}]$

Gravitational constant is $G = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

Let $L \propto {h^x}{c^y}{G^z}$.

$[L] = {[{M^1}{L^2}{T^{ - 1}}]^x}{[{L^1}{T^{ - 1}}]^y}{[{M^{ - 1}}{L^3}{T^{ - 2}}]^z}$

Comparing, we get

$\left. {\matrix{ \hfill {x - z = 0} \cr \hfill {2x + y + 3z = 1} \cr \hfill { - x - y - 2z = 0} \cr } } \right\}$

Solving, we get

$x = z$

$y + 5x = 1$

$ - y - 3x = 0$

or, $2x = 1$

$x = {1 \over 2} = z$

$y = - {3 \over 2}$

Therefore,

$L \propto {h^{1/2}}{C^{ - 3/2}}{G^{1/2}}$

$L \propto \sqrt h $

$L \propto \sqrt G $

Let $M \propto {h^a}{C^b}{G^c}$

$[M] = {[{M^1}{L^2}{T^{ - 1}}]^a}{[{L^1}{T^{ - 1}}]^b}{[{M^{ - 1}}{L^3}{T^{ - 2}}]^c}$

Comparing, we get $a - c = 1$.

$\left. \matrix{ 2a + b + 3c = 0 \hfill \cr - a - b - 2c = 0 \hfill \cr} \right\}a + c = 0$

Therefore,

$2a = 1$

$a = {1 \over 2}$

$c = - {1 \over 2}$

$b = {1 \over 2}$

Therefore,

$M \propto {h^{1/2}}{c^{1/2}}{G^{ - 1/2}}$

$M \propto \sqrt c $

In given Vernier callipers, each 1 cm is equally divided into 8 main scale divisions (MSD). Thus, 1 MSD = ${1 \over 8}$ = 0.125 cm. Further, 4 main scale divisions coincide with 5 Vernier scale divisions (VSD) i.e., 4 MSD = 5 VSD. Thus, 1 VSD = 4/5 MSD = 0.8 $\times$ 0.125 = 0.1 cm. The least count of the Vernier callipers is given by

LC = 1 MSD $-$ 1 VSD = 0.125 $-$ 0.1 = 0.025 cm.

In screw gauge, let l be the distance between two adjacent divisions on the linear scale. The pitch p of the screw gauge is the distance travelled on the linear scale when it makes one complete rotation. Since circular scale moves by two divisions on the linear scale when it makes one complete rotation, we get p = 2l. The least count of the screw gauge is defined as ratio of the pitch to the number of divisions on the circular scale (n) i.e.,

$LC' = {p \over n} = {{2l} \over {100}} = {l \over {50}}$ ....... (1)

If p = 2 LC = 2(0.025) = 0.05 cm, then $l = {p \over 2} = 0.025$ cm. Substitute l in equation (1) to get the least count of the screw gauge

$LC' = {{0.025} \over {50}} = 5 \times {10^{ - 4}}$ cm = 0.005 mm.

If l = 2 LC = 2(0.025) = 0.05 cm then equation (1) gives

$LC' = {{0.05} \over {50}} = 1 \times {10^{ - 3}}$ cm = 0.01 mm.

Relative error in measurement of time,

${{\Delta t} \over t} = {{1\,s} \over {40\,s}} = {1 \over {40}}$

Time period, $T = {{40\,s} \over {20}} = 2\,s$

Error in measurement of time period,

$\Delta T = T \times {{\Delta t} \over t} = 2\,s \times {1 \over {40}} = 0.05\,s$

The time period of simple pendulum is

$T = 2\pi \sqrt {{l \over g}} $

or, ${T^2} = {{4{\pi ^2}l} \over g}$

or, $g = {{4{\pi ^2}l} \over {{T^2}}}$

$\therefore$ ${{\Delta g} \over g} = {{2\Delta T} \over T} = 2 \times {1 \over {40}} = {1 \over {20}}$ ($\because$ ${{\Delta T} \over T} = {{\Delta t} \over t}$)

Percentage error in determination of g is

${{\Delta g} \over g} \times 100 = {1 \over {20}} \times 100 = 5\% $

From coulombs law, we know that

$F = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over {{r^2}}} \Rightarrow {\varepsilon _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {{r^2}F}}$

Therefore, ${\varepsilon _0} = {{[{I^2}{T^2}]} \over {[{L^2}][ML{T^{ - 2}}]}} = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$

Now, $c = {1 \over {\sqrt {{\varepsilon _0}{\mu _0}} }} \Rightarrow {c^2} = {1 \over {{\varepsilon _0}{\mu _0}}}$

Therefore, ${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}} = {1 \over {{{[L{T^{ - 1}}]}^2}[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}]}} = [ML{T^{ - 2}}{I^{ - 2}}]$

Answer (B), (C)

Induction $L = {\phi \over i} = $ weber/Ampere

Also, $e = - L\left( {{{di} \over {dt}}} \right) \Rightarrow L = {{ - e} \over {(di/dt)}} = {{volt - {\mathop{\rm second}\nolimits} } \over {Ampere}}$

Also, $U = {1 \over 2}L{i^2} \Rightarrow L = {{2U} \over {{i^2}}} = {{Joule} \over {{{(Ampere)}^2}}}$

and $U = {1 \over 2}L{i^2} = {i^2}RT \Rightarrow L = RT = $ Ohm-second

Answer (A), (B), (C), (D)