Units & Measurements

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]

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.

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

Time period of a pendulum (T) = $2\pi \sqrt {{l \over g}} $

$ \Rightarrow $ T² = $4{\pi ^2}{l \over g}$

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

Fractional change

$\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 - \left( {2{{dT} \over T}} \right) \times 100$

$ \therefore $ Maximum possible percentage error,

$\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 + \left( {2{{dT} \over T}} \right) \times 100$

Error in time period(dT) = least count of time = 1 second

and T = 30 second

Error in length(dl) = least count of length = 1 mm

and $l$ = 55.0 cm

$ \therefore $ $\left( {{{dg} \over g}} \right) \times 100 =$ $\left( {{{0.1} \over {55}}} \right) \times 100 + 2\left( {{1 \over {30}}} \right) \times 100$ = 6.8%

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] = [S^a I^b h^c]

$ \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 $ ${\left[ {ML{T^{ - 1}}} \right]}$ = [ M^{a + b + c} L^{2b +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] = [S^1/2I^1/2h⁰]

$\sqrt {{{{ \in _0}} \over {{\mu _0}}}} $ = ${{{ \in _0}} \over {\sqrt {{\mu _0}{ \in _0}} }}$ = c $ \times $ ${{ \in _0}}$

$ \therefore $ $\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$ = $\left[ {L{T^{ - 1}}} \right] \times \left[ {{ \in _0}} \right]$

We know, F = ${1 \over {4\pi { \in _0}}}{{{q^2}} \over {{r^2}}}$

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

$ \Rightarrow $ $\left[ {{ \in _0}} \right] = {{{{\left[ {AT} \right]}^2}} \over {\left[ {ML{T^{ - 2}}} \right] \times \left[ {{L^2}} \right]}}$ = $\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$

$ \therefore $ $\left[ {\sqrt {{{{ \in _0}} \over {{\mu _0}}}} } \right]$ = $\left[ {L{T^{ - 1}}} \right]$ $ \times $ $\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$

                 = [A²T³M^–1L^–2]

$\left[ {{\ell \over r}} \right] = $ T

[CV] $=$ AT

So,   $\left[ {{\ell \over {rCV}}} \right]$ = ${T \over {AT}}$ = [A^$-$1]

Least count = ${{Pitch} \over {Number\,\,of\,\,division\,\,on\,\,circular\,scale}}$

5 $ \times $ 10^$-$6 = ${{{{10}^{ - 3}}} \over N}$

N = 200

We know,

Young's modulus (Y) = ${{{F \over A}} \over {{{\Delta l} \over l}}}$

$ \therefore $ [Y] = ${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}$ = [ ML^-1T^-2]

Let [Y] = [V]^x [A]^y [F]^z

$ \therefore $ [ ML^-1T^-2] =

[LT^-1]^x [LT^-2]^y [MLT^-2]^z

$ \Rightarrow $ [ ML^-1T^-2] =

[ M^z L^{x + y + z} T^{-x -2y - 2z}

For dimensional balance, the dimension on both sides should be same.

So, z = 1

x + y + z = -1

$ \Rightarrow $ x + y = -2 ........(1)

and -x -2y - 2z = -2

$ \Rightarrow $ x + 2y = 0 ...........(2)

By solving those two equations we get,

x = -4 and y = 2

$ \therefore $ [Y] = V^$-$4A²F¹

$F = \alpha \beta {e^{\left( {{{ - {x^2}} \over {\alpha KT}}} \right)}}$

$\left[ {{{{x^2}} \over {\alpha KT}}} \right] = {M^o}{L^o}{T^o}$

${{{L^2}} \over {\left[ \alpha \right]M{L^2}{T^{ - 2}}}}$ $=$ ${M^o}{L^o}{T^o}$

$ \Rightarrow $  $\left[ \alpha \right] = {M^{ - 1}}{T^2}$

$\left[ F \right] = \left[ \alpha \right]\left[ \beta \right]$

MLT^$-$2 = M^$-$1T²[$\beta $]

$ \Rightarrow $  [$\beta $] = M²LT^$-$4

Volume of cylinder(V) = $\pi $r²h

= $\pi {{{d^2}} \over 4}h$

= $3.14 \times {{{{\left( {12.6} \right)}^2}} \over 4} \times 34.2$

= 4260

${{\Delta V} \over V} = 2{{\Delta d} \over d} + {{\Delta h} \over h} = 2\left( {{{0.1} \over {12.6}}} \right) + {{0.1} \over {34.2}}$ = 0.0188

$ \therefore $ $\Delta $V = 0.0188 $ \times $ 4260 = 80

Here given that

density of a material in SI units is 128 kg m^–3

And a new unit system is introduced where 1 unit of length = 25 cm and 1 unit of mass = 50 g

You should know that, physical quantity is same in any unit system. And to calculate a physical quantity yo should know two things

(1) numerical value of the physical quantity (n)

(2) unit of the physical quantity (u)

And n $ \times $ u = constant in any unit system.

Here in SI unit system,

n₁ = 128

u₁ = kg/m³

And in new unit system,

n₂ = ?

u₂ = 50gm/(25cm)³

As n₁u₁ = n₂u₂

$ \therefore $ 128 $ \times $ (kg/m³) = n₂ $ \times $ 50gm/(25cm)³

$ \Rightarrow $ 128 $ \times $ ${{1000\,gm} \over {{{\left( {100\,cm} \right)}^3}}}$ = n₂ $ \times $ ${{50\,gm} \over {{{\left( {25\,cm} \right)}^3}}}$

$ \Rightarrow $ n₂ = 128 $ \times $ ${{20} \over {{4^3}}}$ = 40

We know,

Least count (LC) = ${{Pitch} \over {no.\,of\,divisions}}$

$ \therefore $  LC = ${{0.5} \over {100}}$

= 0.5 $ \times $ 10^$-$2 mm

Reading = MSR + CSR $-$ positive error

Given, Main scale reading (MSR) = 5.5 mm

Circular scale reading (CSR)

= 48 $ \times $ 0.5 $ \times $ 10^$-$2 mm

= 0.24

As zero of its circular scale lines 3 division below the mean line, it means error is position error.

$ \therefore $  positive error

= 3 $ \times $ 0.5 $ \times $ 10^$-$2 mm

= 0.015 mm

$ \therefore $  Reading = 5.5 + 0.24 $-$ 0.015

= 5.725 mm

Let t $ \propto $ G^x h^y c^z

$ \therefore $   [t] = [G]^x [h]^y [c]^z    . . . . . (1)

We know,

F = ${{G{M^2}} \over {{R^2}}}$

$ \Rightarrow $  G = ${{F{R^2}} \over {{M^2}}}$

$ \therefore $  [G] = ${{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]} \over {\left[ {{M^2}} \right]}}$

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

Also,

E = hf

$ \therefore $  [h] = ${{[E]} \over {[F]}}$

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

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

[C] = $\left[ {{M^o}L\,{T^{ - 1}}} \right]$

From equation (1) we get,

$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = {\left[ {{M^{ - 1}}\,{L^3}\,{T^{ - 2}}} \right]^x}{\left[ {M\,{L^2}\,{T^{ - 1}}} \right]^y}{\left[ {{M^o}L{T^{ - 1}}} \right]^z}$

$\left[ {{M^o}\,{L^o}\,{T^1}} \right] = \left[ {{M^{ - x + y}}\,{L^{3x + 2yz}}\,{T^{ - 2x - y - z}}} \right]$

By comparing the power of M, L, T

$-$ x + y = 0

$ \Rightarrow $  x = y

3x + 2y + z = 0

$ \Rightarrow $  5x + z = 0 . . . . . (2)

$-$ 2x $-$ y $-$ z = 1

$ \Rightarrow $  $-$ 3x $-$ z = 1  . . . . (3)

By solving (2) and (3), we get,

x = ${1 \over 2}$ = y and z = $-$ ${5 \over 2}$

$ \therefore $  t $ \propto $ $\sqrt {{{Gh} \over {{C^5}}}} $

We know,

$R = {{\rho l} \over A}$

and Volume (V) = A$l$

$ \Rightarrow $   A $=$ ${V \over l}$

$ \therefore $   $R = {{\rho {l^2}} \over v}$

$ \therefore $   ${{\Delta R} \over R} = 2{{\Delta l} \over l}$

$=$   2 $ \times $ 0.5

$=$ 1%

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

[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]

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

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 %

Density of a material (d) = ${M \over {{L^3}}}$

$\therefore$ Error in density,${{\Delta d} \over d} = {{\Delta M} \over M} + 3{{\Delta L} \over L}$

${{\Delta d} \over d} \times 100 = {{\Delta M} \over M} \times 100 + 3{{\Delta L} \over L} \times 100$

$ \Rightarrow {{\Delta d} \over d} \times 100 = 1.5\% + 3\left( 1 \right)\% $ = 4.5 %

Plank length,

$\ell $ = k G^p $\hbar $^q C^r

[ M^o L T^o] = [ M^$-$1 L³ T^$-$2 ]^p [ M L² T^$-$1] ^q [ L T^$-$1]^r

[M^o L T^o ] = [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)

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 $

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.

Given, ratio $r = {{1 - a} \over {1 + a}}$ ..... (1)

Given, error in measurement of a is $\Delta$a ($\Delta$a/a << 1)

Taking natural log of Eq. (1), we get

$\ln r = \ln \left( {{{1 - a} \over {1 + a}}} \right)$

$ \Rightarrow \ln r = \ln (1 - a) - \ln (1 + a)$

${{\Delta r} \over r} = {{\Delta a} \over {1 - a}} + {{\Delta a} \over {1 + a}} = {{(1 + a)\Delta a + (1 - a)\Delta a} \over {(1 - a)(1 + a)}}$

$ \Rightarrow {{\Delta r} \over r} = {{2\Delta a} \over {(1 - a)(1 + a)}}$

$ \Rightarrow \Delta r = r\,.\,{{2\Delta a} \over {(1 - a)(1 + a)}}$

Substituting value of r, we get

$\Delta r = {{1 - a} \over {1 + a}}{{2\Delta a} \over {(1 - a)(1 + a)}} = {{2\Delta a} \over {{{(1 + a)}^2}}}$

Therefore, error $\Delta$r in determining r is ${{2\Delta a} \over {{{(1 + a)}^2}}}$

The law of radioactive decay gives number of undecayed nuclei N at time t as $N = {N_0}{e^{ - \lambda t}}$, where N₀ is number of nuclei at t = 0 and $\lambda$ is the decay constant. The number of nuclei decayed till time t is given by ${N_d} = {N_0} - N$. Thus, the decay constant is given by

$\lambda = {1 \over t}\ln {{{N_0}} \over N} = {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}}$.

Since, N₀ = 3000 and t = 1 s are given without any error, we assume that error in $\lambda$ is due to error in N_d only. Hence, we can write above equation as

$\lambda + \Delta \lambda = {1 \over t}\ln {{{N_0}} \over {{N_0} - ({N_d} + \Delta {N_d})}}$

$ = {1 \over t}\ln {{{N_0}} \over {({N_0} - {N_d})(1 - \Delta {N_d}/({N_0} - {N_d}))}}$

$ = {1 \over t}\ln {{{N_0}} \over {{N_0} - {N_d}}} - {1 \over t}\ln \left( {1 - {{\Delta {N_d}} \over {{N_0} - {N_d}}}} \right)$

$ \approx \lambda + {1 \over t}{{\Delta {N_d}} \over {{N_0} - {N_d}}}$,

which gives $\Delta \lambda = {{\Delta {N_d}} \over {t({N_0} - {N_d})}}$. Substitute $\Delta$N_d = 40, t = 1 s, and N_d = 1000 to get $\Delta$$\lambda$ = 0.02.

Given,

P = a$^{{1 \over 2}}$ b² c³ d^$-$4

Relative error =

${{\Delta P} \over P}$ $ \times $ 100 = (${1 \over 2}$ $ \times $ ${{\Delta a} \over a}$ + 2${{\Delta b} \over b}$ + 3${{\Delta c} \over c}$ + 4${{\Delta d} \over d}$) $ \times $ 100

= ${1 \over 2}$ $ \times $ 2 + 2 $ \times $ 1 + 3 $ \times $ 3 + 4 $ \times $ 5

= 32%

Let,

M $ \propto $ T^x C^y h^z

$\therefore\,\,\,$ [M¹L^oT^o]  =  [T¹]^x  [L¹ T^$-$1]^y  [M¹L²T^$-$1]^z

[M¹L^o T^o]  =  [M^z L^{y + 2z} T^x$-$y$-$z]

By comparing both sides we get,

z = 1

y + 2z = 0

x $-$ y $-$ z = 0

$\therefore\,\,\,$ y = $-$ 2z = $-$ 2

x = y + z = $-$2 + 1 = $-$1

$\therefore\,\,\,$ [M] = [M^$-$1 C^$-$2 h¹]

Surface tension,

T = ${{rhg} \over 2} \times {10^3}N/m$

Relative error,

${{\Delta T} \over T} = {{\Delta r} \over r} + {{\Delta h} \over h}$

Percentage error,

${{\Delta T} \over T} \times 100 = {{\Delta r} \over r} \times 100 + {{\Delta h} \over h} \times 100$

${{\Delta T} \over T} \times 100 = \left( {{{{{10}^{ - 2}} \times 0.01} \over {1.25 \times {{10}^{ - 2}}}} + {{{{10}^{ - 2}} \times 0.01} \over {1.45 \times {{10}^{ - 2}}}}} \right) \times 100$

= (0.8 + 0.689) = 1.489 % = 1.5 %

Let the time taken by stone to reach bottom of will be t₁ and time taken by sound to reach the observer be t₂ then

${t_1} = \sqrt {{{2L} \over g}} $ .... (1_)

and ${t_2} = {L \over {{v_s}}}$ .... (2)

Where L is depth of well

g is acceleration due to gravity

v_s is velocity of sound

Total time taken, $T = {t_1} + {t_2} = {{\sqrt {2L} } \over g} + {L \over {{v_s}}}$ ..... (3)

If $\delta$L is error in depth of well measured and $\delta$T is error is time measured then

$T + \delta T = \sqrt {{{2(L + \delta L)} \over g}} + {{(L + \delta L)} \over {{v_s}}}$

$T + \delta T = \sqrt {{{2L} \over g}\left( {1 + {{\delta L} \over L}} \right)} + {L \over {{v_s}}}\left( {1 + {{\delta L} \over L}} \right)$

expanding ${\left( {1 + {{\delta L} \over L}} \right)^{1/2}}$ using binomial approximation

$T + \delta T = \sqrt {{{2L} \over g}} \left( {1 + {1 \over 2}{{\delta L} \over L}} \right) + {L \over {{V_s}}}\left( {1 + {{\delta L} \over L}} \right)$

$ = \sqrt {{{2L} \over g}} + \sqrt {{{2L} \over g}} \left( {{1 \over 2}{{\delta L} \over L}} \right) + {L \over {{v_s}}} + {L \over {{v_s}}}{{\delta L} \over L}$

$ = \sqrt {{{2L} \over g}} + {L \over {{v_s}}} + \left( {{1 \over 2}\sqrt {{{2L} \over g}} + {L \over {{v_s}}}} \right){{\delta L} \over L}$

given, L = 20 m, g = 10 m/s, v_s = 300 m/s and using equation (3) : $\sqrt {{{2L} \over g}} + {L \over {{v_s}}} = T$, above equation becomes

$T + \delta T = T + \left( {{1 \over 2}\sqrt {{{2 \times 20} \over {10}}} + {{20} \over {300}}} \right){{\delta L} \over L}$

$ \Rightarrow \delta T = \left( {{1 \over 2}\sqrt 4 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {1 + {1 \over {15}}} \right){{\delta L} \over L} = \left( {{{16} \over {15}}} \right){{\delta L} \over L}$

$ \Rightarrow {{\delta L} \over L} = \delta T\left( {{{15} \over {16}}} \right)$

It is given that $\delta$T = 0.01 sec, therefore,

$\left( {{{\delta L} \over L}} \right) \times 100\% = {{15} \over {16}} \times {1 \over {100}} \times 100\% = {{15} \over {16}}\% $

error $ \simeq 1\% $

Answer (B)

We know. resistivity ($\rho $) = ${{RA} \over L}$

and conductivity = ${1 \over \rho }$ = ${1 \over {RA}}$

As   R = ${V \over {\rm I}}$

$ \therefore $   conductivity = ${{L{\rm I}} \over {VA}}$

Also  V = ${\omega \over q}$ = ${\omega \over {it}}$ = ${{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {\rm I} \right]\left[ T \right]}}$ = $\left[ {M{L^2}{T^{ - 3}}{{\rm I}^{ - 1}}} \right]$

$ \therefore $   Conductivity = ${{\left[ {\rm{L}} \right]\left[ {\rm I} \right]} \over {\left[ {M{L^2}{T^{ - 3}}{{\rm{I}}^{ - 1}}} \right]\left[ {{L^2}} \right]}}$

=   $\left[ {{M^{ - 1}}{L^{ - 3}}{T^3}{{\rm I}^2}} \right]$

Least count = ${{0.5} \over {50}}$ = 0.01 mm

Zero error = (45 - 50)$ \times $0.01 mm = - 0.05 mm

Thickness of sheet = (0.5 + 25$ \times $0.01) - (-0.05)

= 0.50 + 0.30 = 0.80 mm

Here t₁ = 90 s, t₂ = 91 s, t₃ = 95 s, t₄ = 92 s

Mean(t) = ${{{t_1} + {t_2} + {t_3} + {t_4}} \over 4}$

= ${{90 + 91 + 95 + 92} \over 4}$ = 92 s

Now mean deviation

= ${{2 + 1 + 3 + 0} \over 4}$ = 1.5 s

Since least count of clock is one second, so reported mean time

= (92 $ \pm $ 2) s

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.

In both calipers C₁ and C₂, 1 cm is divided into 10 equal divisions on the main scale. Thus, 1 division on the main scale is equal to x_m1 = x_m2 = 1 cm/10 = 0.1 cm.

In calipers C₁, 10 equal divisions on the Vernier scale are equal to 9 main scale divisions.
Thus, 1 division on the Vernier scale of C₁ is equal to x_v1 = 9x_m1/10 = 0.09 cm.

In calipers C₂, 10 equal divisions on the Vernier scale are equal to 11 main scale divisions.
Thus, 1 division on the Vernier scale of C₂ is equal to x_v2 = 11x_m2/10 = 0.11 cm.


Let main scale reading be MSR and v^th division of the Vernier scale coincides with m^th division of the main scale (m is counted beyond MSR). The value measured by this calipers is

X = MSR + x = MSR + mx_m $-$ vx_v .........(1)

In calipers C₁, MSR₁ = 2.8 cm, m₁ = 7 and v₁ = 7 and in calipers C₂, MSR₂ = 2.8 cm, m₂ = 8 and v₂ = 7. Substitute these values in equation (1) to get

X₁ = MSR₁ + m₁x_m1 $-$ v₁x_v1

= 2.8 + 7(0.1) $-$ 7(0.09) = 2.87 cm.

X₂ = MSR₂ + m₂x_m2 $-$ v₂x_v2

= 2.8 + 8(0.1) $-$ 7(0.11) = 2.83 cm.

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.

Measured length of rod = 3.50 cm

That means least count of the measuring instrument should be 0.01 cm = 0.1 mm

For vernier scale 1 main scale division = 1 mm

And 9 MSD = 10 VSD

Least count = 1 MSD - 1 VSD

= 1 - 0.9 = 0.1 mm

From Coulomb's law we know,

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

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

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

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

Let's break down the dimensional analysis of each physical quantity:

P. Boltzmann Constant ($k_B$)

The Boltzmann constant relates energy to temperature. Its dimensional formula is derived from the relationship: $E = k_B T$.

Energy ($E$) has dimensions of [ML²T^-2]. Temperature ($T$) has dimensions of [K]. Therefore:

$[k_B] = \frac{[E]}{[T]} = \frac{[ML^2T^{-2}]}{[K]} = [ML^2T^{-2}K^{-1}]$

Q. Coefficient of Viscosity ($\eta$)

Viscosity is a measure of a fluid's resistance to flow. It is defined as the ratio of shear stress to shear rate. Shear stress has dimensions of [ML^-1T^-2] (force per unit area), and shear rate has dimensions of [T^-1] (velocity gradient). Therefore:

$[\eta] = \frac{[ML^{-1}T^{-2}]}{[T^{-1}]} = [ML^{-1}T^{-1}]$

R. Plank Constant ($h$)

Planck's constant relates energy to frequency. Its dimensional formula is derived from the relationship: $E = h\nu$.

Energy ($E$) has dimensions of [ML²T^-2]. Frequency ($\nu$) has dimensions of [T^-1]. Therefore:

$[h] = \frac{[E]}{[\nu]} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$

S. Thermal Conductivity ($k$)

Thermal conductivity is a measure of a material's ability to conduct heat. It is defined as the rate of heat transfer per unit area per unit temperature gradient. Heat transfer rate has dimensions of [ML²T^-3]. Area has dimensions of [L²], and temperature gradient has dimensions of [K/L]. Therefore:

$[k] = \frac{[ML^2T^{-3}]}{[L^2][K/L]} = [MLT^{-3}K^{-1}]$

Matching the Dimensions:

Based on the dimensional analysis, we can match the quantities in List I with the dimensions in List II as follows:

List I	List II
P. Boltzmann Constant ($k_B$)	4. [ML2T-2K-1]
Q. Coefficient of Viscosity ($\eta$)	2. [ML-1T-1]
R. Plank Constant ($h$)	1. [ML2T-1]
S. Thermal conductivity ($k$)	3. [MLT-3K-1]

Therefore, the correct answer is Option C.

From the given data :

• One main scale division (MSD) is $ 5.15 \text{ cm} - 5.10 \text{ cm} = 0.05 \text{ cm} $.


• One Vernier scale division (VSD) is $ \frac{2.45 \text{ cm}}{50} = 0.049 \text{ cm} $.

The least count (LC) of the Vernier calipers is:

$ \text{LC} = 1 \text{ MSD} - 1 \text{ VSD} = 0.05 \text{ cm} - 0.049 \text{ cm} = 0.001 \text{ cm} $

Given the main scale reading (MSR) is 5.10 cm and the Vernier scale reading (VSR) is 24. The diameter $ D $ of the cylinder is calculated as:

$ \begin{aligned} D & = \text{MSR} + \text{VSR} \times \text{LC} \\\\ & = 5.10 \text{ cm} + 24 \times 0.001 \text{ cm} \\\\ & = 5.124 \text{ cm} \end{aligned} $

We know R = ${V \over I}$

$\therefore$ ${{\Delta R} \over R} = {{\Delta V} \over V} + {{\Delta I} \over I}$

Percentage error in R =

${{\Delta R} \over R} \times 100 = {{\Delta V} \over V} \times 100 + {{\Delta I} \over I} \times 100$

$\therefore$ ${{\Delta R} \over R} \times 100 =$ 3% + 3 % = 6%

30 vernier scale divisions coincide with 29 main scale divisions.

Therefore 1 V.S.D = ${{29} \over {30}}$ M.S.D

Least count = 1 M.S.D - 1 V.S.D

= 1 M.S.D - ${{29} \over {30}}$ M.S.D

= ${{1} \over {30}}$ M.S.D

= ${{1} \over {30}}$ $ \times $ 0.5^o

Reading of Vernier = Main Scale Reading + Vernier scale reading $ \times $ Least count

Given that,

Main Scale Reading = 58.5

Vernier scale reading = 09 division

$\therefore$ Reading of Vernier = 58.5^o + 9 $ \times $ ${{0.5^\circ } \over {30}}$

= 58.65^o

To determine Young's modulus using Searle's method, we use the formula:

$ Y = \frac{4MLg}{\pi l d^2} $

Given:
Length of wire, $L = 2 \text{ m}$
Diameter of wire, $d = 0.5 \text{ mm}$
Load, $M = 2.5 \text{ kg}$
Extension in length, $l = 0.25 \text{ mm}$

The quantities $d$ and $l$ are measured using a screw gauge and a micrometer, respectively, both having a pitch of 0.5 mm and 100 divisions on their circular scale. The least count, which is the smallest measurement that can be accurately read, is:

$ \text{Least count} = \frac{\text{Pitch}}{\text{Total number of divisions on the circular scale}} = \frac{0.5 \text{ mm}}{100} = 0.005 \text{ mm} $

Hence, the least count, $\Delta d$ and $\Delta l$, are both 0.005 mm.

Error in measurement of $l$ (extension length) is:

$ \frac{\Delta l}{l} = \frac{0.005}{0.25} = \frac{1}{50} $

Error in measurement of $d$ (diameter) is:

$ \frac{\Delta d}{d} = \frac{0.005}{0.5} = \frac{1}{100} $

Using the formula for Young's modulus, the combined error is:

$ Y = \frac{4MLg}{\pi ld^2} $

The relative error in $Y$ is given by:

$ \frac{\Delta Y}{Y} = \frac{\Delta l}{l} + 2 \frac{\Delta d}{d} $

As can be seen:

$ \frac{\Delta l}{l} = 2 \frac{\Delta d}{d} = \frac{1}{50} $

Therefore, the contributions to the error in the measurement of $Y$ from errors in measuring $l$ and $d$ are the same.

Least count of screw gauge

= ${{Pitch} \over {Number\,of\,division\,on\,circular\,scale}}$

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

= 0.01 mm

Diameter of the wire = M.S.R + C.S.R $ \times $ L.C

= 0 + 52 $ \times $ 0.01

= 0.52 mm

= 0.052 cm