Units & Measurements

To determine the percentage error in the measurement of $ C $, which is related to $ p $, $ q $, $ r $, and $ s $ as:

$ C = \frac{p q^2}{r^3 \sqrt{s}} $

we first express it in terms of powers:

$ C = p^1 q^2 r^{-3} s^{-1/2} $

The percentage error in $ C $ can be calculated using the formula for the propagation of error, which is:

$ \left(\frac{\Delta C}{C}\right)_{\max} = \left|\frac{\Delta p}{p}\right| + 2\left|\frac{\Delta q}{q}\right| + 3\left|\frac{\Delta r}{r}\right| + \frac{1}{2}\left|\frac{\Delta s}{s}\right| $

Given the percentage errors for $ p $, $ q $, $ r $, and $ s $ are $ 1\% $, $ 2\% $, $ 3\% $, and $ 2\% $ respectively, we substitute these values into the formula:

$ \left(\frac{\Delta C}{C}\right)_{\max} = 1\% + 2 \times 2\% + 3 \times 3\% + \frac{1}{2} \times 2\% $

Calculating each part:

Contribution from $ p $: $ 1\% $

Contribution from $ q $: $ 4\% $ (since $ 2 \times 2\% = 4\% $)

Contribution from $ r $: $ 9\% $ (since $ 3 \times 3\% = 9\% $)

Contribution from $ s $: $ 1\% $ (since $ \frac{1}{2} \times 2\% = 1\% $)

Adding these contributions together:

$ 1\% + 4\% + 9\% + 1\% = 15\% $

Thus, the maximum percentage error in the measurement of $ C $ is $ 15\% $.

Given, $Q = {{a{b^4}} \over {cd}}$

$a = (60 \pm 3)\,Pa \Rightarrow a = 60\,Pa,\,\Delta a = 3Pa$

$b = (20 \pm 0.1)\,m \Rightarrow b = 20m,\,\Delta b = 0.1\,m$

$c = (40 \pm 0.2)\,Ns{m^{ - 2}} \Rightarrow c = 40\,Ns{m^{ - 2}},\,\Delta c = 0.2\,Ns{m^{ - 2}}$

$d = (50 \pm 0.1)\,m \Rightarrow \,d = 50\,m,\,\Delta d = 0.1m$

As, $Q = {{a{b^4}} \over {cd}}$

by taking ln on both sides,

$\ln Q = \ln a + u\ln b - \ln c - \ln d$

Now, by differentiating,

${{dQ} \over Q} = {{da} \over a} + 4{{db} \over b} - {{dc} \over c} - {{dd} \over d}$

So, maximum fractional error in Q is given by,

${{\Delta Q} \over Q} = {{\Delta a} \over a} + 4{{\Delta b} \over b} + {{\Delta c} \over c} + {{\Delta d} \over d}$

$ \Rightarrow {{\Delta Q} \over Q} = {3 \over {60}} + 4\left( {{{0.1} \over {20}}} \right) + {{0.2} \over {40}} + {{0.1} \over {40}}$

$ = {1 \over {20}} + {1 \over {50}} + {1 \over {200}} + {1 \over {500}}$

$ \Rightarrow {{\Delta Q} \over Q} = {{50 + 20 + 5 + 2} \over {1000}} = {{77} \over {1000}}$

Hence the % error in $Q = {{\Delta Q} \over Q} \times 100\% $

$ = {{7700} \over {1000}}\% = {x \over {1000}}$ (given)

So, $x = 7700$

Given, L = 5 mm, B = 2.5 mm

We know, least count of a screw gauge,

$L.C. = {\text{Pitch length} \over \text{No. of division on circular scale}}$

$ \Rightarrow L.C. = {{0.75} \over {15}} = 0.05\,mm$

We know, $A = LB$

By taking $ln$ on both sides,

$ \Rightarrow \ln A = \ln L + \ln B$

by differentiating both sides,

$ \Rightarrow {{dA} \over A} = {{dL} \over L} + {{dB} \over B}$

Here, $dL = dB = 0.05\,mm$ (L.C.)

So fractional error,

${{dA} \over A} = {{0.05} \over 5} + {{0.05} \over {2.5}}$

$ = {1 \over {100}} + {2 \over {100}} = {3 \over {100}} = {x \over {100}}$ (given)

Hence, $x = 3$.

The least count (L.C.) of a screw gauge is given by:

$ \text{L.C.} = \frac{\text{Pitch}}{\text{Number of Divisions on Circular Scale}} $

Let the original pitch be $p$ and the original number of divisions be $n$. Then, the original least count is:

$ \frac{p}{n} = 0.01 \, \text{mm} $

After modifications:

The pitch is increased by 75%, so the new pitch is:

$ p_{\text{new}} = p + 0.75p = 1.75p $

The number of divisions is reduced by 50%, so the new number of divisions is:

$ n_{\text{new}} = 0.5n $

The new least count is then:

$ \text{L.C.}_{\text{new}} = \frac{p_{\text{new}}}{n_{\text{new}}} = \frac{1.75p}{0.5n} = \frac{1.75}{0.5} \times \frac{p}{n} = 3.5 \times \frac{p}{n} $

Substitute the original least count:

$ \text{L.C.}_{\text{new}} = 3.5 \times 0.01 \, \text{mm} = 0.035 \, \text{mm} $

Expressed in the form $\times 10^{-3} \, \text{mm}$, this becomes:

$ 0.035 \, \text{mm} = 35 \times 10^{-3} \, \text{mm} $

Thus, the new least count is:

$ 35 \times 10^{-3} \, \text{mm} $

Least count of screw gauge $ = {{0.5} \over {100}}$ mm $ = {{1} \over {200}}$ mm

Zero error of screw gauge $ = +{{6} \over {200}}$ mm $ = +{{3} \over {100}}=0.03$ mm

Reading of screw gauge $ = 4\times0.5+{{46} \over {200}}$ mm

$ = 2+{{23} \over {100}}$ mm $=2.23$ mm

So diameter of wire $=2.23$ mm $-~0.03$ mm

$=2.20$ mm

$=220\times10^{-2}$ mm

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

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

$ = 2 \times {1 \over {50}} \times 100 + {1 \over {100}} \times 100 = 5\% $

1 MSD = 1 mm

10 VSD = 9 MSD

LC = ${1 \over {10}}$ mm

0 + 4$\left( {{1 \over {10}}} \right)$ mm = 0.4 mm

Reading = 41 + 6$\left( {{1 \over {10}}} \right)$

= 41 + 0.6

= 41.6 mm

True reading = 41.2 mm

= 412 $\times$ 10^$-$2 cm

Since zero error is negative, we will add 0.05 cm.

$\Rightarrow$ Corrected reading = 1.7 cm + 5 $\times$ 0.1 mm + 0.05 cm

= 180 $\times$ 10^$-$2 cm

${I_{mean}} = {{1.22 + 1.23 + 1.19 + 1.20} \over 4} = 1.21$

$\Delta {I_{mean}} = {{0.01 + 0.02 + 0.02 + 0.01} \over 4} = 0.015$

So % $I = {{\Delta {I_{mean}}} \over {{I_{mean}}}} \times 100 = {{0.015} \over {1.21}} \times 100 = {{150} \over {121}}\% $

$x = 150$

40 M = 1 cm

$\Rightarrow$ M = 0.025 cm .......... (1)

Also, 50 V = 49 M

$\Rightarrow$ Least count = M $-$ V = M $-$ ${{49} \over {50}}$ M = ${{M} \over {50}}$

$\Rightarrow$ LC = ${{0.025} \over {50}}$ cm

= ${{250} \over {50}}$ $\times$ 10^$-$6 m

$\Rightarrow$ LC = 5 $\times$ 10^$-$6 m

1 MSD = ${1 \over {20}}$ cm

$\because$ 10 VSD = 9 MSD

1 VSD = ${9 \over {10}}$ $\times$ ${1 \over {20}}$ cm = ${9 \over {200}}$ $\times$ 10 mm = 0.45 mm

Now, 1 MSD = ${1 \over {20}}$ $\times$ 10 mm = 0.50 mm

LC = (0.50 $-$ 0.45) mm = 0.05 mm

= 5 $\times$ 10^$-$2 mm

% error in $z = 3 \times 4 + {1 \over 2} \times 12$

$ = 12 + 6 = 18\% $

To understand the measurement of the spherical bob's diameter using Vernier calipers, let's break it down step by step:

Vernier Scale Calculation:

We are given that 9 divisions on the main scale (MSD) are equivalent to 10 divisions on the Vernier scale (VSD).

One main scale division is 1 mm. Therefore, 9 MSD = 9 mm.

Since 9 mm = 10 VSD, one VSD equals $ \frac{9}{10} = 0.9 \, \text{mm} $.

Least Count (LC) of the Vernier Callipers:

The least count is the smallest measurable value and is calculated as:

$ \text{LC} = 1 \, \text{MSD} - 1 \, \text{VSD} = 1 \, \text{mm} - 0.9 \, \text{mm} = 0.1 \, \text{mm} $

Calculate the Initial Reading:

The main scale reading (MSR) is 10 mm.

The 8th division of the Vernier scale coincides with the main scale division, meaning the Vernier scale reading (VSR) is 8.

Total reading before zero error correction = MSR + VSR × LC:

$ 10 \, \text{mm} + 8 \times 0.1 \, \text{mm} = 10.8 \, \text{mm} $

Adjust for the Zero Error:

The Vernier calipers have a positive zero error of 0.04 cm (or 0.4 mm, as 1 cm = 10 mm).

To get the correct measurement, subtract the zero error:

$ 10.8 \, \text{mm} - 0.4 \, \text{mm} = 10.4 \, \text{mm} $

Compute the Radius:

The diameter calculated is 10.4 mm. To find the radius, divide by 2:

$ \text{radius} = \frac{10.4 \, \text{mm}}{2} = 5.2 \, \text{mm} $

Convert the radius from millimeters to centimeters:

$ 5.2 \, \text{mm} = 0.52 \, \text{cm} = 52 \times 10^{-2} \, \text{cm} $

The radius of the spherical bob is $ 52 \times 10^{-2} $ cm.

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

$E = mgl{{{\theta ^2}} \over 2} = m{g^2}{{{T^2}{\theta ^2}} \over {8{\pi ^2}}}$

${{dE} \over E} = 2\left( {{{dg} \over g} + {{dT} \over T}} \right)$

$ = (4 + 3) = 14\% $

Zero error occurs when the screw gauge does not read zero when its jaws are closed. This error needs to be accounted for in the final measurement. The least count (LC) of a screw gauge, calculated as the pitch (p) divided by the number of divisions (N) on the circular scale, helps us determine the value of this error.

The least count of both screw gauges is LC = $p / N=0.1 / 100=0.001 \mathrm{~cm}$.

For screw gauge A, the zero error is calculated by adding the main scale reading (MSR) at zero and the circular scale reading (CSR) at zero, multiplied by the least count. Screw gauge A's MSR at zero is 0 and its CSR at zero is 5. Therefore, its zero error is ( 0 + 5 $\times$ 0.001 = 0.005 ) cm.

For screw gauge B, the zero error involves a negative main scale reading and a circular scale reading. The MSR at zero is -1 (since the main scale is 0.1 cm per division, this corresponds to -0.1 cm) and the CSR at zero is 92. Thus, the zero error for B is ( (-1 $\times$ 0.1) + 92 $\times$ 0.001 = -0.008 ) cm.

To find the difference between the final circular scale readings of screw gauges A and B, we look at the difference in their zero errors. The absolute difference is ( |0.005 - (-0.008)| = |0.005 + 0.008| = 0.013 ) cm. This value represents the absolute difference in the final readings of the two screw gauges, accounting for their individual zero errors.

However, to express this difference in terms of the circular scale divisions (since the least count is 0.001 cm per division), we multiply this difference by 1000 (since 1 cm = 1000 mm and each division represents 0.001 cm) :

$ \text{Difference in Divisions} = 0.013 \text{ cm} \times 1000 $
$ = 13 \text{ divisions} $

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

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

$ \Rightarrow {{\Delta g} \over g} \times 100 = {{\Delta l} \over l} \times 100 + 2{{\Delta T} \over T} \times 100$

$ = \left( {{{\Delta l} \over l} + {{2\Delta T} \over {T}}} \right)100\% $

${E_1} = {{20} \over {64}}\% $

${E_2} = {{30} \over {64}}\% $

${E_3} = {{19} \over {18}}\% $

Given, radius of sphere = (7.50 $\pm$ 0.85) cm

$ \therefore $ r = 7.50 and dr = 0.85

We know, volume of a sphere $v = {4 \over 3}\pi {r^3}$

taking log both sides, we get

$\ln v = \ln {{4\pi } \over 3} + 3\ln r$

Differentiating both sides,

${{dv} \over v} = 0 + 3{{dr} \over r}$

$ \therefore $ Fractional error in volume ${{dv} \over v} = 3{{dr} \over r}$

$ \therefore $ % error in volume,

${{dv} \over v} \times 100 = 3{{dr} \over r} \times 100$

$ = 3 \times {{0.85} \over {7.50}} \times 100$

= 34%

$R = {V \over I}$

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

% error in $R = {2 \over {50}} \times 100 + {{0.2} \over {20}} \times 100$

% error in R = 4 + 1

$ \therefore $ % error in R = 5%

$\rho $ = ${M \over V}$ = ${M \over {{4 \over 3}\pi {{\left( {{D \over 2}} \right)}^3}}}$

$ \Rightarrow $ $\rho $ = ${6 \over \pi }M{D^{ - 3}}$

For maximum error

${{d\rho } \over \rho } \times 100 = {{dM} \over M} \times 100 + 3{{dD} \over D} \times 100$

= 6 + 3 $ \times $ 1.5

= 10.5 %

= $\left( {{{1050} \over {100}}} \right)$ %

$ \therefore $ x = 1050

To determine the maximum percentage error in the volume of the cone, we first need to understand the dependence of the cone's volume on its dimensions: the diameter of the base and its height.

The volume $ V $ of a cone is given by the formula:

$ V = \frac{1}{3} \pi r^2 h $

where $ r $ is the radius of the base and $ h $ is the height.

Given that both the diameter ($ D $) of the base and the height ($ h $) are measured to be $ 20.0 \mathrm{~cm} $, we can find the radius $ r $ as follows:

$ r = \frac{D}{2} = \frac{20.0 \mathrm{~cm}}{2} = 10.0 \mathrm{~cm} $

Now, let's denote the errors in measuring the diameter and height as $ \Delta D $ and $ \Delta h $ respectively.

Since the measurements are taken with a scale having a least count of $ 2 \mathrm{~mm} $, we have:

$ \Delta D = 2 \mathrm{~mm} = 0.2 \mathrm{~cm} $

and

$ \Delta h = 2 \mathrm{~mm} = 0.2 \mathrm{~cm} $

To find the maximum percentage error in the volume, we need to use the formula for the propagation of relative errors. Considering the volume formula $ V = \frac{1}{3} \pi r^2 h $, the relative errors in $ r $ and $ h $ will propagate into the volume as follows:

The relative error in the radius $ \Delta r / r $ is:

$ \left(\frac{\Delta r}{r}\right) = \left(\frac{\Delta D / 2}{D / 2}\right) = \left(\frac{\Delta D}{D}\right) = \frac{0.2 \mathrm{~cm}}{20.0 \mathrm{~cm}} = 0.01 $

The relative error in the height $ \Delta h / h $ is:

$ \left(\frac{\Delta h}{h}\right) = \frac{0.2 \mathrm{~cm}}{20.0 \mathrm{~cm}} = 0.01 $

Since $ V $ is proportional to $ r^2 $ and $ h $, the total relative error in the volume is given by:

$ \left(\frac{\Delta V}{V}\right) = 2 \left(\frac{\Delta r}{r}\right) + \left(\frac{\Delta h}{h}\right) $

Substituting the relative errors, we get:

$ \left(\frac{\Delta V}{V}\right) = 2 (0.01) + 0.01 = 0.02 + 0.01 = 0.03 $

Hence, the maximum percentage error in the volume is:

$ 0.03 \times 100\% = 3\% $

Therefore, the maximum percentage error in the determination of the volume is 3%.

Given :

- Object distance $u = 10.0 \pm 0.1 \, \text{cm}$

- Image distance $v = 20.0 \pm 0.2 \, \text{cm}$

According to the lens formula for a thin lens :

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

So, we can calculate the focal length :

$ f = \frac{1}{\left(\frac{1}{v} - \frac{1}{u}\right)} = \frac{1}{\left(\frac{1}{20 \, \text{cm}} - \frac{1}{-10 \, \text{cm}}\right)} = \frac{1}{0.05 \, \text{cm}^{-1} + 0.1 \, \text{cm}^{-1}} = \frac{1}{0.15 \, \text{cm}^{-1}} = \frac{20}{3} \, \text{cm} $

Next, we need to calculate the error in the determination of the focal length. For that, we find the differential of the lens formula :

Then, the derivative of the equation gives us the change in the focal length $(df)$ in terms of the changes in the object distance $(du)$ and the image distance $(dv)$:

$-\frac{1}{v^2} dv + \frac{1}{u^2} du = \frac{-1}{f^2} df$

For maximum error, we get :

$\frac{1}{f^2} df = \frac{1}{v^2} dv + \frac{1}{u^2} du$

This equation tells us how errors in $(u)$ and $v$ propagate to an error in $f$. Now, when you compute the relative error in the focal length, you get :

$\frac{df}{f} \times 100 = \left(\frac{1}{v^2} dv + \frac{1}{u^2} du\right) \times \frac{f}{1} \times 100$

Plugging in your values of $u = 10 \mathrm{~cm}$, $du = 0.1 \mathrm{~cm}$, $v = 20 \mathrm{~cm}$, $dv = 0.2 \mathrm{~cm}$, and $f = 20/3 \mathrm{~cm}$, you indeed get:

$\frac{df}{f} \times 100 = \left(\frac{0.2}{20^2} + \frac{0.1}{10^2}\right) \times \frac{20}{3} \times 100 = 1\%$

So, the error in the focal length of the lens is indeed 1% (i.e., $n = 1$).

$ \begin{aligned} & \mathrm{B}=\mathrm{e}^\alpha\left(\mathrm{m}_{\mathrm{e}}\right)^\beta \mathrm{h}^\gamma \mathrm{k}^\delta \\\\ & {[\mathrm{~B}]=\left[\mathrm{e}^\alpha\right]\left[\mathrm{m}_{\mathrm{e}}\right]^\beta[\mathrm{h}]^\gamma\left[\mathrm{k}^\delta\right]} \\\\ & {\left[\mathrm{M}^1 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]=[\mathrm{AT}]^\alpha[\mathrm{m}]^\beta\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^\gamma\left[\mathrm{ML}^3 \mathrm{~A}^{-2} \mathrm{~T}^{-4}\right]^\delta} \\\\ & \mathrm{M}^{\mathrm{l}} \mathrm{T}^{-2} \mathrm{~A}^{-1}=\mathrm{m}^{\beta+\gamma+\delta} \mathrm{L}^{2 \mathrm{r}+3 \delta} \mathrm{T}^{\alpha-\gamma-4 \delta} \mathrm{A}^{\alpha-2 \delta} \\\\ & \text {Compare }: \beta+\gamma+\delta=1 ; 2 \gamma+3 \delta=0, \alpha-\gamma-4 \delta=-2, \alpha-2 \delta=-1 \\\\ & \text {On solving } \alpha=3, \beta=2, \gamma=-3, \delta=2 \\\\ & \alpha+\beta+\gamma+\delta=4 \end{aligned} $

$u = ({x_2} - {x_1}) = 75 - 45 = 30$ cm

$\Delta u = \Delta {x_2} + \Delta {x_1} = {1 \over 4} + {1 \over 4} = {1 \over 2}$ cm

$v = ({x_3} - {x_2}) = 135 - 75 = 60$ cm

$\Delta v = \Delta {x_3} + \Delta {x_2} = {1 \over 4} + {1 \over 4} = {1 \over 2}$ cm



$ \therefore $ ${1 \over v} - {1 \over u} = {1 \over f} \Rightarrow {1 \over {60}} + {1 \over {30}} = {1 \over f}$

$ \therefore $ f = 20 cm Also, ${{ - dv} \over {{v^2}}} + {{ - du} \over {{u^2}}} = {{ - df} \over {{f^2}}}$

$ \Rightarrow {{df} \over f} = f\left[ {{{dv} \over {{v^2}}} + {{du} \over {{u^2}}}} \right] = 20\left[ {{1 \over {{{60}^2}}} + {1 \over {{{30}^2}}}} \right]{1 \over 2}$

$ \therefore $ ${{df} \over f} \times 100 = 10\left[ {{1 \over {36}} + {1 \over 9}} \right] = {{50} \over {36}}$ = 1.38 and 1.39 (both)

We know that $\Delta L = {W \over {(YA/L)}}$

where W is weight or load = mg = 1.2 $\times$ 10 = 12 kg m s^$-$2, Y is Young's modulus = 2 $\times$ 10¹¹ N m^$-$2, L is length of wire with load = 1.0 m, A is area of steel wire = $ = \pi {r^2} = {\pi \over 4}{d^2} = {\pi \over 4} \times {(0.5 \times {10^{ - 3}})^2}$

Therefore,

$\Delta L = {{1.2 \times 10} \over {2 \times {{10}^{11}} \times {\pi \over 4}{{(0.5 \times {{10}^{ - 3}})}^2} \times {1 \over {1.0\,m}}}}$

$ = {{1.2 \times 10 \times 4} \over {2 \times {{10}^{11}} \times \pi \times {{(0.5)}^2} \times {{10}^{ - 6}}}}$

$ \Rightarrow \Delta L = 0.3 \times {10^{ - 3}}$ m = 0.3 mm

Now, least count of vernier scale $ = \left( {1 - {9 \over {10}}} \right)$ mm = 0.1 mm

Therefore, Vernier reading $ = {{\Delta L} \over {least\,count}}$

Vernier reading $ = {{0.3\,mm} \over {0.1\,mm}} = 3$

Therefore, 3^rd vernier scale division coincides with the main scale division.

$E(t) = {A^2}{e^{ - \alpha t}}$ ...... (i)

$\alpha$ = 0.2 s^$-$1

$\left( {{{dA} \over A}} \right) \times 100 = 1.25\% $

$\left( {{{dt} \over t}} \right) \times 100 = 1.50$

$ \Rightarrow (dt \times 100) = 1.5t = 1.5 \times 5 = 7.5$

Differentiating on both sides of equation (i), we get

$ d \mathrm{E}=(2 \mathrm{~A} d \mathrm{~A}) e^{-\alpha t}+\mathrm{A}^2 e^{-\alpha t}(-\alpha d t) $

Dividing throughout by $\mathrm{E}=\mathrm{A}^2 e^{-\alpha t}$

$ \frac{d \mathrm{E}}{\mathrm{E}}=\frac{2}{\mathrm{~A}} d \mathrm{~A}+\alpha d t $

(Considering worst possible case)

$\therefore$ $\left( {{{dE} \over E}} \right) \times 100 = 2\left( {{{dA} \over A}} \right) \times 100 + \alpha (dt \times 100)$

$ = 2(1.25) + 0.2(7.5)$

$ = 2.5 + 1.5$

$ = 4\% $

Given $d \propto {\rho ^a}{S^b}{b^c}$

${M^0}L{T^0} \propto {(M{L^{ - 3}})^a}{(M{T^{ - 3}})^b}{({T^{ - 1}})^c}$

${M^0}L{T^0} \propto {M^{(a + b)}}{L^{ - 3a}}{T^{ - 3b - c}}$

Equating the coefficients, we get

$a + b = 0 - 3a = 1 - 3b - c = 0$

$b = - a$ $a = - {1 \over 3} - c = 3b$

$b = {1 \over 3}c = - 3b \Rightarrow c = 1$

Therefore, $b = {1 \over n} = {1 \over 3} \Rightarrow n = 3$.

The difference between the two measurements by Vernier scale gives elongation of the wire caused by the additional load of 2 kg. In the first measurement, main scale reading is MSR = 3.20 $\times$ 10^$-$2 m and Vernier scale reading is VSR = 20. The least count of Vernier scale is LC = 1 $\times$ 10^$-$5 m. Thus, the first measurement by Vernier scale is

L₁ = MSR + VSR $\times$ LC

= 3.20 $\times$ 10^$-$2 + 20(1 $\times$ 10^$-$5)

= 3.220 $\times$ 10^$-$2 m.

In the second measurement, MSR = 3.20 $\times$ 10^$-$2 m and VSR = 45. Thus, the second measurement by Vernier scale is

L₂ = 3.20 $\times$ 10^$-$2 + 45(1 $\times$ 10^$-$5)

= 3.245 $\times$ 10^$-$2 m.

The elongation of the wire due to force F = 2g is

l = L₂ $-$ L₁ = 0.025 $\times$ 10^$-$2 m.

The maximum error in the measurement of l is $\Delta$l = LC = 1 $\times$ 10^$-$5 m. Young's modulus is given by Y = ${{FL} \over {lA}}$. The maximum percentage error in the measurement of Y is

${{\Delta Y} \over Y} \times 100 = {{\Delta l} \over l} \times 100 = {{1 \times {{10}^{ - 5}}} \over {0.025 \times {{10}^{ - 2}}}} \times 100 = 4\% $.

The distance moved on the linear scale when circular scale makes one complete rotation is p = 1 mm (pitch). The number of divisions on the circular scale is N = 100. Thus, one division on the circular scale is LC = ${p \over N} = {1 \over {100}}$ = 0.01 mm. The linear scale reading (LSR) is 1 mm and the circular scale reading (CSR) is 47. Thus, the diameter of the wire is

d = LSR + CSR $\times$ LC

= 1 + 47 $\times$ 0.01 = 1.47 mm = 0.147 cm.

The curved surface area is S = 2$\pi$rL. That is,

$S = 2\pi \left( {{d \over 2}} \right)L$

$S = \pi dL = \pi \left( {{{1.47} \over {10}}} \right)5.6 = $ 2.5848 cm² = 2.6 cm²

which is corrected to two significant digits.

We know that (n + 1) divisions on the vernier scale = n division on the main scale

Therefore,

1 division on the vernier scale = $\left( {{n \over {n + 1}}} \right)$ divisions on the main scale = $\left( {{n \over {n + 1}}} \right)a$ units

Therefore, the least count (LC) of the vernier caliper is

$1(MSD) - 1(VSD) = a - \left( {{n \over {n + 1}}} \right)a = {a \over {n + 1}}$

$\left[ {{a \over {{V^2}}}} \right] = \left[ P \right]$
$\therefore \left[ a \right] = \left[ {P{V^2}} \right]$
$ = {{ML{T^{ - 2}}} \over {{L^2}}}{L^6} = M{L^5}{T^{ - 2}}$

$\begin{array}{l}\begin{array}{l}\left[X\right]\;=\;\left[C\right]\;=\;\left[M^{-1}L^{-2}T^2Q^2\right]\\\left[Z\right]\;=\;\left[B\right]\;=\;\left[MT^{-1}Q^{-1}\right]\\\\\Rightarrow\;\left[Y\right]\;=\;\frac{\;\left[M^{-1}L^{-2}T^2Q^2\right]}{\;\left[MT^{-1}Q^{-1}\right]^2}\;\end{array}\\\\=\;\left[M^{-3}L^{-2}T^4Q^4\right]\\\\\\\\\\\end{array}$

We know E = hv
So $h = {E \over v} = {{\left[ {M{L^2}{T^{ - 2}}} \right]} \over {\left[ {{T^{ - 1}}} \right]}} = \left[ {M{L^2}{T^{ - 1}}} \right]$