Units & Measurements

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