Correct option is (A)

To determine the spring constant $k$ of a spring experimentally, we can use the formula derived from Hooke's Law and the period of oscillation for a mass-spring system:

$T=2\pi \sqrt{\frac{m}{k}}$

Here, $T$ is the period of oscillation, $m$ is the mass, and $k$ is the spring constant. Rearranging the formula to solve for $k$, we get:

$k=\frac{4{\pi }^{2}m}{T^2}$

To find the error in $k$, we have to consider the errors in both the measurements of $T$ and $m$. Let's denote the percentage errors as follows:

$\mathrm{\Delta }T/T\cdot 100\mathrm{\%}=2\mathrm{\%}$ (positive error)

$\mathrm{\Delta }m/m\cdot 100\mathrm{\%}=-1\mathrm{\%}$ (negative error)

According to the rules of error propagation, the relative error in $k$ can be found by adding the relative errors in the measurements, each multiplied by the respective powers to which they affect $k$. Since $T$ is squared in the denominator and $m$ is linear in the numerator, the calculation is as follows:

$\frac{\mathrm{\Delta }k}{k}=|-2\cdot \frac{\mathrm{\Delta }T}{T}|+|1\cdot \frac{\mathrm{\Delta }m}{m}|$

Substituting the percentage errors:

$\frac{\mathrm{\Delta }k}{k}=|-2\cdot 0.02|+|1\cdot (-0.01)|$

$\frac{\mathrm{\Delta }k}{k}=0.04+0.01$

$\frac{\mathrm{\Delta }k}{k}=0.05$

Thus, the percentage error in determining the value of $k$ is:

$\frac{\mathrm{\Delta }k}{k}\cdot 100\mathrm{\%}=5\mathrm{\%}$

The correct answer is Option A: 5%

Correct option is (D)

To calculate the percentage error in the resistivity of the material of the wire, we need to understand the formula for resistivity. The resistivity $\rho$ of a wire is given by:

$\rho =\frac{RA}{l}$

where:

• $R$ is the resistance


• $A$ is the cross-sectional area of the wire


• $l$ is the length of the wire

The cross-sectional area $A$ of the wire with radius $r$ is:

$A=\pi r^2$

We can plug this into the equation for resistivity to get:

$\rho =\frac{R\pi r^2}{l}$

Now, to find the percentage error in resistivity, we need to find the percentage errors in $R$, $r$, and $l$ and then use the following rule for combining errors:

For a given function, $f=f(x,y,z,...)$, where $x,y,z,...$ are the measured quantities with possible errors, the percentage error in $f$, denoted as $(\delta f{)}_{\mathrm{\%}}$, can be approximated by adding the relative percentage errors of the input quantities. If $f$ has the form of a product and quotient of the measured quantities as in our case ($\rho =\frac{R\pi r^2}{l}$), the percentage error in $f$ is given by:

$(\delta f{)}_{\mathrm{\%}}=(\delta x{)}_{\mathrm{\%}}+(\delta y{)}_{\mathrm{\%}}+(\delta z{)}_{\mathrm{\%}}+...$

Where $(\delta x{)}_{\mathrm{\%}}$, $(\delta y{)}_{\mathrm{\%}}$, and $(\delta z{)}_{\mathrm{\%}}$ are the percentage errors in each measured quantity $x,y,z,...$ respectively.

For our case:

• The percentage error in radius $(\delta r{)}_{\mathrm{\%}}$ is given by the error in $r$ divided by the average radius and then multiplied by 100:

$(\delta r{)}_{\mathrm{\%}}=\left(\frac{0.05}{0.35}\right)\times 100$

• The percentage error in resistance $(\delta R{)}_{\mathrm{\%}}$ is:

$(\delta R{)}_{\mathrm{\%}}=\left(\frac{10}{100}\right)\times 100$

• The percentage error in length $(\delta l{)}_{\mathrm{\%}}$ is:

$(\delta l{)}_{\mathrm{\%}}=\left(\frac{0.2}{15}\right)\times 100$

Now let's calculate each:

$(\delta r{)}_{\mathrm{\%}}=\left(\frac{0.05}{0.35}\right)\times 100\approx 14.29\mathrm{\%}$

$(\delta R{)}_{\mathrm{\%}}=\left(\frac{10}{100}\right)\times 100=10\mathrm{\%}$

$(\delta l{)}_{\mathrm{\%}}=\left(\frac{0.2}{15}\right)\times 100\approx 1.33\mathrm{\%}$

However, since the area $A$ is proportional to $r^2$, the percentage error in $A$ will be twice the percentage error in $r$. Thus:

$(\delta A{)}_{\mathrm{\%}}=2\times (\delta r{)}_{\mathrm{\%}}=2\times 14.29\mathrm{\%}\approx 28.58\mathrm{\%}$

Finally, we add the percentage errors to find the percentage error in resistivity:

$(\delta \rho {)}_{\mathrm{\%}}=(\delta R{)}_{\mathrm{\%}}+(\delta A{)}_{\mathrm{\%}}+(\delta l{)}_{\mathrm{\%}}$

$(\delta \rho {)}_{\mathrm{\%}}=10\mathrm{\%}+28.58\mathrm{\%}+1.33\mathrm{\%}\approx 39.91\mathrm{\%}$

This calculation gives us a value close to 39.91%, which means the correct option is closest to this value. Thus, the best answer is:

Option D $39.9\mathrm{\%}$

Correct option is (A)

$\begin{array}{rl}& \mathrm{T}=2\pi \sqrt{\frac{\ell }{\mathrm{g}}}\\ & \mathrm{g}=\frac{4{\pi }^2\ell }{{\mathrm{T}}^2}\\ & \frac{\mathrm{\Delta }\mathrm{g}}{\mathrm{g}}=\frac{\mathrm{\Delta }\ell }{\ell }+\frac{2\mathrm{\Delta }\mathrm{T}}{\mathrm{T}}\\ & =\frac{0.2}{20}+2\left(\frac{1}{40}\right)\\ & =\frac{1.2}{20}\end{array}$

Percentage change $=\frac{1.2}{20}\times 100=6\mathrm{\%}$

F

Correct option is (D)

$\begin{array}{rl}& \mathrm{R}=\frac{\rho \mathrm{L}}{\pi \frac{{\mathrm{d}}^2}{4}}\\ & \frac{\mathrm{\Delta }\mathrm{R}}{\mathrm{R}}=\frac{\mathrm{\Delta }\mathrm{L}}{\mathrm{L}}+\frac{2\mathrm{\Delta }\mathrm{d}}{\mathrm{d}}\\ & \frac{\mathrm{\Delta }\mathrm{L}}{\mathrm{L}}=0.1\mathrm{\%}\text{ and }\frac{\mathrm{\Delta }\mathrm{d}}{\mathrm{d}}=0.1\mathrm{\%}\\ & \frac{\mathrm{\Delta }\mathrm{R}}{\mathrm{R}}=0.3\mathrm{\%}\end{array}$

Correct option is (B)

$\begin{array}{rl}& \mathrm{Q}=\frac{{\mathrm{a}}^4{\mathrm{b}}^3}{{\mathrm{c}}^2}\\ & \frac{\mathrm{\Delta }\mathrm{Q}}{\mathrm{Q}}=4\frac{\mathrm{\Delta }\mathrm{a}}{\mathrm{a}}+3\frac{\mathrm{\Delta }\mathrm{b}}{\mathrm{b}}+2\frac{\mathrm{\Delta }\mathrm{c}}{\mathrm{c}}\end{array}$

$\frac{\mathrm{\Delta }\mathrm{Q}}{\mathrm{Q}}\times 100=4(\frac{\mathrm{\Delta }\mathrm{a}}{\mathrm{a}}\times 100)+3(\frac{\mathrm{\Delta }\mathrm{b}}{\mathrm{b}}\times 100)+2(\frac{\mathrm{\Delta }\mathrm{c}}{\mathrm{c}}\times 100)$

$\begin{array}{rl}\mathrm{\%}\text{ error in }\mathrm{Q}& =4\times 3\mathrm{\%}+3\times 4\mathrm{\%}+2\times 5\mathrm{\%}\\ & =12\mathrm{\%}+12\mathrm{\%}+10\mathrm{\%}\\ & =34\mathrm{\%}\end{array}$