Correct answer is (A)

To find the kinetic energy of the body, we can use the formula:

$KE=\frac{1}{2}m{v}^2$

Given the mass $m=(5\pm 0.5)\text{kg}$ and the velocity $v=(20\pm 0.4)\text{m/s}$, we can find the kinetic energy and its uncertainty by applying the rules for the propagation of errors in multiplication.

For the product of two quantities, the relative error is the sum of the relative errors of the individual quantities:

$\frac{\mathrm{\Delta }(m{v}^2)}{m{v}^2}=\frac{\mathrm{\Delta }m}{m}+\frac{\mathrm{\Delta }v}{v}+\frac{\mathrm{\Delta }v}{v}$

Now, substitute the given values:

$\frac{\mathrm{\Delta }(m{v}^2)}{m{v}^2}=\frac{0.5}{5}+\frac{0.4}{20}+\frac{0.4}{20}$

$\frac{\mathrm{\Delta }(m{v}^2)}{m{v}^2}=0.1+0.02+0.02$

$\frac{\mathrm{\Delta }(m{v}^2)}{m{v}^2}=0.14$

Now, calculate the kinetic energy:

$KE=\frac{1}{2}(5)(20{)}^2=1000\text{J}$

To find the uncertainty in the kinetic energy, multiply the relative error by the calculated kinetic energy:

$\mathrm{\Delta }KE=0.14\times 1000=140\text{J}$

So, the kinetic energy of the body is $(1000\pm 140)\text{J}$.

Correct answer is (B)

The percentage error in a quantity that is a product or quotient of other quantities is given by the sum of the percentage errors in those quantities, each multiplied by the power to which it is raised in the expression for the quantity.

Given the physical quantity P as

$P=\frac{a^2{b}^3}{c\sqrt{d}}$

The percentage error in P, denoted as $\mathrm{\Delta }P/P$, is given by:

$\frac{\mathrm{\Delta }P}{P}=2\left(\frac{\mathrm{\Delta }a}{a}\right)+3\left(\frac{\mathrm{\Delta }b}{b}\right)+\left(\frac{\mathrm{\Delta }c}{c}\right)+\frac{1}{2}\left(\frac{\mathrm{\Delta }d}{d}\right)$

Substituting the given percentage errors for a, b, c, and d:

$\frac{\mathrm{\Delta }P}{P}=2(0.01)+3(0.02)+0.03+\frac{1}{2}(0.04)=0.02+0.06+0.03+0.02=0.13$

Therefore, the percentage error in P is 13%.

Correct answer is (C)

The density of a cylindrical wire is given by the formula:

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

where $m$ is the mass, $r$ is the radius, and $l$ is the length.

The relative error in a calculated quantity is the sum of the relative errors in the quantities it depends on. For the density, this is given by:

$\frac{\mathrm{\Delta }\rho }{\rho }=\frac{\mathrm{\Delta }m}{m}+2\frac{\mathrm{\Delta }r}{r}+\frac{\mathrm{\Delta }l}{l}$

Given that $\mathrm{\Delta }m=0.01$ g, $m=0.4$ g, $\mathrm{\Delta }r=0.03$ mm, $r=6$ mm, $\mathrm{\Delta }l=0.04$ cm, and $l=8$ cm, we can substitute these values into the formula to find the relative error in the density:

$\frac{\mathrm{\Delta }\rho }{\rho }=\frac{0.01}{0.4}+2\frac{0.03}{6}+\frac{0.04}{8}=0.025+0.01+0.005=0.04$

So the relative error in the density is 0.04, or 4%.

Correct answer is (C)

In the problem, we are given two resistances, $R_1$ and $R_2$, each with a certain measurement error, $\mathrm{\Delta }{R}_1$ and $\mathrm{\Delta }{R}_2$. These resistances are connected in parallel, and we are asked to find the percentage error in the equivalent resistance of this combination.

The formula for the equivalent resistance $R$ of two resistors $R_1$ and $R_2$ in parallel is:

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

We want to find the percentage error in $R$, which is given by $(\mathrm{\Delta }R/R)\times 100\mathrm{\%}$.

In order to find $\mathrm{\Delta }R/R$, we differentiate both sides of the above equation with respect to $R$, $R_1$, and $R_2$. This gives us:

$\frac{\mathrm{\Delta }R}{R^2}=\frac{\mathrm{\Delta }{R}_1}{R_1^2}+\frac{\mathrm{\Delta }{R}_2}{R_2^2}$

We can then solve this equation for $\mathrm{\Delta }R/R$:

$\frac{\mathrm{\Delta }R}{R}=(\frac{\mathrm{\Delta }{R}_1}{R_1^2}+\frac{\mathrm{\Delta }{R}_2}{R_2^2})R$

Substituting the given values, $R_1=10\mathrm{\Omega }$, $R_2=15\mathrm{\Omega }$, $\mathrm{\Delta }{R}_1=\mathrm{\Delta }{R}_2=0.5\mathrm{\Omega }$, and $R=R_1{R}_2/(R_1+R_2)=6\mathrm{\Omega }$, we get:

$\frac{\mathrm{\Delta }R}{R}=(\frac{0.5}{100}+\frac{0.5}{225})\times 6=\frac{13}{300}$

Finally, to convert this to a percentage, we multiply by 100, giving:

$\frac{\mathrm{\Delta }R}{R}\times 100=\frac{13}{3}=4.33\mathrm{\%}$

This tells us that the percentage error in the equivalent resistance of the two resistances in parallel is $4.33\mathrm{\%}$.