$A$ physical quantity $y$ is represented by the formula $y = m^{2} r^{-4} g^{x} l^{-\frac{3}{2}}$. If the percentage errors in $y, m, r, l,$ and $g$ are $18, 1, 0.5, 4,$ and $p$ respectively,then find the value of $x$ and $p$.

Quantity $Z$ varies with $x$ and $y$ according to the equation $Z = x^2y - xy^2$,where $x = 3.0 \pm 0.1$ and $y = 2.0 \pm 0.1$. The value of $Z$ is:

$A$ physical quantity $A$ is related to four observables $a, b, c,$ and $d$ as follows: $A = \frac{a^2 b^3}{c \sqrt{d}}$. The percentage errors of measurement in $a, b, c,$ and $d$ are $1\%, 3\%, 2\%,$ and $2\%$ respectively. What is the percentage error in the quantity $A$?

$A$ physical quantity $X$ is related to four measurable quantities $a, b, c$ and $d$ as follows: $X = a^2b^3c^{\frac{5}{2}}d^{-2}$. The percentage errors in the measurement of $a, b, c$ and $d$ are $1\%$,$2\%$,$3\%$ and $4\%$ respectively. What is the percentage error in quantity $X$? If the value of $X$ calculated on the basis of the above relation is $2.763$,to what value should you round off the result?

During Searle's experiment,the zero of the Vernier scale lies between $3.20 \times 10^{-2} \text{ m}$ and $3.25 \times 10^{-2} \text{ m}$ of the main scale. The $20^{\text{th}}$ division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of $2 \text{ kg}$ is applied to the wire,the zero of the Vernier scale still lies between $3.20 \times 10^{-2} \text{ m}$ and $3.25 \times 10^{-2} \text{ m}$ of the main scale,but now the $45^{\text{th}}$ division of the Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is $2 \text{ m}$ and its cross-sectional area is $8 \times 10^{-7} \text{ m}^2$. The least count of the Vernier scale is $1.0 \times 10^{-5} \text{ m}$. The maximum percentage error in the Young's modulus of the wire is:

The length $l$,breadth $b$,and thickness $t$ of a block of wood were measured with the help of a measuring scale. The results with permissible errors are $l = 15.12 \pm 0.01 \; cm$,$b = 10.15 \pm 0.01 \; cm$,and $t = 5.28 \pm 0.01 \; cm$. The percentage error in volume up to proper significant figures is: (in $\%$)