A physical quantity z depends on four observables a, b, c and d , as z =frac{ a ^{2} b ^{frac{2}{3}}

English

Gujarati

Hindi

1.Units, Dimensions and Measurement

medium

A physical quantity $z$ depends on four observables $a,$ $b,$ $c$ and $d ,$ as $z =\frac{ a ^{2} b ^{\frac{2}{3}}}{\sqrt{ c } d ^{3}} .$ The percentage of error in the measurement of $a, b, c$ and $d$ are $2 \%, 1.5 \%, 4 \%$ and $2.5 \%$ respectively. The percentage of error in $z$ is$......\%$

A

$12.5$

B

$14.5$

C

$16.5$

D

$13.5$

(JEE MAIN-2020)

Solution

$\frac{\Delta Z}{Z}=\frac{2 \Delta a }{ a }+\frac{2}{3} \frac{\Delta b }{ b }+\frac{1}{2} \frac{\Delta c }{ c }+\frac{3 \Delta d }{ d }=14.5 \%$

Standard 11

Physics

Share

Similar Questions

A public park, in the form of a square, has an area of $(100 \pm 0.2)\; m ^2$. The side of park is ………… $m$

medium

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

medium

(JEE MAIN-2021)

A body of mass $(5 \pm 0.5) kg$ is moving with a velocity of $(20 \pm 0.4) m / s$. Its kinetic energy will be

medium

(JEE MAIN-2023)

The period of oscillation of a simple pendulum is $T=2\pi \sqrt {\frac{l}{g}} $. Measured value of $L$ is $20.0\; cm$ known to $1\; mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90\ s$ using a wrist watch of $1\; s$ resolution. The accuracy in the determination of $g$ is …….. $\%$

hard

(JEE MAIN-2015)

If there is a positive error of $50\%$ in the measurement of velocity of a body, then the error in the measurement of kinetic energy is ………….. $\%$

medium