The relative error in the determination of the surface area of a sphere is $\alpha $. Then the relative error in the determination of its volume is
JEE MAIN 2018, Diffcult
Download our app for free and get started
Relative eroor in surface area, $\frac{\Delta s}{s}=2 \times \frac{\Delta r}{r}=\alpha$ and
relative error in volume, $\frac{\Delta v}{v}=3 \times \frac{\Delta r}{r}$
$\therefore$ Relative error in volume w. r.t relative error in area,
$\frac{\Delta v}{v}=\frac{3}{2} \alpha$
relative error in volume, $\frac{\Delta v}{v}=3 \times \frac{\Delta r}{r}$
$\therefore$ Relative error in volume w. r.t relative error in area,
$\frac{\Delta v}{v}=\frac{3}{2} \alpha$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1The $SI$ unit of magnetic permeability isView Solution
- 2View SolutionWhich one of the following is not a unit of young's modulus
- 3View SolutionThe dimensions of couple are
- 4View SolutionWhich of the following physical quantities have the same dimensions?
- 5According to Newton, the viscous force acting between liquid layers of area $A$ and velocity gradient $\Delta v/\Delta z$ is given by $F = - \eta A\frac{{\Delta v}}{{\Delta z}}$ where $\eta $ is constant called coefficient of viscosity. The dimension of $\eta $ areView Solution
- 6In the determination of Young's modulus $\left(Y=\frac{4 MLg }{\pi / d ^2}\right)$ by using Searle's method, a wire of length $L=2 \ m$ and diameter $d =0.5 \ mm$ is used. For a load $M =2.5 \ kg$, an extension $\ell=0.25 \ mm$ in the length of the wire is observed. Quantities $d$ and $\ell$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of $0.5 \ mm$. The number of divisions on their circular scale is $100$ . The contributions to the maximum probable error of the $Y$ measurementView Solution
- 7One main scale division of a vernier callipers is $a$ $cm$ and $n ^{\text {th }}$ division of the vernier scale coincide with $( n -1)^{\text {th }}$ division of the main scale. The least count of the callipers in $mm$ isView Solution
- 8$A$ and $B$ possess unequal dimensional formula then following operation is not possible in any case:-View Solution
- 9The position of a particle at time $t$ is given by the relation $x(t) = \left( {\frac{{{v_0}}}{\alpha }} \right)\,\,(1 - {e^{ - \alpha t}})$, where ${v_0}$ is a constant and $\alpha > 0$. The dimensions of ${v_0}$ and $\alpha $ are respectivelyView Solution
- 10View SolutionThe dimensional formula for the modulus of rigidity is