If radius of the sphere is $(5.3 \pm 0.1)\;cm$. Then percentage error in its volume will be
If radius of the sphere is $(5.3 \pm 0.1)\;cm$. Then percentage error in its volume will be
- A
$3 + 6.01 \times \frac{{100}}{{5.3}}$
- B
$\frac{1}{3} \times 0.01 \times \frac{{100}}{{5.3}}$
- C
$\left( {\frac{{3 \times 0.1}}{{5.3}}} \right) \times 100$
- D
$\frac{{0.1}}{{5.3}} \times 100$
Similar Questions
A student measures the time period of $100$ oscillations of a simple pendulum four times. The data set is $90\;s$ ,$91\;s $, $95\;s$ and $92\;s$. If the minimum division in the measuring clock is $1\;s$, then the reported mean time should be
A student measures the time period of $100$ oscillations of a simple pendulum four times. The data set is $90\;s$ ,$91\;s $, $95\;s$ and $92\;s$. If the minimum division in the measuring clock is $1\;s$, then the reported mean time should be
- [JEE MAIN 2016]
A simple pendulum is being used to determine the value of gravitational acceleration $\mathrm{g}$ at a certain place. The length of the pendulum is $25.0\; \mathrm{cm}$ and a stop watch with $1\; \mathrm{s}$ resolution measures the time taken for $40$ oscillations to be $50\; s$. The accuracy in $g$ is ....... $\%$
A simple pendulum is being used to determine the value of gravitational acceleration $\mathrm{g}$ at a certain place. The length of the pendulum is $25.0\; \mathrm{cm}$ and a stop watch with $1\; \mathrm{s}$ resolution measures the time taken for $40$ oscillations to be $50\; s$. The accuracy in $g$ is ....... $\%$
- [JEE MAIN 2020]
The measured value of the length of a simple pendulum is $20 \mathrm{~cm}$ with $2 \mathrm{~mm}$ accuracy. The time for $50$ oscillations was measured to be $40$ seconds with $1$ second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $\mathrm{N} \%$. The value of $\mathrm{N}$ is:
The measured value of the length of a simple pendulum is $20 \mathrm{~cm}$ with $2 \mathrm{~mm}$ accuracy. The time for $50$ oscillations was measured to be $40$ seconds with $1$ second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is $\mathrm{N} \%$. The value of $\mathrm{N}$ is:
- [JEE MAIN 2024]
The maximum percentage errors in the measurement of mass (M), radius (R) and angular velocity $(\omega)$ of a ring are $2 \%, 1 \%$ and $1 \%$ respectively, then find the maximum percenta? error in the measurement of its moment of inertia $\left(I=\frac{1}{2} M R^{2}\right)$ about its geometric axis.
The maximum percentage errors in the measurement of mass (M), radius (R) and angular velocity $(\omega)$ of a ring are $2 \%, 1 \%$ and $1 \%$ respectively, then find the maximum percenta? error in the measurement of its moment of inertia $\left(I=\frac{1}{2} M R^{2}\right)$ about its geometric axis.
In an experiment of simple pendulum time period measured was $50\,sec$ for $25$ vibrations when the length of the simple pendulum was taken $100\,cm$ . If the least count of stop watch is $0.1\,sec$ . and that of meter scale is $0.1\,cm$ then maximum possible error in value of $g$ is .......... $\%$
In an experiment of simple pendulum time period measured was $50\,sec$ for $25$ vibrations when the length of the simple pendulum was taken $100\,cm$ . If the least count of stop watch is $0.1\,sec$ . and that of meter scale is $0.1\,cm$ then maximum possible error in value of $g$ is .......... $\%$