Error in the measurement of radius of a sphere is $0.2\%$. The error in the calculated value of its volume is ......... $\%$
Error in the measurement of radius of a sphere is $0.2\%$. The error in the calculated value of its volume is ......... $\%$
- A
$0.2$
- B
$0.6$
- C
$0.4$
- D
$0.8$
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 ......... $m$
A public park, in the form of a square, has an area of $(100 \pm 0.2) m ^2$. The side of park is ......... $m$
A student performs an experiment for determination of $g \left(=\frac{4 \pi^{2} l }{ T ^{2}}\right), \ell =1 m$ and he commits an error of $\Delta \ell$. For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\Delta T$ and he commits a human error of $0.1 s$ For which of the following data, the measurement of $g$ will be most accurate?
A student performs an experiment for determination of $g \left(=\frac{4 \pi^{2} l }{ T ^{2}}\right), \ell =1 m$ and he commits an error of $\Delta \ell$. For $T$ he takes the time of $n$ oscillations with the stop watch of least count $\Delta T$ and he commits a human error of $0.1 s$ For which of the following data, the measurement of $g$ will be most accurate?
A sliver wire has mass $(0.6 \pm 0.006) \; g$, radius $(0.5 \pm 0.005) \; mm$ and length $(4 \pm 0.04) \; cm$. The maximum percentage error in the measurement of its density will be $......\,\%$
A sliver wire has mass $(0.6 \pm 0.006) \; g$, radius $(0.5 \pm 0.005) \; mm$ and length $(4 \pm 0.04) \; cm$. The maximum percentage error in the measurement of its density will be $......\,\%$
- [JEE MAIN 2022]
The period of oscillation of a simple pendulum is given by $T = 2\pi \sqrt {\frac{l}{g}} $ where $l$ is about $100 \,cm$ and is known to have $1\,mm$ accuracy. The period is about $2\,s$. The time of $100$ oscillations is measured by a stop watch of least count $0.1\, s$. The percentage error in $g$ is ......... $\%$
The period of oscillation of a simple pendulum is given by $T = 2\pi \sqrt {\frac{l}{g}} $ where $l$ is about $100 \,cm$ and is known to have $1\,mm$ accuracy. The period is about $2\,s$. The time of $100$ oscillations is measured by a stop watch of least count $0.1\, s$. The percentage error in $g$ is ......... $\%$
The current voltage relation of diode is given by $I=(e^{1000V/T} -1)\;mA$, where the applied voltage $V$ is in volts and the temperature $T$ is in degree Kelvin. If a student makes an error measuring $ \mp 0.01\;V$ while measuring the current of $5\; mA$ at $300\; K$, what will be the error in the value of current in $mA$ ?
The current voltage relation of diode is given by $I=(e^{1000V/T} -1)\;mA$, where the applied voltage $V$ is in volts and the temperature $T$ is in degree Kelvin. If a student makes an error measuring $ \mp 0.01\;V$ while measuring the current of $5\; mA$ at $300\; K$, what will be the error in the value of current in $mA$ ?