Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$
Assertion $A$ : A spherical body of radius $(5 \pm 0.1)$ $mm$ having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is $4\,\%$.
Reason $R$ : The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.
In the light of the above statements, choose the correct answer from the options given below on :
Both $A$ and $R$ are true but $R$ is NOT the correct explanation of $A$
Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
$A$ is false but $R$ is true
$A$ is true but $R$ is false
A cylindrical wire of mass $(0.4 \pm 0.01)\,g$ has length $(8 \pm 0.04)\,cm$ and radius $(6 \pm 0.03)\,mm$.The maximum error in its density will be $......\,\%$
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
In an experiment to determine the acceleration due to gravity $g$, the formula used for the time period of a periodic motion is $T=2 \pi \sqrt{\frac{7(R-r)}{5 g}}$. The values of $R$ and $r$ are measured to be $(60 \pm 1) \mathrm{mm}$ and $(10 \pm 1) \mathrm{mm}$, respectively. In five successive measurements, the time period is found to be $0.52 \mathrm{~s}, 0.56 \mathrm{~s}, 0.57 \mathrm{~s}, 0.54 \mathrm{~s}$ and $0.59 \mathrm{~s}$. The least count of the watch used for the measurement of time period is $0.01 \mathrm{~s}$. Which of the following statement($s$) is(are) true?
($A$) The error in the measurement of $r$ is $10 \%$
($B$) The error in the measurement of $T$ is $3.57 \%$
($C$) The error in the measurement of $T$ is $2 \%$
($D$) The error in the determined value of $g$ is $11 \%$
The random error in the arithmetic mean of $100$ observations is $x$; then random error in the arithmetic mean of $400$ observations would be
Thickness of a pencil measured by using a screw gauge (least count $0.001 \,cm$ ) comes out to be $0.802 \,cm$. The percentage error in the measurement is ........... $\%$