A packet contains silver powder of mass $20.23 \,g \pm 0.01 \,g$. Some of the powder of mass $5.75 \,g \pm 0.01 \,g$ is taken out from it. The mass of the powder left back is ................
$14.48 \,g \pm 0.00 \,g$
$14.48 \pm 0.02 \,g$
$14.5 \,g \pm 0.1 \,g$
$14.5 \,g \pm 0.2 \,g$
A student determined Young's Modulus of elasticity using the formula $Y=\frac{M g L^{3}}{4 b d^{3} \delta} .$ The value of $g$ is taken to be $9.8 \,{m} / {s}^{2}$, without any significant error, his observation are as following.
Physical Quantity | Least count of the Equipment used for measurement | Observed value |
Mass $({M})$ | $1\; {g}$ | $2\; {kg}$ |
Length of bar $(L)$ | $1\; {mm}$ | $1 \;{m}$ |
Breadth of bar $(b)$ | $0.1\; {mm}$ | $4\; {cm}$ |
Thickness of bar $(d)$ | $0.01\; {mm}$ | $0.4 \;{cm}$ |
Depression $(\delta)$ | $0.01\; {mm}$ | $5 \;{mm}$ |
Then the fractional error in the measurement of ${Y}$ is
A physical parameter a can be determined by measuring the parameters $b, c, d $ and $e $ using the relation $a =$ ${b^\alpha }{c^\beta }/{d^\gamma }{e^\delta }$. If the maximum errors in the measurement of $b, c, d$ and e are ${b_1}\%$, ${c_1}\%$, ${d_1}\%$ and ${e_1}\%$, then the maximum error in the value of a determined by the experiment is
A student performs an experiment to determine the Young's modulus of a wire, exactly $2 \mathrm{~m}$ long, by Searle's method. In a particular reading, the student measures the extension in the length of the wire to be $0.8 \mathrm{~mm}$ with an uncertainty of $\pm 0.05 \mathrm{~mm}$ at a load of exactly $1.0 \mathrm{~kg}$. The student also measures the diameter of the wire to be $0.4 \mathrm{~mm}$ with an uncertainty of $\pm 0.01 \mathrm{~mm}$. Take $g=9.8 \mathrm{~m} / \mathrm{s}^2$ (exact). The Young's modulus obtained from the reading is
Two resistors of resistances $R_1 = (100 \pm 3) \,\Omega $ and $R_2 = (200 \pm 4)$ are connected in series. The maximm absolute error and percentage error in equivalent resistance of the series combination is
The energy of a system as a function of time $t$ is given as $E(t)=A^2 \exp (-\alpha t)$, where $\alpha=0.2 s ^{-1}$. The measurement of $A$ has an error of $1.25 \%$. If the error in the measurement of time is $1.50 \%$, the percentage error in the value of $E(t)$ at $t=5 s$ is