- A$W\,{m^{ - 2}}\,{K^{ - 1}}$
- B$W\,{m^2}\,{K^{ - 4}}$
- ✓$W\,{m^{ - 2}}\,{K^{ - 4}}$
- D$W\,{m^{ - 2}}\,{K^4}$
$⇒$ $\sigma = \frac{E}{{{T^4}}}$
where,$E = $ $\frac{{{\rm{Energy}}}}{{{\rm{Area}} \times {\rm{Time}}}}$ $ = \frac{{{\rm{Watt}}}}{{{m^{\rm{2}}}}}$
$\sigma = \frac{{{\rm{Watt - }}{m^{ - 2}}}}{{{K^4}}} = Watt - {m^{ - 2}}{K^{ - 4}}$
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${F}={A} \cos {Bx}+{C} \sin {Dt}$
The dimensional formula of $\frac{{AD}}{{B}}$ is -

$z \pm \Delta z=\frac{x \pm \Delta x}{y \pm \Delta y}=\frac{x}{y}\left(1 \pm \frac{\Delta x}{x}\right)\left(1 \pm \frac{\Delta y}{y}\right)^{-1} .$
The series expansion for $\left(1 \pm \frac{\Delta y}{y}\right)^{-1}$, to first power in $\Delta y / y$, is $1 \mp(\Delta y / y)$. The relative errors in independent variables are always added. So the error in $z$ will be $\Delta z=z\left(\frac{\Delta x}{x}+\frac{\Delta y}{y}\right)$.
The above derivation makes the assumption that $\Delta x / x \ll<1, \Delta y / y \ll<1$. Therefore, the higher powers of these quantities are neglected.
($1$) Consider the ratio $r =\frac{(1- a )}{(1+ a )}$ to be determined by measuring a dimensionless quantity a.
If the error in the measurement of $a$ is $\Delta a (\Delta a / a \ll<1)$, then what is the error $\Delta r$ in
$(A)$ $\frac{\Delta a }{(1+ a )^2}$ $(B)$ $\frac{2 \Delta a }{(1+ a )^2}$ $(C)$ $\frac{2 \Delta a}{\left(1-a^2\right)}$ $(D)$ $\frac{2 a \Delta a}{\left(1-a^2\right)}$
($2$) In an experiment the initial number of radioactive nuclei is $3000$ . It is found that $1000 \pm$ 40 nuclei decayed in the first $1.0 s$. For $|x|<1$, In $(1+x)=x$ up to first power in $x$. The error $\Delta \lambda$, in the determination of the decay constant $\lambda$, in $s ^{-1}$, is
$(A) 0.04$ $(B) 0.03$ $(C) 0.02$ $(D) 0.01$
Give the answer or quetion ($1$) and ($2$)

Then the velocity as a function of time and the height as a function of time will be