MCQ
Star $A$ has radius $ r$ surface temperature $T$ while star $B$ has radius $4r$ and surface temperature $T/2$ . The ratio of the power of two starts, $P_A : P_B$ is
- A$16:1$
- B$1:16$
- ✓$1:1$
- D$1:4$
==> $\frac{{{P_2}}}{{{P_1}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} \right)^2} \times {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^4}$$ = {\left( {\frac{{4r}}{r}} \right)^2} \times {\left( {\frac{{T/2}}{T}} \right)^4}$ $=1$ .
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The trajectory of particle $1$ with respect to particle $2$ will be
$[\pi=3.14]$


Reason : The fringe width is inversely proportional to the distance between the two slits.