Physics M.C.Q (1 Marks)

$==> T = 10^{7}\, K$

= $ - 2.5 \times 0.6021 = - 1.5$.

Distance of Galaxy from us.

According to Doppler’s effect speed of Galaxy $v = \frac{{c\Delta \lambda }}{\lambda }$

==> $r = \frac{{c\Delta \lambda }}{{H\lambda }} = \frac{{c \times 0.1\lambda }}{{H\lambda }} = \frac{{0.1\, \times 3 \times {{10}^8}}}{{19.3 \times 3 \times {{10}^{ - 3}}}} = 1.6 \times {10^9}\,ly$

 $\therefore S' = \frac{S}{{{{(5.3)}^2}}}$

and magnitude of given star$ = 0$

Now $m_2$ -$m_1$ $= 0 -(-5) = 5$

The brightness ratio is given by

$\frac{{{l_1}}}{{{l_2}}} = {100^{({m_2} - {m_1})/s}} = {100^{5/5}} = 100$

So bright star is $100$ time bright that the dim star.

where $v = 250 \,km/sec = 250 ×10^{3} \,\frac{m}{{\sec }}$

$r = 3 x10^4 ;y = 3 ×10^4 ×9.46 × 10^{12}$ km $\approx 3 ×10^{20}$ m

$\therefore m = \frac{{{{(250 \times {{10}^3})}^2} \times (3 \times {{10}^{20}})}}{{6.6 \times {{10}^{ - 11}}}}$ $\approx 3 × 10^{41} kg$.

Age of the universe, ${t_0} = \frac{1}{H} = \frac{r}{v}$

Taking $r = 430 ×10^6;ly = 430 × 10^6 × 9.46 × 10^{12}$ km

==> ${t_0} = \frac{{430 \times {{10}^6} \times 9.46 \times {{10}^{12}}}}{{8600}}\sec $

$ = \frac{{430 \times {{10}^6} \times 9.46 \times {{10}^{12}}}}{{8600 \times 3600 \times 24 \times 365}} = 1.49 \times {10^{10}}year$