The activity of a radioactive sample falls from $700s^{-1}$ to $500s^{-1}$ in $30$ minutes. Its half life is close to :

From equation of radioactivity we know

$A_t=A_0{e}^{-\lambda t}$

Where $A_t=$ Amount of material at time 't'

$A_0=$ Amount of substance at $t=0$

$\lambda =$ decay constant

$t=$ time.

Taking log we have 

$\ln \left[\frac{A_0}{A_t}\right]=\lambda t$

For half life $A_t=\frac{A_0}{2}$.

$\Rightarrow \ln 2=\lambda t_{1/2}$ __(A) $t_{1/2}=$ half life

From given condition.

$\ln \left[\frac{700}{500}\right]=\lambda (30min)$ __(B)

$(A)-(B)$

$\frac{\ln 2}{\ln [7/5]}=\frac{t_{1/2}}{30}$

$\Rightarrow t_{1/2}=\frac{\ln 2\times 30}{\ln [7/5]}\Rightarrow t_{1/2}=\frac{0.693\times 30}{0.336}$

$\Rightarrow t_{1/2}=61.8min$

option (B) is correct.