The activity of a radioactive sample falls from $700{\text{s}}^{-1}$ to $500{\text{s}}^{-1}$ in $30$ minutes. It's half life is close to :
$\left(\text{Take}\text{ln 2 = 0.693 and ln 1.4 = 0.336}\right)$

The correct option is B $\text{62 min}$
From the law of radioactivity,

$A=A_0{e}^{-\lambda t}$
Here, $A_0=700{\text{s}}^{-1},A=500{\text{s}}^{-1},t=30\text{min}$

$\because \lambda =\frac{\ln 2}{T_{1/2}}$

$\Rightarrow 500=700e^{-30\left(\ln 2/T_{1/2}\right)}$

$\Rightarrow \ln \frac{7}{5}=\frac{30\ln 2}{T_{1/2}}$

$\Rightarrow T_{1/2}=30\frac{\ln 2}{\ln 1.4}=61.8\text{min}$

$\Rightarrow T_{1/2}\approx \text{62 minute}$

Hence, option $\left(B\right)$ is correct.