If a radioactive element with a half-life of 30 min undergoes beta decay. The fraction of the radioactive element that remains undecayed after 90 min is:
For radioactive decay:
1. Number of Half-Lives: - Time elapsed: t = 90 min. - Half-life: T1/2 = 30 min. - Number of half-lives:
\[n = \frac{t}{T_{1/2}} = \frac{90}{30} = 3.\]
2. Remaining Fraction: - Fraction remaining after n half-lives:
\[\frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 = \frac{1}{8}.\]
Final Answer:
\(\boxed{\frac{1}{8}}\)
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

Method used for separation of mixture of products (B and C) obtained in the following reaction is: 