Question

19) The half-life of a certain radioactive isotope is 32 hours. What fraction of the sample would remain after 16 huors?
a) 0.50
b) 0.25
c) 0.62
d) 0.71
20) A potentiometer wire of length 100cm has a resistance of 10Ê. It is connected in series to to a resistance R and a cell of emf 2V and negligible internal resistance. A source of emf of 10mV is balanced by a length of 40cm of the potentiometer wire. What is the value of the resistance R?
a) 200Ê
b) 950Ê
c) 2000Ê
d) 790Ê

Answer 1

Answer on question #61367, Physics / Electromagnetism

19) The half-life of a certain radioactive isotope is 32 hours. What fraction of the sample would remain after 16 hours?

a) 0.50

b) 0.25

c) 0.62

d) 0.71

Solution:

The law radioactive decay is

$$N = N _ {0} \cdot 2 ^ {- \frac {t}{T}}$$

Here $N$ is the quantity at time $t$, and $N_0$ is the initial quantity, i.e. the quantity at time $t = 0$, $T$ is the half-life.

From here,

$$\frac {N}{N _ {0}} = 2 ^ {- \frac {t}{T}}$$ $$\frac {N}{N _ {0}} = 2 ^ {- \frac {16}{32}} = 2 ^ {- \frac {1}{2}} = 0.71$$

Answer: d) 0.71

20) A potentiometer wire of length $100\mathrm{cm}$ has a resistance of $10\Omega$. It is connected in series to a resistance R and a cell of emf 2V and negligible internal resistance. A source of emf of $10\mathrm{mV}$ is balanced by a length of $40\mathrm{cm}$ of the potentiometer wire. What is the value of the resistance R?

a) $200\Omega$

b) $950\Omega$

c) $2000\Omega$

d) $790\Omega$

Solution:

The current in the circuit

$$I = \frac {U}{R + r}$$

Now as the $100\mathrm{cm}$ wire has a resistance of $10\Omega$, the resistance of $40\mathrm{cm}$ of wire will be $40\times (10 / 100) = 4\Omega$.

Potential drop across 40 cm wire will be

$$V = 41$$ $$I = \frac{2}{R + 10}$$

given $V = 10$ mv

$$10 \cdot 10^{-3} = 4 \cdot \frac{2}{R + 10}$$ $$R = 4 \cdot \frac{2}{10 \cdot 10^{-3}} - 10$$ $$R = 790\ \Omega$$

Answer: d) 790 Ω

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