Field of NoThingness - spl BiNal

Given, Potential $V = 300\;V$ Capacitor $C_1 = 3\mu F$ Capacitor $C_2 = 6\mu F$ Energy stored in each capacitor $E = ?$ When the capacitors are connected in parallel, then the equivalent capacitance $C = C_1 + C_2 = (3 + 6)\;\mu F = 9 \; \mu F$ We have, Energy stored in a capacitor ($C_1$); $E_1 = \frac{1}{2}C\;V^2 = \frac{1}{2} * 3 * 10^{-6} * 300^2 = 0.135\;J$ Energy stored in a capacitor ($C_2$); $E_2 = \frac{1}{2}C\;V^2 = \frac{1}{2} * 6 * 10^{-6} * 300^2 = 0.27\;J$ Energy stored in the equivalence capacitance ($C$); $E = \frac{1}{2}C\;V^2 = \frac{1}{2} * 9 * 10^{-6} * 300^2 = 0.405\;J$ Return to Main Menu

Two spheres of radii $5\; cm$ and $10\;cm$ are given charges $100\;C$ and $50\;C$ respectively and then connected by a wire. Calculate the loss of energy after connection.

Given, Radius of $1^{st}$ sphere $(r_1) = 5\;cm$ Radius of $2^{st}$ sphere $(r_2) = 10\;cm$ Charge on 1^{st} sphere $(q_1) = 100\;C$ Charge on $2^{nd}$ sphere $(q_1) = 100\;C$ Loss of energy $(\Delta E) = ?$ Capacitance of first sphere $C_1 = 4\; \pi \; \epsilon_0 \; r_1 = 4 \pi \;*\; 10^{-12}\;*\;0.05 = 5.56 * 10^{-12}\;F$ Capacitance of second sphere $C_2 = 4\; \pi \; \epsilon_0 \; r_2 = 4 \pi \;*\; 10^{-12}\;*\;0.1 = 1.11 \; *\; 10^{-12}\;F$ Then, $V_1 = $ $\frac{q_1}{C_1} = \frac{100}{5.56\;*\;10^{-12}}$ $ = 1.8\;*\;10^{13}\;V$ $V_2 = $ $\frac{q_2}{C_2} = \frac{50}{1.11\;*\;10^{-11}}$ $ = 4.5\;*\;10^{12}\;V$ Total energy before connection, $E_1 = \frac{1}{2}C_1\;V_1^2 + \frac{1}{2}C_2\;V_2^2$ $E_1= \frac{1}{2}* 5.56 * 10^{-12} * (1.8 * 10^{13})^2 + \frac{1}{2}* 1.11 * 10^{-11} * (4.5 * 10^{12})^2$ $= 1.01 * 10^{15}\;J$ Total energy after connection, $E_2 = \frac{1}{2} \frac{(C_1\;V_1\; + \; C_2 \; V_2)^2}{C_1 \; + \; C_2} = \frac{1}{2} \frac{(5....

Find the half-life of $U^{238}$, if one gram of it emits $1.24 * 10^4$ $\alpha - particles$ per second. (Avogadro's Number = $6.023 * 10^{23}$).

Given, $\frac{dN}{dt}$ $ = 1.24 * 10^{4}$ Half Life $(T_{1/2}) = ?$ But, 238 gm of U contains $6.023 * 10^{23}$ atoms 1 gm of U contains $\frac{6.023 \; * \; 10^{23}}{238}$ $ = 2.51 * 10^{21}$ ⇒ $N = 2.51 * 10^{21}$ atoms We have, $\frac{dN}{dt} = \lambda \; N = \frac{ln(2)}{T_{1/2}} * N$ $T_{1/2} = \frac{0.693 \; * \; N}{\frac{dN}{dt}} = \frac{0.693 \; * \; 2.51 \; * \; 10^{21}}{1.24 \; * \; 10^4}$ $ = 1.41 * 10^{17}\; Sec = 4.5 * 10^9\; years$ Return to Main Menu

Calculate the binding energy per nucleon of $_{26}{Fe}^{56}$. Atomic mass of $_{26}{Fe}^{56}$ is $55.9349 \;u$ and that of $_1H^1$ is $1.00783\;u$. Mass of $_0 n^1 = 1.00867\;u$ and $1u = 931\;MeV$.

Given, Atomic mass of $_{26}{Fe}^{56} = A = 55.9349 \;u$ Mass of $_1H^1$ $ = m_p = 1.00783\;u$ Mass of $_0 n^1 $ $ = m_n = 1.00867\;u$ $1u = 931\;MeV$ B.E. per nucleon, $(\overline{BE}) = ?$ Now, $B.E. = [Z\;m_p + (A-Z)\;m_n - M] * 931 \;MeV$ $B.E. = [26 * 1.00783 + (56 - 26) * 1.00867 - 55.9349] * 931 = 492.29418\;MeV$ $\overline{BE} = $ $\frac{B.E.}{A} = \frac{492.29418}{56}$ $ = 8.79096 \; Me V/nucleon = 1.41 * 10^{-12}\;J$. Return to Main Menu

What force is required to stretch a steel wire of cross-sectional area $1\;cm^2$ to double its length?

Given, Area of the cross-section $(A) = 1\;cm^2 = 1 * 10^{-4}\,m$ Young's modulus of steel $(Y) = 2 * 10^{11}\,N/m^2$ Let, length of the wire $(l) = x\;cm$ After extension its length $(L)$ becomes double i.e $2x$ So extension $(e) = \Delta l = L - l = 2x - x = x\;cm$ Force $(F) = ?$ Now we have, $Y = \frac{F/A}{e/l} = \frac{F/A}{x/x} = \frac{F}{A}$ ⇒ $F = Y\,*\,A = 2 * 10^{11} \; * \; 1 \; * \; 10^{-4} = 2 * 10^{7}\;N$ Return to Main Menu

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