c+a-b
c-a-b
a+b+c
a+b-c
we have \(^2_1H+^3_1H\) \(\rightarrow\) \(^4_2He+^1_0n,\)
Energy released, E = (△m)×931MeV
atomic mass unit = 931MeV
\(\triangle\)m=mass of product− mass of reactant
\(\triangle\)m=c−a−b
E=(\(\triangle\)m)×931
i. e ,E =(c−a−b)
The binding energies of \(^2_1H, ^3_1H\) and \(^4_2He\) are respectively a,b and c
Therefore, the correct answer is (B): c-a-b
The current passing through the battery in the given circuit, is:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec:
In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
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