Percentage of Carbon:
The mass of carbon in 0.2 g of CO$_2$ can be calculated using the molar masses: Molar mass of CO$_2$ = 44 g/mol Molar mass of C = 12 g/mol
Mass of C in 0.2 g CO$_2 = \frac{12}{44} \times 0.2 g = 0.0545g$
Percentage of C $= \frac{Mass of C}{Mass of compound} \times 100 = \frac{0.0545}{0.3} \times 100 = 18.18%$
Percentage of Hydrogen:
The mass of hydrogen in 0.1 g of H$_2$O can be calculated using the molar masses: Molar mass of H$_2$O = 18 g/mol Molar mass of H = 1 g/mol (but there are 2 H atoms, thus, 2 g/mol)
Mass of H in 0.1 g H$_2$O $= \frac{2}{18} \times 0.1g = 0.0111g$
Percentage of H $= \frac{\text{Mass of H}}{\text{Mass of compound}} \times 100 = \frac{0.0111}{0.3} \times 100 = 3.70%$
Therefore, the percentage composition of carbon and hydrogen is 18.18% and 3.70%, respectively.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The potential (V) at any axial point, at 2 m distance(r) from the centre of the dipole of dipole moment vector
\(\vec{P}\) of magnitude, 4 × 10-6 C m, is ± 9 × 103 V.
(Take \(\frac{1}{4\pi\epsilon_0}=9\times10^9\) SI units)
Reason R : \(V=±\frac{2P}{4\pi \epsilon_0r^2}\), where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below :
The output (Y) of the given logic gate is similar to the output of an/a :