If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the fundamental frequency n of the original string is given by: A n=n1+n2+n3 B 1/n=1/n1+1/n2+1/n3 C 1/sqrt(n)=1/sqrt(n1)+1/sqrt(n2)+1/sqrt(n3) D sqrt(n)=sqrt(n1)+sqrt(n2)+sqrt(n3)

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are EKx, EKy, EKz. Total K.e = EKx + EKy + EKz. Since the motion of molecule is equally probable in all three directions, therefore EKx = EKy = EKz = 1/3 EK = 1/3 x 3/2 kT = 1/2 kT, where k = R/NA = Boltzmann constant. K. E. = 1/2 kT per molecule or = 1/2 RT per mole.

In vibration motion, molecules possess both kinetic energy as well as potential energy. This means energy of vibration involves two degrees of freedom. Vibration energy = 2 x 1/2 kT = 2 x 1/2 RT [therefore two degrees of freedom per mole]. If the gas molecules have n1 translational degrees of freedom, n2 rotational degrees of freedom and n3 vibrational degrees of freedom, the total energy = n1[kT/2] + n2[kT/2] + n3[kT/2] x 2. Where 'n' is the atomicity of gas. The rotational kinetic energy of H2O molecule is equal to

A 3/2 kT B 1/2 kT C kT D 2 kT

The total energy of molecules is divided equally amongst the various degrees of freedom of a molecule. The distribution of kinetic energy along x, y, z axis are E_Kx, E_Ky, E_Kz. Total K.e = E_Kx + E_Ky + E_Kz Since the motion of the molecule is equally probable in all three directions, therefore E_Kx = E_Ky = E_Kz = (1/3) E_K = (1/3) × (3/2) kT = (1/2) kT, where k = (R / N_A) = Botzman constant. K. E. = (1/2) kT per molecule or = (1/2) RT per mole.

In vibration motion, molecules possess both kinetic energy as well as potential energy. This means the energy of vibration involves two degrees of freedom. Vibration energy = 2 × (1/2) kT = 2 × (1/2) RT [therefore two degrees of freedom per mole] If the gas molecules have n1 translational degrees of freedom, n2 rotational degrees of freedom, and n3 vibrational degrees of freedom, the total energy = n1[(kT / 2)] + n2[(kT / 2)] + n3[(kT / 2)] × 2

Where 'n' is the atomicity of the gas. How many total degrees of freedom are present in H2 molecules in all types of motions?

A. 3 B. 5 C. 6 D. 4

$40 \text{ gm NaOH}, 106 \text{ gm Na}_{2} \text{CO}_{3}$ and $84 \text{ gm NaHCO}_{3}$ are dissolved in water and the solution is made 1 lit. $20 \text{ ml}$ of this stock solution is titrated with $1 \text{ N HCl$. Hence, which of the following statements are correct?

• The burette reading of $\text{HCl}$ will be $40 \text{ ml}$ if phenolphthalein is used as an indicator from the beginning.

• The burette reading of $\text{HCl}$ will be $40 \text{ ml}$ if methyl orange is used as an indicator after the first end point.

• The burette reading of $\text{HCl}$ will be $60 \text{ ml}$ if phenolphthalein is used as an indicator from the beginning.

• The burette reading of $\text{HCl}$ will be $80 \text{ ml}$ if methyl orange is used as an indicator from the very beginning.

Find the sum of the first $n$ terms of the series $3+7+13+21+31+\ldots$.

Or if $a$, $b$ and $c$ are in GP and $x$, $y$ are the arithmetic means of $a, b$ and $b, c$ respectively, prove that $\frac{a}{x} + \frac{c}{y} = 2$ and $\frac{1}{x} + \frac{1}{y} = \frac{2}{b}$.

Consider a source of sound $S$ and an observer $P$. The source emits sound of frequency $N_{0}$. The frequency observed by $P$ is found to be $N_{1}$ if $P$ approaches $S$ at a speed $v$ and $S$ is stationary; $N_{2}$ if $S$ approaches $P$ at a speed $v$ and $P$ is stationary; and $N_{3}$ if each of $P$ and $S$ has speed $v / 2$ towards one another.

• $n_{1}=n_{2}=n_{3}$
• $n_{1}<n_{2}$
• $n_{3}>n_{0}$
• $n_{3}$ lies between $n_{1}$ and $n_{2}$