Physics NCERT Exemplar Solutions Class 11th Chapter Ten

$\Rightarrow \mathrm{}{\stackrel{?}{\mathrm{V}}}_1=2\stackrel{?}{{\mathrm{U}}_{\mathrm{c}\mathrm{m}}}-\stackrel{?}{{\mathrm{U}}_1}\Rightarrow \mathrm{}2\left[\frac{\mathrm{m}/2{\mathrm{V}}_{\mathrm{o}}}{\frac{\mathrm{m}}{2}+\frac{\mathrm{m}}{3}}\right]-{\mathrm{V}}_{\mathrm{o}}$

$\Rightarrow \mathrm{}\frac{6}{5}{\mathrm{V}}_{\mathrm{o}}-{\mathrm{V}}_{\mathrm{o}}\Rightarrow \frac{{\mathrm{V}}_0}{5}$

$\Rightarrow \mathrm{}{\lambda }_{\mathrm{o}}=\frac{\mathrm{h}\mathrm{c}}{\left(\frac{\mathrm{m}}{2}{\mathrm{V}}_{\mathrm{o}}\right)}\mathrm{}{\lambda }_{\mathrm{f}}=\frac{\mathrm{h}\mathrm{c}}{\left(\frac{\mathrm{M}}{2}\frac{{\mathrm{V}}_{\mathrm{o}}}{5}\right)}$

$\Rightarrow \mathrm{\Delta }\lambda =\frac{8\mathrm{h}\mathrm{c}}{{\mathrm{m}\mathrm{V}}_{\mathrm{o}}}$

$K=\frac{\left(\frac{Q}{r}\right)\mathrm{\Delta }x}{\mathrm{\Delta }AT}$

$\begin{array}{ll}\Rightarrow & \frac{\left({\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-2}\right)\left(\mathrm{L}\right)}{\left({\mathrm{L}}^2\right)\left(\theta \right)\left(\mathrm{T}\right)}\\ \Rightarrow \mathrm{}& {\mathrm{M}}^1{\mathrm{L}}^{1-}{\mathrm{T}}^{-3}{\theta }^{-1}\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) Streamline flow is more likely for liquids having low density. We know that greater the coefficient of viscosity of a liquid more will be velocity gradient hence each line of flow can be easily differentiated. Also higher the coefficient of viscosity lower will be Reynolds number, hence flow more like to be streamline.

This is a multiple choice answer as classified in NCERT Exemplar

(c), (d) For liquids coefficient of viscosity $\eta \propto \frac{1}{\sqrt[]{T}}$

i.e with increase in temperature $\eta$ decreases.

For gases coefficient of viscosity $\eta \propto \sqrt[]{T}$

i.e with increase in temperature $\eta$ decreases.

This is a multiple choice answer as classified in NCERT Exemplar

(a), (b) When the coin falls into the water, weight of the (block + coin) system decreases, which was balanced by the upthrust force earlier. As weight of the system decreases, hence upthrust force will also decrease which is only possible when l decreases.

As l decreases volume of water displaced by the block decreases, hence h decreases. As the coin falls into water, it displaces some volume of water which is very less. Hence, we neglect volume of the coin.

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c) Pressure is defined as the ratio of magnitude of component of the force normal to the area and the area under consideration.

As magnitude of component is considered, hence, it will not have any direction. So, Pressure is a scalar quantity.

This is a multiple choice answer as classified in NCERT Exemplar

(b), (d) Consider the diagram where two molecules of a liquid are shown. One is well inside the liquid and other is on the surface. The molecule (A) which is well inside experiences equal forces from all directions, hence net force on it will be zero.

And molecules on the liquid surface have some extra energy as it surrounded by Only lower half side of liquid molecules.

This is a multiple choice answer as classified in NCERT Exemplar

(c) According to the question, the observed meniscus is of convex figure shape. Which is only possible when angle of contact is obtuse. Hence, the combination will be of mercury-glass (140°)

This is a multiple choice answer as classified in NCERT Exemplar

(a) d₁= diameter at 1^st point is 2.5

d₂= diameter at 2^nd  point is 3.75

Applying equation of continuity for cross section A₁ and A₂

A₁V₁=A₂V₂

^{$\frac{V_1}{V_2}=\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2}=\left(\frac{r_2}{r_1}\right)$ 2}

= ${\left(\frac{\frac{3.75}{2}}{\frac{2.5}{2}}\right)}^2$ = ${\left(\frac{3.75}{2.5}\right)}^2=\frac{9}{4}$