Physics MCQs for NEET — Practice Questions with Answers

(b) Increases in surface energy= work done in area$\times$Surface tension 

$\because$Increase in surface area,

 $∆A=\left(5\times 4-4\times 2\right)\times 2 \left(\because flim has two surfaces\right)$

=$$\begin{gathered} \left(20-8\right)\times 2 c{m}^2=24 c{m}^2 \\ =24\times {10}^{-4} m^2 \end{gathered}$$ 

So, work done, W=T.$∆A$

  $$\begin{gathered} 3\times {10}^{-4}=T\times 24\times {10}^{-4} \\ \therefore T=\frac{1}{8}=0.125 N/m \end{gathered}$$

(b) According to ascent formula for capillary tube,

              $$\begin{gathered} h=\frac{2T\cos \theta }{pgr} \\ \therefore \frac{\cos {\theta }_1}{p_1}=\frac{\cos {\theta }_2}{p_2}=\frac{\cos {\theta }_3}{p_3} \\ Thus,\cos \theta \propto \rho \\ \therefore p_1>p_2>p_3 \\ \therefore \cos {\theta }_1>\cos {\theta }_2>\cos {\theta }_3 \\ 0\le {\theta }_1<{\theta }_2<{\theta }_3<\frac{\mathrm{\pi }}{2} \end{gathered}$$

According to Archimedes' principle, the buoyant force on the cylinder is equal to the weight of the liquid displaced by the cylinder. The volume of the liquid displaced is pLπr^2, where r is the radius of the cylinder. The weight of this liquid is pLπr^2ρn. This weight must equal the weight of the cylinder, which is pLπr^2d. Therefore, d = ρn.

Given d=2700 m

ρ=10³kg/m³

Compressibility=45.4 x 10^-11 Pa^-1  per pascal

The pressure at the bottom of ocean is given by

P=ρgd=10³x10x2700=27x10⁶Pa

So, fractional compression=compressibility x pressure

=45.4 x 10^-11x 27 x10⁶

=1.2x10^-2

From Bernoulli's theorem

p₁+$\frac{1}{2}$ρv₁²=p₂+$\frac{1}{2}$ρv₂² 

where, p₁ and p₂ are pressure inside and outside the roof and v₁ v₂ are velocities of wind inside and outside the roof, Neglect the width of the roof.

Pressure difference is

p₁-p₂=$\frac{1}{2}$ρ(v₂²-v₁²)

=$\frac{1}{2}$x1.2(40²-0)=960N/m²

Force acting on the roof is given by

F=(p₁-p₂)A=960 x 250=24x10⁴N=24X10⁵N

As the pressure inside the roof is more than outside to it. So the force will act m the upward direction, i.e. F=24x10⁵N,upward

The liquid flow is incompressible and steady. By the principle of conservation of mass, the volume flow rate must be constant throughout the tube. Therefore, the speed v in the tube is related to the speed v' through the holes by the equation πR^2v = nπr^2v', which gives v' = vR^2/nr^2.

Given pressure=150mm of Hg

Pumping rate of heart of a man=$\frac{dV}{dt}$=$\frac{{\displaystyle 5x{10}^3}}{{\displaystyle 60}}$ m³/s

Power of heart=P.$\frac{dV}{dt}$=ρgh.$\frac{dV}{dt}$ [P=ρgh]

=>(13.6x10³kg/m³) (10x0.15x5x10^-3)/60

=1.70W

In case of insufficient length, angle of contact adjusts itself so that water does not spill

$$\begin{gathered} If the surface area changes, it will change the surface energy as well. As the surface area is decrea\sin g, energy will be released. \\ Change in surface energy=T\times ∆A \\ If there is n no. if small drops; \\ Volume of a large drop=volume of n small drops \\ \frac{4}{3}{\mathrm{\pi R}}^3=\mathrm{n}\times \frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{Change} \mathrm{in} \mathrm{area}; \\ ∆\mathrm{A}=4{\mathrm{\pi R}}^2-\mathrm{n}\times 4{\mathrm{\pi r}}^2=3\left(\frac{4{\mathrm{\pi R}}^3}{3\mathrm{R}}-\mathrm{n}\times \frac{4{\mathrm{\pi r}}^3}{3\mathrm{r}}\right)=3\left(\frac{4{\mathrm{\pi R}}^3}{3\mathrm{R}}-\frac{4{\mathrm{\pi R}}^3}{3\mathrm{r}}\right)=3\left(\frac{\mathrm{V}}{\mathrm{R}}-\frac{\mathrm{V}}{\mathrm{r}}\right)=3\mathrm{V}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right) \\ \mathrm{E}=\mathrm{T}\times ∆\mathrm{A}=3\mathrm{VT}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{r}}\right), \mathrm{released}. \end{gathered}$$

(d) The value of angle of contact determines whether a liquid will spread on the surface