Past NEET Papers

Hint

Density of liquid, \(\rho_1=760 \mathrm{~kg} / \mathrm{m}^3\)
Density of mercury, \(\rho_m=13600 \mathrm{~kg} / \mathrm{m}^3\)
Height of liquid column in mercury barometer,
\(
h_m=76 \mathrm{~cm}=0.76 \mathrm{~m}
\)
If height of liquid in liquid column beh, then
\(
\begin{aligned}
& p_{\text {liquid }}=p_{\text {mercury }} \\
& \Rightarrow \quad h_1 \rho_1 g=h_m \rho_m g \\
& \Rightarrow \quad h_1=\frac{h_m \rho_m}{\rho_1}=\frac{0.76 \times 13600}{760} \\
& =13.6 \mathrm{~m} \\
\end{aligned}
\)

Hint

Given, the density of the small ball is \(d\)
The mass of the small ball is \(M\)
The density of the glycerine is \(d / 2\).
As we know that,
viscous force \(=\) weight – buoyant force
Viscous force \(=V d_1 g-V d_2 g\)
Here, \(V\) is the volume of submerged bodies,
\(g\) is the acceleration due to gravity,
\(d_1\) is the density of the small ball,
\(d_2\) is the density of the glycerine,
\(d_1=d\) and \(d_2=d / 2\)
Substituting the given values in the viscous force expression, we get
Viscous force \(=V d g-V \frac{d}{2} g\)
Viscous force \(=\frac{V d g}{2}=\frac{M g}{2} \quad(\because M=d \times V)\)

Hint

A liquid does not wet the solid surface, if the angle of contact is obtuse i.e., \(\theta>90^2\).

Explanation:

When the contact angle is obtuse (more than \(90^{\circ}\) ), cohesive forces \(>\) adhesive forces, hence the liquid beads up.
If the angle of contact is less than \(90^{\circ}\), the liquid tends to spread at the solid surface. As the angle of contact becomes smaller, the spreading of the liquid over the solid surface increases.

Hint

(b) Force of surface tension balances the weight of water in capillary tube.
\(
F_S=2 \pi r T \cos \theta=m g
\)
Here, \(T\) and \(\theta\) are constant.
So, \(m \propto r\)
Let \(m_1\) and \(m_2\) be the mass of water in two capillary tube.
\(
\begin{aligned}
& \therefore \frac{m_2}{m_1}=\frac{r_2}{r_1} \\
& \Rightarrow \frac{m_2}{5.0}=\frac{2 r}{r} \quad\left(\because r_2=2 r\right) \\
& \Rightarrow m_2=10.0 \mathrm{~g}
\end{aligned}
\)

Hint

According to Pascal’s law “Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel.” In the given situation as shown in the figure below

Pressure due to water column of height \(15 \mathrm{~cm}=\) Pressure due to oil column of height \(20 \mathrm{~cm}\)
\(
\begin{aligned}
& \Rightarrow \quad h_w \rho_w g=h_0 \rho_0 g \\
& 15 \rho_w=20 \rho_0 \\
& \Rightarrow \quad \rho_0=\frac{15}{20} \rho_w
\end{aligned}
\)
\(
\begin{aligned}
\rho_0= & \frac{15}{20} \times 1000 \\
& \quad\left(\because \text { given } \rho_w=1000 \mathrm{~kg} \mathrm{~m}^{-3}\right) \\
& =750 \mathrm{kgm}^{-3}
\end{aligned}
\)

Hint

The rate of liquid flow moving with velocity \(v\) through an area \(A\) is given by Rate \((R)=\) Area \((A) \times\) Velocity \((v)\) Given, area of the hole,
\(
\begin{aligned}
A & =2 \mathrm{~mm}^2 \\
& =2 \times 10^{-8} \mathrm{~m}^2
\end{aligned}
\)
height of tank, \(h=2 \mathrm{~m}\).
The given situation can also be depicted as shown in the figure below.

As the velocity of liquid flow is given as
\(
\begin{aligned}
& v=\sqrt{2 g h} \\
& \therefore R=A v=A \sqrt{2 g h}
\end{aligned}
\)
Substituting the given values, we get
\(
\begin{aligned}
R & =2 \times 10^{-8} \times \sqrt{2 \times 10 \times 2} \\
& =2 \times 10^{-8} \times 6.32=12.64 \times 10^{-8} \mathrm{~m}^3 / \mathrm{s} \\
& =12.6 \times 10^{-8} \mathrm{~m}^3 / \mathrm{s}
\end{aligned}
\)

Hint

The terminal velocity achieved by ball in a viscous fluid is
\(
v_{\mathrm{t}}=\frac{2(\rho-\sigma) r^2 g}{9 \eta}
\)
where, \(\rho=\) density of metal of ball,
\(\sigma=\) density of viscous medium,
\(r=\) radius of ball and
\(\eta=\) coefficient of viscosity of medium
Terminal velocity of first ball,
\(
\begin{aligned}
v_{t_1} & =\frac{2\left(\rho_1-\sigma\right) r_1^2 g}{9 \eta} \\
& =\frac{2\left(8 \rho_2-\sigma\right) r_1^2 g}{9 \eta} \quad \ldots(i)\left[\because \rho_1=8 \rho_2\right]
\end{aligned}
\)
Similarly, for second ball
\(
v_{t_2}=\frac{2\left(\rho_2-\sigma\right) r_2^2 g}{9 \eta} \dots(ii)
\)
From Eq. (i) and (ii), we get
\(
\begin{aligned}
& \frac{v_{\mathrm{t}_1}}{v_{\mathrm{t}_2}}=\frac{2\left(8 \rho_2-\sigma\right) r_1^2 g}{2\left(\rho_2-\sigma\right) r_2^2 g} \times \frac{9 \eta}{9 \eta} \\
&=\left(\frac{8 \rho_2-0.1 \rho_2}{\rho_2-0.1 \rho_2}\right)\left(\frac{r_1}{r_2}\right)^2 \ldots \text { (iii) }\\
& \left[\because \sigma=0.1 \rho_2\right]
\end{aligned}
\)
Here, \(r_1=1 \mathrm{~mm}\) and \(r_2=2 \mathrm{~mm}\)
Substituting these values in Eq. (iii), we get
\(
\begin{aligned}
\Rightarrow \quad \frac{v_{\mathrm{t}_1}}{v_{\mathrm{t}_2}} & =\left(\frac{7.9 \rho_2}{0.9 \rho_2}\right)\left(\frac{1}{2}\right)^2 \\
& =\frac{79}{36}
\end{aligned}
\)

Hint

(b) The pressure inside the soap bubble
\(
=\mathrm{P}_a+\frac{4 \mathrm{~T}}{\mathrm{R}}
\)
And pressure at a point below the surface of the water \(=\rho g Z_0+P_a\)
[where \(\mathrm{P}_a=\) atmospheric pressure]
By equating these pressure we get,
\(
\begin{aligned}
& \mathrm{P}_a+\frac{4 \mathrm{~T}}{\mathrm{R}}=\mathrm{P}_a+\rho \mathrm{g} \mathrm{Z}_0 \\
& \mathrm{Z}_0=\frac{4 \mathrm{~T}}{\mathrm{R} \times \rho \mathrm{g}} \\
& \mathrm{Z}_0=\frac{4 \times 2.5 \times 10^{-2}}{10^{-3} \times 1000 \times 10} \mathrm{~m} \\
& \mathrm{Z}_0=1 \mathrm{~cm}
\end{aligned}
\)

Hint

Key Concept The rate of heat generation is equal to the rate of work done by the viscous force which in turn is equal to its power.
Power = Rate of heat produced, \(\frac{d \emptyset}{d t}=F \times v_T\)
where, \(F\) is the viscous force and \(v_T\) is the terminal velocity.
\(
\begin{aligned}
& \text { As, } \quad F=6 \pi \eta r v_T \\
& \Rightarrow \quad \frac{d \emptyset}{d t}=6 \pi \eta r v_T \times v_T  \\
& =6 \pi \eta r v_T^2 \dots(i)\\
\end{aligned}
\)
From the relation for terminal velocity,
\(
v_T=\frac{2}{9} \frac{r^2(\rho-\sigma)}{\eta} g \text {, we get }
\)
\(
v_T \propto r^2 \dots(ii)
\)
From Eq. (ii), we can rewrite Eq. (i) as
\(
\frac{d \emptyset}{d t} \propto r \cdot\left(r^2\right)^2
\)
or \(\quad \frac{d \emptyset}{d t} \propto r^5\)

Hint

Both ends of the U tube are open, so the pressure on both the free surfaces must be equal.
\(
\text { Here, } \mathrm{h}_{\text {oil }} \times \rho_{\text {oil }} \times \mathrm{g}=\mathrm{h}_{\text {water }} \times \rho_{\text {water }} \times \mathrm{g}
\)
\(
\begin{aligned}
& \rho_0 \mathrm{~g} \times 140 \times 10^{-3}=\rho_{\mathrm{w}} \mathrm{g} \times 130 \times 10^{-3} \\
& \rho_{\text {oil }}=\frac{130}{140} \times 10^3 \approx 928 \mathrm{~kg} / \mathrm{m}^3 \\
& \qquad\left[\because \rho_{\mathrm{w}}=1 \mathrm{kgm}^{-3}\right]
\end{aligned}
\)

Hint

According to the question, the situation can be drawn as follows.

Applying Archimedes principle
Weight of cylinder \(=(\text { upthrust })_1+\) (upthrust) \({ }_2\)
i.e. \(\quad A L d g=(1-P) L A \rho g+(P L A) n \rho g\)
\(\Rightarrow \quad d=(1-P) \rho+\) Pn \(\rho\)
\(=\rho-P \rho+n P \rho\)
\(=\rho+(n-1) P \rho=\rho[1+(n-1) P]\)

Hint

Increase in surface energy= work done in area \(\times\) surface tension
\(\because\) nncrease in surface area,
\(\Delta \mathrm{A}=(5 \times 4-4 \times 2) \times 2\)
( \(\because\) film has two surfaces) \(=(20-8) \times 2 \mathrm{~cm}^2=24 \mathrm{~cm}^2\) \(=24 \times 10^{-4} \mathrm{~m}^2\)

So, work done, \(W=T \cdot \Delta A\)
\(
\begin{array}{ll}
& 3 \times 10^{-4}=T \times 24 \times 10^{-4} \\
\therefore & T=\frac{1}{8}=0.125 \mathrm{~N} / \mathrm{m}
\end{array}
\)

Hint

According to ascent formula for capillary tube,
\(
\begin{aligned}
h & =\frac{2 T \cos \theta}{\rho g r} \\
\therefore \quad \quad \frac{\cos \theta_1}{\rho_1} & =\frac{\cos \theta_2}{\rho_2}=\frac{\cos \theta_3}{\rho_3}
\end{aligned}
\)
Thus, \(\cos \theta \propto \rho\)
\(
\begin{array}{ll}
\therefore \quad & \rho_1>\rho_2>\rho_3 \\
\therefore \quad & \cos \theta_1>\cos _2>\cos _3 \\
& 0 \leq \theta_1<\theta_2<\theta_3<\frac{\pi}{2}
\end{array}
\)

Hint

(b) Compressibility of water, \(\mathrm{K}=45.4 \times 10^{-11} \mathrm{~Pa}^{-1}\)
density of water \(P=10^3 \mathrm{~kg} / \mathrm{m}^3\)
depth of ocean, \(h=2700 \mathrm{~m}\)
We have to find \(\frac{\Delta \mathrm{V}}{\mathrm{V}}=\) ?
As we know, compressibility,
\(
\mathrm{K}=\frac{1}{\mathrm{~B}}=\frac{(\Delta \mathrm{V} / \mathrm{V})}{\mathrm{P}}(\mathrm{P}=\rho \mathrm{gh})
\)
So, \((\Delta \mathrm{V} / \mathrm{V})=\mathrm{K} \rho \mathrm{gh}\)
\(
=45.4 \times 10^{-11} \times 10^3 \times 10 \times 2700=1.2 \times 10^{-2}
\)

Hint

From Bernoulli’s theorem
\(
p_1+\frac{1}{2} \rho v_1^2=p_2+\frac{1}{2} \rho v_2^2
\)
where, \(p_1, p_2\) are pressure inside and outside the roof and \(v_1, v_2\) are velocities of wind inside and outside the roof.
Neglect the width of the roof.
Pressure difference is
\(
\begin{aligned}
p_1-p_2 & =\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \\
& =\frac{1}{2} \times 1.2\left(40^2-0\right) \\
& =960 \mathrm{~N} / \mathrm{m}^2
\end{aligned}
\)
Force acting on the roof is given by
\(
\begin{aligned}
F & =\left(p_1-p_2\right) A=960 \times 250 \\
& =24 \times 10^4 \mathrm{~N}=2.4 \times 10^5 \mathrm{~N}
\end{aligned}
\)
As the pressure inside the roof is more than outside to it. So the force will act in the upward direction,
i.e. \(\mathbf{F}=2.4 \times 10^5 \mathrm{~N}\), upwards.

Hint

During the streamline flow of viscous and incompressible fluid through a pipe varying cross-section, the product of the area of the cross-section and normal fluid velocity \((Av)\) remains constant throughout the flow. Consider a cylindrical tube of a spray pump has radius \(R\), one end having \(n\) fine holes, each of radius \(r\) and speed of liquid in the tube is \(v\) as shown in the figure.

According to equation of continuity,
\(
A v=\text { constant }
\)
where, \(A\) is a cylindrical tube and \(v\) is velocity of liquid in a tube.
Volume in flow rate = volume in outflow rate

\(
\pi R^2 v=n \pi r^2 v^{\prime} \Rightarrow v^{\prime}=\frac{R^2 v}{n r^2}
\)
Thus, speed of the ejection of the liquid through the holes is \(\frac{R^2 v}{n r^2}\).

Hint

(a) Water rises upto the top of capillary tube and stays there without overflowing.

In case of capillary of insufficient length i.e. \(L \lt h\) the liquid will neither overflow from the upper end like a fountain nor it will tickle along the vertical sides of the tube. The liquid after reaching the upper end will increase the radius of its meniscus without changing nature such that
\(\begin{array}{ll}h r=L r^{\prime} & \text { (Here, } r \text { is the intial radius } \\ \because L<h & \text { of tube and } r^{\prime} \text { is the new } \\ \because r^{\prime}>r & \text { radius) }\end{array}\)

Hint

As energy released \(=\left(A_f-A_i\right) T\) where, \(A_i=4 \pi R^2=\frac{3}{3} \times 4 \pi \frac{R^3}{R}=\frac{3 \mathrm{~V}}{R}\) and \(A_f=4 \pi r^2=\frac{V}{\frac{4}{3} \pi r^2} 4 \pi r^2=\frac{3 V}{r}\)
\(\therefore\) Energy released \(=3 V T\left[\frac{1}{r}-\frac{1}{R}\right]\)

Hint

(c) Wetability of a surface by a liquid primarily depends on the angle of contact between the surface and liquid. If the angle of contact is acute liquids wet the solid and vice-versa.

Hint

(b) According to Bernoulli’s theorem, \(P+\frac{1}{2} \rho v^2=\) constant and \(A v=\) constant, If \(A\) is minimum, \(v\) is maximum, \(P\) is minimum.

Hint

(b) \((h \rho g+H \times 1 \times g) \frac{4}{3} \pi r^3=h \rho g \times \frac{4}{3} \pi(2 r)^3\)
This gives \(\mathrm{H}=7 h \rho\)

Hint

(a) \(\mathrm{K}=\frac{1}{\mathrm{~B}}=\frac{\Delta \mathrm{V} / \mathrm{V}}{\mathrm{P}}\). Here, \(\mathrm{P}=100 \mathrm{~atm}\), \(\mathrm{K}=4 \times 10^{-5}\) and \(\mathrm{V}=100 \mathrm{~cm}^3\).
Hence, \(\Delta \mathrm{V}=0.4 \mathrm{~cm}^3\)

Hint

(c) Note that according to Stoke’s law \(6 \pi \eta r v_r=m g\)
Hence, the valid relation is \(v_r \propto m g / r \eta\)

Hint

(a) We know that angle of contact is the angle between the tangent to liquid surface at the point of contact and solid surface inside the liquid. In case of pure water and pure glass, the angle of contact is zero.

Hint

The initial speed of the ball is zero and it finally attains the terminal speed

Hint

Excess pressure inside the bubble \(=\Delta P=\frac{4 T}{R}\)
\(
P_{\text {in }}=P_{\text {out }}+\frac{4 T}{R}
\)
as ‘ \(R\) ‘ increases ‘ \(P\) ‘ decreases

Hint

Venturi-meter works on the Bernoulli’s principle.