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**[IIT-JEE-2009]**

**Sol.**206

$(500-\mathrm{H}) \mathrm{P}_{0}=300\left(\mathrm{P}_{0}-\mathrm{rg} \times 0.2\right)$

$(0.5-\mathrm{H}) \times 10^{5}=0.3\left[10^{5}-10^{4} \times 0.2\right)$

$0.5-\mathrm{H}=0.294$

H = 206 mm

** [IIT-JEE-2009]**

**Sol.**6

**Paragraph for Questions no. 3 to 5**

When liquid medicine of density is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When this force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

[/esquestion]

(A) $2 \pi r T$

(B) $2 \pi \mathrm{RT}$

(C) $\frac{2 \pi r^{2} T}{R}$

(D) $\frac{2 \pi R^{2} T}{r}$

**[IIT-JEE-2010]**

**Sol.**(C)

(A) $1.4 \times 10^{-3} \mathrm{m}$

(B) $3.3 \times 10^{-3} \mathrm{m}$

(C) $2.0 \times 10^{-3} \mathrm{m}$

(D) $4.1 \times 10^{-3} \mathrm{m}$

**[IIT-JEE-2010]**

**Sol.**(A)

(A) $1.4 \times 10^{-6} \mathrm{J}$

(B) $2.7 \times 10^{-6} \mathrm{J}$

(C) $5.4 \times 10^{-6} \mathrm{J}$

(D) $8.1 \times 10^{-6} \mathrm{J}$

**[IIT-JEE-2010]**

**Sol.**(B)

Ans. (B)

$\mathrm{U}=\mathrm{T}\left(4 \pi \mathrm{R}^{2}\right)=0.11 \times 4 \times \pi\left(\sqrt{2} \times 10^{-3}\right)^{2}=2.7 \times 10^{-6} \mathrm{J}$

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.

**[IIT-JEE-2011]**

**Sol.**(D)

Initially position increases while momentum decreases

(A) $\mathrm{E}_{1}=\sqrt{2 \mathrm{E}}_{2}$

(B) $\mathrm{E}_{1}=2 \mathrm{E}_{2}$

(C) $\mathrm{E}_{1}=4 \mathrm{E}_{2}$

(D) $\mathrm{E}_{1}=16 \mathrm{E}_{2}$

**[IIT-JEE-2011]**

**Sol.**(C)

$\Rightarrow \mathrm{E}_{1}=\frac{1}{2} \cdot \mathrm{k}(2 \mathrm{a})^{2}=4 \cdot \frac{1}{2} \mathrm{ka}^{2}=4 \mathrm{E}_{2}$

**[IIT-JEE-2011]**

**Sol.**(B)

Initially position decreases while momentum (–ve) increases and on completion of cycle net KE i.e. is decreased

$(\mathrm{A}) \mathrm{d}_{\mathrm{A}}<\mathrm{d}_{\mathrm{F}}$

(B) $\mathrm{d}_{\mathrm{B}}>\mathrm{d}_{\mathrm{F}}$

(C) $\mathrm{d}_{\mathrm{A}}>\mathrm{d}_{\mathrm{F}}$

(D) $\mathrm{d}_{\mathrm{A}}+\mathrm{d}_{\mathrm{B}}=2 \mathrm{d}_{\mathrm{F}}$

**[IIT-JEE-2011] **

**Sol.**(A,B,D)

(A) more than half-filled if $\rho_{\mathrm{c}}$ is less than 0.5

(B) more than half-filled if $\rho_{\mathrm{c}}$ is more than 1.0

(C) half-filled if $\rho_{\mathrm{C}}$ is more than 0.5

(D) less than half-filled if $\rho_{\mathrm{C}}$ is less than 0.5

**[IIT-JEE-2012]**

**Sol.**(A)

(A) the net elongation of the spring is $\frac{4 \pi \mathrm{R}^{3} \rho \mathrm{g}}{3 \mathrm{k}}$

(B) the net elongation of the spring is $\frac{8 \pi \mathrm{R}^{3} \rho \mathrm{g}}{3 \mathrm{k}}$

(C) the light sphere is partially submerged.

(D) the light sphere is completely submerged.

**[IIT-JEE-2013]**

**Sol.**(A,D)

For system

At equilibrium $2 f_{b}=\frac{4}{3} \pi R^{3} \rho g+\frac{4}{3} \pi R^{3}(3 \rho) g$

$2 \mathrm{f}_{\mathrm{b}}=\frac{16}{3} \pi \mathrm{R}^{3} \rho \mathrm{g}=2 \rho\left(\mathrm{v}_{\text {submerged }}\right) \mathrm{g}$

$\Rightarrow \mathrm{v}_{\text {submerged }}=\frac{8}{3} \pi \mathrm{R}^{3}$

$\Rightarrow$ light sphere is completely submerged

Also for one sphere

$\frac{4}{3} \pi \mathrm{R}^{3}(2 \rho) \mathrm{g}=\mathrm{kx}+\frac{4}{3} \pi \mathrm{R}^{3} \rho \mathrm{g}$

$\Rightarrow \mathrm{x}=\frac{4 \pi \mathrm{R}^{3} \rho \mathrm{g}}{3 \mathrm{k}}$

** [JEE Advanced-2014]**

**Sol.**(D)

**Paragraph for Questions 24 to 25**

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere.

(A) $0.1 \mathrm{ms}^{-1}$

(B) $1 \mathrm{ms}^{-1}$

(C) $2 \mathrm{ms}^{-1}$

(D) $8 \mathrm{ms}^{-1}$

**[JEE Advanced-2014]**

**Sol.**(C)

(A) $\sqrt{\frac{\rho_{\mathrm{a}}}{\rho_{\ell}}}$

(B) $\sqrt{\rho_{\mathrm{a}} \rho_{\ell}}$

(C) $\sqrt{\frac{p_{6}}{\rho_{0}}}$

$(\mathrm{D}) \rho \ell$

**[JEE Advanced-2014]**

**Sol.**(A)

By applying conservation of energy

$\frac{1}{2} \rho_{\mathrm{A}} \mathrm{v}^{2}=\frac{1}{2} \rho_{\mathrm{L}} \mathrm{v}^{2}+\rho_{\mathrm{L}} \mathrm{gH}$

Assuming height of liquid is negligible v $\propto \sqrt{\frac{\rho_{A}}{\rho_{\mathrm{L}}}}$

**[JEE Advanced-2014]**

**Sol.**(C)

Since $\mathrm{g}_{\mathrm{eff}}$ cancelled out therefore distance will be same in case P, Q and R. When lift falls freely no water leaks out of Jar.

**[JEE Advanced-2015]**

**Sol.**(A,D)

Consider a body of density $\mathrm{P}_{\mathrm{b}}$ kept in density $P_{\ell}$ whose viscosity is $\eta$ and terminal velocity V. Then

**[JEE Advanced-2016]**

**Sol.**3

$\frac{V_{\mathrm{TP}}}{V_{\mathrm{TQ}}}=3$

**[JEE Advanced-2017]**

**Sol.**6

(A) For a given material of the capillary tube, h decreases with increase in r

(B) For a given material of the capillary tube, h is independent of $\sigma$.

(C) If this experiment is performed in a lift going up with a constant acceleration, then h decreases.,

(D) h is proportional to contact angle $\theta$.

**[JEE Advanced-2017]**

**Sol.**(A,C)

Field due to straight wire is perpendicular to the wire & radially outward. Hence $\mathrm{E}_{z}=0$ Length, $\mathrm{PQ}=2 \mathrm{R} \sin 60=\sqrt{3} \mathrm{R}$According to Gauss’s law

total flux $=\left[\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{ds}}=\frac{\mathrm{q}_{\mathrm{in}}}{\epsilon_{0}}=\frac{\lambda \sqrt{3} \mathrm{R}}{\epsilon_{0}}\right.$

**Sol.**

Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity $\mathrm{u}_{0}$. Which of the following statements is (are) true ?

(A) The resistive force of liquid on the plate is inversely proportional to h

(B) The resistive force of liquid on the plate is independent of the area of the plate

(C) The tangential (shear) stress on the floor of the tank increases with $\mathrm{u}_{0}$.

(D) The tangential (shear) stress on the plate varies linearly with the viscosity $\eta$ of the liquid.

#tag# **[JEE Advanced-2018]**

#sol# (A,C,D)

Viscous force is given by F = $-\eta \mathrm{A} \frac{\mathrm{dv}}{\mathrm{dy}}$ since h is very small therefore, magnitude of viscous force is given by

$\mathrm{F}=\eta \mathrm{A} \frac{\Delta \mathrm{v}}{\Delta \mathrm{y}}$

$\therefore \mathrm{F}=\frac{\eta \mathrm{Au}_{0}}{\mathrm{h}} \Rightarrow \mathrm{F} \propto \eta \& \mathrm{F} \propto \mathrm{u}_{0} \quad ; \quad \mathrm{F} \propto \frac{1}{\mathrm{h}}, \mathrm{F} \propto \mathrm{A}$

Since plate is moving with constant velocity, same force must be acting on the floor.

Nice stuff

Good collection