At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be: (molar mass of urea $= 60 \mathrm {~g} \mathrm {~mol} ^ { - 1 }$)
(1) 0.028 mm Hg
(2) 0.027 mm Hg
(3) 0.031 mm Hg
(4) 0.017 mm Hg

A solution of sodium sulphate contains $92 g$ of $\mathrm { Na } ^ { + }$ ions per kilogram of water. The molality of $\mathrm { Na } ^ { + }$ ions in that solution in $\mathrm { mol \, kg} ^ { - 1 }$ is:
(1) 16
(2) 4
(3) 12
(4) 8

An element has a face-centered cubic (fcc) structure with a cell edge of $a$. The distance between the centres of two nearest tetrahedral voids in the lattice is
(1) $\frac{a}{2}$
(2) $\frac{3}{2}a$
(3) $a$
(4) $\sqrt{2}\,a$

1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose ebullioscopic constants are in the ratio of $1 : 5$. The ratio of the elevation in their boiling points, $\frac { \Delta \mathrm { T } _ { \mathrm { b } } \mathrm { A } } { \Delta \mathrm { T } _ { \mathrm { b } } \mathrm { B } }$, is: (assuming they have the same molar mass)
(1) $10 : 1$
(2) $1 : 5$
(3) $1 : 0.2$
(4) $5 : 1$

The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in $\text{mol kg}^{-1}$) of the aqueous solution is:
(1) $13.88 \times 10^{-3}$
(2) $13.88 \times 10^{-1}$
(3) $13.88$
(4) $13.88 \times 10^{-2}$

The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of $\mathrm { PbSO } _ { 4 }$ electrolyzed in g during the process is: (Molar mass of $\mathrm { PbSO } _ { 4 } = 303 \mathrm {~g \, mol} ^ { - 1 }$)
(1) 22.8
(2) 15.2
(3) 11.4
(4) 7.6

The following results were obtained during kinetic studies of the reaction. $2 \mathrm {~A} + \mathrm { B } \rightarrow$ product

Experiment	A in mol L ${ } ^ { - 1 }$	B in mol L$^{-1}$	Initial rate of reaction in $\mathrm { mol } \mathrm { L } ^ { - 1 } \mathrm {~min} ^ { - 1 }$
I	0.10	0.20	$6.93 \times 10 ^ { - 3 }$
II	0.10	0.25	$6.93 \times 10 ^ { - 3 }$
III	0.20	0.30	$1.386 \times 10 ^ { - 2 }$

The time (in minutes) required to consume half of A is
(1) 100
(2) 10
(3) 5
(4) 1

For the reaction of $\mathrm { H } _ { 2 }$ with $\mathrm { I } _ { 2 }$, the rate constant is $2.5 \times 10 ^ { - 4 } \mathrm { dm } ^ { 3 } \mathrm {~mol} ^ { - 1 } \mathrm {~s} ^ { - 1 }$ at $327 ^ { \circ } \mathrm { C }$ and $1.0 \mathrm { dm } ^ { 3 } \mathrm {~mol} ^ { - 1 } \mathrm {~s} ^ { - 1 }$ at $527 ^ { \circ } \mathrm { C }$. The activation energy for the reaction, in kJ mol$^{-1}$ is: $\mathrm { R } = 8.314 \mathrm { JK } ^ { - 1 } \mathrm {~mol} ^ { - 1 }$
(1) 166
(2) 59
(3) 72
(4) 150

For any two statement $p$ and $q$, the negative of the expression $p \vee ( \sim p \wedge q )$ is
(1) $\sim p \vee \sim q$
(2) $p \wedge q$
(3) $\sim p \wedge \sim q$
(4) $p \leftrightarrow q$

For each $t \in R$, let $[ t ]$ be the greatest integer less than or equal to $t$. Then, $\lim _ { x \rightarrow 1 ^ { + } } \frac { ( 1 - | x | + \sin | 1 - x | ) \sin \left( [ 1 - x ] \frac { \pi } { 2 } \right) } { | 1 - x | [ 1 - x ] }$
(1) equals 0
(2) equals - 1
(3) does not exist
(4) equal 1

The expression $\sim ( \sim p \rightarrow q )$ is logically equivalent to
(1) $p \wedge \sim q$
(2) $\sim p \wedge \sim q$
(3) $p \wedge q$
(4) $\sim p \wedge q$

Which one of the following statements is not a tautology?
(1) $p \vee q \rightarrow p \vee ( \sim q )$
(2) $p \wedge q \rightarrow ( \sim p \vee q )$
(3) $p \rightarrow p \vee q$
(4) $p \wedge q \rightarrow p$

If $p \Rightarrow ( q \vee r )$ is False, then the truth values of $p , q , r$ are respectively, (where T is True and F is False)
(1) $T , F , F$
(2) $F , T , T$
(3) $F , F , F$
(4) $T , T , F$

The logical statement $[\sim(\sim p \vee q) \vee (p \wedge r)] \wedge (\sim q \wedge r)$ is equivalent to
(1) $(\sim p \wedge \sim q) \wedge r$
(2) $(p \wedge r) \wedge \sim q$
(3) $(p \wedge \sim q) \vee r$
(4) $\sim p \vee r$

A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to: $\left( 1 \mathrm { HP } = 746 \mathrm {~W} , g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 } \right)$
(1) $1.7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(2) $1.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(3) $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
(4) $2.0 \mathrm {~m} \mathrm {~s} ^ { - 1 }$

A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is 25.0 cm and a stopwatch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in g is:
(1) $5.40\%$
(2) $3.40\%$
(3) $4.40\%$
(4) $2.40\%$

Three point particles of masses $1.0 \mathrm {~kg} , 1.5 \mathrm {~kg}$ and 2.5 kg are placed at three corners of a right angle triangle of sides $4.0 \mathrm {~cm} , 3.0 \mathrm {~cm}$ and 5.0 cm as shown in the figure. The centre of mass of the system is at a point:
(1) 0.6 cm right and 2.0 cm above 1 kg mass.
(2) 1.5 cm right and 1.2 cm above 1 kg mass.
(3) 2.0 cm right and 0.9 cm above 1 kg mass.
(4) 0.9 cm right and 2.0 cm above 1 kg mass.

A particle of mass $m$ is fixed to one end of a light spring having force constant $k$ and unstretched length $l$. The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:
(1) $\frac { m l \omega ^ { 2 } } { k - \omega m }$
(2) $\frac { m l \omega ^ { 2 } } { k - m \omega ^ { 2 } }$
(3) $\frac { m l \omega ^ { 2 } } { k + m \omega ^ { 2 } }$
(4) $\frac { m l \omega ^ { 2 } } { k + m \omega }$

A particle moves such that its position vector $\vec{r}(t) = \cos\omega t\,\hat{i} + \sin\omega t\,\hat{j}$ where $\omega$ is a constant and $t$ is time. Then which of the following statements is true for the velocity $\vec{v}(t)$ and acceleration $\vec{a}(t)$ of the particle:
(1) $\vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed away from the origin
(2) $\vec{v}$ and $\vec{a}$ both are perpendicular to $\vec{r}$
(3) $\vec{v}$ and $\vec{a}$ both are parallel to $\vec{r}$
(4) $\vec{v}$ is perpendicular to $\vec{r}$ and $\vec{a}$ is directed towards the origin

The coordinates of the centre of mass of a uniform flag-shaped lamina (thin flat plate) of mass 4 kg. (The coordinates of the same are shown in the figure) are:
(1) $( 1.25 \mathrm {~m} , 1.50 \mathrm {~m} )$
(2) $( 0.75 \mathrm {~m} , 1.75 \mathrm {~m} )$
(3) $( 0.75 \mathrm {~m} , 0.75 \mathrm {~m} )$
(4) $( 1 \mathrm {~m} , 1.75 \mathrm {~m} )$

As shown in figure. When a spherical cavity (centred at $O$) of radius 1 is cut out of a uniform sphere of radius $R$ (centred at $C$), the centre of mass of remaining (shaded part of sphere) is at $G$, i.e., on the surface of the cavity. $R$ can be determined by the equation:
(1) $\left(R^{2} + R + 1\right)(2 - R) = 1$
(2) $\left(R^{2} - R - 1\right)(2 - R) = 1$
(3) $\left(R^{2} - R + 1\right)(2 - R) = 1$
(4) $\left(R^{2} + R - 1\right)(2 - R) = 1$

Consider a uniform rod of mass $M = 4 m$ and length $l$ pivoted about its centre. A mass $m$ moving with velocity $v$ making angle $\theta = \frac { \pi } { 4 }$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is:
(1) $\frac { 3 } { 7 \sqrt { 2 } } \frac { v } { l }$
(2) $\frac { 3 } { 7 } \frac { v } { l }$
(3) $\frac { 3 \sqrt { 2 } } { 7 } \frac { v } { l }$
(4) $\frac { 4 } { 7 } \frac { v } { l }$

A particle of mass $m$ is dropped from a height $h$ above the ground. At the same time another particle of the same mass is thrown vertically upwards from the ground with a speed of $\sqrt{2gh}$. If they collide head-on completely inelastically, the time taken for the combined mass to reach the ground, in units of $\sqrt{\frac{h}{g}}$ is:
(1) $\sqrt{\frac{1}{2}}$
(2) $\sqrt{\frac{3}{4}}$
(3) $\frac{1}{2}$
(4) $\sqrt{\frac{3}{2}}$

In an experiment to verify Stokes law, a small spherical ball of radius $r$ and density $\rho$ falls under gravity through a distance $h$ in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of $h$ is proportional to: (ignore viscosity of air)
(1) $r^4$
(2) $r$
(3) $r^3$
(4) $r^2$

Consider two solid spheres of radii $R _ { 1 } = 1 \mathrm {~m} , R _ { 2 } = 2 \mathrm {~m}$ and masses $M _ { 1 }$ and $M _ { 2 }$, respectively. The gravitational field due to sphere (1) and (2) are shown. The value of $\frac { M _ { 1 } } { M _ { 2 } }$ is:
(1) $\frac { 2 } { 3 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 2 }$
(4) $\frac { 1 } { 3 }$