MikaStars39/nano-eval huggingface dataset:
In $YDSE, D = 1\ m, d = 4\ mm$ and $\lambda = \dfrac {1}{2}mm$. In the figure shown, a parallel beam of light is incident on the plane of the slits of $YDSE$. Light incident on the slit $s_{1}$, passes through a medium of variable refractive index $\mu = (2x + 1)$ (where $x$ is the distance from the plane of slits in $mm$) up to a distance $l = 1\ mm$ before falling on $s_{1}$. Rest of the space filled with air. Then choose the correct option(s).

Options:
A. Position of central bright fringe is at a distance \( y = \dfrac {1}{\sqrt {7}} m \) above the central line
B. Position of central bright fringe is at a distance \( y = \dfrac {1}{\sqrt {6}} m \) above the central line
C. Position of central bright fringe is at a distance $y = \dfrac {1}{\sqrt {3}} m$ above the central line
D. Number of maxima obtaining on the screening is $8$
E. Position of central bright fringe is at a distance $y = \dfrac {1}{2}m$ above the central line
F. Position of central bright fringe is at a distance \( y = \dfrac {1}{\sqrt {8}} m \) above the central line
G. Position of central bright fringe is at a distance \( y = \dfrac {1}{\sqrt {4}} m \) above the central line
H. Number of maxima obtaining on the screening is $7$
I. Position of central bright fringe is at a distance $y = \dfrac {1}{\sqrt {2}} m$ above the central line
J. Position of central bright fringe is at a distance \( y = \dfrac {1}{\sqrt {5}} m \) above the central line
Answer: C
the magnetic needles has magnetic moment $6.7 \times {10^{ - 2}}\;A{m^2}$ and moment of inertia $I = 7.5 \times {10^{ - 6}}\;kg{m^2}$ . It performs 10 complete oscillations in 6.70 s. What is the magnitude of the magnetic field?

Options:
A. 0.1 T
B. 0.06 T
C. 0.01 T
D. 0.12 T
E. 0.02 T
F. 0.2 T
G. 0.04 T
H. 0.03 T
I. 0.05 T
J. 0.08 T
Answer: C
A block of mass m moves along $x-axis$ so as to obey the relation $x=t^3+t^2+1.$ Find the angular momentum of the particle about C as a function of time.

Options:
A. $$m(3t^2+2t+2)h \hat k$$
B. $$m(3t^2+2t)h \hat k$$
C. $$m(3t^2+2t-3)h \hat k$$
D. $$m(3t^2+2t-1)h \hat k$$
E. $$2tmh \hat k$$
F. $$0$$
G. $$m(3t^2+2t+1)h \hat k$$
H. $$m(3t^2+2t-2)h \hat k$$
I. $$3t^2mh \hat k$$
Answer: B
The ionization constant of $NH^+_4$ in water is $ 5.6 \times 10^{-10}$ at $25^0C$. The rate constant for the reaction of $NH^+_4$ and $OH^-$ to form $NH_3$ and $H_2O$ at $25^0C$ is $ 3.4 \times 10^{-10}$ lit $mol^{-1}sec^{}-1$. The rate constant for proton transfer from water to $NH_3$ is :

Options:
A. k=3.03 \times 10^{-7}sec^{-1}
B. k=3.40 \times 10^{-5}sec^{-1}
C. k=3.03 \times 10^{-4}sec^{-1}
D. $$k=6.07\times10^{-6}sec^{-1}$$
E. $$k=6.07\times10^{-5}sec^{-1}$$
F. k=3.03 \times 10^{-6}sec^{-1}
G. k=6.07 \times 10^{-4}sec^{-1}
H. $$k=3.03\times10^{-5}sec^{-1}$$
I. $$k=3.40 \times 10^{-7}sec^{-1}$$
Answer: E
A spring fixed at the bottom end of an incline of inclination $37^o$. A small block of mass 2 kg starts slipping down the incline from a point 4.8 m away from the spring. The block compresses the spring by 20 cm, stops momentarily and then rebounds through a distance of 1 m up the incline and come to rest again. Take $g = 10 m/s^2$.
Work done by gravity up to block come to rest second time :

Options:
A. 43 J
B. 49 J
C. 48 J
D. 50 J
E. 44 J
F. 45 J
G. 47 J
H. None of these
I. 46 J
J. 42 J
Answer: D
The vapour pressure of the solution of two liquids $A(P^{\circ} = 80\ mm)$ and $B(p^{\circ} = 120\ mm)$ is found to be $100\ mm$ when $x_{A} = 0.4$. The result shows that :

Options:
A. Solution will show negative deviations for higher concentration and positive deviations for lower concentrations
B. Solution shows positive deviations
C. Solution will show positive deviations for higher concentration and negative deviations for lower concentrations
D. Solution will show negative deviations for lower concentration and positive deviations for higher concentrations
E. Solution will show positive deviations for lower concentration and negative deviations for higher concentrations
F. Solution will show negative deviations for lower concentration and negative deviations for higher concentrations
G. Solution will show positive deviations for lower concentration and positive deviations for higher concentrations
H. Solution exhibits ideal behaviour
I. Solution will show positive deviations for higher concentration and positive deviations for lower concentrations
J. Solution shows negative deviations
Answer: E
A particle $P$ is sliding down a frictionless hemi-spherical howl. It passes the point $A$ at $t=0$. At this instant of time, the horizontal component of its velocity is $V$ A bead $Q$ of the same mass as $P$ is ejected from $A$ at $t=0$ along the horizontal string $AB$ with speed $v$. Friction between the bead and the string may be neglected. Let $t_{P}$ and $t_{Q}$ be the respective times taken by $P$ and $Q$ to reach the point $B$. Then

Options:
A. $$t_P = t_Q$$
B. $$t_P = \dfrac{1}{3} t_Q$$
C. $$t_P = \dfrac{1}{2} t_Q$$
D. $$t_P < t_Q$$
E. $$\dfrac {t_P}{t_Q}=\dfrac {Length\ of\ arc\ ACB}{Length\ of\ AB} $$
F. $$t_P > t_Q$$
G. $$t_P = 2t_Q$$
H. $$t_P = \dfrac{3}{2} t_Q$$
I. $$t_P = \dfrac{2}{3} t_Q$$
Answer: D
Calculate the change in pH of one litre of buffer solution containing $0.10$ mol each of $NH_3$ & $NH_4Cl$ upon addition of $(i)$ $0.02$ mol of dissolved gaseous $HCl$ $(ii)$ $0.02$ mol of dissolved $NaOH$. Assume no change in solution volume. $(K_b\: for \:NH_3\, =\, 1.8\, \times\, 10^{-5})$

Options:
A. 0.1661, decreases
B. $0.206$, decreases
C. 0.1961, increases
D. 0.1861, increases
E. 0.1761, increases
F. $0.1761$, decreases
G. 0.1961, decreases
H. 0.1861, decreases
I. $0.206$, increases
Answer: F
A mass is being whirled in the horizontal circle with radius 1m and angular velocity, $\omega = (t^{2} + t) rad s^{-1}$. Net acceleration acting on the mass, at $t = 1s$ from start is

Options:
A. $$10 ms^{-2}$$
B. $$3 ms^{-2}$$
C. $$9 ms^{-2}$$
D. $$4 ms^{-2}$$
E. $$8 ms^{-2}$$
F. $$2 ms^{-2}$$
G. $$5 ms^{-2}$$
H. $$6 ms^{-2}$$
I. $$7 ms^{-2}$$
Answer: D
$$\dfrac{1}{2}H_{2(g)} +\dfrac{1}{2}Cl_{2(g)} \rightarrow HCl_{(g)}\Delta H^o =-92.4kJ/mol$$
$$HCl_{(g)} +nH_2O \rightarrow H^+_{(aq)}+cl^-_{(aq)};\Delta H^o=-74.8\,kJ/mol$$
$\Delta H^of$ for $Cl^-_{(aq)}$ is

Options:
A. $$-77.4\,kJ/mole$$
B. $$-167.2\,kJ/mole$$
C. $$-35.2\,KJ/mole$$
D. $$-16.6\,kJ/mole$$
E. $$-17.6\,kJ/mole$$
F. $$-19.6\,kJ/mole$$
G. $$-15.6\,kJ/mole$$
H. $$-52.4\,kJ/mole$$
I. $$-18.6\,kJ/mole$$
J. $$+17.6\,Kj/mole$$
Answer: E
An aqueous solution containing $28\%$ by mass of a liquid A (molecular mass = 140) has a vapour pressure of 160 mm at $37^oC$. Find the vapour pressure of pure liquid A (the vapour pressure of water $37^oC$ is 150 mm).

Options:
A. $$p_A=258.3 mm$$
B. $$p_A=388.3 mm$$
C. $$p_A=368.3 mm$$
D. $$p_A=353.3 mm$$
E. $$p_A=358.3 mm$$
F. $$p_A=378.3 mm$$
G. $$p_A=348.3 mm$$
H. $$p_A=375.3 mm$$
I. $$p_A=363.3 mm$$
J. $$None of these$$
Answer: E
The correct stability order of the following resonance structure is:(I) ${{\text{H}}_{\text{2}}}\mathop {\text{C}}\limits^ -   - \mathop {\text{N}}\limits^ +   \equiv {\text{N}}$(II) ${{\text{H}}_{\text{2}}}{\text{C}} = \mathop {\text{N}}\limits^ +  {\text{ = }}\mathop {\text{N}}\limits^ -  $(III) ${{\text{H}}_{\text{2}}}\mathop {\text{C}}\limits^ +   - {\text{N}} = \mathop {\text{N}}\limits^ -  $(IV) ${{\text{H}}_{\text{2}}}\mathop {\text{C}}\limits^ -   - {\text{N}} = \mathop {\text{N}}\limits^ +  $

Options:
A. (II) > (I) > (IV) > (III)
B. (I) > (IV) > (II) > (III)
C. (I) > (II) > (I) > (III)
D. (I) > (III) > (II) > (IV)
E. (II) > (III) > (IV) > (I)
F. (II) > (IV) > (III) > (I)
G. (II) > (IV) > (I) > (III)
H. (II) > (III) > (I) > (IV)
I. (III) > (I) > (IV) > (II)
J. (II) > (I) > (III) > (IV)
Answer: J
$C(s)+O_2(g)\rightarrow CO_2, (g);$ $\Delta H=-94.3$ kcal/mol
$CO(g)+\dfrac{1}{2}O_2(g)\rightarrow CO_2(g);$ $\Delta H=-67.4$ kcal/mol
$O_2(g)\rightarrow 2O(g);$ $\Delta H=117.4$ kcal/mol
$CO(g)\rightarrow C(g)+O(g);$ $\Delta H=230.6$ kcal/mol.
Calculate $\Delta H$ for $C(s)\rightarrow C(g)$ in kcal/mol.

Options:
A. $$154$$
B. 153.5
C. 146.5
D. $$17.1$$
E. 156.3
F. $$145$$
G. 148
H. $$117$$
I. 147
Answer: F
A weightless rod of length $l$ with a small load of mass $m$ at the end is hinged at point $A$ as shown and occupies a strictly vertical position, touching a body of mass $M$. A light jerk sets the system in motion. Then,

Options:
A. tthe mass ratio $M/m$ for which the rod will form an angle $\alpha = \displaystyle\ \frac{\pi}{6}$ with the horizontal at the moment of the separation from the body is $4$.
B. the velocity $u$ of the body at this moment will be $\displaystyle\ \frac{1}{2} \sqrt{\frac{gl}{2}}$
C. the velocity $u$ of the body at this moment will be $\displaystyle\ \frac{1}{2} \sqrt{ListItemIcon\ \frac{gl}{4}}$
D. the velocity $u$ of the body at this moment will be $\displaystyle\ \frac{1}{2} \sqrt{\frac{gl}{3}}$
E. the mass ratio $M/m$ for which the rod will form an angle $\alpha = \displaystyle\frac{\pi}{6}$ with the horizontal at the moment of the separation from the body is $6$.
F. the velocity \( u \) of the body at this moment will be \(\displaystyle\ \frac{1}{2} \sqrt{\frac{gl}{8}} \)
G. the velocity $u$ of the body at this moment will be $\displaystyle\ \frac{1}{2} \sqrt{\frac{3gl}{4}}$
H. the velocity \( u \) of the body at this moment will be \(\displaystyle\ \frac{1}{2} \sqrt{\frac{gl}{7}} \)
I. the velocity $u$ of the body at this moment will be $\displaystyle\ \frac{1}{2} \sqrt{\frac{gl}{5}}$
J. the velocity $u$ of the body at this moment will be $\displaystyle\ \frac{1}{2} \sqrt{\frac{gl}{6}}$
Answer: B
The potential across a $3 \mu F $ capacitor is $412V$ when it is not connected to anything. It is then connected in parallel with an uncharged $6\mu F$ capacitor. At equilibrium, the charge q on the $3\mu F$ capacitors and the potential difference V across it are

Options:
A. $q=12\mu C$, $V= 6V$
B. $$q=18\mu C, V= 6V$$
C. $q= 12\mu C$, $V= 4V$
D. $q=24\mu C$, $V= 8V$
E. $$q=16\mu C, V= 8V$$
F. $q=12\mu C$, $V= 8V$
G. $$q=20\mu C, V= 8V$$
H. $q=35\mu C$, $V=12V$
I. $$q=30\mu C, V= 10V$$
J. $q=12\mu C$, $V= 10V$
Answer: D
A particle is moving along the x-axis and force acting on it is given by $F={F}_{0}\sin{{\omega}_{x}}N$, where $\omega$ is a constant. The work done by the force from $x=0$ to $x=2$ will be

Options:
A. $$\cfrac { { F }_{ 0 } }{ \omega \left( 1+\cos { \omega } \right) + 1 }$$
B. $$\cfrac { { F }_{ 0 }\sin ^{ 2 }{ \omega } }{ \omega } $$
C. $$\cfrac { { F }_{ 0 } }{ \omega \left( 1+\cos { \omega } \right) }$$
D. $$\cfrac { { F }_{ 0 } }{ \omega \left( 1+\cos { 2\omega } \right) }$$
E. $$\cfrac { { F }_{ 0 } }{ \omega \left( 1-\cos { \omega } \right) } $$
F. $$\cfrac { { F }_{ 0 } }{ 2\omega \left( 1+\cos { \omega } \right) }$$
G. $$\cfrac { { F }_{ 0 } }{ \omega \left( 1+\cos { \omega } \right) - 1 }$$
H. $$\cfrac { { F }_{ 0 } }{ \omega \left( 1-\cos { \omega } \right) + 1 }$$
I. $$\cfrac { { F }_{ 0 } }{ \omega \left( 1-\cos { 2\omega } \right) } $$
J. $$\cfrac { { F }_{ 0 } }{ 2\omega \left( 1-\cos { 2\omega } \right) } $$
Answer: E
The ozone layer in the earth's upper atmosphere is important in shielding the earth from very harmful ultraviolet radiation. The ozone, O$_3$, decomposes according to the equation $2O_3(g)  \rightarrow 3O_2(g)$. The mechanism of the reaction is thought to proceed through an initial fast, reversible step and then a slow second step.Step 1 : Fast, reversible, $O_3(g)  \rightleftharpoons O_2 (g) + O(g)$Step 2 : Slow,                $O_3(g) + O(g)  \rightarrow 2O_2(g)$Then which of the following is correct?

Options:
A. Molecularity is 1 for step 2
B. Molecularity is 3 for step 2
C. Molecularity is 3 for step 1
D. The rate expression for step 2 is $- \dfrac{d}{dt} [O_3] = k [O_3]^2$
E. Molecularity is 1 for step 1
F. The rate expression for step 2 is $- \dfrac{d}{dt} [O_3] = k [O_3][O_2]$
G. Step 2 is rate determining step
H. The rate expression for step 2 is $- \dfrac{d}{dt} [O_3] = k [O_3][O]$
I. Molecularity is 2 for step 2
J. Molecularity is 2 for step 1
Answer: I
For two particles A and B (moving with constant velocities) to collide, relative velocity of A with respect to B should be along AB At time t = 0, particle A is at (1m, 2m) and B is at (5m, 5m). Velocity of B along x-direction is 2 $m s ^ { - 1 }$  and along y-direction is 4  $m s ^ { - 1 }$. Velocity of particle A is $\sqrt { 2 }$ at $45 ^ { \circ }$ with the x-axis. A collides with B. Value of v is ........ $m s ^ { - 1 }$

Options:
A. 4
B. 12
C. 6
D. 8
E. 15
F. 5
G. 3
H. 25
I. 10
J. 7
Answer: F
A copper wire is held at the ends by rigid supports. At $50^{o}C$ the wire is just taut, with negligible tension. If $Y=1.2\times10^{11}N/m^{2},a=1.6\times10^{-5}/C$ and $p=9.2\times10^{3}kgm^{3}$, then the speed of transverse waves in this wire at $30^{o}C$ is

Options:
A. $$19.2m/s$$
B. $$16.2m/s$$
C. $$24.2m/s$$
D. $$32.2m/s$$
E. $$21.2m/s$$
F. $$23.2m/s$$
G. $$64.6m/s$$
H. $$22.2m/s$$
I. $$17.2m/s$$
J. $$20.2m/s$$
Answer: F
At $2$ 'O clock in the afternoon, the hour hand of the clock has a mechanical failure because of which its hour hand (length $= 0.5\, m$) falls freely under gravity with its one end being attached to the clock centre. Calculate the angular velocity of the hand as it goes past the 6 O clock mark. 
($g = 10\, m/s^2$)

Options:
A. $5 \sqrt {10}$ rad/s
B. $2 \sqrt {10}$ rad/s
C. $$3 \sqrt {5}$$
D. $$4 \sqrt {5}$$
E. $$3 \sqrt {10}$$
F. $$4 \sqrt {10}$$
G. $$$\sqrt {10}$ rad/s$$
H. $$7 \sqrt {10}$$
I. $$6 \sqrt {10}$$
J. $$5 \sqrt {5}$$
Answer: E
A particle of mass m executes SHM according to equation $x = A \cos \omega t$ . The average velocity and average kinetic energy over a time interval 0 to $\mathrm { T } / 2$ (T = Time period) are, respectively

Options:
A. $$\dfrac { A \omega } { \pi } , \dfrac { m A ^ { 2 } \omega ^ { 2 } } { 8 }$$
B. $$\dfrac { 2 A \omega } { \pi } , \dfrac { m A ^ { 2 } \omega ^ { 2 } } { 8 }$$
C. $$\dfrac { A \omega } { \pi } , \dfrac { m A ^ { 2 } \omega ^ { 2 } } { 16 }$$
D. $$\dfrac { A \omega } { 2 \pi } , \dfrac { m A ^ { 2 } \omega ^ { 2 } } { 16 }$$
E. $$\dfrac { A \omega } { 2\pi } , \dfrac { m A ^ { 2 } \omega ^ { 2 } } { 4 }$$
F. $$\dfrac { 2 \mathrm { A } \omega } { \pi } , \dfrac { \mathrm { mA } ^ { 2 } \omega ^ { 2 } } { 4 }$$
G. $$\dfrac { A \omega } { \pi } , \dfrac { m A ^ { 2 } \omega ^ { 2 } } { 4 }$$
H. $$0 , \dfrac { m A ^ { 2 } \omega ^ { 2 } } { 2 }$$
I. $$\mathrm { A } \omega , 0$$
J. $$\dfrac { A \omega } { 2 \pi } , \dfrac { m A ^ { 2 } \omega ^ { 2 } } { 8 }$$
Answer: G
The molar heat capacity of solid gold is given by the relation
$$C_{p,m}=25.69-7.32\times 10^{-4}T+4.56\times 10^{-6}T^2$$
in units of $J mol^{1} K^{1}$ . Calculate the entropy change for heating 2.50 moles of gold from $22.0^0 C$ to $1000^0 C$ at constant pressure.

Options:
A. 200.1 J/K
B. 100.9 J/K.
C. 1001.9 J/K.
D. 100.8 J/K.
E. None of the above
F. 100.7 J/K
G. 201.0 J/K.
H. 200.0 J/K.
I. 101.9 J/K
J. 100.1 J/K.
Answer: B
For the cell at $298$K$Ag(s) | AgBr(s) | Br^-(0.01M) | | I^-(0.02M) | AgI (s) | Ag(s)$the correct information is:[ Given : $Ksp (AgBr) = 4 \times 10^{-13}, Ksp (AgI) = 8 \times 10^{-17}, \dfrac{2.303RT}{F} = 0.06V, log 2 = 0.3$]

Options:
A. $$E^0_{cell} = 0.06V$$
B. $$E^0_{cell} = 0.006V$$
C. $$E_{cell} = 0.018V$$
D. $$E^0_{cell} = 0.018V$$
E. $$K_{eq}\ for\ cell\ reaction = 2 \times 10^{-4}$$
F. $$E^0_{cell} = 0.009V$$
G. $$E^0_{cell} = 0$$
H. $$E^0_{cell} = 0.001V$$
I. $$\Delta G^0 for \ cell\ reaction = - 0.06 \times 96500 \times 3.7J$$
J. $$E^0_{cell} = 0.003V$$
Answer: G
A $1^{st}$ order reaction is $50\%$ complete in $30$ minutes at $27^oC$ and in $10$ minutes at $47^oC$. Calculate energy of activation for the reaction.
(Assume log $3$=$0.48$)

Options:
A. $$45.11 \, KJ \, mol^{-1}$$
B. $$45.56 \, KJ \, mol^{-1}$$
C. $$42.56 \, KJ \, mol^{-1}$$
D. $$44.11 \, KJ \, mol^{-1}$$
E. $$43.65 \, KJ \, mol^{-1}$$
F. $$49.20 \, KJ \, mol^{-1}$$
G. $$42.21 \, KJ \, mol^{-1}$$
H. $$46.11 \, KJ \, mol^{-1}$$
I. $$46.87 \, KJ \, mol^{-1}$$
J. $$45.65 \, KJ \, mol^{-1}$$
Answer: D
$\frac{1}{24.63}$ mole of an ideal mono-atomic gas undergoes a reversible process for which $PV^3=C$ where C $(L^3-atm)$ is a constant. The gas expanded from an initial volume of 1 L and an initial temperature of 300 K to a final volume of  2 L then heat lost to be surrounded by gas is

Options:
A. 75.9 J
B. 15.21 J
C. 52.1 J
D. 22.8 J
E. 15.23 J
F. 15.3 J
G. 37.9 J
H. 15.2 J
I. 20.3 J
J. 30.4 J
Answer: H
A Helmholtz galvanometer has coils of radius $\dfrac {11}{\sqrt {5}}cm$ and the number of turns $70$. Calculate the current through the coil which produces a deflection of $45^{o}$. Given magnetic intensity due to earth along horizontal is $H=\dfrac {80}{\pi}A/m$

Options:
A. $$12.5\ mA$$
B. $$20\ mA$$
C. $$7.5\ mA$$
D. $$25\ mA$$
E. $$5\ mA$$
F. $$15\ mA$$
G. $$30\ mA$$
H. $$10\ mA$$
Answer: E
A boy is walking away from a wall towards an observer at a speed of 1 metre/sec and blows a whistle whose frequency is 690 Hz. The number of beats heard by the observer per second is (Velocity of sound in air = 340 metres/sec)

Options:
A. 4
B. Zero
C. 5
D. 1
E. 8
F. 12
G. 7
H. 2
I. 6
J. 3
Answer: B
Solid zinc hydroxide, when dissolved in water, can show the following equilibria.$Zn(OH)_{2(s)}\rightleftharpoons\ Zn(OH)_{2(aq)}$      $k_{1}=10^{-6}M$$Zn(OH)_{2(aq)}\rightleftharpoons\ Zn(OH)_{2(aq)}$    $k_{2}=10^{2}M$$Zn(OH)_{(aq)}^{+}\rightleftharpoons\ Zn^{2+}_{(aq)}+OH_{(aq)}$               $k_{3}=10^{-9}M$What will be the solubility of zinc hydroxide at  $pH$ of $6?$

Options:
A. $$1.2 \times 10^{-4}M$$
B. $$1.2 \times 10^{-8}M$$
C. $$10^{-4}M$$
D. $$1.2 \times 10^{-7}M$$
E. $$10^{-3}M$$
F. $$1.2\times 10^{-5}M$$
G. $$1.2\times 10^{-6}M$$
H. $$10^{-6}M$$
I. $$1.2 \times 10^{-9}M$$
J. $$10^{-5}M$$
Answer: C
A piston of mass $M= 3 kg$ and radius $R=4 cm$ has a hole into which a thin pipe of radius $r= 1 cm$ is inserted. The piston can enter a cylinder tightly and without friction, and initially it is at the bottom of the cylinder. $750gm$ of water is now poured into the pipe so that the piston and pipe are lifted up as show. Then,

Options:
A. the height $h$ of water in the pipe is $\cfrac { 3 }{ \pi }m$
B. the height $h$ of water in the pipe is $\cfrac { 9 }{ 32\pi } m$
C. the height $h$ of water in the pipe is $\cfrac { 5 }{ \pi }m$
D. the height $h$ of water in the pipe is $\cfrac { 4 }{ 32\pi } m$
E. the height $h$ of water in the pipe is $\cfrac { 2 }{ \pi }m$
F. the height $h$ of water in the pipe is $\cfrac { 1 }{ \pi }m$
G. the height $h$ of water in the pipe is $\cfrac { 11 }{ 32\pi } m$
H. the height $H$ of water in the cylinder is $\cfrac { 11 }{ 32\pi } m$
I. the height $h$ of water in the pipe is $\cfrac { 7 }{ 32\pi } m$
J. the height $H$ of water in the cylinder is $\cfrac { 2 }{ \pi }m$
Answer: E
Equal volumes of 1.0$ { M } { KCl }$ and 1.0${ M} { AgNO_{3} }$ , are mixed. The depression of freezing point of the resulting solution will be $(K_{f}{(H_{2}O)}$= $1.86K Kg mol^{-1}$, $1M=1m):$

Options:
A. 0.465 K
B. 0.93 K
C. 1.395 K
D. None of these
E. 5.58 K
F. 3.72 K
G. 1.025 K
H. 1.86 K
I. 1.41 K
J. 2.79 K
Answer: H
Two horizontal disc of different radii are free to rotate about their central vertical axes. One is given some angular velocity, the other is stationary. Their rims are now brought in contact. There is friction between the rims. Then

Options:
A. the force of friction between the rims will disappear when the disc rotate with equal angular speeds
B. the linear momentum of the system will not be conserved
C. the translational kinetic energy of the system will not be conserved
D. the rotational kinetic energy of the system will be conserved
E. the linear momentum of the system will be conserved
F. the rotational kinetic energy of the system will not be conserved
G. the force of friction between the rims will disappear when they have equal linear velocities
H. the translational kinetic energy of the system will be conserved
Answer: F
A water film is formed on a glass-block. A light ray is incident on water film from air at an angle of $\displaystyle { 60 }^{ \circ  }$ with the normal. The angle of incidence on glass slab is
($\displaystyle { \mu  }_{ g } = 1.5,{ \mu  }_{ w } = \frac { 4 }{ 3 } $)

Options:
A. $$\displaystyle { sin }^{ -1 }\left( \frac { 3\sqrt { 3 } }{ 10 } \right)$$
B. $$\displaystyle { sin }^{ -1 }\left( \frac { 3\sqrt { 3 } }{ 8 } \right) $$
C. $$\displaystyle { sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) $$
D. $$\displaystyle { sin }^{ -1 }\left( \frac { 3\sqrt { 3 } }{ 7 } \right)$$
E. $$\displaystyle { sin }^{ -1 }\left( \frac { 7\sqrt { 3 } }{ 16 } \right)$$
F. $$\displaystyle { \sin }^{ -1 }\left( \frac { 5\sqrt { 3 } }{ 12 } \right)$$
G. $$\displaystyle { sin }^{ -1 }\left( \frac { 9\sqrt { 3 } }{ 16 } \right) $$
H. $$\displaystyle { sin }^{ -1 }\left( \frac { 4\sqrt { 3 } }{ 9 } \right) $$
I. $$\displaystyle { sin }^{ -1 }\left( \frac { 2\sqrt { 3 } }{ 7 } \right)$$
Answer: B
The vapour pressure of a solution containing 1 mole benzene and two-mole Toluene is $P=190 + 120 { X }_{ B }\ torr$ at 25 C where ${ X }_{ B }$ is the mole fraction of Benzene. Which of the following statement/(s) is/are correct regarding the above solution?

Options:
A. The bubble point pressure of the above solution is 230 torr
B. The dew point pressure of the above mixture is 208 torr
C. The mole fraction of benzene in the vapour which is equilibrium with mixture of liquids is $\dfrac { 5 }{ 11 }$
D. Toluene is more volatile than benzene
E. The bubble point pressure of the above solution is 218 torr
F. The bubble point pressure of the above solution is 225 torr
G. The bubble point pressure of the above solution is 220 torr
H. The bubble point pressure of the above solution is 216 torr
I. The bubble point pressure of the above solution is 215 torr
J. The bubble point pressure of the above solution is 210 torr
Answer: G
Consider the following complex formation reactions and comment on their formation constant value
(i) $Fe^{2+} (aq) + 6NH_3 \ \ \ {\stackrel{K_f}{ \rightleftharpoons}} \ \ \  [Fe(NH_3)_6]^{2+}$
(i) $Fe^{2+} (aq) + 3en \ \ \ {\stackrel{K'_f}{ \rightleftharpoons}} \ \ \  [Fe(en)_3]^{2+}$

Options:
A. $$K_f > K'_f$$
B. $$K_f = \frac{1}{2}K'_f$$
C. $$K_f = 4K'_f$$
D. $$cannot\ be\ compared$$
E. $$K_f = K'_f$$
F. $$K_f = 2K'_f$$
G. $$K_f = \frac{3}{2}K'_f$$
Answer: A
Three bodies $A,\: B\: and\: C$ have equal surface area and thermal emissivities in the ratio $e_{A}:e_{B}:e_{C}=1: \displaystyle \frac {1}{2}: \displaystyle \frac {1}{4}$. All the three bodies are radiating at same rate. Their wavelengths corresponding to maximum intensity are $ \lambda_{A},\: \lambda_{B}\:and\: \lambda_{C}$ respectively and their temperatures are $T_{A},\: T_{B}\: and\: T_{C}$ on kelvin scale, then select the incorrect statement :

Options:
A. \sqrt {e_{A} \lambda_{A} T_{A} + e_{C} \lambda_{C} T_{C}} = e_{B} \lambda_{B} T_{B}
B. $$\sqrt {T_{A} T_{C}}=T_{B}$$
C. $$\sqrt {\lambda_{A} \lambda_{C}}=\lambda_{B}$$
D. $$\sqrt {e_{A} \lambda_{A} T_{A}.e_{B} \lambda_{B} T_{B}}=e_{C} \lambda_{C} T_{C}$$
E. \sqrt {e_{A} T_{A} + e_{C} T_{C}} = e_{B} T_{B}
F. \sqrt {e_{A} T_{A}^2 \cdot e_{C} T_{C}^2} = e_{B} T_{B}^2
G. $$\sqrt {e_{A} T_{A}} \sqrt {e_{C} T_{C}}=e_{B} T_{B}$$
H. \sqrt {e_{A} \lambda_{A} T_{A}.e_{C} \lambda_{C} T_{C}}=e_{B} \lambda_{B} T_{B}
I. \sqrt {e_{A} T_{A}^2 + e_{C} T_{C}^2} = e_{B} T_{B}
Answer: C
For the equilibrium,${ PCl }_{ 5 }\leftrightharpoons { PCl }_{ 3 }+{ Cl }_{ 2 }$${ K }_{ c }= { a }^{ 2 }/(1-a)V.$temperature remaining constant :

Options:
A. ${ K }_{ c } $ may increases or decrease with the change in volume depending upon its numerical value
B. ${ K }_{ c }$ will not change with the change in pressure
C. ${ K }_{ c }$ will decrease with decrease in volume
D. ${ K }_{ c }$ will decrease with increase in volume
E. ${ K }_{ c }$ will increase with increase in temperature
F. ${ K }_{ c } $ will increase with increase in volume
G. ${ K }_{ c }$ will not change with the change in volume
H. ${ K }_{ c }$ will increases with decrease in volume
I. ${ K }_{ c }$ will not change with the change in temperature
Answer: H
The capacity of a parallel plate condenser with air as a dielectric is $ 2 \mu F $. The space between the plates is filled with a dielectric slab with $K=5$. It is charged to a potential of $200V$ and disconnected from a cell. Work was done in removing the slab from the condenser completely

Options:
A. $$0.096J$$
B. $$0.6J$$
C. $$1.2J$$
D. $$0.032J$$
E. $$0.040J$$
F. $$0.16J$$
G. $$0.064J$$
H. $$0.8J$$
I. $$0.016J$$
J. $$0.048J$$
Answer: D
Carbon $- 11$ decays to $boron - 11$ according to the following formula.
$$^{11}_{6}C \Rightarrow _{5}^{11}B + e^{+} + v_{e} + 0.96MeV$$
Assume that positrons $(e^{+})$ produced in the decay combine with free electrons in the atmosphere and annihilate each other almost immediately. Also assume that the neutrinos $(v_{e})$ are massless and do not intersect with the environment. At $t = 0$ we have $1\mu g$ of $_{6}^{12}C$. If the half-life of the decay process is $t_{0}$, the net energy produced between time $t = 0$ and $t = 2t_{0}$ will be nearly.

Options:
A. $$4 \times 10^{15}MeV$$
B. $$4 \times 10^{17}MeV$$
C. $$4 \times 10^{18}MeV$$
D. $$8\times 10^{18}MeV$$
E. $$8 \times 10^{17}MeV$$
F. $$8 \times 10^{15}MeV$$
G. $$4\times 10^{8}MeV$$
H. $$4\times 10^{16}MeV$$
I. $$8\times 10^{16}MeV$$
Answer: I
A graph is plot between $P$ (in $W$) vs $x^{2}$, where $P$=power and $x$=displacement of particle moving rectilinearly mass of the particle is $1\ kg$ and its speed at $x=0$ is $1\ m/s$. Choose correct options.

Options:
A. speed of particle at \( x=2m \) is \( (14)^{1/3}m/s \)
B. speed of particle at $x=2m$ is $(15)^{1/3}m/s$
C. speed of particle at $x=2m$ is $(13)^{1/3}m/s$
D. speed of particle at $x=2m$ is $(10)^{1/3}m/s$
E. speed of particle at $x=2m$ is $(12)^{1/3}m/s$
F. Relation between speed and displacement will be given as $v^{3}=x^{3}+3x+2$
G. speed of particle at $x=2m$ is $(5)^{1/3}$, /s
H. Relation between speed and displacement will be given as $v^{3}=x^{3}+2x+1$
I. speed of particle at $x=2m$ is $(11)^{1/3}m/s$
Answer: B
$PtC{I_4}.6{H_2}O$ can exist as hydrated complex . $1$ molal aqueous solution has elevation in boiling point ${2.6^ \circ }C$ .Assume $100\% $ ionisation and ${K_b}\left( {{H_2}O} \right) = 0.52\,K\,Kg\,mo{l^{ - 1}},$ then complex is -

Options:
A. $$\left[ {Pt{{\left( {{H_2}O} \right)}_2}C{I_3}} \right].4{H_2}O$$
B. $$\left[ {Pt{{\left( {{H_2}O} \right)}_6}} \right]C{I_4}$$
C. $$\left[ {Pt{{\left( {{H_2}O} \right)}_3}C{I_3}} \right].4{H_2}O$$
D. $$\left[ {Pt{{\left( {{H_2}O} \right)}_2}C{I_4}} \right].5{H_2}O$$
E. $$\left[ {Pt{{\left( {{H_2}O} \right)}_4}C{I_2}} \right]C{I_2}.2{H_2}O$$
F. $$\left[ {Pt{{\left( {{H_2}O} \right)}_2}C{I_4}} \right].4{H_2}O$$
G. $$\left[ {Pt{{\left( {{H_2}O} \right)}_2}C{I_3}} \right].6{H_2}O$$
H. $$\left[ {Pt{{\left( {{H_2}O} \right)}_3}C{I_3}} \right]CI.3{H_2}O$$
I. $$\left[ {Pt{{\left( {{H_2}O} \right)}_3}C{I_4}} \right].4{H_2}O$$
J. $$\left[ {Pt{{\left( {{H_2}O} \right)}_2}C{I_3}} \right].5{H_2}O$$
Answer: F
A sound source emits frequency of $180\ \text{Hz}$ when moving towards a rigid wall with speed $5\ \text{m/s}$ and an observer is moving away from wall with speed $5\ \text{m/s}.$ Both source and observer moves on a straight line which is perpendicular to the wall. The number of beats per second heard by the observer will be $\left[\text{speed of sound} = 355\ \text{m/s}\right]$

Options:
A. $$9\ \text{beats/s}$$
B. $$3\ \text{beats/s}$$
C. $$8\ \text{beats/s}$$
D. $$6\ \text{beats/s}$$
E. $$10\ \text{beats/s}$$
F. $$5\ \text{beats/s}$$
G. $$11\ \text{beats/s}$$
H. $$4\ \text{beats/s}$$
I. $$7\ \text{beats/s}$$
Answer: C
A uniform hollow sphere (of radius r) is set into motion on rough horizontal surface as shown with $\omega=\dfrac{3v}{r}$. Translation velocity of the hollow sphere, after it ceases to slip, is

Options:
A. $\dfrac{2v}{3}$ leftwards
B. $\dfrac{2v}{7} $leftwards
C. $\dfrac{3v}{7}$ leftwards
D. $\dfrac{3v}{5}$ leftwards
E. $\dfrac{3v}{7}$ rightwards
F. $\dfrac{2v}{5}$ rightwards
G. $\dfrac{2v}{5} $leftwards
H. $\dfrac{2v}{3}$ righttwards
Answer: D
A sphere of mass m falls on a smooth hemisphere of mass M resting with its plane face on the smooth horizontal table so that at the moment of impact line joining the centres makes an angle $\alpha $ with the vertical. The velocity of the sphere just before impact is u and e is the coefficient of restitution.

Options:
A. The hemisphere will move with a velocity \( V = \frac{{mu\;\sin \alpha \;\cos \alpha \left( {1 + e} \right)}}{{\left( {M\; + 2m\sin^2 \alpha } \right)}} \)
B. The hemisphere will move with a velocity \( V = \frac{{mu\;\sin \alpha \;\cos \alpha \left( {2 + e} \right)}}{{\left( {M\; + m\sin^2 \alpha } \right)}} \)
C. The hemisphere will move with a velocity \( V = \frac{{mu\;\sin \alpha \;\cos \alpha \left( {1 + 2e} \right)}}{{\left( {M\; + m\sin^2 \alpha } \right)}} \)
D. The hemisphere will move with a velocity \( V = \frac{{mu\;\sin \alpha \;\cos \alpha \left( {1 - e} \right)}}{{\left( {M\; + m\sin^2 \alpha } \right)}} \)
E. The hemisphere will move with a velocity \( V = \frac{{mu\;\sin \alpha \;\cos \alpha \left( {2 - e} \right)}}{{\left( {M\; + m\sin^2 \alpha } \right)}} \)
F. The velocity of hemisphere after impact will be zero.
G. The hemisphere will move with a velocity \( V = \frac{{mu\;\sin \alpha \;\cos \alpha \left( {1 - 2e} \right)}}{{\left( {M\; + m\sin^2 \alpha } \right)}} \)
Answer: F
The top and bottom faces of a rigid insulating box of  dimensions $a \times b\times h$ and mass $m$ are charged with uniform surface charge densities $\sigma$ and $-\sigma$ respectively and the box is placed in a uniform horizontal magnetic field $\bar B =B_0 \hat j$. Assuming free space conditions and $h$ negligibly small as compared to the dimensions $a$ and $b$, find how much velocity will the box acquire after the magnetic field is switched off?

Options:
A. \dfrac{\sigma a bhB_0}{3m}\hat i
B. $$\dfrac{2\sigma a bh B_0}{m}\hat i$$
C. $$\dfrac{2\sigma a b h B_0}{m}\hat j$$
D. $$\dfrac{\sigma a bhB_0}{m}\hat j$$
E. \dfrac{3\sigma a bhB_0}{2m}\hat j
F. \dfrac{\sigma a bh B_0}{4m}\hat i
G. \dfrac{3\sigma a bh B_0}{2m}\hat i
H. $$\dfrac{\sigma a bhB_0}{m}\hat i$$
I. \dfrac{\sigma a bhB_0}{2m}\hat i
J. \dfrac{\sigma a bh B_0}{2m}\hat j
Answer: H
The magnetic field in a region of space varies according to the equation $\hat{B}=B_0[x\hat{i}+y\hat{j}+z\hat{k}]$.The magnetic flux associated with the area bounded by the line $x=0,y=0,x=a$ and $y=b$ in the $x-y$ plane is

Options:
A. $$\dfrac{B_0ab}{4}$$
B. Zero
C. $$\dfrac{4B_0ab}{3}$$
D. $$\dfrac{2B_0ab}{3}$$
E. $$\dfrac{B_0ab}{3}$$
F. $$\dfrac{5B_0ab}{3}$$
G. $${B_0ba}$$
H. $$\dfrac{B_0ab}{2}$$
I. $$\dfrac{7B_0ab}{3}$$
J. $$\dfrac{3B_0ab}{2}$$
Answer: G
For the reaction at 298K $CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g), \Delta H^{o}=178.3kJ,\Delta S^{o}=160Jk^{-1}$Select correct statement$(s)$:

Options:
A. The reaction is spontaneously in forward direction only at temperature above $1114K$
B. The reaction is spontaneously in forward direction onlt at temperature above $1000K$
C. If temperature is decrease forward reaction is fevoured
D. The reaction is spontaneously in forward direction onlt at temperature above $114K$
E. The reaction is spontaneously in forward direction onlt at temperature above $1.114 \times 10^2 K$
F. The reaction is spontaneously in forward direction onlt at temperature above $1.141K$
G. The reaction is sponteneously at this temperature $(298 K)$
H. The reaction is spontaneously in forward direction onlt at temperature above $111.4K$
I. The reaction is spontaneously in forward direction onlt at temperature above $1.114K$
J. The reaction is spontaneously in forward direction only at temperature above $114K
Answer: I
A uniform magnetic field $B = kt$ is passing perpendicular to plane of a ring of resistance $R$. $k$ is a constant and $t$ is time. If resistance of the ring is constant, the current flowing through the ring is:-

Options:
A. clockwise and constant
B. zero
C. clockwise and increasing with time
D. Clockwise and increasing with time
E. clockwise and decreasing with time
F. anticlockwise and constant
G. anticlockwise and decreasing with time
H. Anticlockwise and increasing with time
I. Anticlockwise and constant
Answer: A
Two infinitely long parallel wires carry currents of magnitude $I_1$ and $I_2$ and are at a distance $4$ cm apart. The magnitude of the net magnetic field is found to reach a non-zero minimum value between the two wires and $1$ cm away from the first wire. The ratio of the two currents and their mutual direction is?

Options:
A. $\displaystyle\dfrac{I_2}{I_1}=3$, parallel
B. \(\displaystyle\dfrac{I_2}{I_1}=24\), antiparallel
C. \(\displaystyle\dfrac{I_2}{I_1}=25\), antiparallel
D. $\displaystyle\dfrac{I_2}{I_1}=9$, parallel
E. $\displaystyle\dfrac{I_2}{I_1}=3$, antiparallel
F. \(\displaystyle\dfrac{I_2}{I_1}=27\), antiparallel
G. $\displaystyle\dfrac{I_2}{I_1}=27$, parallel
H. \(\displaystyle\dfrac{I_2}{I_1}=27\), parallel
I. \(\displaystyle\dfrac{I_2}{I_1}=8\), antiparallel
J. $\displaystyle\dfrac{I_2}{I_1}=9$, antiparallel
Answer: J
The figure below shows a glass flask whose hemispherical base is of diameter 0.20 m. The height of the flask is 0.25 m. The flask is filled to the brim with 2.5 litres $(1 \:litre = 10^{-3} m^3)$ of water and sealed with a glass lid.
What is the approximate magnitude of the total vertical force exerted by the water on the curved surface of the flask? (Take the acceleration due to gravity, g, to be $10 \:ms^{-2}$).

Options:
A. 35.5 N
B. 25.0 N
C. 78.5 N
D. 45.5 N
E. 43.5 N
F. 38.5 N
G. 53.5 N
H. 0 N
I. 48.5 N
J. 58.5 N
Answer: G
Find the concentration of $H^+$ ions in an aqueous solution, which is saturated with $H_2S$ ($0.1M$) as well as $H_2CO_3$ ($0.2 M$). [Use data: $K_{1}=10^{-7},K_{2}=10^{-14}$ for $ H_{2}S, K_{1}=4\times10^{-7}, K_{2}=4\times10^{-11} $ for $  H_{2}CO_{3} ]$

Options:
A. $$5 \times 10^{-4}M$$
B. $$3.2 \times10^{-4}M$$
C. $$3.5 \times10^{-4}M$$
D. $$4\times10^{-4}M$$
E. $$2.83\times10^{-4}M$$
F. $$2.5 \times10^{-4}M$$
G. $$3.83\times10^{-4}M$$
H. $$4.83 \times10^{-4}M$$
I. $$3\times10^{-4}M$$
J. None of these
Answer: I
For the reaction:
$$2{ Fe }^{ 3+ }(aq)+{ { (Hg }_{ 2 }) }^{ 2+ }(aq)\rightleftharpoons 2{ Fe }^{ 2+ }(aq)+{ { 2Hg } }^{ 2+ }(aq)$$
${ K }_{ C }=9.14\times { 10 }^{ -6 }$ at 25C. If the initial concentration of the ions are ${ Fe }^{ 3+ }=0.5\,M$, ${ Hg }_{ 2 }^{ 2+ }=0.5\,M$, ${ Fe }^{ 2+ }=0.03\,M$ and ${ Hg  }^{ 2+ }=0.03M$ , what will be the concentration of ions at equilibrium?

Options:
A. ${ Fe }^{ 3+ }=0.4987 M$,${ Hg }_{ 2 }^{ 2+ }=0.4973 M$, ${ Fe }^{ 2+ }=0.0337 M$, ${ Hg }^{ 2+ }=0.0337M$
B. ${ Fe }^{ 3+ }=0.4973 M$,${ Hg }_{ 2 }^{ 2+ }=0.4987 M$, ${ Fe }^{ 2+ }=0.0317 M$, ${ Hg }^{ 2+ }=0.0317M$
C. ${ Fe }^{ 3+ }=0.0327 M$,${ Hg }_{ 2 }^{ 2+ }=0.0327 M$, ${ Fe }^{ 2+ }=0.4973 M$, ${ Hg }^{ 2+ }=0.4973M$
D. ${ Fe }^{ 3+ }=0.4987 M$,${ Hg }_{ 2 }^{ 2+ }=0.4973 M$, ${ Fe }^{ 2+ }=0.0327 M$, ${ Hg }^{ 2+ }=0.0327M$
E. ${ Fe }^{ 3+ }=0.3973 M$,${ Hg }_{ 2 }^{ 2+ }=0.3987 M$, ${ Fe }^{ 2+ }=0.1327 M$, ${ Hg }^{ 2+ }=0.1327M$
F. ${ Fe }^{ 3+ }=0.4973 M$,${ Hg }_{ 2 }^{ 2+ }=0.4987 M$, ${ Fe }^{ 2+ }=0.0327 M$, ${ Hg }^{ 2+ }=0.0327M$
G. ${ Fe }^{ 3+ }=0.4983 M$,${ Hg }_{ 2 }^{ 2+ }=0.4977 M$, ${ Fe }^{ 2+ }=0.0337 M$, ${ Hg }^{ 2+ }=0.0337M$
H. None of these
I. ${ Fe }^{ 3+ }=0.4977 M$,${ Hg }_{ 2 }^{ 2+ }=0.4983 M$, ${ Fe }^{ 2+ }=0.0327 M$, ${ Hg }^{ 2+ }=0.0327M$
J. ${ Fe }^{ 3+ }=0.4987 M$,${ Hg }_{ 2 }^{ 2+ }=0.4973 M$, ${ Fe }^{ 2+ }=0.0317 M$, ${ Hg }^{ 2+ }=0.0317M$
Answer: F
A quantity of $27.6\ g$ of $K_2CO_3$ was treated by a series of reagent so as to convert all of its carbon to $K_2Zn_3 [Fe(CN)_6]_2$. The mass of the product formed is $(K=39, Zn=65.4, Fe=56)$

Options:
A. $$69.6\ g$$
B. $$139.2\ g$$
C. $$27.6\ g$$
D. $$12.6\ g$$
E. $$13.2\ g$$
F. $$21.6\ g$$
G. $$23.2\ g$$
H. $$11.6\ g$$
I. $$13.6\ g$$
Answer: H
The simultaneous solubility of $AgCN(K_{ sp }=2.5\times { 10 }^{ -16 })$ and $AgCl (K_{ sp }=1.6\times { 10 }^{ -10 })$ is $1.0\ M\ { NH }_{ 3 }(aq.)$ are respectively: [Given:${ K }_{ f }({ NH }_{ 3 }{ ) }_{ 2 }^{ + }]={ 10 }^{ 7 }]$

Options:
A. $$5.78\times { 10 }^{ -8 }, 0.027$$
B. $$1.58\times { 10 }^{ -8 }, 1.26\times { 10 }^{ -5 }$$
C. $$4.78 \times { 10 }^{ -8 }, 0.037$$
D. $$0.04, 6.25\times { 10 }^{ -8 }$$
E. $$5.78 \times { 10 }^{ -8 }, 0.047$$
F. $$4.78 \times { 10 }^{ -8 }, 0.027$$
G. $$1.58\times { 10 }^{ -8 }, 0.037$$
H. $$0.037, 5.78\times { 10 }^{ -8 }$$
I. $$5.78\times { 10 }^{ -8 }, 0.037$$
J. $$1.58\times { 10 }^{ -8 }, 0.04$$
Answer: I
A small ball bearing is releases at the top of a long vertical column of glycerine of height $2h$. The ball bearing falls through a height $h$ in a time $t_{1}$ and then the remaining height with the terminal velocity in time $t_{2}$ Let $W_{1}$ and $W_{2}$ be the work done against viscous drag over these height. therefore.

Options:
A. t_{1} = \frac{t_{2}}{2}
B. W_{1}>W_{2}
C. t_{1} = \frac{3t_{2}}{2}
D. t_{1} = t_{2}
E. $$W_{1}=W_{2}$$
F. $$t_{1}> t_{2}$$
G. t_{1} = 2t_{2}
H. $$t_{1}< t_{2}$$
I. $$W_{1}< W_{2}$$
Answer: H
For the reactions ${ N }_{ 2 }(g)+{ O }_{ 2 }(g)\rightleftharpoons NO(g)\Delta H=180\quad KJ{ mol }^{ -1 }$. Which of the following is not true ?

Options:
A. The formation of NO is increased at lower pressure.
B. The formation of NO is decreased at higher temperature.
C. $$\Delta \quad n=0$$
D. The formation of NO is decreased at lower temperature.
E. Change in volume at constant pressure does not affect the equilibrium.
F. Change in pressure at constant temperature does not affect the equilibrium.
G. Change in pressure at constant volume does not affect the equilibrium.
H. The formation of NO is increased at lower temperature.
I. The formation of NO is decreased at higher temperature
J. The formation of NO is increased at higher temperature
Answer: G
$$CO(g)+\cfrac{1}{2}{O}_{2}(g)\rightarrow {CO}_{2}(g)$$
$\Delta H=-67.37K.cal$ at ${25}^{o}C$ the change in entropy accompanying process is $-20.7cal$ ${deg}^{-1}$ ${mol}^{-1}$ then at ${25}^{o}C$ the change in free energy is

Options:
A. $-11K$. ${mol}^{-1}$
B. $-10K.cal$ ${mol}^{-1}$
C. $-15K$. ${mol}^{-1}$
D. $-61.2K$. ${mol}^{-1}$
E. -9K.cal ${mol}^{-1}$
F. $-12K$. ${mol}^{-1}$
G. $-13.2K$. ${mol}^{-1}$
H. $-32K$. ${mol}^{-1}$
I. -18K.cal ${mol}^{-1}$
Answer: B
The solubility of solid silver chromate, $Ag_2CrO_4$ is determined in three solvent $K_{sp}$ of $Ag_2CrO_4$ = 9 x $10^{-12}$
l) Pure water ll) 1 M $AgNO_3$ lll) 0.1 M $Na_2CrO_4$
Predict the relative solubility of $Ag_2CrO_4$ in the three solvent

Options:
A. l = lll < ll
B. ll = lll < l
C. l = ll < lll
D. ll = l < lll
E. l < ll = lll
F. l = ll = lll
G. ll < lll < l
H. l < lll < ll
I. ll < l < lll
J. l < ll < lll
Answer: F
Sphere and cylinder are charged to a same value. ${ E }_{ s }$ is the electrical field produced by the sphere at a distance r from its center. ${ E }_{ c }$ is the electrical field produced by the cylinder at distance r from its axis. If L is the length of the cylinder then ${ E }_{ s }$ and ${ E }_{ c }$ are related as :

Options:
A. $${ E }_{ s }=\dfrac { { E }_{ C }L }{ 3r^2 }$$
B. $${ E }_{ s }=\dfrac { { E }_{ C }L }{ 2r } $$
C. $${ E }_{ s }=\dfrac { { E }_{ C }L }{ 3r }$$
D. $${ E }_{ s }=\dfrac { { E }_{ C } }{ 4r }$$
E. { E }_{ s }=\dfrac { { E }_{ C }L }{ 4r }
F. $${ E }_{ s }=\dfrac { { E }_{ C } }{ 2r } $$
G. $${ E }_{ s }=\dfrac { { E }_{ C } }{ r } $$
H. $${ E }_{ s }=\dfrac { { E }_{ C }L }{ 5r }$$
I. $${ E }_{ s }=\dfrac { { E }_{ C } }{ 3r }$$
J. $${ E }_{ s }=\dfrac { { E }_{ C }L }{ r } $$
Answer: F
A small sphere of mass $1$ kg is rolling without slipping on a rough stationary base with liner speed v = $\sqrt{\dfrac{200}{7}}$m/s. It leaves the inclined plane at point C.
Find its linear speed at point C:-

Options:
A. \sqrt{\dfrac{80}{7}} \text{m/s}
B. \sqrt{\dfrac{90}{7}} \text{m/s}
C. $\sqrt{\dfrac{100}{35}}$m/s
D. $\sqrt{\dfrac{150}{35}} \text{m/s}$
E. $\sqrt{\dfrac{150}{7}} \text{m/s}$
F. $\sqrt{\dfrac{100}{7}}$m/s
G. $\sqrt{\dfrac{50}{7}}$m/s
H. $\sqrt{\dfrac{110}{7}}$m/s
I. $\sqrt{\dfrac{200}{35}}$m/s
J. $\sqrt{\dfrac{120}{7}}$m/s
Answer: F
A man is going in a topless car with a velocity of $10.8  km/h$. It is raining vertically downwards. He has to hold the umbrella at an angle of $53^{o}$ to the vertical to protect himself from rain. The actual speed of the rain is $(cos  53^{o}=\displaystyle \frac{3}{5})$

Options:
A. $$0.75\ ms^{-1}$$
B. $$1.25\ ms^{-1}$$
C. $$3.75\ ms^{-1}$$
D. $$2.25\ ms^{-1}$$
E. $$2.00\ ms^{-1}$$
F. $$2.75\ ms^{-1}$$
G. $$2.50\ ms^{-1}$$
H. $$1.50\ ms^{-1}$$
I. $$1.75\ ms^{-1}$$
J. $$2.05\ ms^{-1}$$
Answer: D
The magnetic field in the region $r\leq R$ and  $B_{o}=\dfrac{\mu_{o} aR}{2e^{2}}$  is

Options:
A. $$\dfrac{B_{o}}{r}(1+e^{2r/R})$$
B. $$\dfrac{B_{o}}{r}(e^{2r/R}+1)$$
C. $$\dfrac{B_{o}}{r}(1-e^{2r/R})$$
D. $$\dfrac{B_{o}}{r}(1-e^{-2r/R})$$
E. $$\dfrac{B_{o}}{r}(e^{-2r/R}+1)$$
F. $$\dfrac{B_{o}}{r}(e^{-2r/R}-1)$$
G. $$\dfrac{B_{o}}{r}(e^{2r/R}+e^{-2r/R})$$
H. $$\dfrac{B_{o}}{r}(1+e^{-2r/R})$$
I. $$\dfrac{B_{o}}{r}(e^{2r/R}-1)$$
Answer: I
Calculate the pH of a buffer prepared by mixing $0.10$ mol of sodium formate and $0.05$ mole of formic acid in $1.0$L of solution. $[HCO_2H : K_a=1.8\times 10^{-4}]$

Options:
A. $$3.64$$
B. $$3.79$$
C. $$5.31$$
D. $$3.44$$
E. $$4.25$$
F. $$3.59$$
G. $$4.05$$
H. $$1.8\times 10^{-4}$$
I. $$3.71$$
J. $$3.95$$
Answer: D
A projectile is thrown with velocity $10\sqrt{2}$ m/s at angle 45 to horizontal. Calculate time interval between the moments when velocity 12.5 m/s. (g = 10 $m/s^2$)

Options:
A. $$2.25s$$
B. $$1.25s$$
C. $$1.0 s$$
D. $$1.75s$$
E. $$0.5s$$
F. $$1.5 s$$
G. $$0.85s$$
H. $$0.75s$$
I. $$4.0s$$
Answer: F
$50mL$ of $0.1M$ ${CH}_{3}COOH$ is treated with $50mL$ of $0.05M$ $NaOH$ and the resultant solution is diluted by adding $100ml$ of water. The final pH of solution is ($p{K}_{a}=4.74$)

Options:
A. $$1.63$$
B. $$1.69$$
C. $$3.74$$
D. $$1.74$$
E. $$1.65$$
F. $$4.74$$
G. $$1.67$$
H. $$2.71$$
I. $$3.72$$
J. $$1.31$$
Answer: J
A reaction $A+B\rightarrow C$, follows first order kinetic with respect to each reactant, with an overall rate constant, $k=2.0\times 10^{-3}M^{-1}s^{-1}$. The initial concentrations of $A$ and $B$ are $0.10M$ and $6.93M$, respectively. Given $ln2=0.693$, choose the correct statement (s).

Options:
A. The half-life for $A$ is about $75$ seconds.
B. The half-life for $A$ is about $25$ seconds.
C. The half-life for $A$ is about $50$ seconds.
D. The half-life for $B$ is about $50$ seconds.
E. The half-life for $A$ is about $35$ seconds.
F. The half-life for $A$ is about $200$ seconds.
G. The half-life for $A$ is about $60$ seconds.
H. The half-life for $A$ is about $100$ seconds.
Answer: C
Three solutions are prepared by adding 'w' gm of 'A' into 1 kg of water, 'w' gm of 'B' into another 1 kg of water and 'w' gm of 'C' in another 1 kg of water (A, B, C, are non electrolytic). Dry air is passed from these solutions in sequence (A $\rightarrow$ B $\rightarrow$ C). The loss in weight of solution A was found to be 2 gm while solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A, B and C is :

Options:
A. $$M_C > M_A > M_B$$
B. $M_A : M_B : M_C$ = 3 : 4 : 5
C. \(M_A : M_B : M_C = \frac{1}{5} : \frac{1}{4} : \frac{1}{3}\)
D. \(M_A : M_B : M_C = 3 : 5 : 2\)
E. \(M_A : M_B : M_C = 3 : 5 : 4\)
F. M_A : M_B : M_C = 5 : 3 : 4
G. $$M_B > M_A > M_C$$
H. $$M_A : M_B : M_C = \frac{1}{4} : \frac{1}{3} : \frac{1}{5}$$
I. \(M_A : M_B : M_C = 5 : 4 : 3\)
J. $M_A : M_B : M_C$ = 4 : 3 : 5
Answer: J
The freezing point depression of a 0.1 M a solution weak acid (HX) is ${ -0.20^{ \circ  } }$. What is the value of equilibrium constant for the reaction ?
$$Hx(aq)\rightleftharpoons { H }^{ + }(aq)+{ X }^{ - }(aq)$$
[Given : ${ K }_{ f }$ for water = 1.8 kg $mol^{ -1 } K$. & Molality =Molarity ]

Options:
A. $$1.46\times { 10 }^{ -4 }$$
B. $$1.46 \times { 10 }^{ -3 }$$
C. $$1.21 \times { 10 }^{ -4 }$$
D. $$1.46 \times { 10 }^{ -5 }$$
E. $$1.35\times { 10 }^{ -4 }$$
F. $$1.35 \times { 10 }^{ -5 }$$
G. $$1.35\times { 10 }^{ -3 }$$
H. $$1.21 \times 10^{-3}$$
I. $$1.21\times { 10 }^{ -2 }$$
Answer: G
CuSO$_4.$5$H_2O$(s) $\rightleftharpoons$ CuSO$_4$.3H$_2O$(s) + 2H$_2O$(g); $k_p$ = 4 $10^{4}$ atm$^2.$ If the vapour pressure of water is 38 torr then percentage of relative humidity is : (Assume all data at constant temperature)

Options:
A. 20
B. 40
C. 50
D. 30
E. 25
F. 10
G. 4
H. 35
I. None of these
J. 60
Answer: B
As shown in the given figure the ball is given sufficient velocity at the lowest point to complete the circle. Length of string is $1m$. Find the tension in the string, when it is at $60^{\circ}$ with vertical position.
(Mass of ball $= 5\ kg$).

Options:
A. $$175\ N$$
B. $$180\ N$$
C. $$215\ N$$
D. $$170\ N$$
E. $$160\ N$$
F. $$195\ N$$
G. $$190\ N$$
H. $$225\ N$$
I. $$210\ N$$
J. $$200\ N$$
Answer: J
The displacement y of a particle from the equilibrium position at a time t is given by $\frac{d^2y}{dt^2} + \frac{2 K dy}{dt} + \omega^2y = 0$
Then the time period of vibration is

Options:
A. $$\frac{2\pi}{\omega}$$
B. $$\frac{2\pi}{K}$$
C. $$\frac{2\pi}{\sqrt{\omega^2 - K^2}}$$
D. $$\frac{2\pi}{\sqrt{\omega^2 - K}}$$
E. $$\frac{2\pi}{\sqrt{\omega^2 - 5K^2}}$$
F. $$\frac{2\pi}{\sqrt{\omega^2 - 3K^2}}$$
G. $$\frac{2\pi}{\sqrt{\omega^2 - 2K^2}}$$
H. $$\frac{2\pi}{\sqrt{\omega^2 - 6K^2}}$$
I. $$\frac{2\pi}{\sqrt{\omega^2 + K^2}}$$
J. $$\frac{2\pi}{\sqrt{\omega^2 - 4K^2}}$$
Answer: A
Let $m_{P}$ be the mass of a proton, $m_{n}$ the mass of a neutron, $M_{1}$ the mass of a $_{10}^{20}Ne$ nucleus and $M_{2}$ the mass of a $_{20}^{40} Ca$ nucleus. Then

Options:
A. $$M_{2} < 2M_{1}$$
B. $$M_{1} = 11(m_{n} + m_{p})$$
C. $$M_{1} < 9(m_{n} + m_{p})$$
D. $$M_{1} < 10(m_{n} + m_{p})$$
E. $$M_{1} > 10(m_{n} + m_{p})$$
F. $$M_{2} > 2M_{1}$$
G. $$M_{2} = 2M_{1}$$
H. $$M_{1} = 9(m_{n} + m_{p})$$
I. $$M_{1} = 10(m_{n} + m_{p})$$
J. $$M_{2} = 2M_{1} + 10m_{p}$$
Answer: D
The freezing point of benzene decrease by ${ 0.45 }^{ 0 }C$ when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be :-
(${ K }_{ f }$ for benzene = 5.12 k kg ${ mol }^{ -1 }$)

Options:
A. 79.4%
B. 94.6%
C. 74.6%
D. 80.4%
E. 69.6%
F. 69.4%
G. 59.6%
H. 54.6%
I. 75.6%
J. 64.6%
Answer: J
If we prepared an aqueous solution by mixing $100$ ml of $0.2$M HCOOH with $100$ ml of $0.2$M NAOH, then its degree of hydrolysis & hydrolysis constant will be? [Given: $K_a(HCOOH)=1.8\times 10^{-4}$]

Options:
A. $$K_h=55.5\times 10^{-10}, h=1.98\times 10^{-5}$$
B. $$K_h=5.55 \times 10^{-10}, h=1.98 \times 10^{-5}$$
C. $$K_h=5.55\times 10^{-10}, h=1.04\times 10^{-5}$$
D. $$K_h=55.5\times 10^{-11}, h=7.44\times 10^{-5}$$
E. $$K_h=5.55 \times 10^{-11}, h=7.44 \times 10^{-5}$$
F. $$K_h=5.55 \times 10^{-11}, h=1.98 \times 10^{-5}$$
G. $$K_h=5.55 \times 10^{-10}, h=7.44 \times 10^{-5}$$
H. $$K_h=5.55 \times 10^{-10}, h=1.67 \times 10^{-5}$$
I. $$K_h=5.55 \times 10^{-11}, h=1.04 \times 10^{-5}$$
J. $$K_h=5.55\times 10^{-11}, h=1.67\times 10^{-5}$$
Answer: C
A long straight wire (radius $= 3.0 \mathrm { mm } )$ carries a constant current distributed uniformly over a cross section perpendicular to the axis of the wire. If the current density is 100$\mathrm { A } / \mathrm { m } ^ { 2 }$ . The magnitudes of the magnetic field at (a) 2.0$\mathrm { mm }$ from the axis of the wire and 4.0$\mathrm { mm }$ from the axis of the wire is

Options:
A. $$4 \pi \times 10 ^ { - 8 } \mathrm { T } , \frac { \pi } { 2 } \times 10 ^ { - 8 } \mathrm { T }$$
B. $$2 \pi \times 10 ^ { - 8 } \mathrm { T } , \frac { 5 \pi } { 2 } \times 10 ^ { - 8 } \mathrm { T }$$
C. $$1.4e-7T$$
D. $$2 \pi \times 10 ^ { - 8 } \mathrm { T } , \frac { 7 \pi } { 2 } \times 10 ^ { - 8 } \mathrm { T }$$
E. $$2 \pi \times 10 ^ { - 7 } \mathrm { T } , \frac { 5 \pi } { 2 } \times 10 ^ { - 8 } \mathrm { T }$$
F. $$2 \pi \times 10 ^ { - 8 } \mathrm { T } , \frac { 11 \pi } { 2 } \times 10 ^ { - 8 } \mathrm { T }$$
G. $$2 \pi \times 10 ^ { - 7 } \mathrm { T } , \frac { 3 \pi } { 2 } \times 10 ^ { - 8 } \mathrm { T }$$
H. $$2 \pi \times 10 ^ { - 8 } \mathrm { T } , \frac { 3 \pi } { 2 } \times 10 ^ { - 8 } \mathrm { T }$$
I. $$\pi 10 ^ { - 4 } \mathrm { T } , \frac { 9 \pi } { 2 } \times 10 ^ { - 8 } \mathrm { T }$$
J. $$4 \pi \times 10 ^ { - 8 } \mathrm { T } , \frac { 9 \pi } { 2 } \times 10 ^ { - 8 } \mathrm { T }$$
Answer: J
For the reaction$3N_{2}O(g)+2NH_{3}(g) \rightarrow 4N_{2}(g)+3H_{2}O(g):\ \Delta H^{o}=-879.6\ kJ$If $\Delta H^{o}_{f}[NH_{3}(g)]=-45.9\ kJ\ mol^{-1}$$\Delta H^{o}_{f}[H_{2}(g)]=-241.8\ kJ\ mol^{-1}$then $\Delta H_{o}^{f}[H_{2}O(g)]$ will be?

Options:
A. $$+82\ kJ$$
B. $$+242\ kJ$$
C. $$-84\ kJ$$
D. $$-86\ kJ$$
E. $$+246\ kJ$$
F. $$-242\ kJ$$
G. $$-246\ kJ$$
H. $$-82\ kJ$$
I. $$-244\ kJ$$
J. $$+86\ kJ$$
Answer: A