MikaStars39/nano-eval huggingface dataset:
Some  statements are given for the following equilibrium.
$$NH_4HS(s) \rightleftharpoons NH_3(g)+H_2S(g); \Delta H=+ve$$
$S_1:$ On increase in temperature,equilibrium pressure of ammonia  increase.
$S_2:$  On increase in volume of container at constant tempperature,equilibrium pressure of ammonia increases.
$S_3:$  On increase in mass of  $NH_4HS(s)$ in the container at constant at constant temperature, equilibrium pressure of ammonia increases.

Options:
A. $$TTT$$
B. $$T T F$$
C. $$F T F$$
D. $$F T T$$
E. $$T F T$$
F. $$F F T$$
G. $$F F F$$
H. $$T F F$$
Answer: A
Two identical balls A and B each of mass 0.1 kg are attached to two identical mass less springs.This spring mass system is constrained to move inside a rigid smooth pipe in the form of a circle as in fig .The pipe is fixed in a horizontal plane. The centers of the ball can move in a circle of radius 0.06 m.Each spring ha sa natural length ${0.06\pi}$ m and force constant 0.1N/m.Initially both the balls are displaced by an angle of ${\theta =\pi/6}$ radian with respect to diameter PQ of the circle and released from rest (a) Calculate the frequency of oscillation  of the ball B. (b)What is the total energy of the system.(c) Find the speed of the ball A when A and B are at the two ends of the diameter PQ.

Options:
A. $\displaystyle{f=\frac{1}{\pi}}$; $ {E=5\pi^{2}\times10^{-5}}J$; ${v=2\pi\times10^{-2}}m/s$
B. $\displaystyle{f=\frac{1}{2\pi}}$; $ {E=5\pi^{2} \times10^{-5}}J$; ${v=4\pi \times10^{-2}}m/s$
C. $\displaystyle{f=\frac{1}{\pi}}$; $ {E=4\pi^{2}\times10^{-5}}J$; ${v=2\pi\times10^{-2}}m/s$
D. $\displaystyle{f=\frac{1}{\pi}}$; $ {E=4\pi^{2}\times10^{-5}}J$; ${v=3\pi\times10^{-2}}m/s$
E. $\displaystyle{f=\frac{1}{2\pi}}$; $ {E=4\pi^{2} \times10^{-5}}J$; ${v=3\pi \times10^{-2}}m/s$
F. $\displaystyle{f=\frac{1}{\pi}}$; $ {E=5\pi^{2} \times10^{-5}}J$; ${v=3\pi \times10^{-2}}m/s$
G. $\displaystyle{f=\frac{1}{2\pi}}$; $ {E=5\pi^{2}\times10^{-5}}J$; ${v=2\pi\times10^{-2}}m/s$
H. $\displaystyle{f=\frac{1}{2\pi}}$; $ {E=4\pi^{2}\times10^{-5}}J$; ${v=2\pi\times10^{-2}}m/s$
I. $\displaystyle{f=\frac{1}{2\pi}}$; $ {E=5\pi^{2} \times10^{-5}}J$; ${v=3\pi \times10^{-2}}m/s$
Answer: C
$CaCO_{3(s)}\rightleftharpoons CaO(s)+CO_{2}(g)$ For this equilibrium it is found that $\log K_{p}=8-\dfrac{6400}{T(K)}$
$T=527^{o}C$ if :

Options:
A. $$K_{p}=7.2\times 10^{-4}$$
B. $$K_{p}=7.2\times 10^{-3}$$
C. $$K_{p}=7.2\times 10^{-6}$$
D. $$K_{p}=13.94\times 10^{3}$$
E. $$K_{p}=7.2\times 10^{-7}$$
F. $$K_{p}=1$$
G. $$K_{p}=7.2\times 10^{-5}$$
H. $$K_{p}=10$$
Answer: G
The molar volume of liquid benzene ($density=0.877 g  ml^{-1}$) increases by a factor of $2750$ as it vaporizes at  $20^{\circ}$C and that of liquid toluene ($density=0.867 g ml^{-1}$)increases by a factor of $7720$ at  $20^{\circ}$C. A solution of benzene and toluene has a vapour pressure of $46.0$ torr. Find the mole fraction of benzene in the vapour above the solution:

Options:
A. $$0.62$$
B. 0.55
C. 0.61
D. 0.51
E. $$0.53$$
F. $$0.47$$
G. 0.60
H. 0.59
I. 0.57
J. $$0.38$$
Answer: A
One mole of monoatomic ideal gas undergoes a cyclic process on PV diagram as shown. It is known that rms speed of molecules at $A =$ mean speed of molecules at $C =$ most probable speed of molecules at B. Temperature at$A = 280 \mathrm { K } .$ Then

Options:
A. $$
\text{Work done by gas from } C \text{ to } A = (280 - 105\pi) \, \mathrm{R}
$$
B. $$\text{Work done by gas from } B \text{ to } C = (420 - 105\pi) \, \mathrm{R} \times \frac{2}{3}$$
C. $$
\text{Work done by gas from } B \text{ to } C = (420 - 105\pi) \, \mathrm{R} \times \frac{3}{2}
$$
D. $$\text{Work done by gas from } B \text{ to } C = (420 - 105\pi) \, \mathrm{R} \times \frac{3}{4}$$
E. $$
\text{Work done by gas from } B \text{ to } C = (420 - 105\pi) \, \mathrm{R} \times \frac{5}{3}
$$
F. $$\text{Work done by gas from } B \text{ to } C = (420 - 105\pi) \, \mathrm{R} \times \frac{3}{5}$$
G. $$\text{Work done by gas from } B \text{ to } C = (420 - 105\pi) \, \mathrm{R} \times \frac{4}{3}$$
H. $$(420 - 105 \pi) \mathrm{R} \times \frac{3}{4}$$
I. $$\frac { \mathrm { P } _ { \mathrm { B } } } { \mathrm { P } _ { \mathrm { A } } } = \frac { 3 \pi } { 8 }$$
Answer: C
In a region of uniform magnetic field $B=10^{-2}T$, a circular coil is rotating at $'\omega'$ rpm about an axis which is perpendicular to the direction of 'B' and which forms a diameter of the coil. The radius of the coil is $30$ cm and resistance $\pi^2$ ohm. If the amplitude of the alternating current induced in the coil is $6$ mA, then value of $'\omega'$ is?

Options:
A. $400$ rpm
B. $15$ rpm
C. 150 rpm
D. 252 rpm
E. 24 rpm
F. 25 rpm
G. 105 rpm
H. $21$ rpm
I. $300$ rpm
J. 42 rpm
Answer: H
Find the potential of a half-cell having reaction, $Ag_{2}S+2e\rightarrow2Ag+S^{2-}$ in a solution buffered
$pH=3$ and which is also saturated with $0.1\ M\ H_{2}S$. For $H_{2}S:\ K_{1}=10^{-8}$ and $K_{2}=2\times 10^{-13},\ K_{sp}(Ag_{2}S)=2\times 10^{-48},\ E_{Ag,ag}^{o}=0.8$

Options:
A. $$-0.144\ V$$
B. $$1.658\ V$$
C. $$-0.134\ V$$
D. $$0.432\ V$$
E. $$-0.124\ V$$
F. $$-0.154\ V$$
G. $$-0.174\ V$$
H. $$-0.245\ V$$
I. $$-0.254\ V$$
J. $$-0.164\ V$$
Answer: A
An electron with initial velocity $v_0\hat i$ is in the region of electric end magnetic fields, $-E_0\hat i$ and $B_0\hat j$ respectively.  The de-Broglie wavelength of the electron at time t is given by :

Options:
A. $$\lambda = \frac{h}{qt(E_0^2 + B_0^2)^{1/2}}$$
B. $$\lambda = \frac{h}{qt(E_0 + B_0)}$$
C. $$\lambda = \frac{h}{qt(E_0^2 + B_0^2)}$$
D. $$\lambda = \frac{h}{qt(E_0^2 + v_0^2 B_0^2)^{1/2}}$$
E. $$\lambda = \frac{h}{mv_0}$$
F. $$\lambda = \frac{h}{qt(E_0^2 - v_0^2 B_0^2)^{1/2}}$$
G. $$\lambda = \frac{h}{qt(E_0)}$$
H. $$\lambda = \frac{h}{qt(E_0^2 - B_0^2)^{1/2}}$$
I. $$\lambda = \frac{h}{qt(E_0^2 + v_0 B_0)}\$$
J. $$\lambda = \frac{h}{qt(E_0^2 - v_0 B_0)}$$
Answer: E
A plane spiral with a great number N of turns wound tightly to one another is located in a uniform magnetic field perpendicular to the spiral's plane. the outside radius of the spiral's turns is equal to a and inner radius is zero. the magnetic induction varies with time as $ B = B_v sin \omega t $ where $ B_0$ and  $\omega $ are constants . find the amplitudes of emf induced the spiral.

Options:
A. $$\frac {1}{6} \pi a^2N \omega B_0$$
B. $$\frac {1}{9} \pi a^2N \omega B_0$$
C. $$ \frac {1}{2} \pi a^2N \omega B_0 $$
D. $$\frac {1}{7} \pi a^2N \omega B_0$$
E. $$ \frac {1}{4} \pi a^2N \omega B_0 $$
F. $$\frac {1}{5} \pi a^2N \omega B_0$$
G. $$ \frac {1}{3} \pi a^2N \omega B_0 $$
H. $$\frac {1}{10} \pi a^2N \omega B_0$$
I. None of these
J. $$\frac {1}{8} \pi a^2N \omega B_0$$
Answer: G
$2NO \Leftrightarrow N_2+O_2;K_c=2.2 \times 10^{-33}$$2N_2O_5 \Leftrightarrow  2N_2+5O_2; K_c=3.8 \times 10^{-32}$$2N_2+O_2 \Leftrightarrow  2N_2O; K_c=4 \times 10^{-32}$ From the above data, the most stable oxide is ______.

Options:
A. $$N_2O_5$$
B. $$NO$$
C. $$N_2O_3$$
D. $$NO_3$$
E. $$N_2O_4$$
F. $$NO_2$$
G. $$NO_4$$
H. $$N_2O$$
Answer: F
Circuit for the measurement of resistance by potentiometer is shown. The galvanometer is first connected at point $A$ and zero deflection is observed at length $PJ=10\ cm$. In second case it is connected at point $C$ and zero deflection is observed at a length $30\ cm$ from $P$. Then the unknown resistance $X$ is

Options:
A. $$2R$$
B. 4R
C. $$\dfrac{R}{6}$$
D. $$3R$$
E. $$\dfrac{R}{2}$$
F. $$\dfrac{3R}{4}$$
G. $$\dfrac{3R}{2}$$
H. $$\dfrac{4R}{3}$$
I. $$\dfrac{R}{3}$$
J. $$\dfrac{R}{4}$$
Answer: A
The Born Haber cycle below represents the energy changes occurring at 298K when $KH$ is formed from its elementsv : ${ \Delta H }_{ atomisation }$ $K = 90 kJ/mol$w : ${ \Delta H }_{ ionisation }$ $K = 418 kJ/mol$x : ${ \Delta H }_{ dissociation }$ $H = 436 kJ/mol$y : ${ \Delta H }_{ electron affinity }$ $H = 78 kJ/mol$z : ${ \Delta H }_{ lattice }$ $KH = 710 kJ/mol$Calculate the value of $\Delta $$H$ showing all your working.

Options:
A. -123 kJ/mol
B. 114 kJ/mol
C. 124 J/mol
D. -124 J/mol
E. -114 kJ/mol
F. -134 kJ/mol
G. None of these
H. 124 kJ/mol
I. 134 kJ/mol
J. -124 kJ/mol
Answer: J
Twelve identical resistors each of value 1 $\Omega$ are connected as shown. Net resistance between C and D (R) is :

Options:
A. \displaystyle R = \frac{4}{5} \Omega
B. \displaystyle R = \frac{6}{5} \Omega
C. $$\displaystyle R = \frac{3}{4} \Omega$$
D. $$\displaystyle R = \frac{4}{3} \Omega$$
E. \displaystyle R = \frac{5}{4} \Omega
F. $$R = 1 \Omega$$
G. \displaystyle R = \frac{3}{5} \Omega
H. \displaystyle R = \frac{2}{3} \Omega
I. $$\displaystyle R = \frac{7}{6} \Omega$$
J. \displaystyle R = \frac{5}{6} \Omega
Answer: C
Given,${NH}_{3}(g)+3{Cl}_{2}(g)\rightleftharpoons N{Cl}_{3}(g)+3HCl(g);-\Delta {H}_{1}$${N}_{2}(g)+3{H}_{2}(g)\rightleftharpoons 2{NH}_{3}(g);-\Delta {H}_{2}$${H}_{2}(g)+{Cl}_{2}(g)\rightleftharpoons 2HCl(g);\Delta {H}_{3}$The heat formation of $N{Cl}_{3}(g)$ in terms of $\Delta {H}_{1},\Delta {H}_{2}$ and $\Delta {H}_{3}$ is:

Options:
A. $$\Delta {H}_{f}=-\Delta {H}_{1}+\cfrac{3}{2}\Delta {H}_{2}-\cfrac{3}{2}\Delta {H}_{3}$$
B. none of these
C. $$\Delta {H}_{f}=-\Delta {H}_{1}-\cfrac{1}{2}\Delta {H}_{2}+\cfrac{3}{2}\Delta {H}_{3}$$
D. $$\Delta {H}_{f}=-\Delta {H}_{1}+\cfrac{3}{2}\Delta {H}_{2}+\cfrac{1}{2}\Delta {H}_{3}$$
E. $$\Delta {H}_{f}=-\Delta {H}_{1}+\cfrac{1}{2}\Delta {H}_{2}-\cfrac{3}{2}\Delta {H}_{3}$$
F. $$\Delta {H}_{f}=-\Delta {H}_{1}+\cfrac{1}{2}\Delta {H}_{2}-\cfrac{1}{2}\Delta {H}_{3}$$
G. $$\Delta {H}_{f}=-\Delta {H}_{1}+\cfrac{1}{2}\Delta {H}_{2}+\cfrac{1}{2}\Delta {H}_{3}$$
H. $$\Delta {H}_{f}=\Delta {H}_{1}-\cfrac{1}{2}\Delta {H}_{2}-\cfrac{3}{2}\Delta {H}_{3}$$
I. $$\Delta {H}_{f}=-\Delta {H}_{1}+\cfrac{1}{2}\Delta {H}_{2}-\cfrac{5}{2}\Delta {H}_{3}$$
J. $$\Delta {H}_{f}=-\Delta {H}_{1}+\cfrac{1}{2}\Delta {H}_{2}+\cfrac{3}{2}\Delta {H}_{3}$$
Answer: B
Two inclined planes (D) and (I) have inclination $ lpha $ and $\beta$ respectively with horizontal, (where $lpha+ \beta = 90 ^ { \circ } )$ intersect each other at point 0 as shown in figure. A particle is projected from point $A$ with velocity $u$ along a direction perpendicular to plane $1$. If the particle strikes plane (I1) perpendicularly at $B$ then

Options:
A. time of flight = \( u'g \cos 2\alpha \)
B. time of flight = $u'g \cos \beta$
C. time of flight = $u'g \sin 2\beta$
D. time of flight = $$u'g \cos\alpha$$
E. time of flight = \( u'g \cos \beta \)
F. time of flight = \( u'g \sin 2\alpha \)
G. distance $OB=$ $u ^ { 3 } / 2 g \sin \beta$
H. time of flight = $u'g \sin \beta$
I. time of flight = $u'g \sin\alpha $
J. distance $OB=$ $u ^ { 3 } / 2 g \sin\alpha $
Answer: H
A particle of mass $m$ moves  in a straight line. If $v$ is the velocity at a distance $x$ from a fixed point on the line and $v^{2}=a-bx^{2}$, where $a$ and $b$ are constant then:

Options:
A. the motion is simple harmonic with amplitude \(\sqrt{\frac{a}{3b}} \)
B. the motion is simple harmonic with amplitude \(\sqrt{\frac{a}{b}} \)
C. the total energy of the particle is $$ma$
D. the particle oscillates with a frequency equal to $\dfrac{\sqrt{b}}{2\pi}$
E. the motion continues along the positive $x-$ direction only
F. the motion is simple harmonic with amplitude \(\sqrt{\frac{a}{2b}} \)
G. the motion is simple harmonic with amplitude \(\sqrt{\frac{a}{5b}} \)
H. the motion is simple harmonic with amplitude \(\sqrt{\frac{a}{4b}}\)
I. the motion is simple harmonic with amplitude \(\sqrt{\frac{2a}{b}}\)
J. the motion is simple harmonic
Answer: J
A cell is connected between the points A and C of a circular conductor ABCD with O as centre and angle $AOC = 60^{o}$. If $B_{1}$ and $B_{2}$ are the magnitudes of the magnetic fields at O due to the currents in ABC and ADC respectively, then ratio $\frac{B_{1}}{B_{2}}$ is

Options:
A. $$\frac{4}{3}$$
B. \( 4 \)
C. $$\frac{3}{2}$$
D. 4
E. $$1$$
F. $$2$$
G. $$5$$
H. 3
I. $$\frac{2}{3}$$
J. $$6$$
Answer: E
The work done during combustion of $9\times 10^{-2}\, Kg$ of ethane, $C_2H_6$ (g) at 300 K is:(Given: R 8.314 J $deg^{-1}, mol^{-1}$, atomic mass C = 12, H = 1)

Options:
A. +6.2362 kJ
B. +6.236 kJ
C. - 6.236 kJ
D. +18.71 kJ
E. +18.7112 kJ
F. +18.711 kJ
G. -18.71 kJ
H. -18.711 kJ
I. +18.7111 kJ
Answer: D
A body A of mass m slides on plane inclined at angle $\theta_1$ to the horizontal and $\mu$ is the coefficient of friction between A and the plane.A is connected by a light string passing over a frictionless pulley to another body B,also of mass m, sliding on a frictionless plane inclined at an angle $\theta_2$ to the horizontal. Which of the following statements are true?

Options:
A. A will just start moving up the plane when $\mu=\dfrac{cos\, \theta_1 - cos\, \theta_2}{sin\, \theta_1}$
B. A will just start moving up the plane when \(\mu = \dfrac{cos\, \theta_2 - sin\, \theta_1}{sin\, \theta_1} \)
C. A will just start moving up the plane when \(\mu = \dfrac{cos\, \theta_2 - cos\, \theta_1}{cos\, \theta_1} \)
D. A will never move up the plane
E. A will just start moving up the plane when $\mu=\dfrac{cos\, \theta_2 - cos\, \theta_1}{sin\, \theta_1}$
F. For A to move up the plane , $\theta_2$ must always be greater than $\theta_1$
G. A will just start moving up the plane when \(\mu = \dfrac{sin\, \theta_2 - sin\, \theta_1}{sin\, \theta_1}\)
H. B will always slide down with constant speed
I. A will just start moving up the plane when $\mu=\dfrac{sin\, \theta_2-sin \theta_1}{cos \theta_1}$
J. A will just start moving up the plane when $\mu=\dfrac{sin\, \theta_1 - sin\, \theta_2}{cos\, \theta_1}$
Answer: D
A volume of 120 $ \mathrm{ml}  $ of drink (half alcohol + half water by mass) originally at a temperature of $  25^{\circ} \mathrm{C} $ is cooled by adding 20 $ \mathrm{gm}  $ ice at $  0^{\circ} \mathrm{C} $ If all the ice melts, find the final temperature(in $ ^{\circ} C )  $ of the drink. (density of drink $ =0.833 \mathrm{gm} / \mathrm{c} c,  $ specific heat of alcohol $ =0.6 \mathrm{cal} / \mathrm{gm}^{\circ} \mathrm{C} ) $

Options:
A. 8^{\circ} C
B. 5^{\circ} C
C. $$ 1^{\circ} C $$
D. $$ 3^{\circ} C $$
E. 7^{\circ} C
F. 10^{\circ} C
G. 9^{\circ} C
H. $$ 4^{\circ} C $$
I. $$ 6^{\circ} C $$
J. $$ 2^{\circ} C $$
Answer: I
$4\ gm$ of $NaOH$ was dissolved in $1$ litre of a solution containing $1$ mole of ${ CH }_{ 3 }COOH$ and $1$ mole of ${ CH }_{ 3 }COONa$. Calculate ${ [H }^{ + }]$ in the  resulting solution. (Given, ${ K }_{ a }$ of $CH_3COOH= 1.8\times { 10 }^{ -5 }$)

Options:
A. $$1.9 \times { 10 }^{ -5 }M$$
B. None of these
C. $$1.6 \times { 10 }^{ -5 }M$$
D. $$1.37 \times { 10 }^{ -5 }M$$
E. $$1.4 \times { 10 }^{ -5 }M$$
F. $$1.8\times { 10 }^{ -5 }M$$
G. $$1.7 \times { 10 }^{ -5 }M$$
H. $$1.47\times { 10 }^{ -5 }M$$
I. $$1.5 \times { 10 }^{ -5 }M$$
J. $$2\times { 10 }^{ -5 }M$$
Answer: F
In the circuit shown, initially there is no charge on capacitors and keys $S_1$ and $S_2$ are open. The values of the capacitors are $C_1=10\mu F$, $C_2=30\mu F$, and $C_3=C_4=80\mu F$. Which statements is/are correct?

Options:
A. At time \( t=0 \), the key \( S_1 \) is closed, the instantaneous current in the closed circuit will be \( 23 \)mA
B. The key $S_1$ is kept closed for long time such that capacitors are fully charged. Now key $S_2$ is closed, at this time the instantaneous current across $3\Omega$ resistor (between points P & Q) will be $0.2A$ (round off to $1^{st}$ decimal place.
C. At time $t=0$, the key $S_1$ is closed, the instantaneous current in the closed circuit will be $25$mA
D. If $S_1$ is kept closed for long time such that capacitors are fully charged, the voltage difference between P and Q will be $10V$
E. At time $t=0$, the key $S_1$ is closed, the instantaneous current in the closed circuit will be $30$mA
F. At time \( t=0 \), the key \( S_1 \) is closed, the instantaneous current in the closed circuit will be \( 22 \)mA
G. At time \( t=0 \), the key \( S_1 \) is closed, the instantaneous current in the closed circuit will be \( 15 \)mA
H. If key $S_1$ is kept closed for long time such that capacitors are fully charged, the voltage across $C_1$ will be $4V$.
Answer: C
Two nucleons are at a separation of $1\ fm$. The net force between them is $F_{1}$ if both are neutrons, $F_{2}$ if both are protons, and $F_{3}$ if one is a proton and the other is a neutron then

Options:
A. $$F_{1} = F_{3} > F_{2}$$
B. F_{1} = F_{2} > F_{3}
C. F_{3} = F_{2} > F_{1}
D. $$F_{1} > F_{2} > F_{3}$$
E. F_{1} = F_{2} = F_{3}
F. F_{3} = F_{2} = F_{1}
G. F_{3} = F_{2} < F_{1}
H. $$F_{3}= F_{1} > F_{2}$$
Answer: H
$34.1$ g $Pb_3O_4$ is dissolved in $500$ mL of $4\;M\;HNO_3$, then (atomic weight of $Pb=206$ amu) :

Options:
A. total $2.1$ moles of $NaOH$ are used after reaction of $Pb_3O_4$ with $HNO_3$
B. there is no residue
C. total $2.0$ moles of $NaOH$ are used after reaction of $Pb_3O_4$ with $HNO_3$
D. total $2.3$ moles of $NaOH$ are used after reaction of $Pb_3O_4$ with $HNO_3$
E. the weight of residue is $11.9$ g
F. $300$ mL of $6\;M\;NaOH$ is required to neutralise excess of $HNO_3$
G. total $1.8$ moles of $NaOH$ are used after reaction of $Pb_3O_4$ with $HNO_3$
H. total $1.2$ moles of $NaOH$ are used after reaction of $Pb_3O_4$ with $HNO_3$
I. total $2.2$ moles of $NaOH$ are used after reaction of $Pb_3O_4$ with $HNO_3$
J. total $1.1$ moles of $NaOH$ are used after reaction of $Pb_3O_4$ with $HNO_3$
Answer: I
The standard reduction potential of the reaction, 
$H_2O + e^- \rightarrow \displaystyle{\frac{1}{2}}H_2 + OH^-$ at 298 K is :

Options:
A. $$\displaystyle{E^o = \frac{RT}{2F} ln K_w}$$
B. $$\displaystyle{E^o = \frac{RT}{F} ln K_w}$$
C. \displaystyle{E^o = \frac{RT}{F} ln [P_{H_2}]^{1/2}}
D. \displaystyle{E^o = \frac{2RT}{F} ln K_w}
E. $$\displaystyle{E^o = \frac{RT}{F} ln \frac{[P_{H_2}]^{1/2}}{[OH^-]}}$$
F. \displaystyle{E^o = \frac{RT}{3F} ln K_w}
G. $$\displaystyle{E^o = \frac{RT}{F} ln [P_{H_2}]^{1/2}[OH^-]}$$
H. \displaystyle{E^o = \frac{RT}{2F} ln \frac{K_w}{2}}
I. \displaystyle{E^o = \frac{RT}{2F} ln 2K_w}
J. \displaystyle{E^o = \frac{RT}{F} ln \frac{[OH^-]}{[P_{H_2}]^{1/2}}}
Answer: A
The Solubility product of $CaSO_{4}$ is water at $25^{\circ}C\ 9.0\times 10^{-6} M^{2}$. Then its solubility is $mg/m$ is [Mol wt of $CaSO_{4} = 136]$.

Options:
A. $$3.0 \times 10^{-2}$$
B. $$136\times 10^{-3}$$
C. $$136$$
D. $$3.0 \times 10^{-4}$$
E. $$2.0 \times 10^{-3}$$
F. $$1.5 \times 10^{-3}$$
G. $$408$$
H. $$1.0 \times 10^{-3}$$
I. $$1.5 \times 10^{-2}$$
J. $$3.0\times 10^{-3}$$
Answer: J
Two transverse sinusoidal waves travel in opposite directions along a string. The speed transverse waves in the string is $0.5cm/s$. Each has an amplitude of $3.0cm$ and wavelength of $6.0cm$. The equation for the resultant wave is

Options:
A. Both are wrong
B. $$y=6\sin{\cfrac{\pi x}{6}}\cos{\cfrac{\pi t}{3}}$$
C. $$y=6\sin{\cfrac{\pi t}{3}}\cos{\cfrac{\pi x}{3}}$$
D. $$y=6\sin{\cfrac{\pi x}{6}}\cos{\cfrac{\pi t}{6}}$$
E. $$y=6\sin{\cfrac{\pi x}{6}}\cos{\cfrac{2\pi t}{3}}$$
F. $$y=6\sin{\cfrac{\pi t}{6}}\cos{\cfrac{\pi x}{3}}$$
G. $$y=6\sin{\cfrac{2\pi t}{3}}\cos{\cfrac{\pi x}{6}}$$
H. $$y=6\sin{\cfrac{\pi x}{3}}\cos{\cfrac{\pi t}{6}}$$
I. $$y=6\sin{\cfrac{\pi t}{3}}\cos{\cfrac{\pi x}{6}}$$
J. Both may be correct
Answer: F
Two particles $A$ and $B$ of equal mass $M$ are moving with the same speed $v$ as shown in the figure. They collide completely inelastically and move as a single particle $C$. The angle $\theta$ that the path of $C$ makes with the X-axis is given by :

Options:
A. $$\tan {\mkern 1mu} \theta = \frac{{1 - \sqrt 2 }}{{\sqrt 2 (1 + \sqrt 3 )}}$$
B. $$\tan {\mkern 1mu} \theta = \frac{{\sqrt 3 - \sqrt 2 }}{{1 + \sqrt 2 }}$$
C. $$\tan {\mkern 1mu} \theta = \frac{{\sqrt 3 + \sqrt 2 }}{{1 + \sqrt 2 }}$$
D. $$\tan {\mkern 1mu} \theta = \frac{{1 - \sqrt 3 }}{{1 + \sqrt 2 }}$$
E. \( \tan {\mkern 1mu} \theta = \frac{{1 - \sqrt 2 }}{{1 + \sqrt 3 }} \)
F. $$\tan {\mkern 1mu} \theta = \frac{{\sqrt 3 + \sqrt 2 }}{{1 - \sqrt 2 }}$$
G. $$\tan {\mkern 1mu} \theta = \frac{{\sqrt 3 - 1}}{{\sqrt 2 - 1}}$$
H. $$\tan {\mkern 1mu} \theta = \frac{{\sqrt 3 - \sqrt 2 }}{{1 - \sqrt 2 }}$$
I. $$\tan {\mkern 1mu} \theta = \frac{{\sqrt 3 + 1}}{{\sqrt 2 + 1}}$$
J. $$\tan {\mkern 1mu} \theta = \frac{{1 + \sqrt 3 }}{{1 - \sqrt 2 }}$$
Answer: F
The elevation of boiling point of water produced by dissolving 1.17 g sodium chloride in 100 g water $\left( { K }_{ b }=0.512 K kg{ mol }^{ -1 } \right) $

Options:
A. 0.315 K
B. 0.312 K
C. 0.410 K
D. 0.319 K
E. 0.309 K
F. 0.308 K
G. 0.103 K
H. 0.205 K
I. 0.310 K
J. 0.311 K
Answer: E
A wooden block is floating in a liquid. About $50\%$ of its volume is inside the liquid when the vessel is stationary. Percentage volume immersed when the vessel moves upwards with acceleration $a=\dfrac{g}{2}$ is

Options:
A. $$25\%$$
B. $$58.33\%$$
C. $$62.5\%$$
D. $$70\%$$
E. $$60\%$$
F. $$83.33\%$$
G. $$50\%$$
H. $$75\%$$
I. $$66.67\%$$
J. $$33.33\%$$
Answer: G
Find the mean free path of a molecule of a van der wall gas at 360 K (the Boyle's temperature for the given gas )& 1 atm. given  the excluded volume for the gas is ${\text{9}}{\text{.6 \times 1}}{{\text{0}}^{{\text{ - 3}}}}\;{\text{d}}{{\text{m}}^{\text{3}}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}},\;\left[ {{\text{use}}\;{\text{\pi }} = 3,\;{{\text{N}}_{\text{A}}} = 6 \times {\text{1}}{{\text{0}}^{{\text{23}}}}{\text{,}}\;\sqrt {\text{2}}  = \frac{{\text{3}}}{{\text{2}}}{\text{,}}\;{\text{boltzmann}}\;{\text{constant}}\;{\text{K}}\; = \;1 \times {{10}^{ - 23}}\;{\text{J/K,}}\;{\text{atm}}\; = \;{\text{1}}{{\text{0}}^{\text{5}}}\;{\text{Pa}}} \right],$

Options:
A. $${3 \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}}$$
B. $$8 \times 10^{-7}$$
C. $${7 \times 10^{-7}}$$
D. $${5 \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}}$$
E. $$6 \times 10^{-7}$$
F. $$9 \times 10^{-7}$$
G. $$1 \times 10^{-7}$$
H. $${2 \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}}$$
I. $$4.5 \times 10^{-7}$$
J. $${4 \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}}$$
Answer: H
A polymeric organic compound has formula $(C_6H_3)(C_3Hg)_nBr_2$. If it contains $7.68$% bromine, then the value of $n$ will be (approximate)
(Atomic mass of $Br = 80 U$)

Options:
A. $$25$$
B. $$21$$
C. $$17$$
D. $$11$$
E. $$13$$
F. $$15$$
G. $$19$$
H. $$27$$
I. $$23$$
J. $$8$$
Answer: G
A bead of weight w can slide on the smooth circular wire in a vertical plane. The bead is attached by a light thread to the highest point ofthe wire and in equilibrium,  the thread is taut and make an angle $\theta $ with the vertical, then the tension of the tread and reaction of the wire on the bead are

Options:
A. $$T = W\,\cos \theta \,\,\,R = 2W\sin \theta$$
B. $$T = 2W\,\sin \theta \,\,\,R = W\,\cot \theta$$
C. $$T = W\,\cos \theta \,\,\,R = 2W\cos \theta$$
D. $$T = W\,\cos 2\theta \,\,\,R = W\,\tan \theta$$
E. $$T = W\,\sin \theta \,\,\,R = W\cot \theta $$
F. $$T = W\,\cos \theta \,\,\,R = W\,\tan \theta $$
G. $$T = W\,\cos \theta \,\,\,R = W\sin \theta $$
H. $$T = W\,\cos 2\theta \,\,\,R = W\sin 2\theta$$
I. $$T = W\,\sin 2\theta \,\,\,R = W\,\cos \theta$$
J. $$T = 2W\,\cos \theta \,\,\,R = W$$
Answer: J
For a particle moving in the $x-y$ plane, the $x$ and $y$ coordinates are changing as $ x=a\, \sin\,\omega t$ and $y=a (1-\cos\, \omega t)$, where $'a'$ and $\omega$ are constants. Then, what can be inferred for the trajectory of the particle?

Options:
A. The trajectory is parabolic
B. The trajectory is elliptical
C. The trajectory is such that the distance moved by the particle is inversely proportional to time.
D. The trajectory is such that the distance moved by the particle is inversely proportional to the square root of time.
E. The trajectory is such that distance moved by the particle is directly proportional to time.
F. The trajectory is such that the distance moved by the particle is proportional to the square of the square root of time.
G. The trajectory is such that the distance moved by the particle is proportional to the cube of time.
H. The trajectory of the particle is equidistant from the origin at every instant of time.
I. The trajectory is such that the distance moved by the particle is proportional to the square of time.
J. The trajectory is such that the distance moved by the particle is proportional to the square root of time.
Answer: E
A uniform rod AB of mass $m$ and length $l$ is are on a smooth horizontal surface. An impulse $J$ is applied to the end $B$, perpendicular to the rod in the horizontal direction. Speed of particle $P$ at a distance $\dfrac{l}{6}$ from the centre towards $A$ of the rod after time $t=\dfrac{\pi m l}{12J}$ is

Options:
A. $\dfrac{2J}{\sqrt{3}m}$
B. $\dfrac{3J}{2m}$
C. $$\dfrac{J}{m}$$
D. $\dfrac{\sqrt{3}J}{2m}$
E. $\dfrac{\sqrt{5}J}{m}$
F. $\dfrac{J}{\sqrt{6}m}$
G. $$\sqrt{2}\dfrac{J}{m}$$
H. $$2\dfrac{J}{m}$$
I. $$\dfrac{J}{\sqrt{2}m}$$
J. $\dfrac{J}{\sqrt{3}m}$
Answer: G
A particle is performing oscillations with acceleration ${a=8\pi^2-4\pi^2x}$ where x is coordinate of the particle w.r.t. the origin. The parameters are in S.I. units. The particle is at rest at x = 2 at t = 0. Find coordinates of the particle w.r.t origin at any time.

Options:
A. ${2-4 \ cos \ 2\pi t}$
B. $${2-2 \ cos 2\pi t}$$
C. $${2+2 \ sin \ 2\pi t}$$
D. $${4-4 \ cos \ 2\pi t}$$
E. $${2+4 \sin  \ 2\pi t}$$
F. $${4-2 \ cos 2\pi t}$$
G. $${2-2 \ cos 4\pi t}$$
H. $${2-4 \sin  \ 2\pi t}$$
I. ${2-2 \ sin \ 2\pi t}$
Answer: E
The instantaneous voltages at three terminals marked X, Y and Z are given by              ${V_x} = {V_0}\sin \omega t$              ${V_y} = {V_0}\sin \left( {\omega t + {{2\pi } \over 3}} \right)and$
                ${V_z} = {V_0}\sin \left( {\omega t + {{4\pi } \over 3}} \right)$An ideal voltmeter is configured to read rms value of the potential difference between its terminals. It is connected between points X and Y and then between Y and Z.The reading (s) of the voltmeter will be

Options:
A. Independent of the choice of the two terminals
B. ${V_{XY}}^{rms} = {V_0}\sqrt {{1 \over 4}}$
C. ${V_{YZ}}^{rms} = {V_0}\sqrt {{2 \over 3}} $
D. ${V_{YZ}}^{rms} = {V_0}\sqrt {{3 \over 4}} $
E. $${V_{XY}}^{rms} = {V_0}\sqrt {{3 \over 2}} $$
F. ${V_{XY}}^{rms} = {V_0}\sqrt {{2 \over 3}} $
G. ${V_{XY}}^{rms} = {V_0}\sqrt {{3 \over 4}} $
H. $${V_{YZ}}^{rms} = {V_0}\sqrt {{1 \over 2}} $$
I. $${V_{YZ}}^{rms} = {V_0}$$
J. ${V_{XY}}^{rms} = {V_0}\sqrt {{1 \over 3}} $
Answer: A
An air bubble of radius 1 mm is formed inside water at a depth 10 m below free surface (where air pressure is ${ 10 }^{ 5 }N{ m }^{ -2 })$ the pressure inside the bubble is (surface tension of water = $2\times { 10 }^{ -7 }N{ m }^{ -1 }$)

Options:
A. $$2.4\times { 10 }^{ 5 }N{ m }^{ -2 }$$
B. $$2.28\times { 10 }^{ 5 }N{ m }^{ -2 }$$
C. $$2.32 \times 10^{5} N m^{-2}$$
D. $$2.16 \times 10^{5} N m^{-2}$$
E. $$2.44 \times 10^{5} N m^{-2}$$
F. $$2.38 \times 10^{5} N m^{-2}$$
G. $$2.36 \times 10^{5} N m^{-2}$$
H. $$2.0028\times { 10 }^{ -5 }N{ m }^{ -2 }$$
I. $$2.24 \times 10^{5} N m^{-2}$$
J. $$2.0014\times { 10 }^{ -5 }N{ m }^{ -2 }$$
Answer: A
A plane progressive wave of frequency $25Hz$, amplitude $2.5\times 10^{-5}M$ and inital phase zero moves along the neagative x-direction with a velocity of $300 ms^{-1}$. $A$ and $B$ are two points 6m apart on the line of propagation of the wave. At any instant the phase difference between $A$ and $B$ is $\theta$. The phase difference and path difference when the displacement of the particle at $A$ and $B$ is $\Delta$, then

Options:
A. \theta = \pi; \Delta = 3 \times 10^{-5} m
B. $$\theta = \pi; \Delta = 0$$
C. \theta = \pi; \Delta = 7.5 \times 10^{-5} m
D. $$\theta = \pi; \Delta = 0 = 5\times 10^{-5} m$$
E. \theta = 0; \Delta = 2.5 \times 10^{-5} m
F. \theta = \pi; \Delta = 5 \times 10^{-5} m
G. $$\theta = 0; \Delta = 0$$
H. $$\theta = 0\, \Delta = 5\times 10^{-5}m $$
I. \theta = \pi; \Delta = 1 \times 10^{-4} m
J. \theta = 0; \Delta = 1 \times 10^{-4} m
Answer: H
A gas of identical hydrogen like atoms has some atoms in ground state and some atoms in a particular excited state and there are no atoms in any other energy level. The atoms of the gas make transition to a higher state by absorbing monochromatic light of wavelength $304 \mathring A $. subsequently, the atoms emit radiation of only six different photon energies. Some of emitted photons have wavelength $304 \mathring A $, some have wavelength more and some have less than $304 \mathring A $ ( Take $hc = 12420 eV - \mathring A $)Find the principal quantum number of the initially excited state.

Options:
A. 3
B. 6
C. 8
D. 4
E. 2
F. 9
G. 7
H. 10
I. 1
J. 5
Answer: I
A cylinder rolls on the planks $A$ and $B$ without relative sliding. If the planks move with velocities $- 2 v \hat { i } , v \hat { i }$; respectively and the plank $A$ has acceleration. $\vec{a}=a\hat{i}$ then the instantaneous axis of rotation of the cylinder

Options:
A. $$\frac { 3 } { 2 } R \text { from } A$$
B. \frac { 2 } { 3 } R \text { from } A
C. \frac { 1 } { 3 } R \text { from } B
D. $$\frac { 3 } { 2 } R \text { from } B$$
E. \frac { 3 } { 4 } R \text { from } A
F. \frac { 1 } { 2 } R \text { from } B
G. \frac { 2 } { 3 } R \text { from } B
Answer: D
Four identical pendulums are made by attaching a small ball of mass $100$g on a $20$cm long thread and suspended from the same point. Now each ball is given charge Q so that balls move away from each other with each thread making an angle of $45^o$ from the vertical. The value of Q is close to ($\displaystyle\frac{1}{4\pi\epsilon_0}=9\times 10^9$ in SI units).

Options:
A. 1.25\mu C
B. 1.4\mu C
C. $1.5\mu$C
D. 1.65\mu C
E. $1\mu$C
F. $2\mu$C
G. $2.5\mu$C
H. 1.35\mu C
I. 1.75\mu C
J. 1.45\mu C
Answer: C
An ice cube of edge length $20 cm$ is floating in a tank of base area $2500$ $cm^2$ filled with water to $22 cm$ height . The displacement of centre of mass of whole system (in cm), when ice cube melts completely will be 
$$(\rho_{ice} = 900 kg/m^3, \rho_{\omega} = 1000 kg/m^3, g = 10m/s^2)$$

Options:
A. 1.08
B. 1.22
C. $$1.5$$
D. $$1$$
E. 1.12
F. $$1.44$$
G. $$1.32$$
H. 1.04
I. 1.2
J. 0.88
Answer: D
In the motorcycle stunt called " the well of death" the track is a vertical cylindrical surface of $18 m$ radius. Take the motorcycle to be a point mass and $\mu= 0.8$. The minimum angular speed of the motorcycle to prevent him from sliding down should be:

Options:
A. None of these
B. $5/6 $ rad/s
C. $15/5$ rad/s
D. 5/3
E. $5/8 $ rad/s
F. $6/5 $ rad/s
G. $25/3 $ rad/s
H. $5/3$ rad/s
I. $4/5$ rad/s
J. $18/5$ rad/s
Answer: E
A bullet of mass m moving with a horizontal velocity u strikes a stationary wooden block of mass M suspended by, a string of length L = 50cm. The bullet emerges out of the block with speed $\frac { u }{ 4 } $ If M = 6m, the minimum value of use that the block can complete the vertical circle
(Take $g={10m/s}^{2}$)

Options:
A. 12 m/s
B. 15 m/s
C. 30 m/s
D. 10 m/s
E. 25 m/s
F. 40 m/s
G. 8 m/s
H. 20 m/s
I. 22 m/s
J. 9 m/s
Answer: D
A uniform circular disc of mass $2m$ and radius $R$ placed freely on a horizontal smooth surface as shown in the figure. A particle of mass $m$ is connected to the circumference of the disc with a massless string. Now an impulse $J$ is applied on the particle in the directions shown by dotted line. The acceleration of centre of mass of the disc just after application of impulse is (If $J=10N-sec,m=\sqrt { 10 } kg,R=25cm$)

Options:
A. $$4m/{ s }^{ 2 }$$
B. $$2m/{ s }^{ 2 }$$
C. $$5m/{ s }^{ 2 }$$
D. 3m/{ s }^{ 2 }
E. 6m/{ s }^{ 2 }
F. $$1m/{ s }^{ 2 }$$
G. 7m/{ s }^{ 2 }
Answer: A
A storage battery contains a solution of $H_{2}SO_{4}$ $38\%$ by weight. At what temperature will the battery contents freeze?
$$(k_{r}=1.86\ mol^{-1}kg)$$

Options:
A. $$238\ k$$
B. 208\ k
C. $$260\ k$$
D. $$215\ k$$
E. 205\ k
F. 223\ k
G. $$244\ k$$
H. 210\ k
I. 212\ k
J. 202\ k
Answer: G
A conducting rod AB of length $\ell=1\ m$ is moving at a velocity $v = 4\ m/s$ making an angle $ 30^o $ with its length. A uniform magnetic field $B = 2T$ existing in a direction perpendicular to the plane of motion. Then -

Options:
A. $$V_B - V_A = 12\ V$$
B. $$V_A - V_B = 6\ V$$
C. $$ V_A - V_B = 4\ V $$
D. $$ V_B - V_A = 8\ V $$
E. $$V_A - V_B = 10\ V$$
F. $$V_A - V_B = 2\ V$$
G. $$ V_B - V_A = 4\ V $$
H. $$ V_A - V_B = 8\ V $$
I. $$V_B - V_A = 6\ V$$
J. $$V_B - V_A = 2\ V$$
Answer: G
A radioactive isotope is being formed at a constant rate $K$. At $t = 0$, the number of active nuclei is $N_{0}$. The decay constant of isotope is $\lambda$. The number of active nuclei.

Options:
A. Goes on decreasing then stabilizes
B. Is $\dfrac {K}{\lambda}$ after a time $t > > \dfrac {1}{\lambda}$
C. Goes on increasing then decreases
D. Is $\dfrac {K}{3\lambda}$ after a time $t > > \dfrac {1}{\lambda}$
E. Is $\dfrac {K}{4\lambda}$ after a time $t > > \dfrac {1}{\lambda}$
F. Goes on increasing
G. First increases then decreases
H. Goes on decreasing
I. Is $\dfrac {K}{2\lambda}$ after a time $t > > \dfrac {1}{\lambda}$
J. Is $\dfrac {K}{\lambda^2}$ after a time $t > > \dfrac {1}{\lambda}$
Answer: G
Potassium having atomic mass $=39.1$ u contains 93.10 atom % $_{}^{39}\textrm{K}$, having atomic mass 38.96371 u; 0.0118 atom % $_{}^{40}\textrm{K}$, which has mass of 40.0 u and is radioactive with $t_{1/2}=1.3\times10^9y,$ and 6.88 atom % $_{}^{41}\textrm{K}$ having a mass of 40.96184 u. Calculate the specific activity of naturally occurring potassium.

Options:
A. None of these
B. Specific activity $=14.06\:dis.\:g^{-1}S^{-1}$
C. Specific activity $=60.06\:dis.\:g^{-1}S^{-1}$
D. Specific activity $=45.09\:dis.\:g^{-1}S^{-1}$
E. Specific activity $=16.03\:dis.\:g^{-1}S^{-1}$
F. Specific activity $=15.09\:dis.\:g^{-1}S^{-1}$
G. Specific activity $=30.69\:dis.\:g^{-1}S^{-1}$
H. Specific activity $=15.06\:dis.\:g^{-1}S^{-1}$
I. Specific activity $=30.03\:dis.\:g^{-1}S^{-1}$
J. Specific activity $=14.03\:dis.\:g^{-1}S^{-1}$
Answer: G
If SR = 100 torr and PR = 350 torr, then which of the following cannot be a mole fraction of A in vapour phase and a mole fraction of A in liquid phase respectively?

Options:
A. $\dfrac{2}{9}$ and $\dfrac{7}{10} $
B. $\dfrac{3}{10}$ and $\dfrac{2}{9} $
C. $\dfrac{2}{9}$ and $\dfrac{5}{10}$
D. $\dfrac{2}{9}$ and $\dfrac{3}{10} $
E. $\dfrac{2}{9}$ and $\dfrac{6}{10}$
F. \dfrac{2}{9} and \dfrac{9}{10}
G. \dfrac{3}{10} and \dfrac{2}{7}
H. $\dfrac{2}{9}$ and $\dfrac{8}{10}$
I. \dfrac{2}{9} and \dfrac{4}{10}
J. $\dfrac{7}{9}$ and $\dfrac{3}{10} $
Answer: A
A long circular tube of length $10 m$ and radius $0.3 m$ carries a current $I$ along its curved surface as shown. A wire-loop of resistance $0.005 \Omega$ and of radius $0.1 m$ is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as $I=I_{0}\cos(300t)$ where $I_{0}$ is constant. lf the magnetic moment of the loop is $N\mu_{0}I_{0}\sin(300t)$, then ` $N$' is

Options:
A. 10
B. 12
C. $$7$$
D. 5
E. $$4$$
F. 3
G. $$9$$
H. $$6$$
I. $$8$$
Answer: H
A sinusoidal voltage $e = 150\ sin (100$ $\pi t) V$ is applied to a series connection of resistance $R$ and inductance $L$.Select TRUE statement/s from the following :

Options:
A. The maximum current in the circuit is $0.5\ A$
B. The maximum current in the circuit is $0.25\ A$
C. The maximum current in the circuit is $0.60\ A$
D. The maximum current in the circuit is $0.70\ A$
E. The maximum current in the circuit is $0.65\ A$
F. The maximum current in the circuit is $0.75\ A$
G. The instantaneous voltages across the resistance and inductance become equal at $t = $ $\dfrac{1}{300}s$
H. Steady state current leads the voltage by $30^{\circ}$
I. The maximum current in the circuit is $0.85\ A$
J. Voltage leads the current by $30^{\circ}$
Answer: F
A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it. The correct statement(s) is(are)

Options:
A. The emf induced in the loop is zero if the current is constant and decreasing at a steady rate.
B. The emf induced in the loop is finite if the current is constant.
C. The emf induced in the loop is zero if the current decreases at a steady rate.
D. The emf induced in the loop is zero if the current increases at a steady rate.
E. The emf induced in the loop is zero if the current is constant and increasing at a steady rate.
Answer: C
What is the aqueous ammonia concentration of a solution prepared by dissolving $0.15$ mole of $NH_{4}^{+}CH3COO^{-}$ in $1 L$ of water? Given: $K_{a} (CH_{3}COOH) = 1.8 \times 10^{-5}$; $K_{b} (NH_{4}OH) = 1.8 \times 10^{-5}$.

Options:
A. $$5.52 \times 10^{-3}M $$
B. $$7.3 \times 10^{-4}M$$
C. $$8.3 \times 10^{-4}M $$
D. $$9.3 \times 10^{-4}M$$
E. $$0.15 M$$
F. $$8.1 \times 10^{-4}M$$
G. $$1.2 \times 10^{-4}M$$
H. $$3.8 \times 10^{-4}M $$
I. $$7.8 \times 10^{-4}M$$
J. $$6.3 \times 10^{-4}M$$
Answer: C
A particle is moving in $xy$ plane on a circular track of radius $2\ m$ with constant speed $v=3\ m/s$ as shown in the figure. If at time $t=0$, particle is at point $A$, then acceleration of the particle ( in $m{s^{ - 2}}$) at time $t = \pi\ s$ will be

Options:
A. \dfrac{9}{2}\hat i - \dfrac{9}{2}\hat k
B. $$\dfrac{9}{2}\hat j$$
C. \dfrac{9}{2}\hat i + \dfrac{9}{2}\hat j
D. \dfrac{9}{2}\hat i + \dfrac{9}{2}\hat k
E. $$ - \dfrac{9}{2}\hat j$$
F. -\dfrac{9}{2}\hat i - \dfrac{9}{2}\hat j
G. -\dfrac{9}{2}\hat i - \dfrac{9}{2}\hat k
H. $$ - \dfrac{9}{2}\hat i + \dfrac{9}{2}\hat j$$
I. $$\dfrac{9}{2}\hat i - \dfrac{9}{2}\hat j$$
Answer: I
Find the moment of external forces $N$ acting on the axle $OO^\prime$ in the process of rotation.

Options:
A. \displaystyle N=\frac{1}{48}ml^2\omega^2\sin^2{\theta}
B. $$\displaystyle N=\frac{1}{12}ml^2\omega^2\sin^2{\theta}$$
C. $$\displaystyle N=\frac{1}{24}ml^2\omega^2\sin^2{\theta}$$
D. $$\displaystyle N=\frac{1}{6}ml^2\omega^2\sin^2{\theta}$$
E. \displaystyle N= mRNA^2\omega^2\sin^2{\theta}
F. \displaystyle N= \frac{1}{27}ml^2\omega^2\sin^2{\theta}
G. \displaystyle N=\frac{1}{18}ml^2\omega^2\sin^2{\theta}
H. $$\displaystyle N=\frac{1}{36}ml^2\omega^2\sin^2{\theta}$$
I. \displaystyle N= \frac{1}{30}ml^2\omega^2\sin^2{\theta}
J. \displaystyle N= \frac{1}{16}ml^2\omega^2\sin^2{\theta}
Answer: C
A particle is performing $S.H.M.$ and at $t = \frac{{3T}}{4}$ is ta position $\frac{{\sqrt {3A} }}{2}$ and moving towards the origin. Equilibrium position of the particle is at $x=0$ . Then what was the position and direction of particle at $t=0$?

Options:
A. -\frac{A}{4} away from mean position
B. $\frac{A}{2}$ towards mean position
C. $\frac{A}{2}$ away from mean position
D. \frac{A}{4} towards mean position
E. -\frac{A}{2} towards mean position
F. $ - \frac{A}{2}$ away from mean position
Answer: B
A simple pendulum made of a wooden block of mass 500 g. A bullet of mass 50 g fired from downwards side of the block it comes into rest after entering in to the block. If the displacement of block is 1.1 m vertically then the speed of bullet ......... (g= 10m${s^{ - 2}}$ )

Options:
A. 49\sqrt {2\,} m{s^{ - 1}}
B. $$50m{s^{ - 1}}$$
C. $$50\sqrt {2\,} m{s^{ - 1}}$$
D. 48\sqrt{2\,} m{s^{ - 1}}
E. 49.5\sqrt{2\,} m{s^{ - 1}}
F. 50.5\sqrt{2\,} m{s^{-1}}
G. 51.5\sqrt {2\,} m{s^{ - 1}}
H. 50.6\sqrt {2\,} m{s^{ - 1}}
I. $$51\sqrt {2\,} m{s^{ - 1}}$$
J. $$51.6m{s^{ - 1}}$$
Answer: C
A pulley in the form of a uniform disc of mass $2\ m$ and radius $r$ free to rotate in a vertical plane about a fixed horizontal axis through its centre. A light inextensible string has one end fastened to a point on the rim of the pulley and is wrapped several times round the rim. The portion of string not wrapped round the pulley id of length $8r$ and carries a particle of mass $m$ at its free end. The particle is held close to the rim of the pulley and level with its centre. If the particle is released from this position, then initial angular velocity of the pulley is $\omega$ and the impulse of the sudden tension in the string when it becomes taut is $J$, then

Options:
A. $$J=m\sqrt{gr}$$
B. $$J=4m\sqrt{gr}$$
C. $$J=2m\sqrt{gr}$$
D. $$\omega=2\sqrt {\dfrac {2g}{r}}$$
E. $$J=2m\sqrt{2gr}$$
F. $$J=m\sqrt{2gr}$$
G. $$J=5m\sqrt{gr}$$
H. $$J=3m\sqrt{gr}$$
I. $$J=6m\sqrt{gr}$$
Answer: C
Consider the situation shown in figure. The elevator is going up with an acceleration of $2.00ms^{-2}$ and the focal length of the mirror is $12.0cm$. All the surfaces are smooth and pulley is light. The mass-pulley system is released from rest (with respect to the elevator) at t=0 when the distance of B from the mirror is 42.0cm. Find the distance between the image of the block B and the mirror at t=0.200s.

Options:
A. 8.77 cm
B. 6.57 cm
C. 7.57 cm
D. 7.77 cm
E. $$2.37 cm$$
F. $$8.57 cm$$
G. 10.57 cm
H. $$4.37 cm$$
I. None of these
J. 9.57 cm
Answer: F
Consider a finite line charge of uniform charge density $\lambda$ and of length $L$ lying along the x-axis from to $x=0$. Find the field at a point along the axis at $x=d$.

Options:
A. $$\dfrac{\lambda d}{4\pi\epsilon_0L(L+d)}$$
B. $$\dfrac{\lambda L}{4\pi\epsilon_0d(L+d)}$$
C. $$\dfrac{\lambda L}{4\pi\epsilon_0d(d-L)}$$
D. $$\dfrac{\lambda d}{4\pi\epsilon_0L}$$
E. $$\dfrac{\lambda d}{4\pi\epsilon_0(L+d)}$$
F. $$\dfrac{\lambda L}{4\pi\epsilon_0(L-d)^2}$$
G. $$\dfrac{\lambda L}{4\pi\epsilon_0(L+d)}$$
H. $$\dfrac{\lambda }{4\pi\epsilon_0(L+d)}$$
I. $$\dfrac{\lambda d}{4\pi\epsilon_0(L-d)}$$
J. $$\dfrac{\lambda d}{4\pi\epsilon_0(L-d)^2}$$
Answer: C
A copper block of mass 1kg slides down on a rough inclined plane of inclination $\dfrac { 5 }{ 9 } $ at a constant speed. Find the increase in the temperature of the block as it slides down through 60 cm assuming that the loss in mechanical energy goes into the copper block as thermal energy. (specific heat of copper = $420J{ kg }^{ -1 }{ K }^{ -1 },g=10{ ms }^{ -2 }$)

Options:
A. 10.6 \times { 10 }^{ -3 }
B. 6.1 \times { 10 }^{ -3 }
C. 9.1 \times { 10 }^{ -3 }
D. $$8.6\times { 10 }^{ -3 }$$
E. $$6.6\times { 10 }^{ -3 }$$
F. 7.6 \times { 10 }^{ -3 }
G. 5.6 \times { 10 }^{ -3 }
H. 7.1 \times { 10 }^{ -3 }
I. $$9.6\times { 10 }^{ -3 }$$
Answer: D
In which of the following solution, the solubility of $AgCN$ will be greater than that in pure water: Given ${K}_{sp}(AgCN=4\times {10}^{16})$;$[{K}_{a}(HCN)=5\times {10}^{-10}]$

Options:
A. a buffer solution of $pH=8$
B. a buffer solution of $pH=12$
C. a buffer solution of $pH=11$
D. a buffer solution of $pH=7$
E. a buffer solution of $pH=9$
F. $0.01M$ $Ag{NO}_{3}$ solution
G. a buffer solution of $pH=6$
H. $0.2M$ ${NH}_{3}$ solution
Answer: B
The rate constant for 1st order decomposition is given  by log k=14.34 - $\frac { 1.25\times { 10 }^{ -4 } }{ T } $ The energy of activation is :

Options:
A. 256 kJ/mol
B. 239 kJ/mol
C. 255 kJ/mol
D. 250 kJ/mol
E. 258 kJ/mol
F. 259 kJ/mol
G. 260 kJ/mol
H. 261 kJ/mol
I. 229 kJ/mol
J. 262 kJ/mol
Answer: B
What is the solubility product of $Al(OH)_{3}$ in water?Given: $Al(OH)_{4}^{-}(aq) \leftrightharpoons Al^{3+}(aq) + 4OH^{-} (aq); K = 1.3 \times 10^{-34}$$Al(OH)_{3}(s) + OH^{-}(aq) \leftrightharpoons Al(OH)_{4}^{-}(aq)$; $K =3.85$

Options:
A. $$5.8 \times 10^{-33}$$
B. $$3.1 \times 10^{-35}$$
C. $$4.9 \times 10^{-33}$$
D. $$5.5 \times 10^{-33}$$
E. $$4.2 \times 10^{-33}$$
F. $$5 \times 10^{-33}$$
G. $$6.1 \times 10^{-33}$$
H. $$7.5 \times 10^{-33}$$
I. $$5 \times 10^{-34}$$
J. $$5.2 \times 10^{-33}$$
Answer: F
Two waves travelling in opposite directions produce a standing wave. The individual wave functions are given by $y_1 = 4 sin (3x - 2t)$ and $y_2 = 4 sin (3x + 2t) cm$, where x and y are in cm

Options:
A. Antinodes are formed at x values given by $$\pi/6, \pi/2, 5\pi/6, 7\pi/6, ...$$
B. Nodes are formed at x values given by
$$0, \pi/3, 2\pi/3, 4\pi/3,...$$
C. \(\pi/12, \pi/3, 5\pi/12, 7\pi/12, \ldots\)
D. $$\pi/4, \pi/2, 3\pi/4, 5\pi/4, ...$$
E. \pi/8, 3\pi/8, 5\pi/8, 7\pi/8, ...
F. The maximum displacement of the motion at t = 2.3 s is 4.63 cm.
G. \(\pi/6, \pi/3, 5\pi/6, 7\pi/6, \ldots\)
H. \(\pi/5, 2\pi/5, 3\pi/5, 4\pi/5, \ldots\)
I. The maximum displacement of the motion at x = 2.3 cm is 4.63 cm.
Answer: A
A free hydrogen atom after absorbing a photon of wavelength $\lambda_a$ gets excited from the state $n = 1$ to the state $n = 4$. Immediately after that the electron jumps to n = m state by emitting a photon of wavelength $\lambda_e$. Let the change in momentum of atom due to the absorption and the emission are $\Delta P_a$ and $\Delta P_e$ respectively. If $\lambda_a / \lambda_e = 1/5$, which of the option (s) is/are correct ?
[Use $hc = 1242 eV$ nm ; $1 nm = 10^{-9} m$, h and c are Plank's constant and speed of light, respectively]

Options:
A. m = 1
B. $$\Delta p_a / \Delta p_e = 1/2$$
C. m = 4
D. $$m = 2$$
E. $$m = 3$$
F. m = 6
G. $$\lambda_e = 418 nm$$
H. m = 5
Answer: D
Two spheres are moving towards each other. Both have same radius but their masses are 2 kg and 4 kg. If the velocities are 4 m/s and 2 m/s respectively and coefficient of restitution is e = 1/3, find the maximum potential energy of deformation.

Options:
A. 19 J
B. 14 J
C. 12 J
D. 17 J
E. 21 J
F. 20 J
G. 16 J
H. 36 J
I. 15 J
J. 18 J
Answer: J
A force 'F' stops a body body of mass 'm' moving with a velocity 'u'  in a distance 's'.The force required to stop a body of double the mass moving with double the velocity in the same distance is:

Options:
A. 8F
B. 4F
C. 3F
D. 7F
E. 2F
F. 12F
G. 10F
H. 9F
I. 5F
J. 6F
Answer: E
A bob of mass m is projected with a horizontal velocity $ v = \sqrt \frac{gl}{2}$ as shown in Fig. 8.224. In consequence, it moves in a circular path in a vertical plane by the inextensible string which passes over the smooth fixed peg. Find the maximum angle that the bob swings in the left hand side.

Options:
A. $cos^{-1}(\frac{1}{9})$
B. $cos^{-1}$ ($\frac{1}{8}$)
C. $cos^{-1}$ ($\frac{1}{2}$)
D. $cos^{-1}$ ($\frac{1}{10}$)
E. $cos^{-1}$ ($\frac{1}{5}$)
F. $cos^{-1}$ ($\frac{1}{6}$)
G. $cos^{-1}$ ($\frac{1}{4}$)
H. $cos^{-1}$ ($\frac{1}{3}$)
I. $cos^{-1}$ ($\frac{1}{7}$)
J. $cos^{-1}(\frac{1}{11})$
Answer: F
A block of mass in placed between two springs of constant K and 2K whose ope ends are fixed. The block is given a velocity ${ v }_{ 0 }$ towards spring 1 at t=0 when it is exactly midway between the springs between the free ends of the springs is $\ell $ There (the surface of contact is frictionless)
Time after which spring (1) is compressed to the maximum for the first time is

Options:
A. $$\dfrac { \ell }{ { v }_{ 0 } } +2\pi \sqrt { \dfrac { m }{ k } }$$
B. $$\dfrac { \ell }{ { v }_{ 0 } } +\dfrac { 5\pi }{ 2 } \sqrt { \dfrac { m }{ k } }$$
C. $$\dfrac { \ell }{ { 2v }_{ 0 } } +2\pi \sqrt { \frac { m }{ k } } $$
D. $$\dfrac { \ell }{ { v }_{ 0 } } +\dfrac { 3\pi }{ 4 } \sqrt { \dfrac { m }{ k } }$$
E. $$\dfrac { \ell }{ { 2v }_{ 0 } } +\dfrac { \pi }{ 2 } \sqrt { \dfrac { m }{ k } } $$
F. $$\dfrac { \ell }{ { v }_{ 0 } } +\dfrac { \pi }{ 2 } \sqrt { \frac { m }{ k } } $$
G. $$\dfrac { \ell }{ { v }_{ 0 } } +\pi \sqrt { \dfrac { m }{ k } } $$
H. $$\dfrac { \ell }{ { 2v }_{ 0 } } +\pi \sqrt { \dfrac { m }{ k } }$$
I. $$\dfrac { \ell }{ { 2v }_{ 0 } } +\dfrac { 3\pi }{ 2 } \sqrt { \dfrac { m }{ k } }$$
J. $$\dfrac { \ell }{ { v }_{ 0 } } +\dfrac { 3\pi }{ 2 } \sqrt { \dfrac { m }{ k } }$$
Answer: G
In the superposition of two coherent waves (at $x = 0$) given by $y_{1} = 4\sin 300\pi t$ and $y_{2} = 2\sin  308\pi t$.

Options:
A. The frequency of beats is $8\ Hz$
B. The amplitude of resulting wave varies at a frequency of $12\ Hz$
C. The amplitude of resulting wave varies at a frequency of $8\ Hz$
D. The amplitude of resulting wave varies at a frequency of $4\ Hz$
E. The amplitude of resulting wave varies at a frequency of $2\ Hz$
F. The frequency of the resulting wave is $152\ Hz$
G. The amplitude of resulting wave varies at a frequency of $10\ Hz$
H. The amplitude of resulting wave varies at a frequency of $5\ Hz$
I. The amplitude of resulting wave varies at a frequency of $6\ Hz$
J. The ratio of maximum to minimum intensity is $9$
Answer: E
The standard Gibbs energy for the decomposition ${ H }_{ 2 }O\left( g \right) \rightleftharpoons { H }_{ 2 }\left( g \right) +\frac { 1 }{ 2 } { O }_{ 2 }\left( g \right) $ is 115.15 kJ/mol at 2000 K. Degree of dissociation ${ H }_{ 2 }O$ at this temperature and 1 atm pressure is: Given: $ R=\frac { 25 }{ 3 } J/mol\ K$ and  $2^{ 1/3 }=1.26$

Options:
A. 0.14
B. 0.13
C. 0.15
D. 0.09
E. 0.07
F. 0.12
G. 0.10
H. 0.01
I. 0.70
J. 0.11
Answer: G
A remote -sensing satellite of earth revolves in a circular orbit at a height of $ 0.25 \times 10^6 $ m above the surface of earth. if earth's radius is $ 0.25 \times 10^6 $ m above the surface of earth. if earth's radius is $ 6.38 \times 10^6 m $ and $ g= 9.8 m/s^2 $, then the orbital speed of the satellite is :

Options:
A. $$9.13 km/s$$
B. 7.12 km/s
C. 7.34 km/s
D. 8.56 km/s
E. $$6.67 km/s$$
F. 6.98 km/s
G. $$7.76 km/s$$
H. 7.04 km/s
I. 7.45 km/s
Answer: E