MikaStars39/nano-eval huggingface dataset:
A saturated solution of AgCl was treated with 1.5 M $\displaystyle Na_{2}CO_{3}$ solution giving $\displaystyle Ag_{2}CO_{3}$. The remaining solution contained 0.0026 gm of Cl per liter  $\displaystyle [K_{sp}ofAg_{2}CO_{3}=8.2\times10^{-12}M^{3}]$ Which of the following statements is/are correct?

Options:
A. $\displaystyle K_{sp}$ of AgCl from the above data is $\displaystyle 1.71 \times 10^{-8}M^{2}$
B. $\displaystyle K_{sp}$ of AgCl from the above data is $\displaystyle 1.71 \times 10^{-9}M^{2}$
C. $\displaystyle K_{sp}$ of AgCl from the above data is $\displaystyle 1.71 \times 10^{-11}M^{2}$
D. $\displaystyle K_{sp}$ of AgCl from the above data is $\displaystyle 1.71\times 10^{-10}M^{2}$
E. \displaystyle K_{sp} of AgCl from the above data is \displaystyle 1.71 \times 10^{-7}M^{2}
F. $\displaystyle K_{sp}$ of AgCl from the above data is $\displaystyle 1.71 \times 10^{-12}M^{2}$
G. $\displaystyle K_{sp}$ of AgCl from the above data is $\displaystyle 1.71 \times 10^{-13}M^{2}$
H. The equilibrium concentration of $\displaystyle CO_{3}^{-2}$ is $\displaystyle 8.2 \times 10^{-12}M$
I. The concentration of NaCl at equilibrium is $\displaystyle 7.32 \times 10^{-1}M$
Answer: D
Three particles A, B. and C each of mass m, are placed at the vertices of an equiangular  triabgle of side  $3$ m. The particle at A is released at $t=0$. The ratio of its acceleration of soon after  release and when it reaches centeroid is

Options:
A. $$\sqrt 3 $$
B. $$\frac{1}{\sqrt{3}}$$
C. $$\frac{2}{3}$$
D. $$\frac{\sqrt{3}}{2}$$
E. $$\frac{1}{3}$$
F. $$\frac{3}{\sqrt{3}}$$
G. $$\frac{1}{2}$$
H. $$\frac{1}{{\sqrt 3 }}$$
I. $$3$$
J. $$\frac{\sqrt{3}}{3}$$
Answer: E
A Block is suspended by an aideal spring of force constant k. If the block is pulled down by applying constant force F and if maximum displacement of block from its initial position of rest is $x_0$ then:

Options:
A. Increase in energy stored in spring is \( \dfrac{4k x^2_0}{3} \)
B. Work done by applied force F is $Fx_o$
C. $$x_o = \dfrac{2F}{k}$$
D. Increase in energy stored in spring is \( \dfrac{3k x^2_0}{2} \)
E. Increase in energy stored in spring is k $x^2_0$
F. Increase in energy stored in spring is \( \dfrac{7k x^2_0}{4} \)
G. Increase in energy stored in spring is \(\dfrac{2k x^2_0}{3}\)
H. Increase in energy stored in spring is \(\dfrac{5k x^2_0}{3}\)
I. $$x_o = \dfrac{3F}{2k}$$
Answer: E
Two aldopentoses A and B give the same osazone derivation. A is oxidised to an optically active aldaric acid by dilute nitric acid. Ruff degradation of B gave a tetrose which was similarly oxidised to an optically active aldaric acid. Assign the structure of A and B from the following list:

Options:
A. $$A=IV, B=III$$
B. $$A=IV, B=II$$
C. $$A=II, B=III$$
D. $$A=III, B=II$$
E. $$A=IV, B=I$$
F. $$A=III, B=I$$
G. $$A=III, B=IV$$
H. $$A=I, B=IV$$
Answer: D
$A$ and $B$ are two isolated spheres kept in close proximity so that they can exchange energy by radiation. The two spheres have identical physical dimensions but the surface of $A$ behaves like a perfectly black body while the surface of $B$ reflects 20% of all the radiations it receives. They are isolated from all other sources of radiation.

Options:
A. If they are in thermal equilibrium and exchange equal amounts of radiation per second, then $T_A = (0.8)^{1/4} T_B$.
B. If they are in thermal equilibrium and exchange equal amounts of radiation per second, then $T_A = (1.2)^{1/4} T_B$.
C. If they are in thermal equilibrium and exchange equal amounts of radiation per second, they will be at the same absolute temperature, $T_A = T_B$.
D. If they are in thermal equilibrium and exchange equal amounts of radiation per second, then $T_A = (0.75)^{1/4} T_B$.
E. If they are in thermal equilibrium and exchange equal amounts of radiation per second, then $T_A = (1.05)^{1/4} T_B$
F. If they are not in thermal equilibrium and are each at $i = 0$ at the same temperature $T_A = T_B = T$, then the sphere $A$ will gain thermal energy and $B$ will lose thermal energy.
G. If they are in thermal equilibrium and exchange equal amounts of radiation per second, then $T_A = (0.95)^{1/4} T_B$.
H. If they are in thermal equilibrium and exchange equal amounts of radiation per second, then $T_A = (1.1)^{1/4} T_B$.
I. If they are in thermal equilibrium and exchange equal amounts of radiation per second, then $T_A = (0.9)^{1/4} T_B$.
J. If they are not in thermal equilibrium and are each at $t = 0$ at the same temperature $T_A = T_B = T$ then sphere $A$ will lose thermal energy and $B$ will gain thermal energy.
Answer: C
Calculate the solubility of AgCN in a buffer solution of $pH=3$. Given $K_{sp}$ of $AgCN = 1.2\times 10^{-16}$ and $K_a$ for $HCN  = 4.8\times 10^{-10}$.

Options:
A. 1.3 \times 10^{-5} g\, L^{-1}
B. 2.5 \times 10^{-5} g\, L^{-1}
C. $$1.5\times 10^{-5} g\, L^{-1}$$
D. none of these
E. 2.2 \times 10^{-5} g\, L^{-1}
F. 3.6 \times 10^{-5} g\, L^{-1}
G. $$2.0 \times 10^{-5} g\, L^{-1}$$
H. $$3.2\times 10^{-4} g\, L^{-1}$$
I. $$4.4\times 10^{-4} g\, L^{-1}$$
J. 1.8 \times 10^{-5} g\, L^{-1}
Answer: C
For the process: $ Cu^{2+}+2e^{-}\rightarrow Cu; \ log [Cu^{2+}]$ vs. $ E_{red}$ graph is shown in the figure, where OA = 0.34 V. The electrode potential of the half-cell of $ Cu/Cu^{2+}$ (0.1 M) will be [2.3303 RT/F = 0.06]

Options:
A. $$-0.28 V$$
B. $$ +0.37 V$$
C. $$-0.34 V$$
D. $$ -0.34+\dfrac{0.0591}{2} V$$
E. $$-0.35 V$$
F. $$+0.34 V$$
G. $$ -0.37 V $$
H. $$-0.33 V$$
I. $$ +0.31 V$$
J. $$-0.30 V$$
Answer: D
If a tuning fork of frequency $\left( {{f_0}} \right)$ 340 Hz and tolerance 1% is used in resonance column method $\left[ {v = 2{f_0}\left( {{l_2} - {l_1}} \right)} \right],$ the first and the second resonance are measured at ${l_1} = 24.0\,\,cm$ and ${l_2} = 74.70\,\,cm.$ Find max permissible error in speed of sound:

Options:
A. 1.8%
B. 1.1%
C. 1.6%
D. 1.5%
E. 1.7%
F. 1.0%
G. 1.2%
H. 1.4%
I. 1.3%
J. 1.35%
Answer: H
In a set of experiments on a hypothetical one-electron atom, the wavelengths of the photons emitted from transitions ending in the ground state $(n=1)$ are shown in the energy diagram above. The possible energy of the atom in $n=3$ cannot be

Options:
A. -0.5125\space eV
B. $$-1.95\space eV$$
C. -0.5425\space eV
D. -1.55\space eV
E. -2.15\space eV
F. -1.75\space eV
G. $$-7.8\space eV$$
H. $$-0.121\space eV$$
I. $$-0.4875\space eV$$
J. -0.5625\space eV
Answer: B
A magnetic needle of magnetic moment $M=0.05$ A-$m^2$ and moment of inertia of $7\times 10^{-6}$ kg$m^2$ is oscillating in a magnetic field of $1$G. Time taken for $20$ oscillation is?

Options:
A. $$128.6 \, \text{s}$$
B. 12.86$s
C. $$18.86 \, \text{s}$$
D. $24.86$s
E. $14.86$s
F. $$148.6 \, \text{s}$$
G. $248.6$s
H. $$24.86 \, \text{s}$$
I. $148.6$s
Answer: H
A train of mass $m=2000\:tons$ moves in the latitude $\varphi=60^\circ$ North. Find in what direction and with what velocity the train should move for the resultant of the inertial forces acting on the train in the reference frame fixed to the Earth to be equal to zero.

Options:
A. $V=\sqrt{(gR)/4}$ east to west
B. $V=\sqrt{(gR)/2}$ east to west
C. $V=\sqrt{(gR)/4}$ north to south
D. $V=\sqrt{(gR)/4}$ east to east
E. $V=\sqrt{(3gR)/4}$ east to west
F. $V=\sqrt{(3gR)/4}$ north to south
G. $V=\sqrt{(gR)/4}$ west to east
H. $V=\sqrt{(gR)/8}$ east to west
I. $V=\sqrt{(gR)/3}$ east to west
J. $V=\sqrt{(3gR)/4}$ west to east
Answer: A
Three simple harmonic vibration, ${X}_{1}=6\sin{\left(\omega t+\dfrac{\pi}{2}\right)}$, ${X}_{2}=4\sin{\left(\omega t+\pi/2 \right)}$, ${X}_{3}=2\sin{\left(\omega t+\dfrac{3\pi}{2}\right)}$ Find resultant amplitude:

Options:
A. $$8\sqrt{3}$$
B. $$2\sqrt{15}$$
C. $$20,\tan^{-1}{2}$$
D. $$4\sqrt{3}$$
E. $$8\sqrt{2}$$
F. $$2\sqrt{13}$$
G. $$4\sqrt{2}$$
H. $$2\sqrt 2 $$
I. $$2\sqrt{10}$$
Answer: D
A rod AB with roller is constrained to move against wall and ground End A starts slide down from rest with constant acceleration $ a_A= 2 m/s^2 $ . find acceleration of B in $ m/s^2 $ when $ \theta = 37^0 $ value of $ \theta $ at t=0 was $ 53^0 $ length of rod AB = 5 m.

Options:
A. $$ \dfrac {15}{42} $$
B. $$ \dfrac {3}{2} $$
C. $$ \dfrac {39}{25} $$
D. $$\dfrac{26}{18}$$
E. $$ \dfrac {49}{16} $$
F. $$\dfrac {27}{18}$$
G. $$\dfrac {23}{18}$$
H. $$\dfrac {29}{18}$$
I. $$\dfrac {25}{18}$$
J. $$\dfrac {21}{18}$$
Answer: B
A reaction takes place in three steps, then rate constants are $k_{1},\ k_{2}$ and $k_{3}$ respectively. The overall rate constant $k=\dfrac{k_{1}^{3/2}k_{3}^{3/2}}{k_{2}^{1/4}}$. If activation of each steps are $50, 500$ and $90\ kJ$ respectively, then overall activation energy of the reaction is:

Options:
A. $$280$$
B. $$320$$
C. $$10$$
D. $$270$$
E. $$620$$
F. $$315$$
G. $$275$$
H. $$310$$
I. $$265$$
J. $$300$$
Answer: C
Following data is given for the reaction
$CaCO_{3}(s)$$\rightarrow$$CaO(s)+CO_{2}(g)$
$$\triangle_{f}H^{0}[CaO(s)]=-635.1 kJ mol^{-1}$$
$$\triangle_{f}H^{0}[CO_{2}(g)]=-393.5.1 kJ mol^{-1}$$
$$\triangle_{f}H^{0}[CaCO_{3}(s)]=-1206.9 kJ mol^{-1}$$
Predicate the effect of temperature on the equilibrium constant of the above reaction.

Options:
A. The temperature will increase.
B. The temperature will decrease.
C. The reaction will proceed in forward direction on increasing temperature.
D. both a & b
Answer: D
$ PCl_{5} $ decomposes as $ PCl_{2}(g) \rightleftharpoons PCl_{3}(g) + Cl_{2}(g) $. If at equilibrium, the total pressure is $P$ and densityof the  gaseous mixture is $d$ at a temperature $T$, then the degree of dissociation $a$ is:(Molecular weight of $PCl_{5}  = M$)

Options:
A. $$a = 1- \dfrac{dRT}{2PM}$$
B. $$a = 1- \dfrac{dRT}{PM}$$
C. $$ a = \dfrac{dRT}{PM} - 1 $$
D. $$ a = 1- \dfrac{PM}{dRT} $$
E. $$a = 1- \dfrac{2PM}{dRT}$$
F. $$a = \dfrac{PM}{dRT} - 1$$
G. $$a = \dfrac{PM}{dRT} - 2$$
H. $$a = \dfrac{dRT}{2PM} - 1$$
Answer: F
Three blocks of masses $m_{1}=2.0,m_{2}=4.0$ and $m_{3}=6.0\ kg$ are connected by strings on a frictionless inclined plane of $60^{o}$, as shown in the figure. A force $F=120\ N$ is applied upwards along the incline to the uppermost block, causing an upward movement of the blocks. The connecting cords are light. The values of tensions $T_{1}$ and $T_{2}$ in the cords are

Options:
A. $$T_{1}=45\ N, T_{2}=50\ N$$
B. $$T_{1}=35\ N, T_{2}=50\ N$$
C. $$T_{1}=40\ N, T_{2}=50\ N$$
D. $$T_{1}=30\ N, T_{2}=50\ N$$
E. $$T_{1}=40\ N, T_{2}=60\ N$$
F. $$T_{1}=20\ N, T_{2}=100\ N$$
G. $$T_{1}=60\ N, T_{2}=60\ N$$
H. $$T_{1}=35\ N, T_{2}=45\ N$$
I. $$T_{1}=20\ N, T_{2}=60\ N$$
J. $$T_{1}=40\ N, T_{2}=45\ N$$
Answer: I
Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents $X$ and $Y$. The molecular weights of the solvents are $M_X$ and $M_{Y'}$ respectively where $M_X=\cfrac{3}{4}M_Y$. The relative lowering of vapour pressure of the solution in $X$ is "$m$" times that the solution in $Y$. Given that the number of moles of solute is very small ion composition to that of solvent, the value of "$m$" is :

Options:
A. $$\cfrac{1}{2}$$
B. $$\cfrac{2}{3}$$
C. $$\cfrac{5}{3}$$
D. $$\cfrac{5}{4}$$
E. $$\cfrac{3}{2}$$
F. $$\cfrac{1}{4}$$
G. $$\cfrac{4}{3}$$
H. $$\cfrac{3}{5}$$
I. $$\cfrac{7}{5}$$
J. $$\cfrac{3}{4}$$
Answer: J
The nuclear reaction $n+_{ 5 }^{ 10 }B\longrightarrow _{ 3 }^{ 7 }\quad Li+_{ 2 }^{ 4 }$He is observed to occure even when very slow moving neutrons $\left( { m }_{ n }=1.0087u \right) $ strick a boron atom at rest. For a particular reaction in which ${ k }_{ n }=0,$ the helium (mHe =4.0026u)is observed to have a speed of 9.30$\times { 10 }^{ 6 }\quad { ms }^{ -1 }.$ Evaluate
The Q-VALUE OF the reaction

Options:
A. 3.16 MeV
B. 4.16 MeV
C. 2.16 MeV
D. 1.32 MeV
E. 2.62 MeV
F. 2.82 MeV
G. 2.32 MeV
H. 0.82 MeV
I. 1.02 MeV
J. 1.62 MeV
Answer: I
A radio nuclide of $^{32}\textrm{P}$ with half-life 14.3 day is produced in a nuclear reactor at a constant rate, $q=2.7\times 10^9$ nuclei per second. Period soon after the beginning of production of that nuclide will its activity be equal to $1.7\times 10^9$ dps is :

Options:
A. $$6 \: day$$
B. $$8 \: day$$
C. $$8.2 \: day$$
D. $$8.7 \: day$$
E. $$9 \: day$$
F. $$None of these$$
G. $$9.5 \: day$$
H. $$8.5 \: day$$
I. $$8.9 \: day$$
J. $$7.5 \: day$$
Answer: G
20 gm of a binary electrolye ( molecular weight  = 100 ) are dissolved in 500 gm of water.  The freezing point of the solution is - $0.74^{\circ}C, K_{f}= 1.86 K $ $molality ^{-1}$  The degree of ioisation of the electrolyte is

Options:
A. 0.74 %
B. 0 %
C. 100 %
D. 0.57 %
E. 0.75 %
F. 0.537 %
G. 75 %
H. 0.73 %
I. 0.67 %
J. 74 %
Answer: B
A conductor $AB$ of length $l$ oriented along x-axis moves in XY plane with velocity $\overrightarrow { v } ={ v }_{ 0 }\left( \hat { i } -\hat { j }  \right) A$ magnetic field $\overrightarrow { B } ={ B }_{ 0 }\left( \hat { i } +\hat { j }  \right) $ exists in the region. The induced emf is

Options:
A. $${B}_{0}/{v}_{0}$$
B. $$\frac{2B_0}{v_0}$$
C. $$\frac{2\sqrt{2}B_0}{3v_0}$$
D. zero
E. $$\frac{3\sqrt{2}B_0}{v_0}$$
F. $${B}_{0}/2{v}_{0}$$
G. $$\sqrt{2}{B}_{0}/{v}_{0}$$
H. $$\frac{\sqrt{2}B_0}{2v_0}$$
I. $$2{B}_{0}/{v}_{0}$$
J. $$$2\sqrt{2}{B}_{0}/{v}_{0}$$$
Answer: G
A body of mass $m$ is attached to a spring of spring constant $k$ which hangs from the ceiling of an elevator at rest in equilibrium. now the elevator starts accelerating upwards with its acceleration varying with time as $a=pt+q$, where $p$ and $q$ are positive constants. In the frame of elevator:

Options:
A. the velocity of the block will vary simple harmonically for all value of $p$ and $q$
B. the block will perform $S.H.M$ provided for all value of $p$ and $q$ expect $p=1$
C. the block will perform $S.H.M$ provided for all value of $p$ and $q$ expect $q=1$
D. the block will perform $S.H.M$ provided for all value of $p$ and $q$ expect $p=0$
E. the block will perform $S.H.M$ provided $q=0$ and any value of $p$
F. the block will perform $S.H.M$ provided for all value of $p$ and $q$ expect $p=q$
G. the block will not perform $S.H.M$ in general for all value of $p$ and $q$ expect $p=0$
H. the block will perform $S.H.M$ for all value of $p$ and $q$
I. the block will perform $S.H.M$ provided for all value of $p$ and $q$ expect $q=0$
J. the block will perform $S.H.M$ provided for all value of $p$ and $q$ except $q=0$
Answer: D
A string of mass $m$ is fixed at both its ends. The fundamental mode of string is excited and it has an angular frequency $\omega $ and the maximum displacement amplitude $A$. Then :

Options:
A. The mean kinetic energy of the string averaged over one periodic time is $ =\dfrac{1}{36}mA^2 \omega^2$
B. The mean kinetic energy of the string averaged over one periodic time is $ =\dfrac{1}{8}mA^2 \omega^2$
C. The mean kinetic energy of the string averaged over one periodic time is $ =\dfrac{1}{16}mA^2 \omega^2$
D. The mean kinetic energy of the string averaged over one periodic time is $ =\dfrac{1}{24}mA^2 \omega^2$
E. The maximum kinetic energy of the string is $E_k=\dfrac{1}{2}mA^2\omega^2$
F. The mean kinetic energy of the string averaged over one periodic time is $ =\dfrac{1}{12}mA^2 \omega^2$
G. The mean kinetic energy of the string averaged over one periodic time is $ =\dfrac{1}{32}mA^2 \omega^2$
H. The mean kinetic energy of the string averaged over one periodic time is $ =\dfrac{1}{10}mA^2 \omega^2$
I. The maximum kinetic energy of the string is $E_k=\dfrac{1}{4}mA^2\omega^2$
J. The mean kinetic energy of the string averaged over one periodic time is $ =\dfrac{1}{4}mA^2 \omega^2$
Answer: B
An alkyl bromide (A) on treatment with Na and ether gives a hydrocarbon (E). (E) on treatment with HBr and peroxidegives $ Br-{(CH_2 )}_6 -Br.$ Compound (A) is:

Options:
A. $$ CH_3-CH=CH_2CH_2Br $$
B. $$ CH_3CH_2CH_2Br $$
C. CH_3-CH=CH_2CH_3
D. CH_3-CH_2-CH=CH_2Br
E. $$ CH_3-CH=CHBr $$
F. CH_3-CH=CH-CH_2Br
G. $$ H_2C=CH-CH_2-Br $$
H. CH_3-CH_2-CH=CHBr
I. CH_3-CH_2-CH_2-CH=CHBr
J. $$ CH_3-CH_2-CH_2-CH_2Br $$
Answer: G
A particle is projected vertically upwards with a speed of 16$ms^{-1}$. After some time, when it again passes through the point of projection, its speed is found to be 8$ms^{-1}$. It is known that the work done by air resistance is same during upward and down ward motion. Then the maximum height attained by the particle is (take g = 10$ms^{-2}$).

Options:
A. $$13.6 m$$
B. $$8 m$$
C. $$17.6 m$$
D. $$6.4 m$$
E. $$4.8 m$$
F. $$9.6 m$$
G. $$10.2 m$$
H. $$12.8 m$$
I. $$16.4 m$$
J. $$3.2 m$$
Answer: B
Calculate pH of 0.1 M $H_{2}S$ solution (Given $K_{a_{i}}=10^{-7}$ and $K_{a_{2}}=10^{-14})$

Options:
A. 4
B. 8
C. 7
D. 3
E. 5
F. None of these
G. 6
H. 1
I. 2
J. 9
Answer: D
If $\oint _{ s }^{  }{ E.dS } =0$ over a surface, then:

Options:
A. The number of flux lines entering the surface is greater than the number of flux lines leaving it.
B. The electric field inside the surface is constant.
C. All charges are uniformly distributed on the surface.
D. The number of flux lines entering the surface must equal to the number of flux lines leaving it.
E. The number of flux lines entering the surface must necessarily equal the number of flux lines leaving it.
F. All charges must necessarily be outside the surface.
G. The electric field inside the surface is necessarily uniform.
H. The electric field inside the surface is uniform.
I. The electric field inside the surface and on it is zero.
J. The electric field inside the surface is zero.
Answer: F
One of the reaction that take place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monooxide to give iron metal and carbon dioxide.
$FeO + CO\left( g \right) \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over
{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} Fe\left( s \right) + C{O_2}\left( g \right)$  $Kp = 0.25 at 1050 k$ 
What are the partial pressure of $CO and C{O_2}$ at $1050 K$ if the initial pressure is ${P_CO}=1.5,{P_C{O_2}}=0.8$?

Options:
A. $$${P_CO}=1.1, {P_C{O_2}} = 0.4$$$
B. $$${P_CO}=1.3, {P_C{O_2}} = 0.4$$$
C. $$${P_CO}=1.25, {P_C{O_2}} = 0.35$$$
D. $${P_CO}=1.84,{P_C{O_2}} = 0.46$$
E. $$${P_CO}=1.22, {P_C{O_2}} = 0.32$$$
F. $${P_CO}=1.2,{P_C{O_2}} = 0.3$$
G. $${P_CO}=1.6,{P_C{O_2}} = 0.4$$
H. $$${P_CO}=1.4,{P_C{O_2}} = 0.5$$$
I. $$${P_CO}=1.35,{P_C{O_2}} = 0.45$$$
J. $${P_CO}=1.5, {P_C{O_2}} = 0.8$$
Answer: F
A gas has molar heat capacity $C = 4.5\ R$ in the process $PT = constant$. Find the number of degrees of freedom (n) of molecules in the gas.

Options:
A. $$n = 4$$
B. $$n = 2$$
C. $$n = 3$$
D. $$n = 5$$
E. $$n = 7$$
F. $$n = 8$$
G. $$n = 6$$
H. $$n = 1$$
I. $$n = 9$$
Answer: D
The conductor ABCDE has the shape shown. It lies in the yz plane, with A and E on the y-axis. When it moves with a velocity $v$ in a magnetic field B, an emf $\varepsilon$ is induced between A and E.

Options:
A. $\varepsilon = 2Bav$, if $v$ is in the y-direction and B is in the x-direction.
B. \(\varepsilon = B \lambda v\), if \(v\) is in the y-direction and B is in the y-direction.
C. $\varepsilon = B \lambda v$, if $v$ is in the z-direction and B is in the x-direction.
D. $\varepsilon = B \lambda v$, if $v$ is in the x-direction and B is in the z-direction.
E. \(\varepsilon = B \lambda v\), if \(v\) is in the z-direction and B is in the y-direction.
F. \(\varepsilon = 2B \lambda v\), if \(v\) is in the x-direction and B is in the z-direction.
G. \(\varepsilon = 2B \lambda v\), if \(v\) is in the y-direction and B is in the z-direction.
H. \(\varepsilon = 2B \lambda v\), if \(v\) is in the z-direction and B is in the x-direction.
I. $\varepsilon = 0$, if $v$ is in the y-direction and B is in the x-direction.
J. \(\varepsilon = B \lambda v\), if \(v\) is in the y-direction and B is in the z-direction.
Answer: D
Daimeter of two limbs of a U-tube are $2 \,mm$ and $5 \,mm$, and surface tension is $70 \,dyne/cm$, if water density is $1000 \,kg/m2$ and $g = 10 \,m/s$ then water level difference in both the tubes will be

Options:
A. $$84 \,cm$$
B. $$8.4 \,cm$$
C. $$0.084 \,m$$
D. $$84 \,m$$
E. $$8.4 \,m$$
F. $$0.0084 \,m$$
G. $$0.84 \,cm$$
H. $$0.84 \,m$$
Answer: E
An electron moves with speed $2\times{10}^{5}m{s}^{-1}$ along the positive x-direction in the presence of a magnetic induction $\overline { B } =\hat { i } +4\hat { j } -3\hat { k } $ tesla. The magnitude of the force experienced by the electrons in newton is (charge on the elextron $=1.6\times{10}^{-19}C$)

Options:
A. $$1.24 \times 10^{-13}$$
B. $$1.12 \times 10^{-13}$$
C. $$1.38 \times 10^{-13}$$
D. $$1.22 \times 10^{-13}$$
E. $$1.02 \times 10^{-13}$$
F. $$1.18\times{10}^{-13}$$
G. $$1.72\times{10}^{-13}$$
H. $$1.44 \times 10^{-13}$$
I. $$1.28\times{10}^{-13}$$
J. $$1.6\times{10}^{-13}$$
Answer: F
The rate law for the reaction; $2H_{2(g)}+2NO_{(g)}\rightarrow N_{2(g)}+2H_2O_{(g)}$ is $\displaystyle\frac{d[N_2]}{dt}=k[H_2][NO]^2$. Which of the following mechanisms is consistent with the rate law?

Options:
A. $$H_2O_{(g)}+NO_{(g)}\overset{k_3}{\rightarrow} H_{2(g)}+NO_{2(g)}$$
B. $$H_{2(g)}+2NO_{(g)}\overset{k_4}{\rightarrow} N_{2(g)}+2H_2O_{(g)}+\displaystyle \frac{1}{3}O_{2(g)}$$
C. $$H_{2(g)}+N_2O_{(g)}\overset{k_2}{\rightarrow} N_{2(g)}+H_2O_{(g)}$$
D. $$ H_{2(g)}+NO_{(g)}\overset{k_1}{\rightarrow} N_2O_{(g)}+H_2O_{(g)} + \displaystyle \frac{1}{2}O_{2(g)}$$
E. $$H_{2(g)}+2NO_{(g)}\overset{k_1}{\rightarrow} N_{2(g)}+2H_2O_{(g)}+\displaystyle \frac{1}{2}O_{2(g)}$$
F. $$H_{2(g)}+2NO_{(g)}\overset{k_4}{\rightarrow} N_{2(g)}+2H_2O_{(g)}+\displaystyle \frac{1}{2}O_{2(g)}$$
G. $$H_{2(g)}+2NO_{(g)}\overset{k_4}{\rightarrow} N_{2(g)}+H_2O_{(g)}+\displaystyle \frac{1}{2}O_{2(g)}$$
H. $$H_{2(g)}+2NO_{(g)}\overset{k_1}{\rightarrow} N_2O_{(g)}+H_2O_{(g)}$$
I. $$H_{2(g)}+2NO_{(g)}\overset{k_1}{\rightarrow} N_{2(g)}+2H_2O_{(g)}+\displaystyle \frac{1}{4}O_{2(g)}$$
J. $$H_{2(g)}+2NO_{(g)}\overset{k_1}{\rightarrow} N_{2(g)}+H_2O_{(g)}+\displaystyle \frac{1}{2}O_{2(g)}$$
Answer: G
Two particles moving initially in the same direction undergo a one dimensional, elastic collision. Their relative velocities before and after collision are $\overrightarrow { { v }_{ 1 } } $ and $\overrightarrow { { v }_{ 2 } } $. Then :

Options:
A. $$\left| \overrightarrow { { v }_{ 1 } } . \overrightarrow { { v }_{ 2 } } \right| = 2 \left| \overrightarrow { { v }_{ 1 } } \right| \left| \overrightarrow { { v }_{ 2 } } \right|$$
B. $$\left| \overrightarrow { { v }_{ 1 } } . \overrightarrow { { v }_{ 2 } } \right| = \frac{1}{4} \left| \overrightarrow { { v }_{ 1 } } \right| \left| \overrightarrow { { v }_{ 2 } } \right|$$
C. $$\left| \overrightarrow { { v }_{ 1 } } . \overrightarrow { { v }_{ 2 } } \right| = \frac{3}{4} \left| \overrightarrow { { v }_{ 1 } } \right| \left| \overrightarrow { { v }_{ 2 } } \right|$$
D. $$\left| \overrightarrow { { v }_{ 1 } } . \overrightarrow { { v }_{ 2 } } \right| = \frac{1}{2} \left| \overrightarrow { { v }_{ 1 } } \right| \left| \overrightarrow { { v }_{ 2 } } \right|$$
E. $\overrightarrow { { v }_{ 1 } } =-\overrightarrow { { v }_{ 2 } } $ only if the two are of equal mass
F. $$\left| \overrightarrow { { v }_{ 1 } } . \overrightarrow { { v }_{ 2 } } \right| = \frac{1}{3} \left| \overrightarrow { { v }_{ 1 } } \right| \left| \overrightarrow { { v }_{ 2 } } \right|$$
G. $$\overrightarrow { { v }_{ 1 } } =\overrightarrow { { v }_{ 2 } }={ \left| \overrightarrow { { v }_{ 1 } } \right| }^{ 2 }$$
H. $$\left| \overrightarrow { { v }_{ 1 } } . \overrightarrow { { v }_{ 2 } } \right| ={ \left| \overrightarrow { { v }_{ 1 } } \right| }^{ 2 }$$ and $$\left| \overrightarrow { { v }_{ 1 } } \right| =\left| \overrightarrow { { v }_{ 2 } } \right| $$ and $$\overrightarrow { { v }_{ 1 } } =-\overrightarrow { { v }_{ 2 } }={ \left| \overrightarrow { { v }_{ 1 } } \right| }^{ 2 }$$
Answer: H
A glass capillary tube of internal radius r=0.25 mm is immersed in water. The top end of the tube projects by 2 cm above the surface of the water. At what angle does the liquid meet the tube? Surface tension of water=$0.07 N m^{-1}$ and its contact angle is 0.

Options:
A. $$sin^{-1} 5/13$$
B. $$cos^{-1} 5/14$$
C. $$sin^{-1} 6/14$$
D. $$cos^{-1} 6/14$$
E. $$sin^{-1} (0.5)$$
F. $$cos^{-1} (0.5))$$
G. $$cos^{-1} 4/13$$
H. $$sin^{-1} 5/14$$
I. $$cos^{-1} 7/14$$
J. $$cos^{-1} 4/15$$
Answer: B
A uniform rod of M =5 kg and length 8 m supported horizontally by a string of length 10 m and breaking tension 52 N. A ball of mass 1 kg starts rolling at t =0 from end A with constant speed of 50 cm/s towards end B. Find the time when the string breaks ${ g=10m/s }^{ 2 }$

Options:
A. $11.42$ seconds
B. $8.92$ seconds
C. $7.92$ seconds
D. $10.92$ seconds
E. $9.72$ seconds
F. $9.42$ seconds
G. $11.92$ seconds
H. $9.92$ seconds
Answer: D
A particle of specific charge (charge\mass) starts moving from origin under action of an electric field $\overrightarrow E  = {E_0}\widehat i$ and magnetic field  $\overrightarrow B  = {B_0}\widehat {k.}$ its velocity at $({X_0},{Y_0},0)\;is\;(4\widehat i - 3\widehat j)$ . The value of ${X_0}$ is

Options:
A. $$\frac{{5\alpha }}{{2{B_0}}}$$
B. $$\frac{{125}}{{4\alpha {E_0}}}$$
C. $$\frac{{25}}{{2\alpha {E_0}}}$$
D. $$\frac{{25\alpha {E_0}}}{{2{B_0}}}$$
E. $$\frac{{13\alpha {E_0}}}{{2\alpha {B_0}}}$$
F. $$\frac{{16\alpha {B_0}}}{{{E_0}}}$$
G. $$\frac{{25}}{{2\alpha {B_0}}}$$
H. $$\frac{{13\alpha {E_0}}}{{2{B_0}}}$$
I. $$\frac{{125}}{{2\alpha {E_0}}}$$
J. $$\frac{{125}}{{2\alpha {B_0}}}$$
Answer: C
At what angle should a body be projected with a velocity $24m{s}^{-1}$ just to pass over the obstacle $14m$ high at a distance of $24m$. (Take $g=10{ms}^{-2}$)

Options:
A. $$\tan{\theta}=1.5$$
B. $$\tan{\theta}=2.5$$
C. $$\tan{\theta}=1.1$$
D. $$\tan{\theta}=0.8$$
E. $$\tan{\theta}=3.2$$
F. $$\tan{\theta}=1.2$$
G. $$\tan{\theta}=1$$
H. $$\tan{\theta}=3.8$$
I. $$\tan{\theta}=2$$
J. $$\tan{\theta}=2.2$$
Answer: H
An ideal monoatomic gas undergoes a reversible process: $\cfrac{P}{V}=$ constant, from $(2bar,273K)$ to $4$ bar. The value of $\cfrac{\Delta U}{w}$ for this process is:

Options:
A. $$+3.0$$
B. $$-2.0$$
C. $$-1.0$$
D. $$-3.0$$
E. $$-1.5$$
F. $$-2.5$$
G. $$+1.5$$
Answer: D
A black body (heat capacity = C) at absolute temperature T is kept in a surrounding of absolute temperature T/2. Its rate of cooling is :($\sigma $ is stefan's constant)

Options:
A. $$\dfrac{15 \sigma A T^{4}}{C 16}$$
B. $$\dfrac{\sigma A}{C} \left( \dfrac{3T}{4} \right)^{4}$$
C. $$\dfrac{15 \sigma A T^{4}}{C 2}$$
D. $$\dfrac{15\sigma A }{C}\left ( \dfrac{T}{4} \right )^{4}$$
E. $$\dfrac{\sigma A}{C}\left ( \dfrac{T}{2} \right )^{4}$$
F. $$\dfrac{\sigma A}{C} \left( \dfrac{T}{3} \right)^{4}$$
G. $$\dfrac{\sigma A}{C} \left( \dfrac{T}{4} \right)^{4}$$
H. $$\dfrac{15\sigma A }{C}\left ( \dfrac{T}{2} \right )^{4}$$
I. $$\dfrac{\sigma AT^{4}}{C 2}$$
J. $$\dfrac{15\sigma A }{C}\left ( \dfrac{T}{3} \right )^{4}$$
Answer: H
The activation energies of two reactions are ${E}_{1}$ and ${E}_{2}$ with ${E}_{1}> {E}_{2}$. If the temperature of the reacting system is increased from ${T}_{1}$ (rate constants are ${K}_{1}$ and ${K}_{2}$) to ${T}_{2}$(rate constants are ${K}_{1}'$ and ${K}_{2}'$) predict which of the following alternative is correct?

Options:
A. $$\cfrac { { K }_{ 1 }' }{ { K }_{ 1 } } <\cfrac { { K }_{ 2 }' }{ { K }_{ 2 } } $$
B. $$\cfrac { { K }_{ 1 }' }{ { K }_{ 1 } } =\cfrac { { K }_{ 2 }' }{ { K }_{ 2 } } $$
C. \cfrac { { K }_{ 1 }' }{ { K }_{ 1 } } >\cfrac { { K }_{ 2 } }{ { K }_{ 2 }' }
D. $$\cfrac { { K }_{ 1 }' }{ { K }_{ 1 } } >\cfrac { { K }_{ 2 }' }{ { K }_{ 2 } }$$
E. $$\cfrac { { K }_{ 1 } }{ { K }_{ 1 }' } = \cfrac { { K }_{ 2 } }{ { K }_{ 2 }' }$$
F. $$\cfrac { { K }_{ 1 } }{ { K }_{ 1 }' } >\cfrac { { K }_{ 2 } }{ { K }_{ 2 } '} $$
G. $$\cfrac { { K }_{ 1 } }{ { K }_{ 1 }' } <\cfrac { { K }_{ 2 } }{ { K }_{ 2 } '} $$
H. $$\cfrac { { K }_{ 1 }' }{ { K }_{ 2 } } <\cfrac { { K }_{ 1 } }{ { K }_{ 2 }' }$$
I. $$\cfrac { { K }_{ 1 } }{ { K }_{ 1 }' } = \cfrac { { K }_{ 2 } }{ { K }_{ 2 } }$$
Answer: F
Two identical adiabatic vessels are filled with oxygen at pressure ${ P }_{ 1 }$ and ${ P }_{ 2 }$ $({ P }_{ 1 }>{ P }_{ 2 })$ . the vessels are inter connected with each other by a non-conducting pipe. If $V{ 0 }_{ 1 }$ and $V{ 0 }_{ 2 }$ denote initially internal energies of oxygen and $V{ f }_{ 1 }$ and $V{ f }_{ 2 }$ denote final internal energies in first and second vessel respectively, then

Options:
A. $$\frac { V{ o }_{ 1 } }{ V{ o }_{ 2 } } =\frac { { P }_{ 1 } }{ { P }_{ 2 } } ,{ Vf }_{ 1 }<{ Vf }_{ 2 }$$
B. \(\frac{V{ o }_{ 1 }}{V{ o }_{ 2 }} = \frac{P_{2}}{P_{1}} , V{ f }_{ 1 } < V{ f }_{ 2 }\)
C. \(\frac{V{ o }_{ 1 }}{V{ o }_{ 2 }} = \frac{P_{1}}{P_{2}} , V{ f }_{ 1 } = V{ f }_{ 2 }\)
D. \(\frac{V{ o }_{ 1 }}{V{ o }_{ 2 }} = \frac{P_{2}}{P_{1}} , V{ f }_{ 1 } = V{ f }_{ 2 }\)
E. \(\frac{V{ o }_{ 1 }}{V{ o }_{ 2 }} = \frac{P_{1}}{P_{2}} , V{ f }_{ 1 } > V{ f }_{ 2 }\)
F. \(\frac{V{ o }_{ 1 }}{V{ o }_{ 2 }} = \frac{P_{2}}{P_{1}} , V{ f }_{ 1 } > V{ f }_{ 2 }\)
G. $$\frac { V{ o }_{ 1 } }{ V{ o }_{ 2 } } =\frac { { P }_{ 2 } }{ { P }_{ 1 } } ,{ Vf }_{ 1 }<{ Vf }_{ 2 }$$
H. $$\frac { V{ o }_{ 1 } }{ V{ o }_{ 2 } } =\frac { { P }_{ 2 } }{ { P }_{ 1 } } ,{ Vf }_{ 1 }>{ Vf }_{ 2 }$$
I. \(\frac{V{ o }_{ 1 }}{V{ o }_{ 2 }} = \frac{P_{1}}{P_{2}} , V{ f }_{ 1 } < V{ f }_{ 2 }\)
J. $$\frac { V{ o }_{ 1 } }{ V{ o }_{ 2 } } =\frac { { P }_{ 1 } }{ { P }_{ 2 } } ,{ Vf }_{ 1 }>{ Vf }_{ 2 }$$
Answer: H
The electric potential in a certain region of space is given by $V=3x^{2}+2x$, where $x$ is in meter and $V$ is in volt, in this region the equipotential surface are

Options:
A. Plane parallel to the YZ plane
B. Concentric sphere concentrated at y-axis.
C. Plane parallel to $XZ$ plane
D. Concentric sphere concentrated at origin.
E. Concentric cylinder with axis as $x-$axis
F. Plane parallel to the XZ plane
G. Concentric cylinder with axis as y-axis
H. Concentric sphere concentrated at z-axis.
I. Plane parallel to $XY$ plane
J. Plane parallel to $YZ$ plane
Answer: D
As per Hardy-Schulze formulation, the flocculation value of the following for ferric hydroxide sol are in the order :

Options:
A. K$_3$[Fe(CN)$_6$] $<$ K$_2$CrO$_4$ $<$ KBr $<$ $KNO_3$ $<$ $AlCl_3$
B. AlCl$_3 >$ K$_3$[Fe(CN)$_6$] $>$ K$_2$CrO$_4 >$ KBr$=$KNO$_3$
C. K$_3$[Fe(CN)$_6$] $<$ K$_2$CrO$_4 $<$ KBr $=$ KNO$_3 $>$ AlCl$_3
D. K$_3$[Fe(CN)$_6$] $<$ K$_2$CrO$_4$ $<$ AlCl$_3$ $<$ KBr $<$ KNO$_3$
E. K$_3$[Fe(CN)$_6$] $<$ K$_2$CrO$_4 <$ KBr $ = \  $KNO$_3 = \ $AlCl$_3$
F. K$_3$[Fe(CN)$_6$] $<$ AlCl$_3 $<$ K$_2$CrO$_4 $<$ KBr $<$ KNO$_3
G. K$_3$[Fe(CN)$_6$] $>$ AlCl$_3$ $>$ K$_2$CrO$_4$ $>$ KBr $>$ KNO$_3$
H. K$_3$[Fe(CN)$_6$] $<$ K$_2$CrO$_4$ $<$ AlCl$_3$ $<$ KBr $=$ KNO$_3$
Answer: A
A minute spherical air bubble is rising slowly through a column of mercury contained in a deep jar. If the radius of the bubble at a depth of 100 cm is 0.1 mm, calculate its depth where its radius is 0.126 mm given that the surface tension of mercury is 567 dyne/cm. Assume that the atmospheric pressure is 76 cm of mercury

Options:
A. $$\displaystyle h_{2}=10.48\: cm$$
B. $$\displaystyle h_{2}=19.48\: cm$$
C. $$\displaystyle h_{2}=15.48\: cm$$
D. $$\displaystyle h_{2}=9.48\: cm$$
E. $$\displaystyle h_{2}=18.48\: cm$$
F. $$h_{2}=17.48\: cm$$
G. $$h_{2}=9.48\: m$$
H. $$\displaystyle h_{2}=9.48\: m$$
I. $$\displaystyle h_{2}=16.48\: cm$$
J. $$h_{2}=18.48\: cm$$
Answer: D
The concentration of \( N_2O_5 \) in liquid bromine varied with time as follows:
| \( t \) (s) | 0 | 100 | 300 | 500 |
|-------------|--------|-------|-------|-------|
| \([N_2O_5]\) (M) | 0.12 | 0.088 | 0.049 | 0.026 |
What will be the rate constant and order of the reaction?

Options:
A. $k=2.06 \times 10^{-3}s^{-1}$, order$=0th$
B. $k=3.06\times 10^{-3}s^{-1}$, zero
C. $k=2.06 \times 10^{-3}s^{-1}$, order$=2nd$
D. $k=4\times 10^{-3}s^{-1}$, zero
E. $k=2\times 10^{-3}s^{-1}$, order$=1st$
F. None of these
G. $k=2.06 \times 10^{-3}s^{-1}$, order$=1st$
H. $k=3.06 \times 10^{-3}s^{-1}$, order$=1st$
Answer: F
A coil of $50$ turns and $10cm$ diameter has resistance of $10\, ohm$. What must be potential difference across the coil so as to nullify the earth's magnetic field $B=0.314G$ at the centre of the coil.

Options:
A. 1.75 volt
B. $1.5$ volt
C. 1.25 volt
D. $2.5$ volt
E. $1.0$ volt
F. 0.9 volt
G. $0.5$ volt
H. 1.1 volt
I. 0.75 volt
J. 1.2 volt
Answer: E
Ice at $0^oC$ is added to 200g of water initially at $70^oC$ in a vacuum flask. When 50g of ice has been added and has all melted the temperature of the flask and contents is $40^oC$. When a further 80 g of ice has been added and has all melted, the temperature of the whole is $10^oC$. Calculate the specific latent heat of fusion of ice. [Take $S_w=1 cal/gm^oC$]

Options:
A. $$3.0 \times 10^4 J/kg$$
B. $$2.4 \times 10^5 J/kg$$
C. $$3.5 \times 10^4 J/kg$$
D. $$9.0 \times 10^4 J/kg$$
E. $$1.2 \times 10^4 J/kg$$
F. $$3.2 \times 10^4 J/kg$$
G. $$3.4 \times 10^4 J/kg$$
H. $$3.3 \times 10^4 J/kg$$
I. $$4.2 \times 10^4 J/kg$$
J. $$3.6 \times 10^4 J/kg$$
Answer: D
The energy released by the fission of 1g of $^{ 235 }{U}$ in joule, given that the energy released per fission is 200 MeV. 
(Avogadro's number = 6.023$\times { 10 }^{ 23 }$)

Options:
A. $$8.202 \times 10^{13}$$
B. $$8.202\times { 10 }^{ 12 }$$
C. $$8.202\times { 10 }^{ 14 }$$
D. $$8.202\times { 10 }^{ 8 }$$
E. $$8.202\times { 10 }^{ 10 }$$
F. $$8.202\times { 10 }^{ 11 }$$
G. $$8.202 \times { 10 }^{ 13 }$$
H. $$8.202\times { 10 }^{ 15 }$$
I. $$8.202 \times 10^{11}$$
Answer: E
The electric potential at a certain distance from a point charge $Q$ is $810 V$ and electric field is $300 N/C$. The minimum speed with which a particle of charge $-Q$ and mass $m = 6 \times 10^{-16} kg$ should be projected from that point so that it moves into the field free region is:

Options:
A. $$9 \times 10^{3} m/s$$
B. $$9 \times 10^{5} m/s$$
C. $$81 \times 10^{3} m/s$$
D. $$81 \times 10^{5} m/s$$
E. $$9 \times 10^{4} m/s$$
F. $$81 \times 10^{6}m/s$$
G. $$81 \times 10^{4} m/s$$
H. $$9 \times 10^{6}m/s$$
Answer: G
The equation of state for one mole of a gas is $PV=RT+BP$, where $B$ is a constant, independent of temperature. The internal energy of fixed amount of gas is function of temperature only. If one mole of the above gas is isothermally expanded from $12L$ to $22L$ at a constant external pressure of $1$ bar at $400K$, then the change in enthalpy of the gas is approximate $(B=2L/mol)$

Options:
A. $$0$$
B. $$-664J$$
C. $$-132J$$
D. $$-332J$$
E. $$-66.4J$$
F. $$-133J$$
G. $$-33.2J$$
H. $$-3.32J$$
I. $$-166J$$
Answer: A
The region between two concentric spheres of radii $'a'$ and $'b'$ , respectively( see figure), has volume charge density $\rho = \dfrac {A}{r}$, where $A$ is a constant and $r$ is the distance from the centre. At the center of the spheres is a point charge $Q$. The value of $A$ such that the electric field in the region between the spheres will be constant is:

Options:
A. \dfrac {Q}{2\pi (a^{2} - b^{2})}
B. \dfrac {Q}{4\pi a^{2}}
C. $$\dfrac {2Q}{\pi a^{2}}$$
D. \dfrac {Q}{2\pi (a^{2} + b^{2})}
E. \dfrac {Q}{2\pi b^{2}}
F. $$\dfrac {Q}{2\pi (b^{2} - a^{2})}$$
G. \dfrac {2Q}{\pi b^{2}}
H. \dfrac {Q}{4\pi (b^{2} - a^{2})}
I. $$\dfrac {2Q}{\pi (a^{2} - b^{2})}$$
J. $$\dfrac {Q}{2\pi a^{2}}$$
Answer: J
$2$g of brass containing $Cu$ and $Zn$ only reacts with $3$M $HNO_3$ solution. Following are the reactions taking place.$Cu(s)+HNO_3(aq)\rightarrow Cu^{2+}(aq)+NO_2(g)+H_2O(l)$$Zn(s)+H^+(aq)+{NO}^-_3(aq)\rightarrow NH^+_4(aq)+Zn^{2+}(aq)+H_2O(l)$The liberated $NO_2(g)$ was found to be $1.04$L at $25^o$C and $1$ atm $[Cu =63.5, Zn=65.4]$.How many grams of $NH_4NO_3$ will be obtained in the reaction?

Options:
A. $0.405$g
B. $0.0428$g
C. 0.1852g
D. 0.1894g
E. 0.1985g
F. 0.1518g
G. 0.2058g
H. 0.1952g
I. $0.358$g
J. $0.2018$g
Answer: J
Four point charges q, q, q and -3q are placed at the vertices of a regular tetrahedron of side L. The work done by electric force in taking all the charges to the centre of the tetrahedron is (where $k=\frac {1}{4\pi \epsilon_0})$ :

Options:
A. $$\dfrac {-9kq^2}{L}$$
B. $$\dfrac {-3kq^2}{L}$$
C. $$\dfrac {3kq^2}{L}$$
D. $$\dfrac {12kq^2}{L}$$
E. $$\dfrac {6kq^2}{L}$$
F. $$\dfrac {-15kq^2}{L}$$
G. $$\dfrac {-12kq^2}{L}$$
H. $$\dfrac {-18kq^2}{L}$$
I. $$\dfrac {9kq^2}{L}$$
J. $$\dfrac {-6kq^2}{L}$$
Answer: J
On dissolving $10.8\ g$ of glucose $(m.w.=180)$ in $240\ g$ of water, its boilig point increases by $0.13^{o}C$. Then the molal elevation constant of water is

Options:
A. 104.11
B. 222.1
C. 111.0
D. 104.22
E. 100
F. 222.2
G. 111.2
H. 104.23
I. 104.1
J. 111.1
Answer: J
$$AgNO_3 \overset{\Delta }{\rightarrow} (W) + (X) + O_2$$
$$(X) + H_2O \rightarrow HNO_2 + HNO_3$$
$$(W) + HNO_3 \rightarrow Y + NO + H_2O$$
$$(Y) + Na_2S_2O_3 (excess) \rightarrow (z) + NaNO_3$$
Identify (W) to (Z).

Options:
A. $$W = Ag_2 X = N_2O_2 Y = AgNO_3 Z = Na_3[Ag(S_2O_3)_2]$$
B. $$W = Ag X = N_2O_3 Y = AgNO_3 Z = Na_2[Ag(S_2O_3)_2]$$
C. $$W = Ag X = N_2O Y = AgNO_3 Z = Na_3[Ag(S_2O_3)_2]$$
D. $$W = Ag X = N_2O Y = AgNO_3 Z = Na_2[Ag(S_2O_3)_2]$$
E. $$W = Ag_2O X = N_2O Y = AgNO_3 Z = Na_3[Ag(S_2O_3)_2]$$
F. $$W = Ag_2O X = N_NO Y = AgNO_3 Z = Na_3[Ag(S_2O_3)_2]$$
G. $$W = Ag X = N_2O_4 Y = AgNO_3 Z = Na_3[Ag(S_2O_3)_2]$$
H. $$W = Ag X = N_NO_2 Y = AgNO_3 Z = Na_3[Ag(S_2O_3)_2]$$
I. $$W = Ag X = N_2O_2 Y = AgNO_3 Z = Na_2[Ag(S_2O_3)_2]$$
J. $$W = Ag_2O X = N_2 Y = AgNO_3 Z = Na[Ag(S_2O_3)_2]$$
Answer: H
The displacement x of a particle in motion is given in terms of time by $x (x - 4) = 1 - 5 \cos \omega t$

Options:
A. motion of the particle is neither oscillator; nor simple harmonic
B. The particle executes oscillatory motion which is not SHM
C. The particle executes oscillatory motion which is SHM but with a different amplitude and phase
D. The particle executes oscillatory motion which is SHM but with a different phase
E. The particle is not acted upon by a force when it is at x = 4
F. The particle executes oscillatory motion which is SHM but with a different amplitude
G. The particle executes oscillatory motion which is SHM but with a different frequency
H. The particle executes oscillatory motion which is SHM but with a different phase and frequency
I. The particle executes SHM
J. The particle executes oscillatory motion which is SHM but with a different period
Answer: I
The moment of inertia of thin square plate ABCD of uniform thickness about an axis passing through the center O and perpendicular to the plane is $I$.which of the following is false?

Options:
A. $$I=I_1+I_2+I_3$$
B. $$I=I_2+I_3+I_4$$
C. $$I=I_2+I_3$$
D. $$I=I_1+I_2$$
E. $$I=I_1+I_2+I_3+I_4$$
F. $$I=I_1+I_3+I_2$$
G. $$I=I_1+I_4$$
H. $$I=I_1+I_3+I_4$$
I. $$I=I_4+I_2$$
J. $$I=I_1+I_3$$
Answer: D
Two moles of a liquid $A(p_A^\circ =100 torr)$ and 3 moles of liquid $B(p_B^{\circ}=150 torr)$ form a solution having vapour pressure of 120 torr. Based upon this observation one acn conclude?

Options:
A. Interaction between like molecules < those between unlike molecules
B. Interactions between like molecules > those between unlike molecules
C. Interaction between like molecules = those between unlike molecules
D. Interactions between like molecules ≤ those between unlike molecules
E. Interaction between like molecules ≈ those between unlike molecules
Answer: B
Two spheres of volume $250\ cc$ each but of relative densities $0.8$ and $1.2$ are connected by a string and the combination is immersed in a liquid. Find the tension in the string is:
$$(g=10\ m/s^{2})$$

Options:
A. $$2\ N$$
B. $$1\ N$$
C. 6.0\ N
D. $$0.5\ N$$
E. 4.5\ N
F. 3.5\ N
G. 3.0\ N
H. 2.5\ N
I. $$1.5\ N$$
J. 4.0\ N
Answer: D
2 moles of a diatomic gas undergoes the process $PT^2/V$ = constant. Then the molar heat capacity of the gas during the process will be equal to

Options:
A. $$2R$$
B. $$\dfrac{11R}{2}$$
C. $$6R$$
D. $$R$$
E. $$\dfrac{9R}{2}$$
F. $$3R$$
G. $$\dfrac{13R}{2}$$
H. $$\dfrac{7R}{2}$$
I. $$\dfrac{5R}{2}$$
J. $$4R$$
Answer: J
Electric field and magnetic field in a region of space is given by $\vec{E}={E}_{o}\hat{j}$ and $\vec{B}={B}_{o}\hat{j}$.A particle of specific charge $\alpha$ is released from origin with velocity $\vec{v}={v}_{o}\hat{i}$.Then path of particle. Note- ${E}_{o},{B}_{o}$ and ${v}_{o}$ are constant values.

Options:
A. is a helix with uniform pitch
B. is a helix with constant pitch
C. is a circle
D. is a helix with varying radius
E. is cycloid
F. is a helix with non-constant pitch
G. is a parabola
H. is a helix with varying pitch
I. is a helix with constant radius
J. is a helix with non-uniform pitch
Answer: J
A particle of mass $m$ is projected upwards with velocity $v = \frac { v _ { e } } { 2 } \left( V _ { e } =\right.$ escape velocity \right$)$ . At highest point the potential energy of the particle is

Options:
A. $$- \frac { \mathrm { GMm } } { 7 \mathrm { R } }$$
B. $$-\frac { 5 \mathrm { GMm } } { 8 \mathrm { R } }$$
C. $$- \frac { \mathrm { GMm } } { 2 \mathrm { R } }$$
D. $$- \frac { 3 \mathrm { GMm } } { 4 \mathrm { R } }$$
E. $$- \frac { \mathrm { GMm } } { 6 \mathrm { R } }$$
F. $$- \frac { 2 \mathrm { GMm } } { 3 \mathrm { R } }$$
G. $$- \frac { \mathrm { GMm } } { 4 \mathrm { R } }$$
H. $$-\frac { \mathrm { GMm } } { 8 \mathrm { R } }$$
I. $$- \frac { 3 \mathrm { GMm } } { 2 \mathrm { R } }$$
J. $$-\frac { \mathrm { GMm } } { 3 \mathrm { R } }$$
Answer: D
A 0.01 M weak acid \( HA \) is 1% ionized. Calculate the concentration of hydroxide ions (\([OH^-]\)) in the solution, given \( K_w(H_2O) = 10^{-14} \).

Options:
A. $$10^{-6}\ M$$
B. $$10^{-11}\ M$$
C. $$10^{-10}\ M$$
D. $$10^{-89}\ M$$
E. $$10^{-12}\ M$$
F. $$10^{-8}\ M$$
G. $$10^{-7}\ M$$
H. $$10^{-9}\ M$$
I. $$10^{-14}\ M$$
J. $$10^{-13}\ M$$
Answer: F
The order of stability of the following tautomeric compounds is:$CH_2=\overset{OH}{\overset{|}{C}-}CH_2-\overset{O}{\overset{||}{C}}-CH_3 \rightleftharpoons CH_3-\overset{O}{\overset{||}{C}}-CH_2-\overset{O}{\overset{||}{C}}-CH_3 \rightleftharpoons CH_3-\overset{OH}{\overset{|}{C}=}CH-\overset{O}{\overset{||}{C}}-CH_3$             \n          $ I$                                                   $II$                                                      $III$

Options:
A. $$III> I> II$$
B. $$III> II> I$$
C. $$II> I> III$$
D. $$I> II> III$$
E. $$I> III> II$$
F. $$II> III> I$$
Answer: B
A current 2.0 A is passed for 5 hours through a moltal salt deposits 22.2 g of metal(At wt=177).The oxidation state of the metal in the metal salt is:

Options:
A. +4
B. +3
C. +7
D. +5
E. +2
F. +9
G. +1
H. +6
I. +10
J. +8
Answer: A
The equilibrium constant for the reaction, $A\rightarrow B$ was measured over a range of temperature and can be represented by the equation, $ln K_{p} = 4.814 - (2059/ T)$
Then $\triangle G^{\circ}, \triangle H^{\circ}, \triangle S^{\circ}$ for the reaction at $298\ K$ will be

Options:
A. \(\triangle G^{\circ} = 5.72\ kJ, \triangle H^{\circ} = 17.98\ kJ, \triangle S^{\circ} = 44\ Jk^{-1} mole^{-1}\)
B. $$\triangle G^{\circ} = 6.25\ kJ, \triangle H^{\circ} = 18.29\ kJ, \triangle S^{\circ} = 45\ Jk^{-1} mole^{-1}$$
C. \(\triangle G^{\circ} = 6.25\ kJ, \triangle H^{\circ} = 18.29\ kJ, \triangle S^{\circ} = 48\ Jk^{-1} mole^{-1}\)
D. $$\triangle G^{\circ} = 8.729\ kJ, \triangle H^{\circ} = 20.22\ kJ, \triangle S^{\circ} = 51.23\ Jk^{-1} mole^{-1}$$
E. $$\triangle G^{\circ} = 5.185\ kJ, \triangle H^{\circ} = 17.11\ kJ, \triangle S^{\circ} = 450\ Jk^{-1} mole^{-1}$$
F. \(\triangle G^{\circ} = 6.50\ kJ, \triangle H^{\circ} = 18.54\ kJ, \triangle S^{\circ} = 47\ Jk^{-1} mole^{-1}\)
G. \(\triangle G^{\circ} = 7.12\ kJ, \triangle H^{\circ} = 18.65\ kJ, \triangle S^{\circ} = 46.5\ Jk^{-1} mole^{-1}\)
H. \(\triangle G^{\circ} = 6.25\ kJ, \triangle H^{\circ} = 17.85\ kJ, \triangle S^{\circ} = 44.5\ Jk^{-1} mole^{-1}\)
I. \(\triangle G^{\circ} = 7.05\ kJ, \triangle H^{\circ} = 18.94\ kJ, \triangle S^{\circ} = 46\ Jk^{-1} mole^{-1}\)
J. $$\triangle G^{\circ} = 7.323\ kJ, \triangle H^{\circ} = 19.37\ kJ, \triangle S^{\circ} = 48\ Jk^{-1} mole^{-1}$$
Answer: B
The temperature of a diatomic gas is T. The total kinetic energy of the gas is given as $E = { K }_{ 1 }$ where ${ K }_{ 1 }$ is constant. Find out the total number of molecules of the gas in the sample.  (K = Boltzaman's constant)

Options:
A. $$3K_{ 1 }/4K$$
B. $$4K_{ 1 }K$$
C. None of these.
D. $$5K_{ 1 }/4K$$
E. $$3K_{ 1 }/2K$$
F. $$3K_{ 1 }K$$
G. $$2K_{ 1 }/4K$$
H. $$4K_{ 1 }/2K$$
I. $$5K_{ 1 }/2K$$
J. $$2K_{ 1 }/5K$$
Answer: F
A voltmeter and an ammeter are connected in series across an ideal cell. When certain resistance is connected in parallel to voltmeter, reading of ammeter increases three times and reading of voltmeter reduces to one third. The ratio of resistance of voltmeter to that of ammeter is

Options:
A. 3
B. 1/3
C. 1/9
D. 1/6
E. 1
F. 9
G. 6
H. 4
I. 2
J. 5
Answer: A
$A$: tetracyanomethane  $B$: carbon dioxide$C$: benzene   $D$: 1,3-buta-di-eneRatio of $\sigma$ and $\pi$ bonds is in order:

Options:
A. $$A = B < D = C$$
B. $$A = B < C < D$$
C. $$A = B < C = D$$
D. $$C < D < A < B$$
E. $$A < B < C < D$$
F. $$A = B = C = D$$
G. $$A = B < D < C$$
Answer: B
Find the solubility of $Ag_2C_ro_4$ $\left( {ksp = 1.1 \times {{10}^{ - 12}}} \right)$ in a $sol^n$ having $0.1$m $AgNO_3$

Options:
A. $${1.1 \times 10^{-16}}$$
B. $${1.1 \times 10^{-8}}$$
C. $${1.1 \times 10^{-14}}$$
D. $${1.1 \times {{10}^{ - 9}}}$$
E. $${1.1 \times {{10}^{ - 11}}}$$
F. $${1.1 \times 10^{-13}}$$
G. $${1.1 \times 10^{-12}}$$
H. $${1.1 \times {{10}^{ - 10}}}$$
I. $${1.1 \times 10^{-15}}$$
J. $${1.1 \times {{10}^{ - 12}}}$$
Answer: J
The trajectory of a projectile near the surface of the earth is given as $y=2x-9x^2$. If it were launched at an angle $\theta_0$ with speed $v_0$ then $(g = 10ms^{-2})$:

Options:
A. $$\theta_0 = \cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right) and v_0 = \dfrac{4}{3}ms^{-1}$$
B. $$\theta_0 = \cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right) and v_0 = \dfrac{2}{3}ms^{-1}$$
C. $\theta_0 = \cos^{-1}\left(\dfrac{1}{\sqrt{5}}\right)$ and $v_0 = \dfrac{5}{3}ms^{-1}$
D. $$\theta_0 = \cos^{-1}\left(\dfrac{1}{\sqrt{5}}\right) and v_0 = \dfrac{2}{3}ms^{-1}$$
E. $$\theta_0 = \sin^{-1}\left(\dfrac{1}{\sqrt{5}}\right) and v_0 = \dfrac{4}{3}ms^{-1}$$
F. $\theta_0 = \sin^{-1}\left(\dfrac{2}{\sqrt{5}}\right)$ and $v_0 = \dfrac{5}{3}ms^{-1}$
G. $$\theta_0 = \cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right) and v_0 = \dfrac{5}{3}ms^{-1}$$
H. $$\theta_0 = \sin^{-1}\left(\dfrac{2}{\sqrt{5}}\right) and v_0 = \dfrac{4}{3}ms^{-1}$$
I. $\theta_0 = \sin^{-1}\left(\dfrac{1}{\sqrt{5}}\right)$ and $v_0 = \dfrac{5}{3}ms^{-1}$
Answer: C
An ideal gas with the adiabatic exponent $\gamma$ goes through a process: $P={ P }_{ 0 }-\alpha V$, where ${P}_{0}$ and $\alpha$ are positive constants and $V$ is the volume. At what volume will the gas entropy have the maximum value?

Options:
A. $$\cfrac { \gamma { P }_{ 0 } }{ >>>>m<<pha \left( \gamma -2 \right) }$$
B. $$\cfrac { \gamma { P }_{ 0 } }{ 2\alpha \left( \gamma +1 \right) }$$
C. $$\cfrac { \gamma { P }_{ 0 } }{ \alpha \left( \gamma +1 \right) } $$
D. $$\cfrac { \gamma { P }_{ 0 } }{ \alpha \left( \gamma -1 \right) } $$
E. $$\cfrac { \gamma { P }_{ 0 } }{ 2\alpha \left( \gamma -1 \right) }$$
F. $$\cfrac {\alpha { P }_{ 0 } }{ \left( \gamma +1 \right) } $$
G. $$\cfrac { \gamma { P }_{ 0 } }{ 3\alpha \left( \gamma -1 \right) }$$
H. $$\cfrac { \gamma { P }_{ 0 } }{ 4\alpha \left( \gamma -1 \right) }$$
I. $$\cfrac { \gamma { P }_{ 0 } }{ 3\alpha \left( \gamma +1 \right) }$$
J. $$\cfrac {\alpha { P }_{ 0 } }{ \left( \gamma -1 \right) } $$
Answer: C
There is 50% dimer formation of benzoic acid $(C_6H_5COOH)$ in benzene solution.$2C_6H_5COOH\rightleftharpoons (C_6H_5COOH)_2$,Hence the abnormal molecular weight of benzoic acid ( theoretical value = 122 g $mol^{-1}$) is:

Options:
A. 81 g $mol^{-1}$
B. 205 g $mol^{-1}$
C. 204 g $mol^{-1}$
D. 61 g $mol^{-1}$
E. 203 g $mol^{-1}$
F. 162 g $mol^{-1}$
G. 122 g $mol^{-1}$
H. 163 g $mol^{-1}$
I. 184 g $mol^{-1}$
J. 244 g $mol^{-1}$
Answer: H
In the reaction $A+3B\rightleftharpoons 2C$, initially $10$ moles each of $A$ and $B$ were taken. If equilibrium pressure is $8$ atm and $\propto =20$% then $K_{P}$ will be :

Options:
A. $$\dfrac {1}{4}$$
B. $$\dfrac {1}{8}$$
C. $$\dfrac {1}{16}$$
D. $$\dfrac{1}{256}$$
E. $$\dfrac {1}{32}$$
F. $$\dfrac {1}{64}$$
G. $$\dfrac {1}{2}$$
H. $$\dfrac{1}{128}$$
Answer: G
A cuboid \( ABCDEFGH \) has vertices \( A(0, 0, 0) \), \( B(0, a, 0) \), \( C(a, a, 0) \), \( D(a, 0, 0) \), \( E(0, 0, b) \), \( F(a, 0, b) \), \( G(a, a, b) \), and \( H(0, a, b) \), is anisotropic with linear expansion coefficients \( \alpha_x = 1 \times 10^{-5}/^\circ \mathrm{C} \), \( \alpha_y = 2 \times 10^{-5}/^\circ \mathrm{C} \), and \( \alpha_z = 3 \times 10^{-5}/^\circ \mathrm{C} \); determine the coefficients of superficial expansion for faces \( ABCD \) (\( xy \)-plane), \( ADEH \) (\( xz \)-plane), and \( BCFG \) (\( yz \)-plane).

Options:
A. $$\beta _{CDEH}=4 \times 10^{-5}/ ^\circ C$$
B. $$\beta _{BCGH}=4 \times 10^{-5}/ ^\circ C$$
C. $$\beta _{CDEH}=6 \times 10^{-5}/ ^\circ C$$
D. $$\beta _{CDEH}=2 \times 10^{-5}/ ^\circ C$$
E. $$\beta _{CDEH}=3 \times 10^{-5}/ ^\circ C$$
F. $$\beta _{ABCD}=5 \times 10^{-5}/ ^\circ C$$
G. $$\beta _{EFGH}=2 \times 10^{-5}/ ^\circ C$$
H. $$\beta _{CDEH}=5 \times 10^{-5}/ ^\circ C$$
Answer: E