MikaStars39/nano-eval huggingface dataset:
Two communicating vessel's contain mercury. The diameter of one vessel is n times larger than the diameter of the other. A column of water of height h is poured into the left vessel. The mercury level will rise in the right-hand vessel(s= relative density of mercury and p=density of water) by

Options:
A. $$\frac { { n }^{ 2 }h }{ { (n }^{ 2 }-1)^{ 2 }s }$$
B. $$\frac { { n }^{ 2 }h }{ { (n }^{ 2 }+1)^{ 3 }s }$$
C. $$\frac { n^{ 2 }h }{ (n^{ 2 }-1)(n^{ 2 }+1)s }$$
D. $$\frac { n^{ 2 }h }{ (n^{ 2 }+1)s }$$
E. $$\frac { { n }^{ 2 }h }{ { (n }^{ 2 }+1)^{ 2 }s } $$
F. $$\frac { h }{ { n }^{ 2 }s } $$
G. $$\frac { h }{ { (n }+1)^{ 2 }s } $$
H. $$\frac { h }{ { (n }^{ 2 }+1)s } $$
I. $$\frac { { n }^{ 2 }h }{ { (n }^{ 2 }+1)^{ 1/2 }s }$$
J. $$\frac{ h }{ (n^{ 2 }-1)s }$$
Answer: E
For 1 mole of an ideal monoatomic gas on moving from one state to other, the temperature is doubled hut pressure becomes $\sqrt { 2 }$  times. Then entropy change in the process will be $R = 2 \mathrm { Cal } / \mathrm { mol } - \mathrm { K } )$

Options:
A. $$3 R \ln 4$$
B. $$3 R \ln 3$$
C. $$2 \mathrm { R } \ln 4$$
D. $$2 \mathrm{R} \ln 2$$
E. $$R \ln 2$$
F. $$4 R \ln 2$$
G. $$2 R \ln 2$$
H. $$4 R \ln 4$$
I. $$\frac { R } { 2 } \ln 2$$
J. $$3 \mathrm { R } \ln 2$$
Answer: E
The compound $CH\equiv CCH_{2}CH_{2}CH=CHCH_{3}$ is subjected to oxidation with $CrO_{3}/CH_{3}COOH$. The products formed are:

Options:
A. $$CO_{2},\:HO_{2}CCH_{2}CH_{2}CH=CHCH_{3}$$
B. $$CO_{2},\:HCO_{2}H,\:CH_{3}CO_{2}H$$
C. $$CH\equiv CCH_{2}CH_{2}CO_{2}H,\:HCO_{2}H$$
D. $$CO_{2},\:HCO_{2}H,\:CH\equiv CCH_{2}CH_{3}$$
E. $$CO_{2},\:HO_{2}CCH_{2}CH_{2}CO_{2}H,\:CH_{3}CO_{2}H$$
F. $$HCO_{2}H,\:HO_{2}CCH_{2}CH_{2}CO_{2}H,\:CH_{3}CO_{2}H$$
G. $$HCO_{2}H,\:CH\equiv CCH_{2}CH_{2}CO_{2}H,\:CH_{3}CO_{2}H$$
H. $$CO_{2},\:HO_{2}CCH_{2}CH_{2}CH_{3}$$
I. $$CH\equiv CCH_{2}CH_{2}CO_{2}H,\:CH_{3}CO_{2}H$$
J. $$CO_{2},\:HCO_{2}H,\:CH\equiv CCH_{2}CH_{2}CO_{2}H$$
Answer: I
$$CuSo_{4},5H_{2}O(s)\rightleftharpoons CuSO_{4}(s)+5H_{2}O(g);K_{p}=10^{-10}$$
$10^{-2} moles of CuSO_{4},5H_{2}O(s)$ is taken in a 2.5 L container at $27^oC$ then at equilibrium(take $R=\frac { 1 }{ 12 } $ L -atm $mole^{-1}K^{-1}$

Options:
A. mole of $CuSO_{4}.5H_{2}O$ left in the container is $9.2 \times10^{-3}$
B. mole of $CuSO_{4}$ at equilibrium in the container is $2\times10^{-4}$
C. mole of $CuSO_{4}$ at equilibrium in the container is $10^{-3}$
D. mole of $CuSO_{4}.5H_{2}O$ left in the container is $9.4 \times10^{-3}$
E. mole of $CuSO_{4}.5H_{2}O$ left in the container is $8.5 \times10^{-3}$
F. mole of $CuSO_{4}.5H_{2}O$ left in the container is $8.9 \times10^{-3}$
G. mole of $CuSO_{4}.5H_{2}O$ left in the container is $9.5 \times10^{-3}$
H. mole of $CuSO_{4}.5H_{2}O$ left in the container is $9.8\times10^{-3}$
I. mole of $CuSO_{4}.5H_{2}O$ left in the container is $9.3 \times10^{-3}$
J. mole of $CuSO_{4}.5H_{2}O$ left in the container is $9\times10^{-3}$
Answer: H
Work performed when a point charge $2\times 10^{-8}$ is transformed from infinity to a point at a distance of $1\ cm$ from the surface of the ball with a radius of $1\ cm$ and a surface charge density $\sigma =10^{-8}C/cm^{2}$ is:

Options:
A. $$0.11\times 10^{-4}J$$
B. $$11\times 10^{-4}J$$
C. $$11 \times 10^{-5}J$$
D. $$1.13 \times 10^{-5}J$$
E. $$1.1 \times 10^{-4}J$$
F. $$0.1 \times 10^{-4}J$$
G. $$1.13 \times 10^{-4}J$$
H. $$1.13 \times 10^{-3}J$$
I. $$113\times 10^{-4}J$$
Answer: B
A man applying a force $F$ upon a stretched spring is stationary in a compartment moving with constant speed $v$. If the compartment covers a distance $L$ in some time $t$, then

Options:
A. The work done by friction acting on man with respect to ground is, (w) $= FL$
B. The work done by friction acting on man with respect to ground is, (w) $= \frac{FL}{2}$
C. The work done by friction acting on man with respect to ground is, (w) $= \frac{3FL}{4}$
D. The work done by friction acting on man with respect to ground is, (w) $= \frac{5FL}{4}$
E. The man acting with force F on spring does the work (w) $= FL$
F. The work done by friction acting on man with respect to ground is, (w) $= \frac{3FL}{2}$
G. The work done by friction acting on man with respect to ground is, (w) $= \frac{FL}{4}$
H. The total work done by man with respect to ground is, (w) $= FL$
I. The total work performed by man on the compartment with respect to ground is zero
J. The work done by friction acting on man with respect to ground is, (w) $= 2FL$
Answer: A
Refer to the Carnot cycle of an ideal gas shown in the figure Let  $W _ { a \rightarrow b } , W _ { b \rightarrow c } , W _ { c \rightarrow d }$  and  $W _ { d \rightarrow a }$  represent the work done by the system during the processes  $a \rightarrow b , b \rightarrow c, c \rightarrow d$  and  $d \rightarrow a$  respectively. Consider the following relations
(1) $W _ { a \rightarrow b } + W _ { b \rightarrow c } + W _ { c \rightarrow d } + W _ { d \rightarrow a } > 0$
(2) $W _ { a \rightarrow b } + W _ { b \rightarrow c } + W _ { c \rightarrow d } + W _ { d \rightarrow a } < 0$
(3) $W _ { a \rightarrow b } + W _ { c \rightarrow d } > 0$
(4) $W _ { b \rightarrow c } + W _ { d \rightarrow a } = 0$
Which of the above relation is/are true

Options:
A. $1,3$ and $4$ only
B. $3$ and $4$ only
C. $1$ and $4$ only
D. $1, 2, 3$ and $4$ only
E. $1, 2, 4$ only
F. 1, 2, and 3 only
G. $1$ and $3$ only
H. $2,3$ and $4$ only
Answer: H
For a given sample of water containing the following impurities$Mg{ \left( H{ CO }_{ 3 } \right)  }_{ 2 }=73mg/L;\quad Ca{ \left( H{ CO }_{ 3 } \right)  }_{ 2 }=162mg/L;\quad Ca{ SO }_{ 4 }=136mg/L$$Mg{ Cl }_{ 2 }=95mg/L;\quad Ca{ Cl }_{ 2 }=111mg/L;\quad NaCl=100mg/L$Then the total hardness (temporary and permanent) of above water sample is?

Options:
A. $$375ppm$$
B. $$500ppm$$
C. $$350ppm$$
D. $$300ppm$$
E. $$315ppm$$
F. $$450ppm$$
G. $$310ppm$$
H. $$400ppm$$
I. $$425ppm$$
J. $$325ppm$$
Answer: D
A wave travels on a light string. The equation of the wave is $\displaystyle Y=A\sin \left ( kx-\omega t+30^{\circ} \right )$. It is reflected from a heavy string tied to an end of the light string at x = 0. If 64% of the incident energy is reflected, then the equation of the reflected wave is

Options:
A. $$\displaystyle Y=0.8A\sin \left ( kx+\omega t+30^{\circ} \right )$$
B. \displaystyle Y=0.8A\sin \left ( kx+\omega t+120^{\circ} \right )
C. \displaystyle Y=0.8A\sin \left ( kx-\omega t-30^{\circ}+180^{\circ} \right )
D. $$\displaystyle Y=0.8A\sin \left ( kx+\omega t+30^{\circ}+180^{\circ} \right )$$
E. $$\displaystyle Y=0.8A\sin \left ( kx-\omega t+30^{\circ}+180^{\circ} \right )$$
F. $$\displaystyle Y=0.8A\sin \left ( kx+\omega t+150^{\circ} \right )$$
G. \displaystyle Y=0.8A\sin \left ( kx-\omega t+150^{\circ} \right )
H. \displaystyle Y=0.8A\sin \left ( kx-\omega t-30^{\circ} \right )
I. \displaystyle Y=0.8A\sin \left ( kx+\omega t-150^{\circ} \right )
J. $$\displaystyle Y=0.8A\sin \left ( kx+\omega t-30^{\circ} \right )$$
Answer: J
A small ball of mass m and charge +q tied with a string of length $l$, rotating in a vertical circle under gravity and a uniform horizontal electric field E as shown. The tension in the string will be minimum for:

Options:
A. $$\displaystyle \theta = tan^{-1} \left(\frac{qE}{mg}\right)$$
B. \(\displaystyle \theta = \pi - tan^{-1} \left( \frac{mg}{qE} \right) \)
C. \(\displaystyle \theta = \frac{\pi}{2} + tan^{-1} \left( \frac{qE}{mg} \right)\)
D. \(\displaystyle \theta = \pi - tan^{-1} \left( \frac{qE}{mg} \right) \)
E. $$\displaystyle \theta = \pi + tan^{-1} \left( \frac{mg}{qE} \right)$$
F. $$\theta = \pi$$
G. \(\displaystyle \theta = \frac{\pi}{2} + tan^{-1} \left( \frac{mg}{qE} \right)\)
H. $$\displaystyle \theta = \pi + tan^{-1} \left(\frac{qE}{mg}\right)$$
I. \(\displaystyle \theta = \pi + tan^{-1} \left( \frac{mg}{2qE} \right) \)
J. $$\theta = o^{circ}$$
Answer: H
What is the volume of Z (compressibility  factor) for 0.5 mole of real gas at 1 atm pressure ?
Assuming the size of gas molecule is negligible 
$$\left[ {{\text{given}}\;{\text{a}}\; = \;{\text{100}}\;{\text{atm}}\;{{\text{L}}^{\text{2}}}{\text{,}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}{\text{,}}\;RT\; = \;20\;{\text{L}}{\text{.atm}}{\text{.}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right]$$

Options:
A. 5
B. 20
C. 27.5
D. 22.5
E. 12.5
F. 25
G. 15
H. 0.25
I. 10
J. 0.5
Answer: F
A spring balance reads $W_{1}$ when ball is suspended from it. A weighing machine reads $W_{2}$ when a tank of liquid, is kept on it. When the ball is immersed in the liquid without touching the tank, the spring balance reads $W_{3}$ and weighing machine reads $W_{4}$.

Options:
A. W_{1} + W_{2} < W_{3} + W_{4}
B. W_{1} < W_{3}
C. $$W_{2} > W_{4}$$
D. W_{1} = W_{3}
E. $$W_{1} > W_{3}$$
F. $$ W_{1} + W_{2} > W_{3} + W_{4}$$
G. W_{1} + W_{2} < W_{3}
Answer: E
A charged object of mass $1\ kg$ is projected at $t = 0$ from ground with a velocity of $10 \sqrt{2}\ m/s$ making an angle $45^o$ with horizontal. In the entire region, an electric field is applied at $t = 0$, due to which a time varying force $F= 12t\ N$, where $t$ is in seconds acts on the object in horizontally forward direction. At how much distance from the starting point, the object strikes the ground? ($g = 10\ m/s^2$)

Options:
A. $$88\ m$$
B. $$80\ m$$
C. $$96\ m$$
D. $$72\ m$$
E. $$68\ m$$
F. $$60\ m$$
G. $$64\ m$$
H. $$10\ m$$
I. $$48\ m$$
J. $$20\ m$$
Answer: E
$5.1g \space NH_4SH$ is introduced in 3.0 L evacuated flask at $327^{0}C$, 30% of the solid $NH_4SH$ decomposed to $NH_3$ and $H_2S as gases. The $K_p$ of the reaction at $327^{0}C$ is ($ R = 0.082 \space L \space atm \space mol^{-1}K^{-1}$, Molar mass of $S =  32 \space 32 \space mol^{01}$, molar mass of $N = 14g \space mol^{-1}$)      [JEE MAINS - ONLINE - 2019]

Options:
A. $$4.9 \times 10^{-4} \space atm^{2}$$
B. $$4.9 \times 10^{-3} \space atm^{2}$$
C. $$5.9 \times 10^{-4} \space atm^{2}$$
D. $$1.23 \times 10^{-3} \space atm^{2}$$
E. $$2.45 \times 10^{-3} \space atm^{2}$$
F. $$4.9 \times 10^{-2} \space atm^{2}$$
G. $$5.9 \times 10^{-3} \space atm^{2}$$
H. $$1 \times 10^{-4} \space atm^{2}$$
I. $$0.242 \times 10^{-4} \space atm^{2}$$
J. $$0.242 atm^{2}$$
Answer: B
Orbital speed of a satellite in a circular orbit is given by  ${v_0} = \sqrt {\frac{{GM}}{r}} $, where $r$ is the radius of circular path and $M$ is the mass of planet. Two satellite each of mass m revolving in circular orbits of radius $5R$ in opposite direction. They collide perfectly in inelastically and fall to the ground. The speed of combined mass on striking the ground will be

Options:
A. $$\sqrt{4v_0}$$
B. $${{v_0}}$$
C. $$2\sqrt {2{v_0}} $$
D. $$\sqrt{5v_0}$$
E. $$\sqrt {2{v_0}}$$
F. $$\sqrt {3{v_0}}$$
G. $$\sqrt{3v_0}$$
H. $$\ {2{v_0}} $$
Answer: H
Calculate the enthalpy of combustion of methyl alcohol at 298 K from the following dataBond$\displaystyle C-H$$\displaystyle C-O$$\displaystyle O-H$$\displaystyle O=O$$\displaystyle C=O$Bond Enthalpy ( kJ $\displaystyle mol^{-1}$414351.5464.5494711Resonance energy of $\displaystyle CO_{2}=-143  kJ  mol^{-1}$Latent heat of vaporisation of methyl alcohol $\displaystyle =35.5 kJ  mol^{-1}$Latent heat of vaporisation of water $\displaystyle =40.6  kJ  mol^{-1}$

Options:
A. $$\displaystyle -136.7 kJ mol^{-1}$$
B. $$\displaystyle -166.7 kJ mol^{-1}$$
C. $$-156.7 kJ mol^{-1}$$
D. $$\displaystyle -696.9 kJ mol^{-1}$$
E. $$-138.7 kJ mol^{-1}$$
F. $$-146.7 kJ mol^{-1}$$
G. $$-151.7 kJ mol^{-1}$$
H. $$-139.7 kJ mol^{-1}$$
I. $$-166.7 kJ mol^{-1}$$
J. $$\displaystyle -659.7 kJ mol^{-1}$$
Answer: A
A charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the middle of a uniformly charged spherical region of total charge $Q$ and radius $R, q$ and $Q$ have opposite signs. The spherically charged region is not free to move. The value of $K_0$ is such that the particle will just reach the boundary of the spherically charged region. How much time does it take for the particle to reach the boundary of the region.

Options:
A. $$t = \dfrac{\pi}{6} \sqrt{\dfrac{2 \pi \varepsilon_0 m R^3}{qQ}}$$
B. $$t = \dfrac{\pi}{4} \sqrt{\dfrac{4 \pi \varepsilon_0 m R^3}{qQ}}$$
C. $$t = \dfrac{\pi}{3} \sqrt{\dfrac{4 \pi \varepsilon_0 m R^3}{qQ}}$$
D. $$t = \dfrac{\pi}{4} \sqrt{\dfrac{2 \pi \varepsilon_0 m R^3}{qQ}}$$
E. $$t = \dfrac{\pi}{2} \sqrt{\dfrac{4 \pi \varepsilon_0 m R^3}{qQ}}$$
F. $$t = \dfrac{\pi}{6} \sqrt{\dfrac{4 \pi \varepsilon_0 m R^3}{qQ}}$$
G. t = \dfrac{\pi}{5} \sqrt{\dfrac{4 \pi \arepsilon_0 m R^3}{qQ}}
H. t = \dfrac{\pi}{7} \sqrt{\dfrac{4 \pi \arepsilon_0 m R^3}{qQ}}
I. $$t = \dfrac{\pi}{2} \sqrt{\dfrac{2 \pi \varepsilon_0 m R^3}{qQ}}$$
Answer: B
A meter scale of uniform thickness has a mass of $5\ kg$. It is being lifted upwards with an acceleration of $2\ m/s^{2}$ by applying a force at the zero marked end. The tension in the rod at $20\ cm$ mark is $(g=10\ m/s^{2})$

Options:
A. $$44\ N$$
B. $$60\ N$$
C. $$54\ N$$
D. $$50\ N$$
E. $$40\ N$$
F. $$58\ N$$
G. $$42\ N$$
H. $$46\ N$$
I. $$52\ N$$
J. $$48\ N$$
Answer: B
A 10 pF parallel plate capacitor is charged with a 4.0 V battery. While the capacitor is
still connected to the battery, a dielectric slab (κ = 5.0) is inserted between the plates
to completely fill the gap. How much electric potential energy is stored in the
capacitor after inserting the dielectric?

Options:
A. $$4.5 × 10^{-10} J$$
B. $$5.3 × 10^-10 J $$
C. $$4.8 × 10^{-10} J$$
D. $$4.7 × 10^{-10} J$$
E. $$4.9 × 10^{-10} J $$
F. $$4.2 × 10^{-10} J$$
G. $$4.4 × 10^{-10} J$$
H. $$4.0 × 10^-10 J $$
I. $$5.6 × 10^{-10} J $$
J. $$4.6 × 10^{-10} J $$
Answer: J
Conductor ABC consist of two quarter circular path of radius R lies in X-Y plane and carries current I as shown. A uniform magnetic field $ \overrightarrow { B }  $ is switched on in the region that exert force $ \overrightarrow { F } =\sqrt { 2 } IRB_{ 0 }\hat { k }  $ on conductor ABC. $ \overrightarrow { B } $ can be:

Options:
A. $$\dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \left( -\hat { i } - \hat { j } \right)$$
B. $$\dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \left( \hat { i } + \hat { j } + \hat { k } \right)$$
C. $$ \dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \hat { i } $$
D. $$\dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \hat { j }$$
E. $$\dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \left( \hat { i } + 2\hat { j } \right)$$
F. $$ \dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \left( -\hat { i } \right) $$
G. $$\dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \left( 2\hat { i } + \hat { j } \right)$$
H. $$\dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \left( -\hat { j } \right)$$
I. $$\dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \left( \hat { i } - \hat { j } \right)$$
J. $$ \dfrac { { B }_{ 0 } }{ \sqrt { 2 } } \left( \hat { i } +\hat { j } \right) $$
Answer: C
A particle of specific charge (charge/mass) $lpha $ starts moving from the origin under the action of electric field $ ec E  =ec E_o\hat { i }$ and magnetic field $ec B = ec B_o\hat{ k}$. Its velocity at $ \left( x_ o,0,0 \right) is\left( 4\hat { i } +3\hat { j }  \right)  $. The value of $ x_o $ is :

Options:
A. $$\dfrac { 125 }{ 6 \alpha E_ \circ }$$
B. $$ \dfrac { 13lpha E_ \circ }{ 2 B_ o } $$
C. $$\dfrac { 25 }{ 3 \alpha E_ \circ }$$
D. $$\dfrac { 25 }{ 4 lpha E_ \circ }$$
E. $$\dfrac { 25 }{ 4 \alpha E_ \circ }$$
F. $$ \dfrac { 25 }{ 2 lpha E_ \circ } $$
G. $$\dfrac { 50 }{ 4 \alpha E_ \circ }$$
H. $$ \dfrac { 5lpha }{ 2B_ \circ } $$
I. $$ \dfrac { 16 lpha B_ \circ }{ E_ o } $$
J. $$\dfrac { 25 }{ 3 lpha E_ \circ }$$
Answer: F
10 mL of $\frac{M}{5}CH_3COONa$ solution is titrated with $\frac{M}{5} HCl$ solution. The pH value at equivalence point is $(pK_a(CH_3COOH) = 4.76)$

Options:
A. 3.76
B. 1.88
C. 2.76
D. 0.7
E. 1.76
F. 2.46
G. 4.76
H. 3.88
I. 1
J. 2.88
Answer: D
A wire of length I m  and radius ${ 10 }^{ -3 }$ m is carrying a heavy current and is assumed to radiate as a black body at equilibrium. its temperature is 900 K while that of surrounding is 300 k. the resistivity of the materials of the wire at 300 k is ${ \pi  }^{ 2 }\times { 10 }^{ -5 }$ ohm-m and its temperature coefficient of resistance is 7.8$\times { 10 }^{ -3 }{ I }^{ o }C$ ( stefan constant $\sigma =5.68\times { 10 }^{ -8 }\quad W/{ m }^{ 2 }{ K }^{ 4 }$ 
the resistivity of wire at 900 k is nearly

Options:
A. $$1.2\times { 10 }^{ 7 }ohm-m$$
B. $$2.0 \times { 10 }^{ -7 }ohm-m$$
C. $$2.2 \times { 10 }^{ -7 }ohm-m$$
D. $$1.8 \times { 10 }^{ -7 }ohm-m$$
E. $$2.4\times { 10 }^{ 7 }ohm-m$$
F. $$2.4\times { 10 }^{ -7 }ohm-m$$
G. $$3.6\times { 10 }^{ -7 }ohm-m$$
H. $$1.2 \times { 10 }^{ -7 }ohm-m$$
I. $$1.3 \times { 10 }^{ -7 }ohm-m$$
J. $$1.6 \times { 10 }^{ -7 }ohm-m$$
Answer: G
The van der Waal's equation for a real gas may be rearranged to give:
$$V_m^{3}-\left ( b+\dfrac{RT}{P} \right )V_m^{2}+\dfrac{a}{p}V_m-\dfrac{ab}{p}=0$$
Where $V_m$ is molar volume of the gas. Which of the following is correct?

Options:
A. At any temperature there are three value of $V_m$, one real two imaginary.
B. At temperature greater than $T_c$, there are three values of $V_m$, one real and two imaginary.
C. At temperature greater than $T_c$, there are three real values of $V_m$.
D. At temperature greater than $T_c$, there are three real values of $V_m$, but two are identical.
E. At temperature $T_c$, there are three real values of $V_m$, but one is distinct from the others.
F. At temperature greater than $T_c$, there are three values of $V_m$, all real and distinct.
G. At temperature $T_c$, there are three real values of $V_m$, but they are not identical.
Answer: B
Chromium plating can involve the electrolysis of an electrolyte of an acidified mixture of chromic acid and chromium sulphate. If during electrolysis the article being plated increases in mass by 2.6 g and $0.6\, dm^{3}$ of oxygen are evolved at an inert anode, the oxidation state of chromiumions being discharged must be : (assuming atomic weight of Cr = 52 and 1 mole of gas at room temperature and pressure occupies a volume of 24 $dm^{3}$)

Options:
A. $$+3$$
B. $$+1$$
C. $$-1$$
D. $$+2$$
E. $$+5$$
F. $$+6$$
G. $$Zero$$
H. $$+4$$
I. $$+7$$
Answer: B
A charged capacitor is allowed to discharge through a resistance $2\Omega$ by closing the switch S at the instant $t=0$. At time $t=ln 2$ms, the reading of the ammeter falls half of its initial value. The resistance of the ammeter equal to:

Options:
A. $$0$$
B. $$1\Omega$$
C. $$1M\Omega$$
D. $$4\Omega$$
E. $$8\Omega$$
F. $$\infty$$
G. $$2\Omega$$
H. $$2M\Omega$$
I. $$4M\Omega$$
Answer: A
a moles of $PCl_5$ are heated in a closed container to equilibrate , $PCl_{5(g)} \rightleftharpoons  PCl_{3(g)} + Cl_{2(g)}$ at a pressure of P atm. If x moles of $PCl_5$ dissociate at equilibrium, then

Options:
A. $$\frac{x}{a} = \lgroup \frac{K_p}{P} \rgroup^{1/2}$$
B. $$\frac{x}{a} = \lgroup \frac{K_p + P}{K_p } \rgroup^{1/2}$$
C. $$\frac{x}{a} = \lgroup \frac{K_p }{K_p + P} \rgroup^{1/2}$$
D. $$\frac{x}{a} = \frac{K_p - P}{K_p}$$
E. $$\frac{x}{a} = \frac{P}{K_p - P}$$
F. $$\frac{x}{a} = \left( \frac{K_p + P}{K_p - P} \right)^{1/2}$$
G. $$\frac{x}{a} = \left( \frac{P}{K_p + P} \right)^{1/2}$$
H. $$\frac{x}{a} = \frac{K_p}{K_p + P}$$
I. $$\frac{x}{a} = \frac{K_p - P}{K_p + P}$$
J. $$\frac{x}{a} = \frac{K_p + P}{K_p}$$
Answer: H
A particle of $1\ kg$ mass is moving in $xy-$plane whose position is described by position vector $\vec {r}=t^{2}\hat {i}+4t\hat {j}$

Options:
A. The magnitude of tangent of acceleration at $t=1\ sec$ is $\sqrt {\dfrac {16}{5}}\ m/s^{2}$
B. $$\sqrt{\dfrac{32}{5}}\ m/s^{2}$$
C. The magnitude of tangent of acceleration at \( t=1\ sec \) is \(\sqrt {\dfrac {25}{5}}\ m/s^{2} \)
D. The magnitude of tangent of acceleration at $t=1\ sec$ is $\sqrt {\dfrac {4}{5}}\ m/s^{2}$
E. $$\sqrt{\dfrac{20}{5}}\ m/s^{2}$$
F. the radius of curvature at $t=1 \ sec\ is\  5\sqrt {5}$
G. $$\sqrt{\dfrac{36}{5}}\ m/s^{2}$$
H. The magnitude of tangent of acceleration at \( t=1\ sec \) is \(\sqrt {\dfrac {9}{5}}\ m/s^{2} \)
I. The minimum radius of curvature is $8\ m$
J. The magnitude of tangent of acceleration at \( t=1\ sec \) is \(\sqrt {\dfrac {18}{5}}\ m/s^{2} \)
Answer: A
An aqueous solution of liquid $'X' (Mol. Wt = 56), 28\%$ by weight has a vapour pressure $150\ mm$. Find the vapour pressure of $'X'$ if vapour pressure of water is $155\ mm$ of $Hg$.

Options:
A. $$110\ mm$$
B. $$150\ mm$$
C. $$180\ mm$$
D. $$210\ mm$$
E. $$160\ mm$$
F. $$230\ mm$$
G. $$215\ mm$$
H. $$171\ mm$$
I. $$200\ mm$$
J. $$220\ mm$$
Answer: A
Particles move with decreasing speed along the circle of the radius R  so that at any moment of the time its tangent and centripetal accelerations are equal in magnitude the initial moment, t=0, its speed is u. 1. The time after which the speed of particle reduce to half of its initial value is

Options:
A. $${R \over u}$$
B. $${{3R} \over {2u}}$$
C. $${{3R} \over {4u}}$$
D. $${{R \over {4u}}}$$
E. $${R \over {2u}}$$
F. $${{5R} \over {4u}}$$
G. $${{R} \over {3u}}$$
H. $${{2R} \over u}$$
I. $${{5R} \over {3u}}$$
Answer: H
Current in an $AC$ circuit is given by $i=3\sin \omega t+4\cos \omega t$, then

Options:
A. If voltage applied is $V=V_m\cos \omega t,$ then the circuit may contain resistance and inductance only.
B. If voltage applied is $V=V_m\cos \omega t,$ then the circuit may contain resistance and capacitance only.
C. rms value of current is $5A$
D. If voltage applied is $V=V_m\sin \omega t,$ then the circuit may contain resistance and inductance only.
E. If voltage applied is \( V=V_m\cos \omega t, \) then the circuit may contain resistance and capacitance only.
F. mean value of this current in positive one-half period will be $\displaystyle \frac{6}{\pi}$
G. If voltage applied is $V=V_m\sin \omega t$ then the circuit must be containing resistance and capacitance.
H. If voltage applied is $V=V_m\cos \omega t,$ then the circuit may contain resistance and capacitance and inductance.
I. If voltage applied is $V=V_m\cos \omega t,$ then the circuit may contain inductance only.
J. If voltage applied is \( V=V_m\cos \omega t, \) then the circuit may contain capacitance and inductance only.
Answer: A
Consider the situation as shown in the figure. The system is released from rest and the block of mass $ 1\ kg$ is found to have a speed $ 0.3 \displaystyle {m}/{s}$ after it has descended through a distance of $ 1\ m$. The coefficient of kinetic friction between the block and the table is (All pulleys are massless and smooth and strings are inextensible and light acceleration due to gravity $=10  m/s^2$.):

Options:
A. $$0.18$$
B. $$0.15$$
C. $$0.07$$
D. $$0.13$$
E. $$0.5$$
F. $$0.11$$
G. $$0.14$$
H. $$0.12$$
I. $$0.08$$
J. $$0.2$$
Answer: H
A mixture of ${n_1}$ moles of monoatomic gas and ${n_2}$ moles of diatomic gas has $\frac{{{C_y}}}{{{C_y}}} = Y = 1.5$ :-

Options:
A. $$2{n_1} = {n_2}$$
B. $$3{n_1} = {n_2}$$
C. $${n_1} = \frac{3}{2}{n_2}$$
D. $$3{n_1} = 4{n_2}$$
E. $$2{n_1} = 3{n_2}$$
F. $${n_1} = 2{n_2}$$
G. $${n_1} = {n_2}$$
H. $$3{n_1} = 2{n_2}$$
I. $$2{n_1} = 4{n_2}$$
Answer: G
Fire is caught at height of $125\;m$ from the fire brigade. To extinguish the fire, water is coming out from the pipe of cross section $6.4\;cm$ with rate of $950\;litre/min$. Find out minimum velocity of water exiting from fire brigade tank $(g=10m/s^2)$

Options:
A. $$30\;m/s$$
B. $$15\;m/s$$
C. $$40\;m/s$$
D. $$35\;m/s$$
E. $$25\;m/s$$
F. $$10\;m/s$$
G. $$18\;m/s$$
H. $$20\;m/s$$
I. $$5\;m/s$$
J. $$50\;m/s$$
Answer: I
A bead of mass m is located on a parabollic wire with its axis vertical and vertex at the origin as shown and whose equation is $ x^2 = 4ay $. the wire is fixed and the bead can slide on it without friction.it is released from the point y= 4A on the wire frame from rest. the tangential acceleration of the bead reaches to position given by y=a is :

Options:
A. $$ \frac {g}{\sqrt 2} $$
B. $$ \frac {g}{2} $$
C. $$\frac{g}{\sqrt{6}}$$
D. $$ \frac {g}{\sqrt 1} $$
E. $$\frac{\sqrt{2} g}{2}$$
F. $$\frac{g}{\sqrt{3}}$$
G. $$\frac{g}{\sqrt{4}}$$
H. $$\frac{g}{\sqrt{5}}$$
I. $$ \frac {\sqrt3 g}{2} $$
J. $$\frac{g}{3}$$
Answer: A
For gaseous decomposition of ${PCI}_{5}$ in a closed vessel the degree of dissociation '$\alpha $', equilibrium pressure 'P' & ${'K}_{p}'$ are related as

Options:
A. $$\alpha =\sqrt { { K }_{ p }+P } $$
B. $$\alpha =\frac { 1 }{ \sqrt { { K }_{ p }+P } } $$
C. \(\alpha =\sqrt { \frac { P }{ { K }_{ p }-1 } } \)
D. $$\alpha =\sqrt { \frac { P }{ { K }_{ p }-2 } }$$
E. \(\alpha =\sqrt { \frac { P }{ { K }_{ p }+1 } } \)
F. $$\alpha =\sqrt { \frac { { K }_{ p } }{ P } } $$
G. $$\alpha =\sqrt { \frac { P }{ { K }_{ p }+3 } }$$
H. $$\alpha =\sqrt { \frac { P }{ { K }_{ p }+2 } }$$
I. $\alpha =\sqrt { \frac { P }{ { K }_{ p } } }$
J. $$\alpha =\sqrt { \frac { { K }_{ p }+P }{ { K }_{ p } } } $$
Answer: F
The volume V of water passing through a point of a uniform tube during t seconds is related to the cross-sectional area A of the tube and velocity u of water by the relation$V\propto A^{\alpha}u^{\beta}t^{\gamma}$  Which one of the following will be true?

Options:
A. $$ \alpha $$
B. $$\alpha \neq \beta \neq \gamma$$
C. $$\alpha = \beta = \gamma$$
D. $$ \alpha \neq \beta = \gamma $$
E. $$ \alpha = \beta \neq \gamma $$
Answer: D
A clock with pendulum made of invar has a period of 0)5sec at $20 ^ { \circ } \mathrm { C }$  If the clock is used in a climate where the temperature increases to $30 ^ { \circ } \mathrm { C }$ , how much time does the clock loose -1 in each oscillation. For invar $\alpha = 9 \times 10 ^ { - 7 \cdot 0 } C ^ { - 1 }$

Options:
A. $$1.5 \times 10 ^ { - 7 } \mathrm { sec }$$
B. $$3.25 \times 10 ^ { - 7 } \mathrm { sec }$$
C. $$5 \times 10 ^ { - 6 } \mathrm { sec }$$
D. $$3.0 \times 10 ^ { - 7 } \mathrm { sec }$$
E. $$3.75 \times 10 ^ { - 7 } \mathrm { sec }$$
F. $$3.5 \times 10 ^ { - 7 } \mathrm { sec }$$
G. $$2.75 \times 10 ^ { - 7 } \mathrm { sec }$$
H. $$1.125 \times 10 ^ { - 6 } \mathrm { sec }$$
I. $$5 \times 10 ^ { - 7 } \mathrm { sec }$$
J. $$2.25 \times 10 ^ { - 6 } \mathrm { sec }$$
Answer: C
A frog sits on the end of a long board of length $L=10cm$. The board rests on frictionless horizontal table. The frog wants to jump to the opposite end of the board. What is minimum  take of speed $v$ in $ms^{-1}$ relative to the ground that the frog follows to do the tricks ? [Assume that the board and frog have equal masses?]

Options:
A. $$0.5\sqrt{10}ms^{-1}$$
B. $$10\sqrt{2}ms^{-1}$$
C. $$0.5\sqrt{3}ms^{-1}$$
D. $$2\sqrt{5}ms^{-1}$$
E. $${5}ms^{-1}$$
F. $$1.0\sqrt{2}ms^{-1}$$
G. $$0.5\sqrt{2}ms^{-1}$$
H. $$1.5\sqrt{2}ms^{-1}$$
I. $$0.5\sqrt{5}ms^{-1}$$
J. $$1.5\sqrt{5}ms^{-1}$$
Answer: G
If $\vec{E} = E_0 \cos(kz) \cos(\omega t) \hat{i}$ then $\vec{B}$ for electromagnetic wave is:

Options:
A. $$\vec{B} = \dfrac{E_0}{C} \cos (kz) \cos (\omega t) \hat{i}$$
B. $$\vec{B} = \frac{E_0}{c} \sin(kz) \sin(\omega t) \hat{i}$$
C. $$\vec{B} = \dfrac{E_0}{C} \cos (kz) \cos (\omega t) \hat{j}$$
D. $$\vec{B} = \frac{E_0}{c} \hat{k}$$
E. $$\vec{B} = \dfrac{E_0}{C} \sin (kz) \cos (\omega t) \hat{i}$$
F. $$\vec{B} = \dfrac{E_0}{C} \cos (kz) \sin (\omega t) \hat{j}$$
G. $$\vec{B} = \frac{E_0}{c} \sin(kz) \cos(\omega t) \hat{j}$$
H. $$\vec{B} = \dfrac{E_0}{C} \sin (kz) \sin (\omega t) \hat{j}$$
I. $$\vec{B} = \frac{E_0}{c} \cos(kz) \sin(\omega t) \hat{j}$$
J. $$\vec{B} = \dfrac{E_0}{C} \cos (kz) \sin (\omega t) \hat{i}$$
Answer: B
A quantity of $5.08\ g$ of iodine held in suspension in water is slowly acted upon by $460\ ml$ of $H_2S$ measured at $0^oC$ and $1$ atm. What weight of sulphur will be liberated? $(I=127)$

Options:
A. $$0.017\ g$$
B. $$0.64\ g$$
C. $$0.668\ g$$
D. $$0.657\ g$$
E. $$0.665\ g$$
F. $$0.667\ g$$
G. $$0.659\ g$$
H. $$1.297\ g$$
I. $$0.658\ g$$
J. $$0.660\ g$$
Answer: D
A mass m starting from A reaches B of a frictionless track. On reaching B, it pushes the track with a force equal to x times its weight, then the applicable relation is :

Options:
A. $$h = \dfrac{(5 + x)}{2}R$$
B. $$h = (\dfrac{x + 1}{2})R$$
C. $$h = \dfrac{(6 - x)}{2}R$$
D. $$h = \dfrac{x}{2}R$$
E. $$h = \dfrac{(4 - x)}{2}R$$
F. $$h = \dfrac{(3 - x)}{2}R$$
G. $$h = \dfrac{(7 - x)}{2}R$$
H. $$h = \dfrac{(5 - x)}{2}R$$
I. h = R
J. $$h = \dfrac{(4 + x)}{2}R$$
Answer: A
$lpha -$ particles of $6 MeV$ energy are scattered back from a silver foil. Calculate the maximum volume in which the entire positive charge of the atom is supposed to be concentrated. $[(Z \,for \,Ag = 47), k = 9.0\times  10^9 Nm^2C^{-2}]$.

Options:
A. $$2.25\times 10^{-14}m^3$$
B. $$4.8\times 10^{41} m^3$$
C. $$1.1\times 10^{-41} m^3$$
D. 3.6 \times 10^{-41} m^3
E. 5.8 \times 10^{-41} m^3
F. 3.8 \times 10^{-40} m^3
G. 5.8 \times 10^{-40} m^3
H. 3.8 \times 10^{-41} m^3
I. $$4.8\times 10^{-41} m^3$$
J. 3.8 \times 10^{-42} m^3
Answer: I
An ideal gas us thermally insulated vessel at internal pressure $P_1$, volume, $V_1$ and absolute temperature $T_1$ expands irreversibly against zero external pressure as shown in the diagram.The final internal pressure, volume and temperature of the gas are $P_2, V_2$ and $T_2$ respectively. For this expansion

Options:
A. $$P_2V_2^2=P_1V_1^2$$
B. $$P_2V_2^{\gamma} = P_1V_1^{\gamma-1}$$
C. $$P_2V_2^{\gamma}=P_1V_1^{\gamma}$$
D. $$P_2V_2^{\gamma-1}=P_1V_1^{\gamma}$$
E. $$P_2V_2=P_1V_1$$
F. $$P_2V_2^{\gamma-1}=P_1V_1^2$$
G. $$P_2V_2^{\gamma-1}=P_1V_1^{\gamma-1}$$
H. $$P_2V_2^{\gamma-1}=P_1V_1$$
Answer: E
$n$ cells of each of EMF $E$ and internal resistance $r$ send the same current $R$ whether the cells are connected in series or parallel, then :

Options:
A. $$R = \sqrt{n} \dfrac{E}{2r}$$
B. $$R = \sqrt{n} \dfrac{E}{3r}$$
C. $$R = nr$$
D. $$r = nR$$
E. $$R = \sqrt{n} \dfrac{E}{r + 1}$$
F. $$ R = \sqrt {n}  \dfrac{E}{r}$$
G. $$R = \sqrt{n} \dfrac{E}{r^3}$$
H. $$R = r$$
I. $$R = \sqrt{n} \dfrac{E}{2r^2}$$
J. $$R = \sqrt{n} \dfrac{E}{r^2}$$
Answer: H
A particle of mass $m$ moves along a circular path of radius $r$ with a centripetal acceleration $a_n$ changing with time $t$ as $a_n = kt^2,$ where $k$ is a positive constant. The average power developed by all the forces acting on the particle during the first $t_0$ seconds is:

Options:
A. $$\dfrac{mkrt_0^2}{7}$$
B. $$\dfrac{mkrt_0^2}{8}$$
C. $$\dfrac{mkrt_0^2}{3}$$
D. $$\dfrac{mkrt_0} {4}$$
E. $$\dfrac {mkrt_0^2} {2}$$
F. $$mkrt_0$$
G. $$\dfrac{mkrt_0^2}{5}$$
H. $$\dfrac{mkrt_0^2}{6}$$
I. $$\dfrac{mkrt_0^2}{4}$$
J. $$\dfrac{mkrt_0} {2}$$
Answer: E
Block A has a mass of 2 kg and block B has a mass of 20 kg. If the coefficient of kinetic friction between block B and the horizontal surface is 0.1 and B is accelerating towards the right with $a = 2 m/s^2$, then the mass of the block C will be

Options:
A. 10.7 kg
B. 8.7 kg
C. 15 kg
D. 10.3 kg
E. 9.5 kg
F. 10.5 kg
G. 11.3 kg
H. 12.5 kg
I. 5.7 kg
J. 11.5 kg
Answer: F
$100$ mL of $0.02\  M$ benzoic acid ($pKa = 4.20$) is titrated using $0.02\ M$ $NaOH$. $pH$ after $50$ mL and $100$ mL of $NaOH$ have been added are:

Options:
A. $$4.2, 8.35$$
B. $$4.2, 7$$
C. $$4.2, 8.2$$
D. $$4.2, 7.5$$
E. $$3.50, 7$$
F. $$4.2, 8.15$$
G. $$4.2, 8.1$$
H. $$4.2, 7.75$$
I. $$4.2, 8.05$$
J. $$4.2, 8.25$$
Answer: B
Consider the following reaction at $ { 600 }^{ \circ  }C$
$$ 2S{ O }_{ 2 }(g)+{ O }_{ 2 }(g)\rightleftharpoons 2S{ O }_{ 3 }(g);{ K }_{ c }=4.00$$
A mixture was  prepared with $ [S{ O }_{ 3 }]=0.50M,[{ O }_{ 2 }]=0.35M\quad and\quad [S{ O }_{ 2 }]=0.0M$
Thus, equilibrium concentration of $O_2$ is

Options:
A. $$0.045\ M$$
B. $$0.0450\ M$$
C. $$0.35\ M$$
D. $$0.0400\ M$$
E. $$0.0300\ M$$
F. $$0.45\ M$$
G. $$0.050\ M$$
H. $$0.040\ M$$
I. $$0.030\ M$$
J. $$0.035\ M$$
Answer: J
A solid sphere of radius R and mass m is released from point A on  a vertical circular surface of radius 5R which has sufficient friction in part AB. Part BCD is smooth.h is the maximum height attained by the centre of the sphere from point C in  part CD. Total energy and kinetic energy at this height is T and K respectively (assume lowest point C of the circular path as reference point for potential energy) Then

Options:
A. $$h=\cfrac {7}7 R$$
B. $$h=\cfrac {31}7 R$$
C. $$h=\cfrac {6}7 R$$
D. $$h=\cfrac {5}7 R$$
E. $$h=\cfrac {3}7 R$$
F. $$T=5mgR$$
G. $$h=\cfrac {5}8 R$$
H. $$h=\cfrac {4}7 R$$
I. $$h=\cfrac {2}7 R$$
J. Friction force on the sphere just before point $B=\cfrac {2 \sqrt {3mg}}5$
Answer: H
A train is moving with a constant speed of 10 m/s in a circle of radius $\displaystyle \frac{16}{\pi }$ m. The plane of the circle lies in horizontal. x-y plane. At time t=0, train is at point $P(\dfrac{16}{\pi},0 ,0)$ and moving in counter-clockwise direction. At this instant a stone is thrown from the train with speed 10 m/s relative to train towards negative x-axis at an angle of $\displaystyle 37^{\circ}$ with vertical z-axis. Find the velocity of particle relative to train at the highest point of its trajectory.

Options:
A. $$\left ( 4\hat{i}+7\hat{j} \right )ms^{-1}$$
B. $$\left ( -4\hat{i}+7\hat{j} \right )ms^{-1}$$
C. $$\left ( -5\hat{i}+7\hat{j} \right )ms^{-1}$$
D. $$\left ( -4\hat{i}+10\hat{j} \right )ms^{-1}$$
E. $$\left ( 4\hat{i}+8\hat{j} \right )ms^{-1}$$
F. $$\displaystyle \left ( 4\hat{i}+10\hat{j} \right )ms^{-1}$$
G. $$\displaystyle \left ( -3\hat{i}+5\hat{j} \right )ms^{-1}$$
H. $$\left ( 4\hat{i}+11\hat{j} \right )ms^{-1}$$
I. $$\displaystyle \left ( -3\hat{i}+4\hat{j} \right )ms^{-1}$$
J. $$\displaystyle \left ( -6\hat{i}+8\hat{j} \right )ms^{-1}$$
Answer: F
A capillary tube of radius $0.25mm$ is submerged vertically in water so that $25mm$ of its length is outside water. The radius of curvature of the meniscus will be ($T=7.5\times {10}^{-3}N/m$)

Options:
A. 0.9\ mm
B. 0.65\ mm
C. $$0.8\ mm$$
D. $$0.6\ mm$$
E. 0.7\ mm
F. 0.2\ mm
G. 0.1\ mm
H. 0.5\ mm
I. $$0.4\ mm$$
J. $$0.3\ mm$$
Answer: D
Acetyl salicylic acid $(mol.\ wt.\ =180)$ called aspirin is a pain killer with $p{ K }_{ a }=6$. It two tablets each of $0.09\ gm$ mass containing aspirin are dissolved in $100\ mL$ solution. Its pH will be:

Options:
A. $$0.5$$
B. $$2.0$$
C. $$2.5$$
D. $$2.3$$
E. $$1.5$$
F. $$1.0$$
G. $$4.0$$
H. $$2.2$$
I. $$3.5$$
J. $$3.0$$
Answer: G
At $t = 0$, capacitor A has no charge while capacitor B is charged to 3 volts with shown polarity. Find power supplied by battery at $t= 0$

Options:
A. $$\dfrac{5}{6}W$$
B. $$\dfrac{5}{3}W$$
C. $$\dfrac{5}{12}W$$
D. $$\dfrac{5}{4}W$$
E. $$\dfrac{5}{9}W$$
F. $$\dfrac{5}{5}W$$
G. $$\dfrac{5}{2}W$$
H. $$\dfrac{5}{8}W$$
I. $$\dfrac{5}{7}W$$
Answer: D
Bromopahenol blue is an indicator with vale of $K_{ a }=5.84\times 10^{ -5 }$ Calculate the % of this indicator in these basic form at a pH of 4.84.

Options:
A. 80%
B. 40%
C. 60%
D. 30%
E. 70%
F. 90%
G. 20%
H. 15%
I. 96%
J. 10%
Answer: G
Suppose the electron enters the electric field at time $t = 0$. What is the velocity $ec v$ of the electron at time $t_1$ when it leaves the plates?

Options:
A. $$\vec v = v_0 \hat i + (\dfrac{eE_yL_1}{3mv_0}) \hat j$$
B. $$\vec v = (\dfrac{eE_yL_1}{mv_0}) \hat i + v_0 \hat j $$
C. $$\vec v = v_0 \hat i + \left(\dfrac{eE_yL_1}{6mv_0}\right) \hat j$$
D. $$\vec v = v_0 \hat i + \left(\dfrac{eE_yL_1}{4mv_0}\right) \hat j$$
E. $$\vec v = v_0 \hat i - (\dfrac{eE_yL_1}{mv_0}) \hat j$$
F. $$\vec v = v_0 \hat i + \left(\dfrac{eE_yL_1}{7mv_0}\right) \hat j$$
G. $$\vec v = -(\dfrac{eE_yL_1}{mv_0}) \hat i + v_0 \hat j$$
H. $$\vec v = v_0 \hat i + (\dfrac{eE_yL_1}{2mv_0}) \hat j$$
I. $$\vec v = v_0 \hat i + \left(\dfrac{eE_yL_1}{5mv_0}\right) \hat j$$
J. $$\vec v = v_0 \hat i + (\dfrac{eE_yL_1}{mv_0}) \hat j$$
Answer: J
$Cu^{+}$ is not stable and undergoes disproportionation. $E^{o}$ for $Cu^{+}$ disproportionation is : $(E^{o}_{Cu^{2+}/Cu^{+}}=+0.153V,\ E{^{o}}_{Cu^{+}/Cu} =0.53V)$

Options:
A. +0.337 V
B. +0.683 V
C. -0.367 V
D. +0.345 V
E. +0.3415 V
F. +0.383 V
G. +0.377 V
H. +0.3315 V
I. +0.4015 V
J. +0.3715 V
Answer: E
A solution is prepared by mixing 100 mL 0.50 M hydrazoic acid $({ HN }_{ 3 })$, whose ${ K }_{ a }=3.6\times { 10 }^{ -4 }$, with 400 mL of 0.10M cyanic acid (HOCN), whose ${ K }_{ a }=8\times { 10 }^{ -4 }$. Which of the following is (are) true regarding

Options:
A. $$[{ H }^{ + }]={ 5.2 \times 10 }^{ -3 }M$$
B. $$[{ H }^{ + }]={ 9.5 \times 10 }^{ -3 }$$
C. $$[{ H }^{ + }]={ 4.5 \times 10 }^{ -3 }M$$
D. $$[{ OH }^{ - }]={ 7.14\times 10 }^{ -13 }$$
E. $$[{ H }^{ + }]={ 10 }^{ -2 }M$$
F. $$[{ H }^{ + }]={ 6.5 \times 10 }^{ -3 }M$$
G. $$[{ H }^{ + }]={ 7.2 \times 10 }^{ -3 }$$
H. $$[{ H }^{ + }]={ 8.2 \times 10 }^{ -3 }$$
I. $$[{ H }^{ + }]={ 3.6\times 10 }^{ -3 }$$
J. $$[{ OCN }^{ - }]={ 4.57\times 10 }^{ -3 }$$
Answer: E
A projectile is projected at an angle  $\left( \alpha > 45 ^ { \circ } \right)$  with an initial velocity  $u.$  The time  $t,$  at which its magnitude of horizontal velocity will equal the magnitude of vertical velocity is:

Options:
A. $$t = \dfrac { u } { g } ( \sin \alpha - \cos \alpha )$$
B. $$t = \dfrac { u } { 2 g } ( \cos \alpha + \sin \alpha )$$
C. $$t = \dfrac { u } { g } ( \cos \alpha - 2 \sin \alpha )$$
D. $$t = \dfrac { u } { g } ( 2 \cos \alpha - \sin \alpha )$$
E. $$t = \dfrac { u } { g } \left( \sin ^ { 2 } \alpha - \cos ^ { 2 } \alpha \right)$$
F. $$t = \dfrac { u } { g } ( \cos \alpha - \sin \alpha )$$
G. $$t = \dfrac { u } { g } ( \sin \alpha - 2 \cos \alpha )$$
H. $$t = \dfrac { u } { g } ( \cos \alpha + \sin \alpha )$$
I. $$t = \dfrac { u } { g } ( \sin \alpha + \cos \alpha )$$
J. $$t = \dfrac { u } { g } ( \cos \alpha + 2 \sin \alpha )$$
Answer: A
Calculate the pH of C M $Na_2S$ solution. $(K_1, K_2 of \, H_2S are \, 5.0 \, \times 10^{-5}, \, 2.5 \times 10{-11})$.

Options:
A. $$pH \, = \, 7 \, + \, \frac{pK_2}{2} \, + \, \frac{1}{2} log C$$
B. $$pH \, = \, 7 \, + \, \frac{pK_1 \times pK_2}{2pK_2} \, + \, \frac{1}{2} log C$$
C. $$pH \, = \, 7 \, + \, \frac{pK_1 + pK_2}{2pK_2} \, + \, \frac{1}{2} log C$$
D. $$pH \, = \, 7 \, + \, \frac{pK_1pK_2}{2} \, + \, \frac{1}{2} log C$$
E. $$pH \, = \, 7 \, + \, \frac{pK_1}{2pK_2} \, + \, \frac{1}{2} log C$$
F. $$pH \, = \, 7 \, + \, \frac{pK_1 - pK_2}{2} \, + \, \frac{1}{2} log C$$
G. $$pH \, = \, 7 \, + \, \frac{pK_1 + pK_2}{2} \, + \, \frac{1}{2} log C$$
H. $$pH \, = \, 7 \, + \, \frac{pK_1 + 2pK_2}{2} \, + \, \frac{1}{2} log C$$
I. $$pH \, = \, 7 \, + \, \frac{pK_1 + pK_2}{2} \, - \, \frac{1}{2} log C$$
J. $$pH \, = \, 7 \, + \, \frac{pK_1}{2} \, + \, \frac{1}{2} log C$$
Answer: G
Three identical conducting large plates are fixed at separation of $d$ from each other as shown. The area of each plate is $A$. Plate $1$ is given charge $+Q$ while plate $2\;and\;3$ are neutral and are connected to each other through coil of inductances $L$ and switch $S$. If resistance of all connected wires is neglected the maximum current flow, through coil after closing switch is $(C=arepsilon_0\,A/d)$ (neglect fringe effect)

Options:
A. \displaystyle \frac{Q_0}{3\sqrt{LC}}
B. $$\displaystyle\frac{2Q_0}{\sqrt{LC}}$$
C. $$\displaystyle\frac{Q_0}{\sqrt{LC}}$$
D. $$\displaystyle \frac{Q_0}{\sqrt{3LC}}$$
E. $$\displaystyle \frac{3Q_0}{2\sqrt{LC}}$$
F. $$\displaystyle \frac{Q_0}{4\sqrt{LC}}$$
G. $$\displaystyle\frac{Q_0}{2\sqrt{LC}}$$
H. $$\displaystyle\frac{Q_0}{\sqrt{2LC}}$$
I. $$\displaystyle \frac{Q_0}{\sqrt{4LC}}$$
J. $$\displaystyle \frac{Q_0}{5\sqrt{LC}}$$
Answer: G
A silver wire dipped in 0.1 M HCI solution saturated with $AgCI$ develops oxidation potential of -0.209 V. If $E_{Ag/g^+}^0$ = -0.799 V, the $K_{sp}$ of $AgCI$ in pure water will be

Options:
A. $$8 x 10^{-11}$$
B. $$10^{-11}$$
C. $$7 x 10^{-11}$$
D. $$5 x 10^{-11}$$
E. $$6 x 10^{-11}$$
F. $$2 x 10^{-11}$$
G. $$9 x 10^{-11}$$
H. $$3 x 10^{-11}$$
I. $$4 x 10^{-11}$$
Answer: H
$CH_3 NH_2$ (0.1 mole , $K_b$ = 5 $\times 10^{-4}$) is added to 0.08 mole of HCl and solution is diluted to one litre, resulting hydrogen ion concentration is

Options:
A. $$1.6 \times 10^{-13}$$
B. $$1.6 \times 10^{-12}$$
C. $$8 \times 10^{-2}$$
D. $$1.6 \times 10^{-10}$$
E. $$1.6 \times 10^{-11}$$
F. 8 $\times 10^{-11}$
G. $$8 \times 10^{-12}$$
H. $$1.6 \times 10^{-9}$$
I. $$8 \times 10^{-10}$$
J. $$5 \times 10^{-5}$$
Answer: E
A compound X $\left( {{C_g}{H_g}} \right)$  react with ammonincal $AgN{o_3}$ to  give a white precipitate, and on  oxidation with hot alkaline $KMn{O_4}$ gives the acid ${\left( {C{H_3}} \right)_2}CHCOOH.$ therefore, X is

Options:
A. $$C{H_2} = CHCH = CHC{H_3}$$
B. $${\left( {C{H_3}} \right)_2}CH - C = CH$$
C. $${\left( {C{H_3}_{}} \right)_2}C = C = C{H_2}$$
D. $$C{H_3}{\left( {C{H_2}} \right)_2}C = CH$$
E. $$CH_2 = CH - CH = CH - CH_3$$
F. $$C{H_2} = CHC{H_2} = CHC{H_3}$$
G. $${\left( {C{H_3}} \right)_2}CH - CH = CH$$
H. $$CH_2 = CH - CH = CH - CH_2 - CH_3$$
I. $$CH_2 = CH - CH_2 - CH = CH_2$$
J. $$C{H_2} = CHC{H_2} = C{H_2}$$
Answer: A
A charged cork ball having mass $1$ gram and charge $q$ is suspended on a light string in a uniform electric field as shown in the figure. The ball is in equilibrium at $\theta = 37^o$, when value of electric field is $\overrightarrow{E} = (3\hat{i} + 5\hat{j}) \times 10^5 N?C$. Assume $T$ as tension in the string.
Which of the following is correct? (Given: $\sin 37^o = 0.60$ and $g = 10 m/s^2$)

Options:
A. $$q = 6.55 \times 10^{-9}C$$
B. $$T = 4.55 \times 10^{-3}N$$
C. $$T = 6.55 \times 10^{-3}N$$
D. $$q = 5.55 \times 10^{-9}C$$
E. $$q = 11 \times 10^{-9}C$$
F. $$T = 5.55 \times 10^{-2}N$$
G. $$q = 12 \times 10^{-9}C$$
H. $$q = 4.55 \times 10^{-3}N$$
I. $$T = 5.55 \times 10^{-4}N$$
J. $$T = 5.55 \times 10^{-3}N$$
Answer: J
An inductance $L$, a capacitance $C$ and a resistance $R$ may be connected to an $AC$ source of angular frequency $\omega$, in three different combinations of $RC, RL$ and $LC$ in series. Assume that $\displaystyle \omega L = \frac{1}{\omega C}$. The power drawn by the three combinations are $P_1, P_2, P_3$ respectively. Then,

Options:
A. $$\displaystyle P_1>P_2>P_3$$
B. $$\displaystyle P_1 = P_2 < P_3$$
C. $$\displaystyle P_1 > P_2 = P_3$$
D. $$\displaystyle P_1 < P_2 < P_3$$
E. $$\displaystyle P_1 < P_2 = P_3$$
F. $$\displaystyle P_1=P_2=P_3$$
G. $$\displaystyle P_1=P_2>P_3$$
Answer: F
For the following cell $Zn| Zn^{2+} // Cd^{2+}// Cd $
$E_{cell}=0.30V $ and $E^0_{cell}=0.36V.$ Then the value of $ $[Cd^{2+}]/ [Zn^{2+}]$ is

Options:
A. $$0.2$$
B. $$100$$
C. $$0.05$$
D. $$20$$
E. $$0.1$$
F. $$0.02$$
G. $$0.5$$
H. $$0.01$$
I. $$1$$
J. $$10$$
Answer: J
For a weak acid $HA$ of concentration $C(mol \ I^{-}$ and degree of dissociation $(\alpha)$ Ostwald's ilution law is represented by the equation

Options:
A. $$K_{a}=\dfrac {C^{2}\alpha}{(1-\alpha)}$$
B. $$K_{a}=\dfrac {C^{2}\alpha}{(1+\alpha)}$$
C. $$K_{a}=\dfrac {C^{2} \alpha}{(1 - 2\alpha)}$$
D. $$K_{a}=\dfrac {C^{2} \alpha}{(1+ \alpha)}$$
E. $$K_{a}=C \alpha$$
F. $$K_{a}=\dfrac {C \alpha^{2}}{(1-2\alpha)}$$
G. $$K_{a}=\dfrac {C\alpha^{2}}{(1-\alpha)}$$
H. $$K_{a}=\dfrac {C \alpha^{2}}{(1+\alpha)}$$
I. $$K_{a}=\dfrac {\alpha^{2}C}{(1-\alpha)}$$
Answer: G
Which of the following aqueous solutions produce the same osmotic pressure

Options:
A. 1.0 \,g \,of \,a \,non - electrolyte \,solute \,(X) \,in \,200 \,ml \,solution \,(Molar \,mass \,of \,X = 200)
B. $1.0 \,g$ of a non - electrolyte solute (X) in $150 \,ml$ solution (Molar mass of $X = 200$)
C. 1.0 \,g \,of \,a \,non - electrolyte \,solute \,(X) \,in \,125 \,ml \,solution \,(Molar \,mass \,of \,X = 200)
D. $1.0 \,g$ of a non - electrolyte solute (X) in $50 \,ml$ solution (Molar mass of $X = 200$)
E. $0.1 \,M \,NaCl$ solution
F. 1.0 \,g \,of \,a \,non - electrolyte \,solute \,(X) \,in \,60 \,ml \,solution \,(Molar \,mass \,of \,X = 200)
G. $1.0 \,g$ of a non - electrolyte solute (X) in $100 \,ml$ solution (Molar mass of $X = 200$)
H. $1.0 \,g$ of a non - electrolyte solute (X) in $75 \,ml$ solution (Molar mass of $X = 200$)
Answer: D
A magnet of total magnetic moment $ 10^{-2} \hat i A-m^2 $ placed in a time varying magnetic field , $ B \hat i ( cos \omega t ) $where B= tesla and $ \omega  =  rad/s . $ the work done for reversing the direction of the magnetic moment at t=  second, is

Options:
A. 0.01 J
B. 0.007 J
C. 0.012 J
D. 0.005 J
E. 0.015 J
F. 0.02 J
G. 0.004 J
H. 0.011 J
I. 0.028 J
J. 0.014 J
Answer: F
A ball of mass $m$ falls vertically from a height $h$ and collides with a block of equal mass $m$ moving horizontally with a velocity $v$ on a surface. The coefficient of kinetic friction between the block and the surface is ${ \mu  }_{ k }=0.2$, while the coefficient of resittution ($e$) between the ball and the block is $0.5$. There is no friction acting between the ball and the block. The velocity of the block just after the collision decreases by

Options:
A. $$0$$
B. $$0.6\sqrt{2gh}$$
C. can't be said
D. $$0.4\sqrt{2gh}$$
E. $$0.7\sqrt{2gh}$$
F. $$0.1\sqrt{2gh}$$
G. $$0.3\sqrt{2gh}$$
H. $$0.5\sqrt{2gh}$$
I. $$0.8\sqrt{2gh}$$
J. $$0.2\sqrt{2gh}$$
Answer: C
The Haber's process for production of ammonia involves the equilibrium:
$${N}_{2}(g)+3{H}_{2}(g)\rightleftharpoons  2{NH}_{3}(g)$$
Assuming $\Delta {H}^{o}$ and $\Delta {S}^{o}$ for the reaction do not change with temperature, which of the statements is true?
($\Delta {H}^{o}=-95kJ$ and $\Delta {S}^{o}=-190J{ K }^{ -1 }$)

Options:
A. Ammonia dissociates spontaneously below $500K$
B. Ammonia dissociates spontaneously above $450K$
C. Ammonia dissociates spontaneously above $525K$
D. Ammonia dissociates spontaneously above $425K$
E. Ammonia dissociates spontaneously above $400K$
F. Ammonia dissociates spontaneously above $500K$
G. Ammonia does not dissociate at any temperature
H. Ammonia dissociates spontaneously below $450K$
I. Ammonia dissociates spontaneously above $475K$
J. Ammonia dissociates at all temperatures
Answer: A
A steel drill is making 180 revolutions per minute under a constant couple of 5 Nm. If it drills a hole in 7 seconds in a steel block of mass 600 gm, the rise in temperature of the block is: (S=0.1 cal/gm/K)

Options:
A. $$48^{0}C$$
B. $$3^{0}C$$
C. $$5.2^{0}C$$
D. $$1.5^{0}C$$
E. $$45^{0}C$$
F. $$43^{0}C$$
G. $$46^{0}C$$
H. $$44^{0}C$$
I. $$1.3^{0}C$$
J. $$47^{0}C$$
Answer: G
In the above question, the intensity of the waves reaching a point P far away on the x-axis from each of the four sources is almost the same and equal to $I_0.$ Then,

Options:
A. If $d=\lambda /6,$ the intensity at P is $3I_0.$
B. If $d=\lambda /6,$ the intensity at P is $5I_0.$
C. If $d=\lambda /4,$ the intensity at P is $4I_0.$
D. If $d=\lambda /6,$ the intensity at P is $4I_0.$
E. If $d=\lambda /2,$ the intensity at P is $3I_0.$
F. None of these is true
G. If $d=\lambda /6,$ the intensity at P is $2I_0.$
H. If $d=\lambda /6,$ the intensity at P is $7I_0.$
I. If $d=\lambda /6,$ the intensity at P is $8I_0.$
J. If $d=\lambda /6,$ the intensity at P is $6I_0.$
Answer: A
The dissociation constant of diprotic weak acid $Ka_1$ $(H_2A)=10^{-7}$ and $Ka_2$ $(H_2A)=10^{-11}$. The equilibrium constant $H_2A+2HO^-\leftrightharpoons A^{-2}+2H_2O$ is

Options:
A. $$10^{-10}$$
B. $$10^{-24}$$
C. $$10^{-17}$$
D. $$10^{-18}$$
E. $$10^{-20}$$
F. $$10^{18}$$
G. $$10^{-7}$$
H. $$10^{-16}$$
I. $$10^{10}$$
J. $$10^{-14}$$
Answer: D
There are three elements A, B, and C. Their atomic number is $Z_1, Z_2$, and $Z_3$ respectively. If $Z_1-Z_2=2$ and $\dfrac{Z_1+Z_2}{2}=Z_3-2$ end the electronic configuration of element A is $[Ar]3d^64s^2$,then correct order of magnetic momentum is/are :

Options:
A. $$B^{2+} > A^{2+} >C^{3+}$$
B. B^{2+} > A > C
C. B > A^{2+} > C^{2+}
D. $$B=A^{3+} > C^{3+}$$
E. B > A^{2+} > C
F. $$B^{3+} > A > C^{2+}$$
G. $$B > C^{2+} > A$$
H. $$B > A >C^{2+}$$
Answer: H
The bond energies (in kJ/mol) at $25^0C$ are $C-C = 346, C-H = 413, H-H= 437, C= C; 611$. From these data, the value of $\Delta H $ at $25^0C$ for the above reaction is ______________.

Options:
A. $$248\ kJ\ mol^{-1}$$
B. $$144\ kJ\ mol^{-1}$$
C. $$124\ kJ\ mol^{-1}$$
D. $$-289\ kJ\ mol^{-1}$$
E. $$-134\ kJ\ mol^{-1}$$
F. $$-124\ kJ\ mol^{-1}$$
G. $$289\ kJ\ mol^{-1}$$
H. $$134\ kJ\ mol^{-1}$$
I. $$-248\ kJ\ mol^{-1}$$
Answer: C
Consider the following hypothetical equilibrium,$2B(g)$$\rightleftharpoons$$B_{2}(g)$
If d is observed vapour density and D is the theoretical vapour density, then degree of association ($\alpha$) will be

Options:
A. $$\alpha = 2\left( \frac{D + d}{d} \right)$$
B. $$\alpha = 2\left( \frac{D - d}{D} \right)$$
C. $$\alpha =\frac { 2D-d }{ D } $$
D. $$\alpha = 2\left( \frac{D - d}{2d} \right)$$
E. $$\alpha = 2\left( \frac{D - d}{2D} \right)$$
F. $$\alpha =2\left( \frac { D-d }{ d } \right) $$
G. $$\alpha = 2\left( \frac{D - 2d}{d} \right)$$
H. $$\alpha =\frac { 2D }{ D-d } $$
I. $$\alpha =2-\frac { 2D }{ d } $$
J. $$\alpha = 2\left( \frac{D + d}{D} \right)$$
Answer: F
A block B is tied to one end of a uniform rope $R$ as shown. The mass of the block is $2 kg$ and that of rope is 1 kg. A force $F=15$ is applied at angle $37^o$ with vertical. The tension at the mid-point of rope is

Options:
A. $$1.2 N$$
B. $$1.6 N$$
C. $$1.8 N$$
D. $$3.5 N$$
E. $$4.5 N$$
F. $$2.5 N$$
G. $$2 N$$
H. $$3 N$$
I. $$2.2 N$$
J. $$1.5 N$$
Answer: J