MikaStars39/nano-eval huggingface dataset:
Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the list
{$en={ H }_{ 2 }NC{ H }_{ 2 }C{ H }_{ 2 }N{ H }_{ 2 }$; atomic numbers: Ti=22, Cr=24; Co=27; Pt=78
List-IList-IIP$[Cr(N{ H }_{ 3 }{ ) }_{ 4 }C{ I }_{ 2 }]CI$1.Paramagnetic and exhibits ionization isomerismQ$[Ti({ H }_{ 2 }O{ ) }_{ 5 }CI]({ NO }_{ 3 }{ ) }_{ 2 }$2.Diamagnetic and exhibits cis-trans isomerismR.$[Pt(en)(N{ H }_{ 3 })CI]{ NO }_{ 3 }$3.Paramagnetic and exhibits cis-trans isomerismS.$[Co(N{ H }_{ 3 }{ ) }_{ 4 }(N{ O }_{ 3 }{ ) }_{ 2 }]{ NO }_{ 3 }\ $4.Diamagnetic and exhibits ionization esomerism

Options:
A. P Q R S
B. 3 1 4 2
C. P Q R S
4 2 3 1
D. 2 1 3 4
E. P Q R S
1 3 4 2
F. P Q R S
1 4 2 3
G. P Q R S
2 1 3 4
H. 1 4 2 3
I. 1 3 4 2
J. P Q R S
3 1 4 2
Answer: C
For the reaction $A(g)+2B(g)\rightleftharpoons 3C(g)+D(g);{ K }_{ p }=0.05atm$ at 1000 K . The value of ${ K }_{ e }$ is is represents by .

Options:
A. $$\dfrac { 5 \times { 10 }^{ -5 } }{ R^2 }$$
B. $$5 \times { 10 }^{ -4 }R^2$$
C. $$\dfrac { 5\times { 10 }^{ -5 } }{ R } $$
D. $$5\times { 10 }^{ -4 }R$$
E. $$\dfrac { 5\times { 10 }^{ -4 } }{ R } $$
F. $$5 \times { 10 }^{ -4 }R^{-1}$$
G. $$5 \times { 10 }^{ -5 }R$$
H. $$5 \times { 10 }^{ -4 }R^3$$
I. $$5 \times { 10 }^{ -3 }R$$
Answer: D
An ideal diatomic gas is expanded so that the amount of heat transferred to the gas is equal to the decrease in its internal energy.
The process can be represented by the equation $TV^n$= constant where the value of n is:

Options:
A. $$n=\frac {1}{5}$$
B. $$n=\frac {3}{5}$$
C. $$n=\frac {7}{5}$$
D. $$n=\frac{4}{5}$$
E. $$n=\frac{1}{3}$$
F. $$n=\frac {2}{5}$$
G. $$n=\frac{1}{4}$$
H. $$n=\frac{2}{3}$$
I. $$n=\frac {3}{2}$$
J. $$n=\frac{3}{4}$$
Answer: A
In which of the following ways the concentration of hydrazine, $N_{2}H_{4}$, could be increased in an equilibrium described by the equation  :$N_{2}(g)+2H_{2}(g)\rightleftharpoons N_{2}H_{4}(g)\: \Delta H=95kJ$

Options:
A. increase the temperature
B. heat the reaction
C. decrease the temperature
D. decrease the temperature and add \( H_2 \)
E. cool the reaction
F. decrease the pressure
Answer: B
A particle of mass $m$ is moving along the side of a square of side $'a'$. with a uniform speed $
u $ in the $x-y$ plane as shown in the figure:
Which of the following statement is false for the angular momentum $L$ about the origin?

Options:
A. $L=mv \left[ \dfrac { R }{ \sqrt { 2 } } -a \right] \hat{k}$ when the particle is moving from $C$ to $D$
B. $L=\dfrac { mv }{ \sqrt { 2 } } R\hat{k}$ when the particle is moving from $A$ to $B$
C. L=\dfrac { mv }{ \sqrt { 3 } } R\hat{k}
D. $L=\dfrac { mv }{ \sqrt { 2 } } (R - a)\hat{k}$ when the particle is moving from $A$ to $B$
E. $L=\dfrac { mv }{ \sqrt { 2 } } (R + a)\hat{k}$ when the particle is moving from $A$ to $B$
F. $L=\dfrac { mv }{ \sqrt { 2 } } (R + a)\hat{k}$ when the particle is moving from $B$ to $C$
G. $L=mv \left[ \dfrac { R }{ \sqrt { 2 } } +a \right] \hat{k}$ when the particle is moving from $B$ to $C$
H. $L=\dfrac { mv }{ \sqrt { 2 } } (R + 2a)\hat{k}$ when the particle is moving from $A$ to $B$
I. L=\dfrac { mv }{ \sqrt { 2 } } (R - a)\hat{k} when the particle is moving from $D$ to $A$
J. $L=-\dfrac { mv }{ \sqrt { 2 } } R\hat{k}$ when the particle is moving from $D$ to $A$
Answer: B
What will b the result if $100ml{\text{ of }}0.06{\text{ M }}Mg{\left( {N{O_3}} \right)_2}$  is added to $50ml{\text{ of 0}}{\text{.06M }}N{a_2}{C_2}{O_4}?$ $\left[ {{K_{sp}}{\text{ of }}Mg{C_2}{O_4} = 8.6{\text{ x }}{{10}^{ - 5}}} \right]$

Options:
A. A precipitate will form and an excess of $Mg{^{2 + }}$ ions will remain in the solution
B. A precipitate will form and an excess of $Na^+$ and ${C_2}{O_4}^{2 - }$ ions will remain in the solution
C. A precipitate will form and an excess of $Na^+$ ions will remain in the solution
D. A precipitate is form but neither ion is present in excess
E. A precipitate will form and an excess of $Na^+$ and $NO_3^-$ ions will remain in the solution
F. A precipitate will form and an excess of $Na^+$ and $Mg^{2+}$ ions will remain in the solution
G. A precipitate will form and an excess of ${C_2}{O_4}^{2 - }$ ions will remain in the solution
H. A precipitate will form and an excess of $Mg^{2+}$ ions will remain in the solution
I. A precipitate will form and an excess of $NO_3^-$ ions will remain in the solution
J. A precipitate will not be formed
Answer: G
The correct order of increasing basicity of the given conjugate bases (${R}=$ $CH_3$) is:

Options:
A. $\mathrm{H}\mathrm{C} ={\mathrm{C}}^-<\mathrm{R}\mathrm{C}\mathrm{O}{\mathrm{O}}^-<{\mathrm{R}}^-<{\mathrm{N}}\mathrm{H}_{2}^-$
B. $\mathrm{R}\mathrm{C}\mathrm{O}\mathrm{O}^-< \mathrm{H}\mathrm{C}$ $\equiv {\mathrm{C}}^-<\mathrm{N}\mathrm{H}_{2}^-<\mathrm{R}^-$
C. $\mathrm{R}\mathrm{C}\mathrm{O}{\mathrm{O}}^-<$ $\mathrm{H} \mathrm{C} ={\mathrm{C}}^-<{\mathrm{R}}^-<{\mathrm{N}}\mathrm{H}_{2}^-$
D. $\mathrm{R}\mathrm{C}\mathrm{O}\mathrm{O}^-<\mathrm{N}\mathrm{H}_{2}^-<$ HC $\equiv {\mathrm{C}}^-<\mathrm{R}^-$
E. $\mathrm{R}^-<$ HC $\equiv {\mathrm{C}}^-<\mathrm{R}\mathrm{C}\mathrm{O}\mathrm{O}^-<\mathrm{N}\mathrm{H}_{2}^-$
F. $$\mathrm{R}\mathrm{C}\mathrm{O}\mathrm{O}^-<\mathrm{N}\mathrm{H}_{2}^-<\mathrm{R}^-< \mathrm{H}\mathrm{C} \equiv {\mathrm{C}}^-$$
Answer: B
An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown.
The radius of the loop is $a$ and distance of its center from the wire is $d(d > > a)$. If the loop applies a force $F$ on
the wire then

Options:
A. $$F \propto \left(\dfrac{a}{d}\right)^3$$
B. $$F \propto \left(\dfrac{a^{2}}{d^{3}}\right)$$
C. $$F \propto \left(\dfrac{a}{d}\right)^{2}$$
D. $$F \propto \left(\dfrac{a}{d^2}\right)$$
E. $$F \propto \left(\dfrac{a}{d^3}\right)$$
F. $$F \propto \left(\dfrac{a^2}{d^2}\right)$$
G. $$F \propto \left(\dfrac{a}{d}\right)$$
H. $$F = 0$$
I. $$F \propto \left(\dfrac{a^2}{d^3}\right)$$
J. $$F \propto \left(\dfrac{a^{3}}{d^{3}}\right)$$
Answer: C
The equation of a wave on a string of linear mass density $ 0.04 k g m ^ { - 1 } \text { is given by } y = 0.02 ( m ) \left[ 2 \pi \left[ \frac { 1 } { 0.04 ( s ) } \frac { x } { 0.50 ( m ) } \right) \right] $ The tension in the string is

Options:
A. 3.0 N
B. 4.5 N
C. 0.5 N
D. 12.5 N
E. 4 N
F. 2.0 N
G. 3.5 N
H. 8.0 N
I. 6.25 N
J. 3.2 N
Answer: I
A force of $10 \,N$ is applied on the right block of a system of two blocks of $1 \,kg$ mass each, on a rough horizontal plane. coefficient of friction is $\mu = 0.2$ for both blocks. The spring is initially in relaxed state. Acceleration of centre of mass at initial instant will be $(g = 10 \,m/s^2)$

Options:
A. $$5 \,m/s^2$$
B. $$3.25 \,m/s^2$$
C. $$3.5 \,m/s^2$$
D. $$6 \,m/s^2$$
E. $$2.5 \,m/s^2$$
F. $$4 \,m/s^2$$
G. $$2 \,m/s^2$$
H. $$3 \,m/s^2$$
I. $$5.5 \,m/s^2$$
J. $$3.75 \,m/s^2$$
Answer: F
When $0.05$ M dimethyl amine is dissolve in $0.1$M NaOH solution then the percentage dissociation of dimethyl amine is? $(K_b)_{(CH_2)_2NH}=5\times 10^{-4}$.

Options:
A. $$5 \times 10^{-10}$$
B. $$5\times 10^{-3}$$
C. $$5\times 10^{-1}$$
D. $$5 \times 10^{-8}$$
E. $$5\times 10^{-2}$$
F. $$5\times 10^{-5}$$
G. $$5 \times 10^{-4}$$
H. $$5 \times 10^{-7}$$
I. $$5 \times 10^{-6}$$
J. $$5 \times 10^{-9}$$
Answer: C
Determine the maximum horizontal force F that may be applied to the plank of mass m for which the solid sphere does not slip as it begins to roll on the plank. The sphere has a mass M and radius R. The coefficient of static and kinetic friction between the sphere and the plank are $\mu_s$ and $\mu_k$, respectively

Options:
A. The linear acceleration for sphere will be $\mu_{s}g \left (1 + \dfrac{1}{2} \right)$
B. The maximum horizontal force, $F = \mu_{s}g \left (M + \dfrac{3}{2}m \right )$
C. The maximum horizontal force is \( F = \mu_{s}g \left (M + \dfrac{13}{2}m \right) \)
D. The maximum horizontal force is $F =\mu_{s}g\left ( M + \dfrac{7}{2}m\right )$
E. None of the above
F. The linear acceleration for sphere will be $\mu_{s}g \left (1 + \dfrac{3}{2} \right)$
G. The maximum horizontal force is $F = \mu_{s}g \left (M + \dfrac{5}{2}m \right)$
H. The maximum horizontal force is $F = \mu_{s}g \left (M + \dfrac{11}{2}m \right)$
I. The maximum horizontal force is $F = \mu_{s}g \left (M + \dfrac{9}{2}m \right)$
J. The linear acceleration for sphere will be $\mu_{s}g$
Answer: J
A car is moving on a plane inclined at $30^0$ to the horizontal with an acceleration of $9.8 m/s^2$ parallel to the plane upward. A bob is suspended by s string from the roof. The angle in degrees which the string makes with the vertical is : (assume that the bob does not move relative to car) [ g = 0=9.8 $m/s^2$]

Options:
A. $$30^0$$
B. $$10^0$$
C. $$35^0$$
D. $$45^0$$
E. $$50^0$$
F. 23^0
G. $$60^0$$
H. $$20^0$$
I. $$15^0$$
J. $$25^0$$
Answer: H
Consider the Hofmann ammonolysis reaction:If $R = {Me}_{3}C-$ ($t-butyl$), the correct order of basic character of the above amines in aqueous medium is :

Options:
A. $$N{H}_{3} > 1 > 2 > 3$$
B. $$1 > N{H}_{3} > 2 > 3 $$
C. $$1 > N{H}_{3} > 3 > 2$$
D. $$1 > 2 > 3 > N{H}_{3}$$
E. $$1 > 2 > N{H}_{3} > 3$$
F. $$3 > 1 > 2 > N{H}_{3}$$
G. $$3 > N{H}_{3} > 1 > 2$$
H. $$3 > 1 > N{H}_{3} > 2$$
I. $$3 > 2 > 1 > N{H}_{3}$$
Answer: A
The period of a particle executing SHM is $8\ s$. At $t = 0$ it is at the mean position. The ratio of the distances covered by the particle in the $1st$ second to the $2nd$ second is

Options:
A. $$\sqrt {2} + 1$$
B. $$\dfrac{1}{\sqrt{3} + 2}$$
C. $$\dfrac {1}{\sqrt {2}}$$
D. \dfrac {1}{\sqrt {3}}
E. $$\dfrac {1}{2\sqrt {2}}$$
F. $$\dfrac{1}{\sqrt{2} + 1}$$
G. $$\sqrt {2}$$
H. $$\dfrac {1}{\sqrt {3} + 1}$$
I. $$\dfrac {1}{2}$$
Answer: C
The freezing point of a 0.1M formic acid aqueous solution is ${ -0.2046 }^{ 0 }C$. Find equilibrium constant of the reaction - 
$${ HCOO }^{ - }(aq.)+{ H }_{ 2 }O(\ell )\quad \rightleftharpoons \quad HCOO(aq.)+{ OH }^{ - }(aq.)$$
Given : ${ K }_{ f }({ H }_{ 2 }O)\quad=\quad 1.86\quad K-kg\quad { mole }^{ -1 }$
Assume solution to be very dilute

Options:
A. $$9\times { 10 }^{ -12 }\quad M$$
B. $$1.1 \times 10^{-10} \quad M$$
C. $$9 \times 10^{-11} \quad M$$
D. $$1.1 \times 10^{-11} \quad M$$
E. $$1.1\times { 10 }^{ -11 }\quad M$$
F. $$1.1\times { 10 }^{ -3 }\quad M$$
G. $$1.1 \times 10^{-14} \quad M$$
H. $$1.1 \times 10^{-13} \quad M$$
I. $$9\times { 10 }^{ -13 }\quad M$$
J. $$1.1 \times 10^{-12} \quad M$$
Answer: A
The equation of the wavelength representation a traveling wave is y=$\[\frac{1}{{1 + {x^2}}}\]$at=$0$ After $0.2$ s, the equation of waveform changes to y=$\[\frac{1}{{{x^2} + 2\left( {x + 1} \right)}}\]$ The velocity of the wave is

Options:
A. 4 m/s towards +ve x-axis
B. 3 m/s towards +ve x-axis
C. 3 m/s towards -ve x-axis
D. $5$ m/s towards-ve x-axis
E. $1$ m/s towards-ve x-axis
F. $2$ m/s towards +ve x-axis
G. $1$ m/s towards +ve x-axis
H. 2 m/s towards -ve x-axis
I. 4 m/s towards -ve x-axis
J. $5$ m/s towards +ve x-axis
Answer: J
At $1200^{o}C$, mixture of $Cl_{2}$ and $Cl$ atoms (both in gaseous state) effuses 1.16 times faster than krypton effuses under identical conditions. The fraction of chlorine molecules dissociated into atoms is approximate: [MW of Kr $=$ 84]

Options:
A. 17%
B. 87%
C. 22%
D. 34%
E. 41%
F. 13%
G. 19%
H. 16%
I. 44%
J. 38%
Answer: F
For the reaction,${\text{2N}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \rightleftharpoons {\text{2NO}}\left( {\text{g}} \right){\text{ + }}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right),\;\;\;{K_C} = 1.8 \times {10^{ - 6}}\;\;at\;{\text{18}}{{\text{5}}^{\text{o}}}{\text{C}}{\text{.}}\;{\text{At}}\;{\text{18}}{{\text{5}}^{\text{o}}}{\text{C}}\;$ the value of ${K_C}$ for the reaction -
$${\text{NO}}\left( {\text{g}} \right) + \frac{1}{2}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \rightleftharpoons {\text{N}}{{\text{O}}_2}\left( {\text{g}} \right)$$

Options:
A. $$1.8 \times {10^{-3}}$$
B. $$1.8 \times {10^3}$$
C. $$1.9 \times {10^{-6}}$$
D. $$7.5 \times {10^2}$$
E. $$1.8 \times {10^6}$$
F. $$0.9 \times {10^{ - 3}}$$
G. $$1.9 5\times {10^3}$$
H. $$1.9 \times {10^{-3}}$$
I. $$0.9 \times {10^6}$$
J. $$0.9 \times {10^{ - 6}}$$
Answer: I
A particle with charge $+Q$ and mass $m$ enters a magnetic field of magnitude B existing only to the right of the boundary YZ. The direction of the motion of the particle is perpendicular to the direction of $B$ let $T=\pi\dfrac{m}{QB}$.The time spent by the particle in the field will be

Options:
A. $$T\theta$$
B. $$T\left(\dfrac{\pi-2\theta}{2\pi}\right)$$
C. $$T\left(\dfrac{\pi + 2\theta}{\pi}\right)$$
D. $$T\left(\dfrac{\pi + 2\theta}{5\pi} \right)$$
E. $$2T\theta$$
F. $$T\left(\dfrac{\pi - 2 \theta}{\pi} \right)$$
G. $$T\left(\dfrac{2\pi + 2\theta}{\pi}\right)$$
H. $$T\left(\dfrac{\pi +2 \theta }{2\pi}\right)$$
I. $$T\left(\dfrac{\pi + 2\theta}{3\pi} \right)$$
J. $$T\left(\dfrac{\pi + 2\theta}{4\pi}\right)$$
Answer: H
Y,K,n represent the Young's modulus, bulk modulus and rigidity modulus of a body respectively. If rigidity modulus is twice the bulk modulus, then

Options:
A. $$Y=9 n/18$$
B. $$Y=10 n/18$$
C. $$Y=9K/5$$
D. $$Y=9 n/5$$
E. $$Y=5 n/9$$
F. $$Y=5 n/10$$
G. $$Y=18 n/9$$
H. $$Y=9 n/10$$
I. $$Y=5K/18$$
J. $$Y=18 K/5$$
Answer: J
In an electrical circuit two resistances $\mathrm{R}_{1}=(4\pm 0.5)$ and $\mathrm{R}_{2}=(12\pm 0.5)$ are connected in parallel. What will be the equivalent resistance with percentage error?
( Given the effective resistance$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$ )

Options:
A. $16\pm 18.6$%
B. $3\pm 19.6$%
C. $3\pm 22.92$%
D. $16\pm 22.92$%
E. $3\pm 20.92$%
F. $3\pm 16.6$%
G. $3\pm 17.6$%
H. $3\pm 21.92$%
I. $3\pm 18.6$%
J. $3\pm 21.6$%
Answer: C
The following observations were taken for determining surface tension $T$ of water by the capillary method : 
The diameter of capillary,  $D=1.25 \times 10^{-2}\, m$  rise of water,  $h=1.45 \times 10^{-2}\, m$. Using  $g=9.80\, m/s^2$  and the simplified relation  $T = \frac{{rhg}}{2} \times {10^3}N/m$,  the possible error in surface tension is closest to :

Options:
A. $2.4$%
B. $15$%
C. $0.15$%
D. 0.25%
E. 1.75%
F. 0.20%
G. $1.5$%
H. 0.10%
I. 0.25
Answer: C
Calculate the pH at which an acid indicator with ${ K }_{ a }-1.0\times { 10 }^{ 5 }\ $ changes color when the indicator is $1.00\times { 10 }^{ 4 }$ M.

Options:
A. pH = 3.00
B. pH = 4.00
C. pH = 5.00
D. pH = 6.00
E. pH = 1.00
F. pH = 2.00
G. pH = 2.50
H. pH = 2.25
I. pH = 1.50
J. pH = 1.75
Answer: F
For the hemisphere of radius 'R' shown in the figure the surface charge density $'\sigma '$ varies with '$\theta$ ' (shown in figure) as $\sigma  = {\sigma _0}\cos \theta .$ The total charge on the sphere is

Options:
A. $$\dfrac{{{\sigma _0}\pi {R^2}}}{6}$$
B. $$\dfrac{{{\sigma _0}\pi {R^2}}}{4}$$
C. $$\dfrac{{7{\sigma _0}\pi {R^2}}}{4}$$
D. $$\dfrac{{{\sigma _0}\pi {R^2}}}{3}$$
E. $$\dfrac{{{\sigma _0}\pi {R^2}}}{2}$$
F. $$3{\sigma _0}\pi {R^2}$$
G. $$\dfrac{{5{\sigma _0}\pi {R^2}}}{4}$$
H. $$2{\sigma _0}\pi {R^2}$$
I. $${{\sigma _0}\pi {R^2}}$$
J. $$\dfrac{{3{\sigma _0}\pi {R^2}}}{4}$$
Answer: H
Let $T_1$ be surface tension between solid and air, $T_2$ the surfaced tension between solid and liquid and T the surface tension between liquid and air. Then in equilibrium, for a drop of liquid on clean glass plate the relation is $(\theta$ is angle of contact $)$

Options:
A. $$cos\theta=\dfrac{T_1-T_2}{T}$$
B. $$cos\theta=\dfrac{T_1-T}{T_2-T}$$
C. $$cos\theta=\dfrac{T_1+T_2}{T}$$
D. $$cos\theta=\dfrac{T_2-T}{T_1-T}$$
E. $$cos\theta=\dfrac{T_1-T}{T_2}$$
F. $$cos\theta=\dfrac{T_2-T_1}{T_1-T}$$
G. $$cos\theta=\dfrac{T}{T_1-T_2}$$
H. $$cos\theta=\dfrac{T}{T_2-T_1}$$
I. $$cos\theta=\dfrac{T_2-T_1}{T}$$
J. $$cos\theta=\dfrac{T}{T_1+T_2}$$
Answer: A
Calculate the volume occupied by 16 gram $O_2$ at 300K and 8.31 MPa if $\dfrac{P_cV_c}{RT_c}$ = 3/8 and $\dfrac{P_rV_r}{T_r}$=2.21.
(Given : R = 8.314 J/ K-mol)

Options:
A. 124.32 mL
B. 250.62 mL
C. 124.31 mL
D. 125.32 mL
E. 126.62 mL
F. 249.62 mL
G. None of these
H. 248.62 mL
I. 126.31 mL
J. 125.31 mL
Answer: C
The work done in the expansion of a real gas ($n$ moles) from volume $\displaystyle { V }_{ 1 }$ to $\displaystyle { V }_{ 2 }$, obeying Van der Waals' equation at temperature T where volume occupied by the molecules may be taken as negligible in comparison to the total volume of gas is :

Options:
A. $$\displaystyle =-nRT\log { \frac { { V }_{ 1 } }{ { V }_{ 2 } } } +{ n }^{ 2 }a\left[ \frac { 1 }{ { V }_{ 2 } } -\frac { 1 }{ { V }_{ 1 } } \right]$$
B. $$\displaystyle =-nRT\log { \frac { { V }_{ 2 } }{ { V }_{ 1 } } } -{ n }^{ 2 }a\left[ \frac { 1 }{ { V }_{ 1 } } -\frac { 1 }{ { V }_{ 2 } } \right]$$
C. $$\displaystyle =-nRT\log { \frac { { V }_{ 2 } }{ { V }_{ 1 } } } +{ n }^{ 2 }a\left[ \frac { 1 }{ { V }_{ 2 } } -\frac { 1 }{ { V }_{ 1 } } \right]$$
D. $$=-nRT\log { \frac { { V }_{ 1 } }{ { V }_{ 2 } } } -{ n }^{ 2 }a\left[ \frac { 1 }{ { V }_{ 1 } } -\frac { 1 }{ { V }_{ 2 } } \right]$$
E. $$\displaystyle =-nRT\log { \frac { { V }_{ 2 } }{ { V }_{ 1 } } } +{ n }^{ 2 }a\left[ \frac { 1 }{ { V }_{ 1 } } -\frac { 1 }{ { V }_{ 2 } } \right] $$
F. $$\displaystyle =-nRT\log { \frac { { V }_{ 2 } }{ { V }_{ 1 } } } -{ n }^{ 2 }a\left[ \frac { 1 }{ { V }_{ 2 } } -\frac { 1 }{ { V }_{ 1 } } \right] $$
G. $$\displaystyle =-nRT\log { \frac { { V }_{ 1 } }{ { V }_{ 2 } } } -{ n }^{ 2 }a\left[ \frac { 1 }{ { V }_{ 2 } } -\frac { 1 }{ { V }_{ 1 } } \right]$$
Answer: F
A semi circular current carrying wire having radius R is placed in x-yplane with its centre at origin 'O'. There is non-uniform magnetic field \[ \vec{B} = \frac{B_0 x}{2R} \hat{k} \](here $B_o$ is $+$ve constant) is existing in the region. The magnetic force acting on semi circular wire will be along.

Options:
A. -\hat{i}
B. $+$x-axis
C. $-\hat{j}$
D. $-2x$-axis
E. $-y$-axis
F. $+\hat{j}$
G. $+$y-axis
H. $-$x-axis
Answer: H
8 g weak acid HX (molecular mass = 80 u; $K_a = 10^{-4}$) is dissolved in 1-00 mL water. If it is titrated with 0.25 M NaOH, find pH at equivalence point:

Options:
A. 8.85
B. 9
C. 8.95
D. 9.15
E. 8.7
F. 9.25
G. 9.45
H. 8.65
I. 9.35
J. 9.2
Answer: I
Minimum energy of proton to ionize He+ ion (which is in ground state and is at rest) is $(consider\,\, mass\, m_{He+} = 4 m_p)$

Options:
A. 70.4 eV
B. 54.4 eV
C. 76.8 eV
D. 102 eV
E. 51.2 eV
F. 108.8 eV
G. 72 eV
H. 136 eV
I. 68 eV
J. 52.4 eV
Answer: B
The standard enthalpy and entropy changes for the reaction in equilibrium for the forward reaction are given below. $CO(g)+H_{2}O(g)\rightleftharpoons CO_{2}(g)+H_{2}(g) $$\Delta H^{^{\circ}}_{300K}=-41.16 kJ mol^{-1} $$\Delta S^{^{\circ}}_{300K}=-4.25 X 10^{-2} kJ mol^{-1} $$\Delta H^{^{\circ}}_{1200K}=-32.96 kJ mol^{-1} $Then, the incorrect statement is?

Options:
A. The reaction proceeds in'the forward direction at 300 K
B. At 300 K, the products will be favored more than reactants at equilibrium
C. At 300 K, the products will be favored equally to the reactants at equilibrium
D. At 1200 K, the products will be favored more than reactants at equilibrium
E. At 1200 K, $ K_{p} $> 1
F. At 300 K, the reactants will be favored more than the products at equilibrium
G. At 300 K, the reactants will be favored more than products at equilibrium
H. At 300 K, the reactants will be favored equally to the products at equilibrium
I. At 300 K, $ K_{p} $< 1
J. At 1200 K, reaction proceeds in the reverse direction
Answer: B
For the reaction in euilibrium :$2NOBr(g)$$\rightleftharpoons$$2NO(g)+Br_{2}(g)$ if $P_{br2}$=p/9 at equilibrium and Pis the total pressure.The ratio $K_{p}/P$ is equal to-

Options:
A. $$\frac{1}{9}$$
B. $$\frac{1}{45}$$
C. $$\frac{1}{81}$$
D. $$\frac{1}{36}$$
E. $$\frac{1}{72}$$
F. $$\frac{1}{18}$$
G. $$\frac{1}{54}$$
H. $$\frac{1}{27}$$
I. $$\frac{1}{24}$$
J. $$\frac{1}{25}$$
Answer: A
Two projectiles A and B are projected with same speed at angles $30^0$ and $60^0$ to horizontal, then choose the wrong statement? (symbols have their usual meaning)

Options:
A. $$H_A = \frac{1}{3}H_B$$
B. $$H_A = 3H_B$$
C. $$\sqrt{3} T_B = T_A$$
D. H_A = \frac{1}{2}H_B
E. All of these
F. $$R_A = R_B$$
G. $$H_A = 4H_B$$
H. $$H_A = \frac{1}{4}H_B$$
I. $$H_B = 3H_A$$
J. $$H_A = 2H_B$$
Answer: C
A long straight wire carries a current along the $x-$axis. Consider the points $A(0, 1, 0),\ B(0, 1, 1),\  C(1, 0, 1)$ and $D(1, 1, 1)$. If the current is $i$ and the magnetic field at $D$ has magnitude $B,$  then,

Options:
A. B makes an angle of $60^o$ with the $xy$ plane
B. B makes an angle of $15^o$ with the $xy$ plane
C. $$B=\dfrac {\mu_0i}{2\pi \sqrt {3}}$$
D. B makes an angle of $50^o$ with the $xy$ plane
E. B makes an angle of $75^o$ with the $xy$ plane
F. $B$ makes an angle of $55^o$ with the $xy$ plane
G. $$B=\dfrac {\mu_0i}{2\sqrt {2}\pi}$$
H. B makes an angle of $30^o$ with the $xy$ plane
I. $B$ is parallel to the $z-$axis
J. $B$ makes an angle of $45^o$ with the $xy$ plane
Answer: J
A spring lies along the x-axis attached to a wall at one end and a block at the other end. The block rests on a friction less surface at x = 0. A force of constant magnitude F is applied to the block that begins to compress the spring, until the block comes to a maximum displacement  $x_{max}$.
During the first half of the motion, applied force transfers more energy to the

Options:
A. Depends upon the spring constant and the mass of the block
B. Kinetic energy and potential energy are equal during the first half of the motion
C. Equal to both
D. Depends upon the spring constant
E. Potential energy
F. Depends upon mass of the block
G. Depends upon the magnitude of the force
H. Kinetic energy and potential energy are equal
I. Kinetic energy
J. Potential energy is more during the first half of the motion
Answer: C
Separation between the plates of a parallel plate capacitor connected to a battery (zero resistance) of constant e.m.f. is increased with constant (very slow) speed by external forces During the process W is the work done by external forces $\displaystyle \Delta U $ is the change in potential energy of the capacitor, $\displaystyle W_{b}$ is the work done by the battery and H is the heat loss in the circuit Then

Options:
A. $$W = H$$
B. $$W = o$$
C. $$\displaystyle H=\Delta U$$
D. $$W = H + W_{b}$$
E. $$W = \Delta U + W_{b}$$
F. $$W = \Delta U$$
G. $$\displaystyle H
eq 0$$
H. $$W = \Delta U - H$$
I. $$\displaystyle W+W_{b}=\Delta U$$
J. $$W = \Delta U - W_{b}$$
Answer: B
A capacitor C is connected to two equal resistances as shown in the figure. Consider the following statements:
(i) At the time of charging of capacitor time constant of the circuit is 2CR
(ii) At the time of discharging of the capacitor the time constant of the circuit is CR
(iii) At the time of discharging of the capacitor the time constant of the circuit is 2CR
(iv) At the time of charging of capacitor the time constant of the circuit is CR

Options:
A. Statements (i) and (iii) only are correct
B. Statements (iii) and (iv) only are correct
C. Statements (i) and (iii) and (iv) only are correct
D. Statements (i) and (ii) only are correct
E. Statements (ii) and (iii) only are correct
F. Statements (ii) and (i) only are correct
G. Statements (ii) and (iii) and (iv) only are correct
H. Statements (i) and (iv) only are correct
I. Statements (i) and (ii) and (iv) only are correct
J. Statements (ii) and (iv) only are correct
Answer: B
Two moles of a monoatomic gas follows the process $PV^{\frac {4}{3}}$= constIf this is process is possible:

Options:
A. its molar heat capacity is positive
B. its molar heat capacity is \(\frac{2}{3}\)
C. its molar heat capacity is \(\frac{1}{4}\)
D. molar heat capacity can't be defined until volume or pressure is constant
E. its molar heat capacity is \(\frac{3}{4}\)
F. its molar heat capacity is zero
G. its molar heat capacity is \(\frac{1}{3}\)
H. its molar heat capacity is negative
I. its molar heat capacity is \(\frac{1}{2}\)
J. its molar heat capacity is \(\frac{3}{2}\)
Answer: F
A reaction can take place by two paths. $k_1$ and $k_2$ are rate constants for the two paths & $E_1$ and $E_2$ are their respective activation energies. At temperature $T_a : k_1 > k_2, E_1 < E_2$Its temperature is raised to $T_b$, the rate constants change to $k_1'$ & $k_2 '$. Which relation is correct between $k_1, k_2, k_1'$ & $k_2'$ (considering activation energy does not change with temperature)?

Options:
A. $$\displaystyle \frac{k_1'}{k_1} < \frac{k_2'}{k_2}$$
B. $$\frac{k_1'}{k_1} < \frac{k_2}{k_2}$$
C. $$\displaystyle \frac{k_1'}{k_1} > \frac{k_2'}{k_2}$$
D. $$\displaystyle \frac{k_1'}{k_1} = \frac{k_2'}{k_2}$$
E. $$\displaystyle \frac{k_1'}{k_1} < \frac{k_1}{k_2}$$
F. \frac{k_1'}{k_1} = \frac{k_2}{k_1}
G. \frac{k_1'}{k_1} > \frac{k_2}{k_2'}
H. \displaystyle \frac{k_1'}{k_1} > \frac{k_2}{k_1}
I. $$\displaystyle \frac{k_1'}{k_2'} > \frac{k_1}{k_2}$$
Answer: A
A gaseous reaction ${ A }_{ (g) }\rightleftharpoons 2{ B }_{ (g) }+{ C }_{ (g) }$ found to be of first order. If the reaction is started with $\rho_{A}=90mm Hg$. the total pressure after $10 min$ is found to be $180 mm Hg$. The rate constant of the reaction is

Options:
A. $5.4x10^{-3} s^{-1}$
B. $6.9x10^{-3} s^{-1}$
C. $4.8$x$10^{-3}$ $s^{-1}$
D. $7.1$x$10^{-3}$ $s^{-1}$
E. $5.2$x$10^{-3}$ $s^{-1}$
F. $3.8$x$10^{-3}$ $s^{-1}$
G. $4.6$x$10^{-3}$ $s^{-1}$
H. $4.90$x$10^{-3}$ $s^{-1}$
I. $1.15$x$10^{-3}$ $s^{-1}$
J. $6.9$x$10^{-2}$ $s^{-1}$
Answer: G
In the space shown a non-uniform magnrtic field $ \overrightarrow { B } =B_{ 0 }\left( I+x \right) \left( -\hat { k }  \right)  $ tesla is present A closed loop of small resistance, placed in the xy plane is given velocity $  V_0 $ The force due to magnetic field on the loop is

Options:
A. Along +x direction
B. along -v direction
C. zero
D. along + y direction
E. along -x direction
F. along -y direction
G. along -z direction
H. along + v direction
Answer: E
Optical axis of a thin equi-convex lens  is the X-axis. The co-ordinate of a point object and its image are (-20 cm, 1 cm) and (25 cm, -2 cm) respectively:-

Options:
A. The focal length of the lens is 8 cm
B. The focal length of the lens is 10 cm
C. The focal length of the lens is 12 cm
D. The focal length of the lens is 13 cm
E. The lens is located at X = 5 cm
F. The focal length of the lens is 9 cm
G. The focal length of the lens is 15 cm
H. The focal length of the lens is 14 cm
I. The lens is located at X = -5 cm
J. The focal length of the lens is 11 cm
Answer: I
Two nucleons are at a separation of $1\times { 10 }^{ -15 } m$ . The net force between them is ${ F }_{ 1 }$, if  both protons are neutrons, ${ F }_{ 2 }$ if both are protons  and ${ F }_{ 3 }$ if one is a proton and other is a neutron. In such a case

Options:
A. ${ F }_{ 1 }$>${ F }_{ 3 }$>${ F }_{ 2 }$
B. ${ F }_{ 2 }$>${ F }_{ 1 }$>${ F }_{ 3}$
C. ${ F }_{ 2 }$=${ F }_{ 1 }$>${ F }_{ 3}$
D. ${ F }_{ 2 }$>${ F }_{ 3 }$>${ F }_{ 1 }$
E. ${ F }_{ 1 }$>${ F }_{ 2 }$>${ F }_{ 3 }$
F. ${ F }_{ 1 }$=${ F }_{ 2 }$>${ F }_{ 3 }$
G. ${ F }_{ 1 }$=${ F }_{ 3 }$>${ F }_{ 2 }$
H. ${ F }_{ 3 }$>${ F }_{ 2 }$>${ F }_{ 1 }$
I. ${ F }_{ 1 }$=${ F }_{ 2 }$=${ F }_{ 3}$
J. ${ F }_{ 3 }$>${ F }_{ 1 }$>${ F }_{ 2 }$
Answer: G
A solid sphere of radius $R$ is set into motion on a rough horizontal surface with a linear speed ${v}_{0}$ in forward direction and an angular velocity ${ \omega  }_{ 0 }={ v }_{ 0 }/2R$ in counter coefficient of friction is $\mu$, then find the time after which sphere starts pure rolling.

Options:
A. $${ t }_{ 0 }=\cfrac { 10{ v }_{ 0 } }{ 7\mu g }$$
B. $${ t }_{ 0 }=\cfrac { 9{ v }_{ 0 } }{ 7\mu g }$$
C. $${ t }_{ 0 }=\cfrac { 12{ v }_{ 0 } }{ 7\mu g }$$
D. $${ t }_{ 0 }=\cfrac { 8{ v }_{ 0 } }{ 7\mu g }$$
E. $${ t }_{ 0 }=\cfrac { 4{ v }_{ 0 } }{ 7\mu g } $$
F. $${ t }_{ 0 }=\cfrac { 5{ v }_{ 0 } }{ 7\mu g }$$
G. $${ t }_{ 0 }=\cfrac { 6{ v }_{ 0 } }{ 7\mu g }$$
H. $${ t }_{ 0 }=\cfrac { 2{ v }_{ 0 } }{ 7\mu g } $$
I. $${ t }_{ 0 }=\cfrac { { v }_{ 0 } }{ 7\mu g } $$
J. $${ t }_{ 0 }=\cfrac { 3{ v }_{ 0 } }{ 7\mu g } $$
Answer: J
A current I flows in a long straight wire with its cross-section having the form of a this half ring of radius R. O is the centre of its mid-section perpendicular to its length. Assume the wire extends to infinite length on either side of the mid-section. Then the magnetic induction field B at the centre O will be :

Options:
A. $$B=\dfrac { { \mu }_{ 0 }i }{ \pi ^{ 2 }R } $$
B. $$B=\dfrac { { \mu }_{ 0 }i }{ 5\pi R }$$
C. $$B=\dfrac { { \mu }_{ 0 }i }{ 3\pi R }$$
D. $$B=\dfrac { { \mu }_{ 0 }i }{ 4\pi R }$$
E. $$B=\dfrac { { \mu }_{ 0 }i }{ 2R } $$
F. $$B=\dfrac { { \mu }_{ 0 }i }{ \pi R } $$
G. $$B=\dfrac { { \mu }_{ 0 }i }{ \sqrt{\pi} R }$$
H. $$B=\dfrac { { \mu }_{ 0 }i }{ 3\pi ^{ 2 }R }$$
I. $$B=\dfrac { { \mu }_{ 0 }i }{ 2\pi ^{ 2 }R }$$
J. $$B=\dfrac { { \mu }_{ 0 }i }{ 2\pi R } $$
Answer: A
The E$^o$ for reduction of HNO$_2$ to NO is 0.99 V and E$^o$ for reduction of NO$_3^{-}$ to HNO$_2$ is 0.94 V. The equilibrium constant at 25$^o$C for the disproportionation of 3 mole of aqueous HNO$_2$ to yield NO and $NO_{3}^{-}$ ions is:

Options:
A. $$K_c = 336.97$$
B. $$K_c = 351.64$$
C. $$K_c = 349.64$$
D. $$K_c = 354.64$$
E. $$K_c = 341.97$$
F. $$K_c = 339.97$$
G. $$K_c = 344.64$$
H. $$K_c = 639.73$$
I. $$K_c = 359.64$$
J. None of these
Answer: A
If Hund's rule is followed, magnetic moment of ${ Fe }^{ 2+ }$, ${ Mn }^{+ }$ and Cr all having 24 electrons will be in order-

Options:
A. $$F{ e }^{ 2+ } = { Cr } < { Mn }^{ + }$$
B. $$Mn^{ + } < { Cr } < { Fe }^{ 2+ }$$
C. $$Mn^{ 2+ }=Cr<{ Fe }^{ 2+ }$$
D. $$F{ e }^{ 2+ } = { Mn }^{ + } < Cr$$
E. $$Mn^{ + } < Cr < { Fe }^{ 2+ }$$
F. $$F{ e }^{ 2+ } < { Mn }^{ + } < Cr$$
G. $${ Fe }^{ 2+ } < { Cr }=Mn^{ + }$$
H. $$Mn^{ + } < { Fe }^{ 2+ } < Cr$$
I. $$F{ e }^{ 2+ } < { Cr } < { Mn }^{ + }$$
Answer: A
Figure shows two parallel and coaxial loops.  the smaller loop (radius r) is above the larger loop (radius R) by distance x>>R. The magnetic field due to current i in the larger loop is nearly constant throughout the smaller loop. Suppose that x is increasing at a constant rate of $dx/dt=v$
Determine the magnetic flux through the smaller loop as a function of x

Options:
A. $$\dfrac {\mu_0\, i\, R^2 \pi\, r^2}{3x^3}$$
B. $$\dfrac {3\mu_0\, i\, R^2 \pi\, r^2}{2x^3}$$
C. $$\dfrac {2\mu_0\, i\, R^2 \pi\, r^2}{x^3}$$
D. $$\dfrac {\mu_0\, i\, R^2 \pi\, r^2}{2x^3}$$
E. $$\dfrac {\sqrt {2}\mu_0\, i\, R^2 \pi\, r^2}{x^3}$$
F. $$\dfrac {\mu_0\, i\, R^2 \pi\, r^2}{6x^3}$$
G. $$\dfrac {\mu_0\, i\, R^2 \pi\, r^2}{7x^3}$$
H. $$\dfrac {\mu_0\, i\, R^2 \pi\, r^2}{5x^3}$$
I. $$\dfrac {\mu_0\, i\, R^2 \pi\, r^2}{x^3}$$
J. $$\dfrac {\mu_0\, i\, R^2 \pi\, r^2}{4x^3}$$
Answer: I
$(+)-2-Butanol$ has $[\theta]_{\lambda}^{25}=+13.9^{o}$. A sample of $2-butanol$ containing both the enantiomers was found to have a specific rotation value of $-3.5^{o}$ under simialr condition. THe percentage of the $(+)$ and $(-)$ enatiomers present in the sample are, respectively :

Options:
A. 38.4\% and 61.6\%
B. $37.4\%$ and $62.6\%$
C. 34.5\% and 65.5\%
D. 55.5\% and 44.5\%
E. $42.4\%$ and $57.8\%$
F. 57.8\% and 42.4\%
G. $62.6\%$ and $37.4\%$
H. $35.5\%$ and $64.5\%$
Answer: B
A body is moving down a long inclined plane of  inclination $ 45^0 $ with horizontal . the coefficient of friction between the body and the plane varies as $ \mu = x/ 2 , $ where x is the distance moved down the plane. initially x  =  0 & v = 0

Options:
A. The body's velocity remains constant after reaching a certain value
B. The body's velocity decreases after reaching a maximum
C. The body's velocity decreases all the time
D. The body's velocity remains constant throughout the motion
E. The body first accelerates and then.decelerates
F. The instantaneous acceleration of the body down the plane is $ \frac { g( x - 1 ) }{ 2 \sqrt {2} } $
G. The instantaneous acceleration of the body down the plane is $ \frac { g( x - 2 ) }{ 2 \sqrt {2} } $
H. The velocity of the body increases all the time
Answer: E
A car is accelerating on a horizontal road with acceleration = 20 $m/s^2$. A box that is placed inside the car, of mass m = 10 kg is put in contact with the vertical wall as shown. The friction coefficient between the box and the wall is $\mu$ = 0.6, then which one is true?

Options:
A. The friction force acting on the box will be 120 N
B. The friction force acting on the box will be 90 N
C. The friction force acting on the box will be 110 N
D. The friction force acting on the box will be 130 N
E. The net contact force between the vertical wall and the box is only of electromagnetic in nature.
F. The friction force acting on the box will be 100 N
G. The friction force acting on the box will be 115 N
H. The friction force acting on the box will be 105 N
I. The acceleration (with respect to ground ) of the box will be 20 $m/sec^2$
J. The contact force between the vertical wall and the box will be $100\sqrt{5}$ N
Answer: F
For the reaction $A(g)+B(g)\rightleftharpoons C(g)$ the equilibrium partial pressures $P_{A}=0.15$ atm and $P_{C}=0.30$ atm. When the volume of the reaction vessel was reduced to such an extent that on re-establishing the equilibrium, the partial pressure of $A$ and $B$ were doubled. The new patrial pressure of $C$ would be

Options:
A. $$0.30\ atm$$
B. $$0.45\ atm$$
C. $$0.50\ atm$$
D. $$1.50\ atm$$
E. $$1.80\ atm$$
F. $$0.75\ atm$$
G. $$1.20\ atm$$
H. $$0.90\ atm$$
I. $$0.60\ atm$$
Answer: A
Two straight segments of wire $ab$ and $bc$ each carrying current $I$, are parallel as shown in fig. The cube edge is 50 cm and magnetic field is uniform along Y-axis having magnitude $0.4\ T$. If $I=3A\ $, the force experienced by wire abc in the presence of magnitude field is :

Options:
A. $$0.6(\sqrt 2\hat i-\hat k)$$
B. $$0.6(\hat i + \hat k)$$
C. $$1.2(\hat i+\hat k)$$
D. $$0.6(\hat i - \hat k)$$
E. $$0.6\hat i$$
F. $$0.6(\sqrt 2\hat i + \hat j)$$
G. $$0.6(\sqrt 2\hat i - \hat j + \sqrt 2\hat k)$$
H. $$0.6(\sqrt 2\hat i - \sqrt 2\hat k)$$
I. $$0.6(\sqrt 2\hat i + \hat k)$$
J. $$0.6(\sqrt 2\hat i+\hat j-\sqrt 2\hat k)$$
Answer: E
A small ring of mass $m$ is attached at one end of a light string of length $l = 0.6\ cm.$ The other end of the string is tied to a small block $B$ of mass $2\ m.$ The ring is free to move on a fixed smooth horizontal rod. The block $B$ is released from rest from the position shown. The velocity (in $ms^{-1}$) of the ring when the string becomes vertical is:

Options:
A. $$3$$
B. $$2.7$$
C. $$4$$
D. $$3.6$$
E. $$3.4$$
F. $$3.5$$
G. $$2.5$$
H. $$3.3$$
I. $$3.2$$
Answer: F
a moles of an ideal mono atomic gas undergoes a process in which the temperature change with volume as $r={ kv }^{ 2 }$ if the temperature of the gas change from ${ T }_{ 0 }$ to ${ 4T }_{ 0 }$ then

Options:
A. heat supplied to the gas is ${ 4nRT }_{ 0 }$
B. ${ 5nRT }_{ 0 }$
C. 8nRT_0
D. non of these
E. ${ 7nRT }_{ 0 }$
F. work done by the gas is ${ 3nRT }_{ 0 }$
G. heat supplied to by the gas is ${ 6nRT }_{ 0 }$
H. ${ 2nRT }_{ 0 }$
I. ${ 3nRT }_{ 0 }$
J. work done on the gas is$\frac { 1 }{ 2 } { nRT }_{ 0 }$
Answer: D
The salt $Al(OH)_3$ is involved in the following two equilibria.
$$Al(OH)_3(s)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq); K_{sp}$$
$$Al(OH)_3(s)+OH(aq)\rightleftharpoons Al(OH)^-_4(aq); K_c$$
Which of the following relationship is correct when the solubility is minimum?

Options:
A. $$[OH^-]=\left(\displaystyle\frac{K_{sp}}{K_C}\right)^{1/5}$$
B. $$[OH^-]=\left(\displaystyle\frac{K_C}{K_{sp}}\right)^{1/3}$$
C. None of these
D. $$[OH^-]=\left(\displaystyle\frac{K_{sp}}{K_C}\right)^{1/2}$$
E. $$[OH^-]=\sqrt{\left(\displaystyle\frac{K_{sp}}{K_C}\right)^{1/4}}$$
F. $$[OH^-]=\left(\displaystyle\frac{K_C}{K_{sp}}\right)^{1/6}$$
G. $$[OH^-]=\left(\displaystyle\frac{K_{sp}}{K_C}\right)^{1/3}$$
H. $$[OH^-]=\left(\displaystyle\frac{K_{sp}}{K_C}\right)^{1/4}$$
I. $$[OH^-]=\left(\displaystyle\frac{K_C}{K_{sp}}\right)^{1/2}$$
J. $$[OH^-]=\left(\displaystyle\frac{K_C}{K_{sp}}\right)^{1/4}$$
Answer: C
Phosphorous pentachloride decomposes to phosphorous trichloride and chlorine gas at elevated temperatures by the following reaction:
$$PCL_{5}(g) \leftrightharpoons PCL_{3} + Cl_{2}(g) $$
If K$_{c}$ = 1 . 8 at 250$^{0}$C what is the value of K$_{p}$ at the same temperature ?

Options:
A. $$7 . 7 \times 10^{-4}$$
B. $$7 . 7 \times 10^{-2}$$
C. $$6 . 7 x 10^{-2}$$
D. 77
E. $$6 . 7 x 10^{-3}$$
F. 67
G. $$7 . 7 \times 10^{-1}$$
H. $$7.7 \times 10^{-3}$$
I. $$6 . 7 \times 10^{-1}$$
Answer: F
A source of frequency $n$ and an observer are moving on a straight line with velocity $x$ and $y$ respectively. If the source is ahead of the observer and the medium is also moving in the same direction with velocity $Z$, then the apparent frequency of the sound heard by the observer will be:($V$ = velocity of sound)

Options:
A. $$\displaystyle \frac{V-Z-y}{V-Z-x}n$$
B. $$\frac{V+Z-y}{V-Z+x}n$$
C. $$\frac{V-Z+y}{V+Z-x}n$$
D. $$\frac{V-Z-y}{V+Z-x}n$$
E. $$\displaystyle \frac{V+Z-y}{V-Z-x}n$$
F. $$\displaystyle \frac{V+Z+y}{V+Z+x}n$$
G. $$\frac{V+Z+y}{V-Z+x}n$$
H. $$\frac{V+Z+y}{V+Z-x}n$$
I. $$\displaystyle \frac{V-Z+y}{V-Z+x}n$$
J. $$\displaystyle \frac{V+Z-y}{V+Z-x}n$$
Answer: I
The average $O-H$ bond energy in $H_{2}O$ with the help of following data :(1) $H_{2}O(l) \rightarrow H_{2}O(g); \triangle H = + 40.6\ KJ\ mol^{-1}$(2) $2H(g) \rightarrow H_{2}(g); \triangle H = + 435.0\ KJ\ mol^{-1}$(3) $O_{2}(g) \rightarrow 2O(g); \triangle H = + 489.6\ KJ\ mol^{-1}$(4) $2H_{2}(g) + O_{2}(g)\rightarrow 2H_{2}O(l); \triangle H = -571.6\ KJ\ mol^{-1}$

Options:
A. $$463.5\ KJ\ mol^{-1}$$
B. $$461.5\ KJ\ mol^{-1}$$
C. $$925\ KJ\ mol^{-1}$$
D. $$445.8\ KJ\ mol^{-1}$$
E. $$231.3\ KJ\ mol^{-1}$$
F. $$345.6\ KJ\ mol^{-1}$$
G. $$489.9\ KJ\ mol^{-1}$$
H. $$584.9\ KJ\ mol^{-1}$$
I. $$279.8\ KJ\ mol^{-1}$$
J. $$462.5\ KJ\ mol^{-1}$$
Answer: J
In a region, electric field varies as $E = 2x^{2} - 4$ where $x$ is distance in $S.I.$ from origin along x-axis. A positive charge of $1\mu C$ is released with minimum velocity from infinity for crossing the origin, then

Options:
A. The kinetic energy at $x = \sqrt {3}m$ must be zero
B. The kinetic energy at $x = 2\sqrt{3}m$ must be zero
C. The kinetic energy at $x = 2m$ must be zero
D. The kinetic energy at the origin must be zero
E. The kinetic energy at $x = 2\sqrt{2}m$ must be zero
F. The kinetic energy at $x = \sqrt {2}m$ must be zero
G. The kinetic energy at $x = \sqrt {2}m$ may be zero
H. The kinetic energy at $x = 2m$ may be zero
I. The kinetic energy at $x = \sqrt {3}m$ may be zero
J. The kinetic energy at the origin may be zero
Answer: F
A real gas obeying Vander Waal equation has $a = 6.72 atm -lit^2 \,mol^{-2}$ and $b = 0.273 \,lit \,mol^{-1}$. At what temperature substances will exist only in gaseous state irrespective of pressure.

Options:
A. $$350 \,K$$
B. $$300 \,K$$
C. $$250 \,K$$
D. $$375 \,K$$
E. $$200 \,K$$
F. $$150 \,K$$
G. $$325 \,K$$
H. $$500 \,K$$
I. $$90 \,K$$
J. $$400 \,K$$
Answer: J
A thin straight vertical conductor has $10 amp$ current flows vertically upwards. it is present at a place where $B_H = 4 \times 10^{-6}\ T$. Arrange the net magnetic induction at the following points in descending order.(a) at $0.5m$ on south of conductor(b) at $0.5m$ on west of conductor(c) at $0.5m$ on east of conductor (d) at $0.5m$ on north-east of conductor

Options:
A. $$a, b, d, c$$
B. $$b, a, d, c$$
C. $$a, b, c, d$$
D. $$b, d, a, c$$
E. $$c, b, d, a$$
F. $$c, a, d, b$$
G. $$b, a, c, d$$
H. $$c, b, a, d$$
I. $$a, c, b, d$$
Answer: B
A electron is projected normally from the surface of a sphere with speed $v_0$ in a uniform magnetic filed perpendicular to the plane of the paper such that its strikes symmetrically opposite on the sphere with respect to the x-axis. Radius of the sphere is $'a'$ and the distance of its centre from the wall is $'b'$. What should be magnetic field such that the charge particle just escapes the wall

Options:
A. $$B = \frac{2bmv_0}{(a^2+b^2)e}$$
B. $$B = \frac{2bmv_0}{(a^2-b^2)q}$$
C. $$( B = \frac{2bmv_0}{(a^2+b^2)e} )$$
D. $$B = \frac{2bmv_0}{(a^2-b^2)e}$$
E. $$B = \frac{(a^2 -b^2)}{2bmv_0}$$
F. $$B = \frac{2bmv_0}{(b^2 + a^2)e}$$
G. $$B = \frac{2bmv_0}{(b^2-a^2)e}$$
H. $$B = \frac{2bmv_0}{(b^2 + a^2)q}$$
I. $$B = \frac{(a^2 + b^2)}{2bmv_0}$$
J. $$B = \frac{(b^2 -a^2)}{2bmv_0}$$
Answer: G
1.0 g sample of substance A at ${ 100 }^{ \circ  }$ C is added to 100 mL of ${ H }_{ 2 }O$ at ${ 25 }^{ \circ  }C$ Using separate 100 mL portions of ${ H }_{ 2 }O,$ the procedure is repeater with substance B and then with substance C. How will the final temperature of the water compare?
$$\begin{matrix} Substance & Specificheat \ A & 0.60{ Jg }^{ -1 }^{ \circ  }{ { C }^{ -1 } } \ B & 0.40Jg^{ -1 }^{ \circ  }{ { C }^{ -1 } } \ C & 0.20Jg^{ -1 }^{ \circ  }{ { C }^{ -1 } } \end{matrix}$$

Options:
A. $${ T }_{ C }>{ T }_{ A }>{ T }_{ B }$$
B. $${ T }_{ A }={ T }_{ B }={ T }_{ C }$$
C. $${ T }_{ A }>{ T }_{ C }>{ T }_{ B }$$
D. $${ T }_{ A }>{ T }_{ B }>{ T }_{ c }$$
E. $${ T }_{ B }>{ T }_{ A }>{ T }_{ C }$$
F. $${ T }_{ B }>{ T }_{ C }>{ T }_{ A }$$
G. $${ T }_{ A }>{ T }_{ B }>{ T }_{ C }$$
H. $${ T }_{ c }>{ T }_{ B }>{ T }_{ A }$$
Answer: H
In Resonance tube experiment, if $400\ Hz$ tuning fork is used, the first resonance occurs when length of air column in the tube is $19\ cm$. If the $400\ Hz$. tuning fork is replaced by $1600\ Hz$ tuning fork then to get resonance, the water level in the tube should be further lowered by (take end correction $= 1\ cm)$.

Options:
A. $$10\ cm$$
B. $$25\ cm$$
C. $$5\ cm$$
D. $$20\ cm$$
E. $$12\ cm$$
F. $$8\ cm$$
G. $$7\ cm$$
H. $$3\ cm$$
Answer: C
A balloon starts rising from the ground with an acceleration of $1.25\,\,m/{s^2}$. After $8s$  a stone is released from the balloon. The stone will talking $g = 10\,m{s^{ - 2}}$

Options:
A. Have a displacement of $50\,m$
B. Reach the ground in $5s$
C. Reach the ground in $4s$
D. Reach the ground in $3.5s$
E. Begin to move down after being released
F. Cover a distance of $42m$ in reaching the ground
G. Cover a distance of $45m$ in reaching the ground
H. Cover a distance of $41m$ in reaching the ground
I. Reach the ground in 3s
J. Cover a distance of $40m$ in reaching the
ground
Answer: C
In the preparation of $CaO$ from $CaCO_{3}$ using the equilibrium$CaCO_{3}(s)\rightleftharpoons CaO(s) + CO_{2}(g), K_{p}$ is expressed as$\log K_{p} = 7.282 - \dfrac {8500}{T}$For the complete decomposition of $CaCO_{3}$, the temperature in Celsius to be used is__________.

Options:
A. $$1050$$
B. $$1097$$
C. $$8500$$
D. $$1067$$
E. $$950$$
F. $$1167$$
G. $$894$$
H. $$1127$$
I. $$850$$
Answer: G
Calculate the voltage E in V, of the cell
$$Ag(s)\:AgIO_{3}(s)\left | Ag^{+}(x\:M),\:HIO_{3}(0.300\:M) \right |\left | Zn^{2+}(0.175\:M) \right |Zn (s)$$
if $K_{SP}=3.02\times 10^{-8}$ for $AgIO_{3}(s)$ and $K_{a}=0.162$ for $HIO_{3},\:E^{\circ}(Zn^{2+}/Zn)=-0.76\:V,\:E^{\circ}(Ag/Ag^{+}=-0.8\:V)$

Options:
A. $$0.47$$
B. $$0.55$$
C. $$0.35$$
D. $$0.32$$
E. $$1.88$$
F. $$0.50$$
G. $$1.45$$
H. $$0.45$$
I. $$0.42$$
J. $$1.52$$
Answer: H
The electric field in a region is given by  $E=(4axy\sqrt { z } )\hat { i } +\left( 2ax^{ { 2 } }\sqrt { z }  \right) \hat { j } +\left( ax^{ { 2 } }/\sqrt { z }  \right) \hat { k } ,$  where  $a$  is a positive constant. The equation of an equipotential surface will be of the form :

Options:
A. $z =$ constant $/ \left[ x ^ { 3 } y ^ { 3 } \right]$
B. $$z = \text{constant} / \left[ x ^ { 3 } y ^ { 1 } \right]$$
C. $$z = \text{constant} / \left[ x ^ { 4 } y ^ { 3 } \right]$$
D. $z =$ constant $/ \left[ x ^ { 4 } y ^ { 2 } \right]$
E. $$z = \text{constant} / \left[ x ^ { 2 } y ^ { 3 } \right]$$
F. $$z = \text{constant} / \left[ x ^ { 2 } y ^ { 2 } \right]$$
G. $z =$ constant $/ \left[ x ^ { 3 } y ^ { 1 } \right]$
H. $z =$ constant $/ \left[ x y ^ { 2 } \right]$
I. None of these
Answer: D
${(CH_3)}_2C=C{(CH_3)}_2$ gives "$X$" on a reductive ozonolysis and "$Y$" on oxidative cleavage with $KMnO_4/H^+$. $X,Y$ respectively

Options:
A. $$CH_3CH_2OH,CH_3COOH$$
B. $$CH_3COCH_3,CH_3CHO$$
C. $$CH_3COOH,CH_3COCH_3$$
D. $$CH_3COOH,CH_3CHO$$
E. $$CH_3COCH_3,CH_3COOH$$
F. $$CH_3CH_2OH,CH_3CHO$$
G. $$CH_3CHO,CH_3COCH_3$$
H. $$CH_3COOH,CH_3CH_2OH$$
I. $$CH_3CH_2OH,CH_3COCH_3$$
Answer: G
The space between two plates of a condenser is filled with two dielectric media of thickness $t_1$ and $t_2$ and dielectric constant $k_1$ and $k_2$ respectively. The capacity of the condenser is given by?

Options:
A. \(\epsilon_0 A \{ \dfrac{K_1}{t_1 + t_2} + \dfrac{K_2}{t_2} \} \)
B. $$C = \frac{\varepsilon_0 A}{\left( \frac{t_1}{K_1} + \frac{t_2}{k_2} \right)}$$
C. $${ { \epsilon  }_{ 0 }A }\{ \dfrac { { K }_{ 1 } }{ { t }_{ 2 } } +\dfrac { { K }_{ 2 } }{ { t }_{ 1 } } \}$$
D. $$(K_1-K_2)arepsilon_0\dfrac{A}{(k_1+K_2)}\cdot (t_1+t_2)$$
E. \( \epsilon_0 A \{ \dfrac{K_1}{t_2} + \dfrac{K_2}{t_1} \} \)
F. $$C = \frac{(K_1 + K_2) \varepsilon_0 A}{t_1 k_1 + t_2 k_2}$$
G. \(\epsilon_0 A \{ \dfrac{K_1}{t_1} - \dfrac{K_2}{t_2} \} \)
H. ${ { \epsilon  }_{ 0 }A }\{ \dfrac { { K }_{ 1 } }{ { t }_{ 1 } } +\dfrac { { K }_{ 2 } }{ { t }_{ 2 } } \} $
I. \(\epsilon_0 A \{ \dfrac{K_1}{t_1} + \dfrac{K_2}{t_2} \} \)
J. \(\epsilon_0 A \{ \dfrac{K_1}{t_1} + \dfrac{K_2}{t_2 + t_1} \} \)
Answer: H
A $40 \,kg$ boy swinging on a swing, the power delivered by gravity force is, when the swinging is making an angle $30^0$ with horizontal and the boy moving with a velocity of $8\,m/s$ upward $(g=10\,m/s)$

Options:
A. $-1200 \sqrt{3} \,W$
B. $-500$W
C. -1200 W
D. -1400 W
E. $-1000$W
F. $-1000 \sqrt{3} \,W$
G. $-1600 \sqrt{3}$ W
H. -900 \,W
I. -800 W
J. $-1600$W
Answer: G
A uniform solid cylinder A, of mass $m_1$, can freely rotate about horizontal axis fixed to a mount B of mass $m_2$ (fig). A constant horizontal force F is applied to the end K of a light thread tightly wound on the cylinder. The friction between the mount and the supporting horizontal plane is assumed to be absent. Find:
(a) the acceleration of the point K
(b) the kinetic energy of this system t seconds after the beginning of motion.

Options:
A. $a_k= \dfrac{F(3m_1+2m_2)}{m_1(m_1+2m_2)}$; $K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+2m_2)}{m_1(m_1+m_2)}t^2$
B. $a_k= \dfrac{F(3m_1+2m_2)}{m_1(m_1+m_2)}$; $K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+2m_2)}{m_1(m_1+m_2)}t^2$
C. $a_k= \dfrac{F(3m_1+2m_2)}{m_1(m_1+2m_2)}$; $K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+2m_2)}{2m_1(m_1+m_2)}t^2$
D. $a_k= \dfrac{F(3m_1+2m_2)}{2m_1(m_1+m_2)}$; $K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+2m_2)}{m_1(m_1+m_2)}t^2$
E. $a_k= \dfrac{F(3m_1+2m_2)}{m_1(m_1+m_2)}$; $K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+2m_2)}{2m_1(m_1+m_2)}t^2$
F. $a_k= \dfrac{F(3m_1+2m_2)}{2m_1(m_1+m_2)}$; $K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+m_2)}{m_1(m_1+m_2)}t^2$
G. $a_k= \dfrac{F(3m_1+2m_2)}{m_1(m_1+2m_2)}$; $K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+2m_2)}{m_1(m_1+2m_2)}t^2$
H. $a_k= \dfrac{F(3m_1+2m_2)}{m_1(m_1+m_2)}$; $K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+m_2)}{m_1(m_1+m_2)}t^2$
I. $a_k= \dfrac{F(3m_1+2m_2)}{2m_1(m_1+m_2)}$; $K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+2m_2)}{2m_1(m_1+m_2)}t^2$
J. a_k= \dfrac{F(3m_1+2m_2)}{m_1(m_1+3m_2)}; K.E = \dfrac{1}{2} \dfrac{F^2(3m_1+2m_2)}{m_1(m_1+3m_2)}t^2
Answer: B
A body of mass 1 kg falls freely from a height of 100m, on a platform of mass 3 kg which is mounted on a spring having spring constant k = 1.25 x ${ 10 }^{ 6 }$ N/m. The body stick to the platform and the spring's maximum compression is found to be x. Given that g = 10 ${ ms }^{ -2 },$ the value of x will be close to :

Options:
A. 40 cm
B. 12 cm
C. 15 cm
D. 4 cm
E. 2.5 cm
F. 8 cm
G. 10 cm
H. 6 cm
I. 2 cm
J. 80 cm
Answer: I
Given the cell:$Cd(s)|Cd(OH)_2(s)|NaOH(aq,\space 0.01\space M)|H_2(g, \space 1\space bar)|Pt(s)$
with $E_{cell} = 0.0\space V$. If $E^{\small\circ}_{Cd^{2+}|Cd} = -0.39\space V$, then $K_{sp}$ of $Cd(OH)_2$ is:

Options:
A. $$10^{-12}$$
B. 10^{-13}
C. 10^{-10}
D. $$10^{-15}$$
E. 10^{-11}
F. $$10^{-17}$$
G. $$10^{-8}$$
H. 10^{-16}
I. $$10^{-14}$$
Answer: D