MikaStars39/nano-eval huggingface dataset:
A circular disc of radius R rolls without slipping on a rough horizontal surface. At the instant shown its linear velocity is V, linear acceleration a, angular velocity $\omega$ and angular acceleration $R \alpha$. Four points A, B, C and D lie on its circumference such that the diameter AC is vertical & BD horizontal then choose the correct options. Then choose the correct option(s) from the given four options.

Options:
A. $$a_D =\sqrt{(\alpha + \omega ^2R )^2+(R\alpha)^2}$$
B. $$a_D =\sqrt{(a + \omega^2 R )^2 + (R\alpha)^2} - 2\omega R$$
C. $$V_C = V + R\omega$$
D. $$a_D =\sqrt{(a + \omega^2 R )^2 + (2R\alpha)^2}$$
E. $$a_D =\sqrt{(a - \omega^2 R )^2 + (R\alpha)^2}$$
F. $$a_A =\sqrt{(a - R \alpha )^2+(\omega ^2R)^2}$$
Answer: A
One mole of monoatomic ideal gas at a temperature $ T_1 K $ expands according to the law $ \frac {P}{V} = const $. Its final temperature is $ T_2K$, The work done by the gas is:-

Options:
A. W = \frac { 3R\left( T_ 2+T_ 1 \right) }{ 4 }
B. W = \frac { 2R\left( T_ 1-T_ 2 \right) }{ 3 }
C. $$ W = \frac { R\left( T_ 2-T_ 1 \right) }{ 2 } $$
D. $$ W = \frac { R\left( T_ 1-T_ 2 \right) }{ 2 } $$
E. \( W = \frac { R\left( T_ 1+T_ 2 \right) }{ 2 } \)
F. W = \frac { 3R\left( T_ 2-T_ 1 \right) }{ 2 }
G. $$ W = \frac { 3R\left( T_ 1-T_ 2 \right) }{ 2 } $$
H. W = \frac { R\left( T_ 1+T_ 2 \right) }{ 3 }
I. W = \frac { R\left( T_ 2+T_ 1 \right) }{ 3 }
J. $$ W = \frac { 3R\left( T_ 1+T_ 2 \right) }{ 2 } $$
Answer: C
Formation of a solution from two components can be considered as(i) Pure solvent $\rightarrow $ separated solute molecules, $\triangle { H }_{ 1 }$(ii) Pure solute $\rightarrow $ separated solute molecules, $\triangle { H }_{ 2 }$solution so formed will be ideal if(iii) Separated solvent and solute molecules $\rightarrow $ solution, $\triangle { H }_{ 3 }$Then:

Options:
A. $$\triangle { H }_{ soln }=\triangle { H }_{ 1 }+\triangle { H }_{ 3 }-\triangle { H }_{ 2 }$$
B. $$\triangle { H }_{ soln }=\triangle { H }_{ 2 }+\triangle { H }_{ 3 }-\triangle { H }_{ 1 }$$
C. $$\triangle { H }_{ soln }=\triangle { H }_{ 1 }+\triangle { H }_{ 2 }-\triangle { H }_{ 3 }$$
D. $$riangle { H }_{ soln }=\triangle { H }_{ 2 }-\triangle { H }_{ 3 }-\triangle { H }_{ 1 }$$
E. $$riangle { H }_{ soln }=\triangle { H }_{ 2 }+\triangle { H }_{ 1 }+\triangle { H }_{ 3 }$$
F. $$riangle { H }_{ soln }=\triangle { H }_{ 3 }+\triangle { H }_{ 1 }-\triangle { H }_{ 2 }$$
G. $$riangle { H }_{ soln }=\triangle { H }_{ 2 }-\triangle { H }_{ 1 }+\triangle { H }_{ 3 }$$
H. $$\triangle { H }_{ soln }=\triangle { H }_{ 1 }-\triangle { H }_{ 2 }-\triangle { H }_{ 3 }$$
I. $$\triangle { H }_{ soln }=\triangle { H }_{ 1 }+\triangle { H }_{ 2 }+\triangle { H }_{ 3 }$$
Answer: I
Assume that decomposition of $HNO_{3}$ can be represented by the following equation $4HNO_{3}(g)\rightleftharpoons 4NO_{2}(g)+2H_{2}O(g)+O_{2}(g)$and the reaction approaches equilibrium at $400\ K$ temperature and $30\ atm$ pressure. At equilibrium partial pressure of $HNO_{3}$ is $2\ atm$.Calculate $K_{c}$ in ${(mol/L)}^{3}$ at $400\ K$ :(Use : $R=0.08\ atm-L/mol-K$)

Options:
A. $$22.8$$
B. $$16$$
C. 17.8
D. $$8$$
E. 17.2
F. 18.4
G. 19.2
H. 21.8
I. 21.4
J. $$4$$
Answer: A
To find the resistance of a galvanometer by the half deflection method the following circuit is used with resistances $R_1=9970 \Omega, \, R_2=30 \Omega$ and $R_3=0$. The deflection in the galvanometer is $d$. With $R_3=107 \Omega$ the deflection changed to $\frac {d}{2}$. The galvanometer resistance is approximately:

Options:
A. $$107 \Omega$$
B. $$137 \Omega$$
C. $$77 \Omega$$
D. $$107 /2 \Omega$$
E. $$117 \Omega$$
F. $$87 \Omega$$
G. $$67 \Omega$$
H. $$127 \Omega$$
I. $$97 \Omega$$
Answer: C
If circular race track is banked at angle $\theta$ ($\theta> $ angle of repose),radius of track is $r$ and $\mu$ is coefficient of friction between car and road. Calculate minimum speed on car on inclined plane to avoid slipping?

Options:
A. $$v={ \left[ rg\left( \cfrac { \tan { \theta -\mu } }{ 1-\mu \tan { \theta } } \right) \right]^{1/2} }$$
B. $$v={ \left[ rg\left( \cfrac { \tan { \theta +\mu } }{ 1+\mu \tan { \theta } } \right) \right]^{1/2} }$$
C. $$v={ \left[ rg\left( \cfrac { \tan { \theta -\mu } }{ 1+\mu \tan { \theta } } \right) \right] }^{ 1/2 }$$
D. $$v={ \left[ rg\left( \cfrac { \tan { \theta - \mu } }{ 1 - \mu \tan { \theta } } \right) \right] }^{ 1/2 }$$
E. $$v={ \left[ rg\left( \cfrac { \tan { \theta + \mu } }{ 1 - \mu \tan { \theta } } \right) \right] }^{ 1/2 }$$
F. $$v={ \left[ rg\left( \cfrac { \tan { \theta +\mu } }{ 1+\mu \tan { \theta } } \right) \right] }^{ 1 }$$
G. $$v=\left[ rg\left( \cfrac { \tan { \theta -\mu } }{ 1+\mu \tan { \theta } } \right) \right] $$
H. $$v={ \left[ rg\left( \cfrac { \tan { \theta +\mu } }{ 1+\mu \tan { \theta } } \right) \right] }^{ 1/2 }$$
I. $$v={ \left[ rg\left( \cfrac { \tan { \theta +\mu } }{ 1-\mu \tan { \theta } } \right) \right]^{1/2}  }$$
Answer: I
A particle of mass m is under the influence of a force F which varies with the displacement x according to the relation $F =  - kx + {F_0}$ in which $k\;and\;{F_0}$ are constants.The particle when disturbed will oscillate

Options:
A. $$about\;x = {F_0}/k,\,with\;\omega = \sqrt {k/m} $$
B. $$about\;x = {F_0}/(3k),\,with\;\omega = \sqrt {k/m}$$
C. $$about\;x = {F_0}/(6k),\,with\;\omega = \sqrt {k/m}$$
D. $$about\;x = {F_0}/k,\,with\;\omega$$
E. $$about\;x = 0,\,with\;\omega
e \sqrt {k/m} $$$
F. $$about\;x = {F_0}/(7k),\,with\;\omega = \sqrt {k/m}$$
G. $$about\;x = 0,\,with\;\omega = \sqrt {k/m} $$
H. $$about\;x = {F_0}/(5k),\,with\;\omega = \sqrt {k/m}$$
I. $$about\;x = {F_0}/(2k),\,with\;\omega = \sqrt {k/m}$$
J. $$about\;x = {F_0}/(4k),\,with\;\omega = \sqrt {k/m}$$
Answer: G
A point P moves in counter clockwise direction on a circular path as shown in the figure. The movement og 'P; is such that it sweeps out a length $s={ t }^{ 2 }+5$, where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when $t=5\sqrt { \frac { 3 }{ 10 }  } $ seconds is nearly :

Options:
A. $$1.6m/{ s }^{ 2 }$$
B. $$3m/{ s }^{ 2 }$$
C. $$2.2m/{ s }^{ 2}$$
D. $$1.9m/{ s }^{ 2}$$
E. $$1.5m/{ s }^{ 2 }$$
F. $$2.5m/{ s }^{ 2 }$$
G. $$2m/{ s }^{ 2 }$$
H. $$1.8m/{ s }^{ 2 }$$
I. $$1.7m/{ s }^{ 2 }$$
Answer: G
The magnetic field in a region is given by $\overrightarrow B  = {B_0}\overrightarrow i $.A charge particle of  charge $\left( {{q_0}} \right)$ and mass $m$ is given a velocity $\overrightarrow v  = \frac{{{v_0}}}{2}\widehat i + \frac{{{v_0}\sqrt 3 }}{2}\widehat j$. Find the pitch of the helical path on the particle  on which the particle moves:

Options:
A. $$\(\frac{{\pi m{v_0}\sqrt 3 }}{{q{B_0}}}\)$$
B. $$\(\frac{{2\pi m{v_0}\sqrt 3 }}{{3q{B_0}}}\)$$
C. $$\frac{{\pi m{v_0}\sqrt 3 }}{{2q{B_0}}}$$
D. $$\(\frac{{\pi m{v_0}}}{{4q{B_0}}}\)$$
E. $$\frac{{\pi m{v_0}}}{{q{B_0}}}$$
F. $$\frac{{\pi m{v_0}}}{{qB}}$$
G. $$\(\frac{{2\pi m{v_0}}}{{3q{B_0}}}\)$$
H. $$\(\frac{{\pi m{v_0}}}{3q{B_0}}\)$$
I. $$\frac{{\pi m{v_0}}}{{2q{B_0}}}$$
J. $$\(\frac{{\pi m{v_0}}}{{3q{B_0}}}\)$$
Answer: I
A train passes an observer standing on a platform. The first carriage of the train passes the observer in time $  t_{1}=1  $ s and the second carriage in $  t_{2}=1.5 s $. Find its acceleration assuming it to be constant. The length of each carriage is $  l=12 \mathrm{m} $

Options:
A. $$ -2.4 \mathrm{m} / \mathrm{s}^{2} $$
B. $$-2.2 \mathrm{m} / \mathrm{s}^{2}$$
C. $$-2.6 \mathrm{m} / \mathrm{s}^{2}$$
D. $$-2.8 \mathrm{m} / \mathrm{s}^{2}$$
E. $$-2.3 \mathrm{m} / \mathrm{s}^{2}$$
F. $3.3$ $ \mathrm{m} / \mathrm{s}^{2} $
G. $$-3.4 \mathrm{m} / \mathrm{s}^{2}$$
H. $$-2.5 \mathrm{m} / \mathrm{s}^{2}$$
I. $$ -3.2  \mathrm{ m} / \mathrm{s}^{2} $$
J. $2.4$  $ \mathrm{m} / \mathrm{s}^{2} $
Answer: A
For the reaction $N_2O_4\rightleftharpoons 2NO_2$, equilibrium mixture contains $NO_2$ at $P = 1.1 atm$ & $N_2O_4$ at $P = 0.28 atm$ at 350 K. The volume of the container is doubled. Calculate the equilibrium pressures of the two gases when the system reaches new equilibrium :

Options:
A. $$P_{NO_2}= 0.64 atm, P_{N_2O_4} = 0.095 atm$$
B. $$P_{NO_2}= 0.32atm, P_{N_2O_4} = 0.049atm$$
C. $$P_{NO_2}= 0.64 atm, P_{N_2O_4} = 0.06 atm$$
D. $$P_{NO_2}= 0.64 atm, P_{N_2O_4} = 0.05 atm$$
E. $$P_{NO_2}= 0.64 atm, P_{N_2O_4} = 0.12 atm$$
F. $$P_{NO_2}= 0.32 atm, P_{N_2O_4} = 0.095 atm$$
G. $$P_{NO_2}= 0.64 atm, P_{N_2O_4} = 0.14 atm$$
H. $$P_{NO_2}= 0.64 atm, P_{N_2O_4} = 0.049 atm$$
I. $$P_{NO_2}= 0.32 atm, P_{N_2O_4} = 0.10 atm$$
J. $$P_{NO_2}= 0.32 atm, P_{N_2O_4} = 0.05 atm$$
Answer: D
Two carbon discs, $1.0\ g$ each, are $1.0\ cm$ apart have equal and opposite charges. If force of attraction between them is $10^{-5}N$, the ratio of excess electrons to the total atoms on the negatively charge disc is $(N_A=6\times 10^{23})$

Options:
A. $$10^{-14}:2.4$$
B. $$2.4:10^{12}$$
C. $$10^{-12}:2.4$$
D. $$10^{12}:2.4$$
E. $$10^{-14}:2.2$$
F. $$10^{-13}:2.5$$
G. $$10^{-13}:2.3$$
H. $$10^{-12}:2.5$$
I. $$2.4\times 10^{-12}:1$$
J. $$10^{-14}:2.5$$
Answer: E
Hydrogen is produced industrially by the steam-hydrocarbon re-forming process as given by:${H_2}O\left( g \right) + C{H_4}\left( g \right) \to CO\left( g \right) + 3{H_2}\left( g \right),If{K_c} = 3.8 \times {10^{ - 3}}at1000K$ what is the value of ${K_p}$ at the same temperature?

Options:
A. $$36.8$$
B. $$22.8$$
C. 27.8
D. $$37.8$$
E. 35.8
F. 28.8
G. $$9.8$$
H. 42.8
I. $$25.61$$
Answer: D
A thin uniform wire to bent to form the two equal sides AB and AC triangle ABC. where AB = AC = $5\ cm$. the third side is  BC, of length $6\ cm,$ is made from uniform wire of twice the density of first. the center of mass is :

Options:
A. $$(2.4,2.2)\ cm$$
B. $$(2.58,1.8)\ cm$$
C. $$(2.4,1.8)\ cm$$
D. $$(2.88,3.3\ )cm$$
E. $$(2.48,1.7) cm$$
F. $$(2.78,1.4)\ cm$$
G. $$(2.48,1.8)\ cm$$
H. $$(2.4,1.6)\ cm$$
I. $$(2.48,2.1)cm$$
J. $$(2.5,1.9)\ cm$$
Answer: C
A particle of mass $3$ gram is moving along x-axis according to relation $x = 3 4t^{ 2 } + t^{ 3 }$. The work done in first $4s$ will be:

Options:
A. $$416 mJ$$
B. $$360 mJ$$
C. $$432 mJ$$
D. $$168 mJ$$
E. $$192 mJ$$
F. $$144 mJ$$
G. $$180 mJ$$
H. $$none of these$$
I. $$384 mJ$$
Answer: I
A ray of sunlight enters a spherical water droplet $(n = \dfrac{4}{3})$ at an angle of incidence $53^\circ$ measured with respect to the normal to the surface. It is reflected from the back surface of the droplet and re-enters into air. The angle between the incoming and outgoing ray is [Take $\sin \ 53^\circ = 0.8$]

Options:
A. $$125^\circ$$
B. $$130^\circ$$
C. $$132^\circ$$
D. $$128^\circ$$
E. $$127^\circ$$
F. $$138^\circ$$
G. $$15^\circ$$
H. $$129^\circ$$
I. $$134^\circ$$
J. $$30^\circ$$
Answer: E
If the time period $t$ of the oscillating of a drop of liquid of density $d$, radius $r$, vibrating under surface tension $s$ is given by the formula $t=\sqrt {r^{2b}.s^{c}.d^{a/2}}$. It is observed that the time period is directly proportional to $\sqrt {\dfrac {d}{s}}$. The value of $b$ should therefore br

Options:
A. \dfrac{5}{2}
B. $$\dfrac {3}{2}$$
C. $$\sqrt {3}$$
D. \dfrac{1}{2}
E. $$\dfrac {3}{4}$$
F. \dfrac{1}{4}
G. \dfrac{5}{3}
H. \dfrac{5}{4}
I. $$\dfrac {2}{3}$$
J. \dfrac {1}{3}
Answer: B
Two straight conducting rails from a right angle where their ends are joined. A conducting bar in contact with the rails starts at the vertex at time t=0 and moves with constant velocity v along them as shown in figure. magnetic fields $\overrightarrow{B}$ is directed into the page. The induced emf in the circuit at any time is proportional to

Options:
A. $$v^4$$
B. $$v^3$$
C. $$t^3$$
D. $$t^0$$
E. $$v$$
F. $$v^2$$
G. $$t^2$$
Answer: F
Best indicator to delect and point in complete titration of $0.2\ M\ K_{2}\ CO_{3}$ against $0.1\ M\ HCl$ is:- 
[For $H_{2}CO_{3}:\ K_{a_{1}}=4\times 10^{-7}; K_{a_{2}}=5\times 10^{-11}$]

Options:
A. $$Thymol blue $(pK_{in} \simeq 2)$$$
B. $$Thymol blue (pK_{in} \simeq 8)$$
C. $$Thymol blue (pK_{in} \simeq 10)$$
D. $$Methyl orange$$
E. $$Malachite green $(pK_{in} \simeq 1)$$$
F. $$Thymolphthalein (pK_{in} \simeq 10)$$
G. $$Phenolpthalein$$
H. $$Thymol blue (pK_{in} \simeq 9)$$
I. $$Thymolphthalein (pK_{in} \simeq 9)$$
J. $$Thymolphthalein (pK_{in} \simeq 11)$$
Answer: G
Under what pressure conditions is $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O$ efflorescent at 25C? How good a drying agent is $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ at the same temperature? Given
$$Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)\rightleftharpoons Cu{ SO }_{ 4 }.3{ H }_{ 2 }O(s)+2{ H }_{ 2 }O(v)$$
${ K }_{ P }=1.086\times { 10 }^{ -4 }{ atm }^{ 2 }$ at 25C. Vapour pressureof water at 25C is 23.8 mm of Hg.

Options:
A. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will efflorescent but on the contrary $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will absorb moisture from the. atmosphere under given conditions.
B. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will efflorescent but on the contrary $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will absorb moisture from the atmosphere under given conditions.
C. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will efflorescent and $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will not absorb moisture from the atmosphere under given conditions.
D. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will not act as efflorescent but on the contrary $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will absorb moisture from the atmosphere under given conditions.
E. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will efflorescent but on the contrary $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will not absorb moisture from the atmosphere under given conditions.
F. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will not act as efflorescent and $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will also not absorb moisture from the atmosphere under given conditions.
G. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will not act as efflorescent and $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will absorb moisture from the atmosphere under given conditions.
H. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will not act as efflorescent and $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will not absorb moisture from the atmosphere under given conditions.
I. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will efflorescent and $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will absorb moisture from the atmosphere under given conditions.
J. $Cu{ SO }_{ 4 }.5{ H }_{ 2 }O(s)$ will not act as efflorescent and $Cu{ SO }_{ 4 }.3{ H }_{ 2 }O$ will also absorb moisture from the atmosphere under given conditions.
Answer: D
A uniform disc of radius r and mass m is charged uniformly withe charge q. this disc is placed flat on a rough horizontal surface having coefficient of friction $ \mu $ A uniform magnetic field is present in a circular region of radius a (> r) but varying as $ kt^3 $. find the time in second after which the disc begins to rotate. (Given r=1 m, m=18 kg, q=1 C, $ \mu = 0.1 $, K=4, $ g=10 m/s^2) $

Options:
A. 12
B. 14
C. 2
D. 16
E. 18
F. 4
G. 6
H. 8
I. 10
J. none of these
Answer: H
An electron of mass $m$ and charge $e$ leaves the lower plate of a parallel plate capacitor of length $L$, with an initial velocity $v_{0}$ making an angle $\alpha$ with the plate and come out of the capacitor making an angle $\beta$ to the plate. The electric field intensity between the plates.

Options:
A. None of the above
B. $$E = \dfrac{m v_0^2 \cos^2 \alpha}{e L} \left( 2 \tan \alpha + \tan \beta \right)$$
C. $$E = \dfrac{m v_0^2 \cos^2 \alpha}{e L} \left( \tan \beta - \tan \alpha \right)$$
D. $$E = \dfrac{m v_0^2 \cos^2 \alpha}{e L} \left( 2 \tan \alpha - \tan \beta \right)$$
E. $$E = \dfrac{m v_0^2 \cos^2 \alpha}{e L} \left( \tan \alpha - \tan \beta \right)$$
F. E = \dfrac {mv_{0}^{2}\cos^{2}\alpha}{eL} (\tan \alpha + \tan \beta)
G. $$E = \dfrac{m v_0^2 \cos^2 \alpha}{e L} \left( 2 \tan \beta - \tan \alpha \right)
$$
H. $$E = \dfrac{m v_0^2 \cos^2 \alpha}{e L} \left( 2 \tan \beta - 2 \tan \alpha \right)$$
I. $$E = \dfrac {mv_{0}^{2}\cos^{2}\alpha}{eL} (\tan 2\alpha - \tan \beta)$$
J. $$E = \dfrac{m v_0^2 \cos^2 \alpha}{e L} \left( \tan \beta + \tan \alpha \right)$$
Answer: E
Due to a point isotropic sonic source, loudness at a point is $ L = 60 \, dB $  If density of air is $ \rho = (15/11) \, kg/m^3 $  and velocity of sound in air is $ v = 33 \, m/s $ , the pressure oscillation amplitude at the point of observation is $ [ I_0 = 10^{-12} \, W/m^2] $

Options:
A. $$3 \times 10^{-5} \, N/m^2$$
B. $$3 \times 10^{-8} \, N/m^2$$
C. $$ 3 \times 10^{-3} \,N/m^2 $$
D. $$ 0.03\,N/m^2 $$
E. $$3 \times 10^{-7}\,N/m^2$$
F. $$0.3\,N/m^2$$
G. $$3 \times 10^{-6}\,N/m^2$$
H. $$3 \times 10^{-4} \, N/m^2$$
Answer: D
$Co|{ Co }^{ 2+ }({ C }_{ 2 })\parallel { Co }^{ 2+ }({ C }_{ 1 })|Co$ for this cell, $\triangle$G is negative if :

Options:
A. $C_2$ > $2C_1$
B. $C_1$ < $C_2$
C. $C_1$ = $C_2$
D. unpredictable
E. C_1 = 2C_2
F. $C_1$ > $C_2$
G. $C_2$ = 2$C_1$
Answer: F
A tank full of water has a small hole at its bottom. Let $t_1$ be the time taken to empty first half of the tank and $t_2$ be the time needed to empty rest half of the tank, then-

Options:
A. $$t_1 = \frac{t_2}{3}$$
B. $$t_1 = \frac{3t_2}{2}$$
C. $$t_1 = 0.414 t_2$$
D. $$\( t_1 = \frac{t_2}{2} \)$$
E. $$t_1 = \frac{t_2}{4}$$
F. $$t_1 > t_2$$
G. t_1 = 2t_2
H. $$t_1 = t_2$$
I. $$t_1 = \frac{2t_2}{3}$$
J. $$t_1 = \frac{t_2}{\sqrt{2}}$$
Answer: H
A disc of radius $R$ rolls on a horizontal surface with linear velocity $v$ and angular velocity $\omega $. A point $P$ on the circumference  of the disc at $\theta $ has a vertical velocity. Then

Options:
A. $$\theta = \pi + {\cos ^{ - 1}}\left( {\frac{v}{{R\omega }}} \right)$$
B. $$\theta = \frac{3\pi}{2} - {\cos^{-1}}\left( \frac{v}{{R\omega}} \right)$$
C. $$\theta = \pi - {\sin ^{ - 1}}\left( {\frac{v}{{R\omega }}} \right)$$
D. $$\theta = \pi - {\cos ^{ - 1}}\left( {\frac{v}{{R\omega }}} \right)$$
E. $$\theta = \frac{\pi }{2} - {\sin ^{ - 1}}\left( {\frac{v}{{R\omega }}} \right)$$
F. $$\theta = 2\pi - {\cos ^{ - 1}}\left( {\frac{v}{{R\omega }}} \right)$$
G. $$\theta = \pi + {\sin ^{ - 1}}\left( {\frac{v}{{R\omega }}} \right)$$
H. $$\theta = 2\pi + {\cos^{-1}}\left( \frac{v}{{R\omega}} \right)$$
I. $$\theta = 2\pi - {\sin ^{ - 1}}\left( {\frac{v}{{R\omega }}} \right)$$
J. $$\theta = \pi + 2{\sin ^{ - 1}}\left( {\frac{v}{{R\omega }}} \right)$$
Answer: D
The elastic collision between two bodies, A and B, can be considered using the following model. A and B are free to move along a common line without friction. When their distance is greater than d = $1$m, the interacting force is zero; when their distance is less than d, a constant repulsive F = $6$ N is present. The mass of body A is $m_A$ = $1$kg and it is initially at rest; the mass of body B is $m_B$ = $3$kg and it is approaching body A head-on with a speed $v_o$ = $2$m/s. Find the minimum distance between A and B

Options:
A. 2.50m
B. $0.75$m
C. 1.50m
D. 1.25m
E. 1.75m
F. $2$m
G. $0.25$m
H. 1.00m
I. 2.25m
J. $0.50$m
Answer: F
A long plank of mass 5m is placed on a rough horizontal surface of $\mu =0.3.$ block of mass m is placed on the surface of the plank for which $\mu =0.2.$ The block is connected to a fixed point by a light inextensible string which is parallel to the horizontal. Then The minimum force F applied on the plank to make it move is

Options:
A. 1.2 mg
B. 1.8 mg
C. 1.6 mg
D. 1.9 mg
E. 1.3 mg
F. 1.4 mg
G. 1.5 mg
H. 2 mg
I. 1.1 mg
J. 1.7 mg
Answer: G
How much power is required by an electric heater to melt $5$ kg of ice at $\displaystyle { 0 }^{ \circ  }C$ in $2$ mins if $60\%$ heat is supplied by the heater? ($L = 336$ J/g)

Options:
A. 90kW
B. 89kW
C. 59kW
D. 102kW
E. 87kW
F. 98kW
G. 28kW
H. 95kW
I. 93kW
J. 94kW
Answer: I
The standard reduction -potential for the half-cell:$NO_3^- (aq) + 2H^+ + e^- \rightarrow NO_2(g) + H_2O$  is  0.78V.$[\text{Take}\ \frac {2.303RT}{F} = 0.06, log 2 = 0.3]$Select correct options.

Options:
A. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 0.40 volt.
B. The reduction potential of a half-cell in 8 M H$^+$ (considering other species to be at unit concentration) is 0.834 volt.
C. The reduction potential of a half-cell in 8 M H$^+$ ( (considering other species to be at unit concentration) is 0.672 volt.
D. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 0.54 volt.
E. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 0.38 volt.
F. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 0.36 volt.
G. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 0.42 volt.
H. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 1.62 volt.
I. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 0.32 volt.
J. The reduction potential of a half-cell in a neutral solution (considering other species to be at unit concentration) is 0.48 volt.
Answer: F
The equation of the standing wave in a string clamped at both ends, vibrating in its third harmonic is given by$y=0.4\sin(0.314x)\cos(600\pi t)$  where, $x$ and $y$ are in cm and $t$ in sec.a) the frequency of vibration is $300 Hz$.b) the length of the string is $30 cm$.c) the nodes are located at  $x = 0, 10 cm, 30 cm$

Options:
A. Only a, c are true
B. a, b, c are false
C. Only c is true
D. Only b is true
E. Only b, c are true
F. a, b, c are true
G. Only a is true
H. Only a, b are true
Answer: F
A certain amount of idea mono atomic gas undergoes, process given by $U{V^{\frac{1}{2}}} = C$ where $U$ is the internal energy of the gas. The molar specific heat of the gas for the process will be

Options:
A. $$3R$$
B. $$\frac{5R}{3}$$
C. $$\frac{{5R}}{2}$$
D. $$\frac{2R}{2}$$
E. $$\frac{R}{2}$$
F. $$\frac{3R}{3}$$
G. $$\frac{3R}{2}$$
H. $$\frac{{ - R}}{2}$$
I. $$\frac{4R}{2}$$
J. $$\frac{6R}{3}$$
Answer: A
Consider a L - C oscillation circuit. Circuit elements has zero resistance Initially at $t = 0$ all the energy is stored in the form of electric field and plate-1 is having positive charge :At time $t = t_{1}$ plate-2 attains half of the maximum $+ve$ change for the first time Value of $t_{1}$ is :

Options:
A. $$\dfrac{2\pi}{3}\sqrt{LC}$$
B. $$\dfrac{\pi}{3}\sqrt{LC}$$
C. $$\dfrac{7\pi}{4}\sqrt{LC}$$
D. $$\dfrac{8\pi}{9}\sqrt{LC}$$
E. $$\dfrac{3\pi}{4}\sqrt{LC}$$
F. $$\dfrac{5\pi}{6}\sqrt{LC}$$
G. $$\pi \sqrt{LC}$$
H. $$\dfrac{7\pi}{6}\sqrt{LC}$$
I. $$\dfrac{4\pi}{3}\sqrt{LC}$$
J. $$\dfrac{5\pi}{4}\sqrt{LC}$$
Answer: I
pH of a solution obtained by mixing equal volume of 0.2 M NaOH & 0.2 M ${ CH }_{ 3 }COOH\quad \left( { K }_{ a }={ 10 }^{ -5 } \right)$ is :

Options:
A. 6.5
B. 7
C. 8
D. 9.5
E. 8.5
F. 7.25
G. 6.75
H. 9
I. 7.5
J. 5
Answer: B
A solid cylinder having radius 0.4 m, initially rotating (at t = 0) with $\omega_0$ = 54 rad/sec is placed on a rough inclined plane with $\theta = 37^0$ having friction coefficient $\mu$ = 0.5. The time taken by the cylinder to start pure rolling is:

Options:
A. 1.6 sec
B. 1.1 sec
C. 1.7 sec
D. 1.3 sec
E. 1.5 sec
F. 1.8 sec
G. 1.2 sec
H. 5.4 sec
I. 1.4 sec
J. 1.9 sec
Answer: G
Given that:
$$\Delta G_{r}^{\circ} (CuO) = -30.4 \text{ kcal/mole}$$
$$\Delta G_{r}^{\circ} (Cu_2O) = -34.98 \text{ kcal/mole} \;\;\;\;\;\;\;\;\; T = 298\ K$$
Now on the basis of above data which of the following predictions will be most appropriate under the standard conditions and reversible reaction.

Options:
A. Finely divided form of $CuO$ kept in excess $O_2$ would be completely converted to mixture of $CuO$ and $Cu_2O$ (having equal amounts of both)
B. Finely divided form of $CuO$ kept in excess $O_2$ would be partially converted to $Cu_2O$ (but not completely)
C. Finely divided form of $Cu_2O$ kept in excess $O_2$ would be partially converted to $CuO$ (but not completely)
D. Finely divided form of $CuO$ kept in excess $O_2$ would be completely converted to mixture of $CuO$ and $Cu_2O$ (having more of $Cu_2O$)
E. Finely divided form of $CuO$ kept in excess $O_2$ would be completely converted to $Cu_2O$
F. Finely divided form of $CuO$ kept in excess $O_2$ would be partially converted to $Cu_2O$ (with a slight predominance of $CuO$)
G. Finely divided form of $CuO$ kept in excess $O_2$ would be completely converted to mixture of $CuO$ and $Cu_2O$ (having more of $CuO$ )
H. Finely divided form of $Cu_2O$ kept in excess $O_2$ would be completely converted to $CuO$
Answer: E
Which of the following regents would carry out the following transformation? $(D=^{ 2 }{ H })$

Options:
A. $NaBD_{4}$ in $H_{2}O $
B. $NaBD_{4}$ in $CH_{3}OH$
C. $NaBD_{4}$ in $D_{2}O $
D. $LiAlH_{4}$, then $D_{2}O$
E. $NaBD_{4}$ in $CH_{3}OD $
F. $LiAlH_{4}$, then $CH_{3}OH $
G. $LiAlD_{4}$, then $D_{2}O$
Answer: B
A solid sphere of radius $  R  $ is placed on a smooth horizontal surface.$ A$ horizontal force $F$ is applied at height $  h  $ from the lowest point. For the maximum acceleration of centre of mass, which is correct?

Options:
A. $$ h=R $$
B. $$ h=0 $$
C. $$h=\frac{2R}{3}$$
D. $$ h=2 R $$
E. $$h=\frac{R}{3}$$
F. h=3R
G. $$ h=\frac{R}{2} $$
H. $$h=4R$$
I. $Centre$ $of$ $mass$ $has$ $same$ $acceleration$ $in$ $each$ $case$
J. $$ h=\frac{3R}{2} $$
Answer: A
Consider the two idealised systems.
(i) a parallel plate capacitor with large plates and small separation and
(ii) a long solenoid of length L $> >$R, radius of cross-section.
In (i) $ec{E}$ is ideally treated as a constant between plates and zero outside. In (ii) magnetic field is constant inside the solenoid and zero outside. These idealised assumptions, however, contradict fundamental laws as given.

Options:
A. $$Case (ii) contradicts Ampère's circuital law$$
B. $$Case (i) contradicts Gauss's law for magnetic fields$$
C. Case (ii) contradicts $\oint ec{H}\cdot ec{dl}=I_{en}$
D. $$Case (i) contradicts Gauss's law for electrostatic fields$$
E. $$Case (i) contradicts Faraday's law of induction$$
F. Case (i) agree with $\oint ec{E}\cdot ec{dl}=0$
G. $$Case (i) contradicts Ampère's circuital law$$
H. $$Case (ii) agrees with $\oint \vec{H}\cdot \vec{dl}=I_{en}$$$
I. $$Case (ii) contradicts Gauss's law for magnetic fields$$
J. $$Case (ii) contradicts Faraday's law of induction$$
Answer: I
A flat coil, C, of n turns , area Aa and resistance R, is placed in a uniform magnetic field of magnitude B. The plane of the coil is initially perpendicular to B. The coil is rotated by an angle $\theta$ about a diameter and charge of amount Q flows through it. Choose the correct Alternative.

Options:
A. $$\theta =90 ^{\circ}\, Q=(BAn/3R)$$
B. $$\theta =90 ^{\circ}\, Q=(BAn/4R)$$
C. $$\theta =90 ^{\circ}\, Q=(BAn/2R)$$
D. $$\theta =90 ^{\circ}\, Q=(3BAn/5R)$$
E. $$ \theta =180 ^{\circ}\, Q=(BAn/R)$$
F. $$\theta =90 ^{\circ}\, Q=(BAn/5R)$$
G. $$ \theta =90 ^{\circ}\, Q=0$$
H. $$\theta =90 ^{\circ}\, Q=(2BAn/R)$$
Answer: G
Sulphur trioxide is prepared by the following two reactions.$S_8(s) + 8O_2(g) \rightarrow 8SO_2(g); \Delta H = -2374. kJ $$2SO_2(g) + O_2(g) \rightarrow 2SO_3(g); \Delta H = -198. kJ $Calculate the net change in enthalpy for the formation of one mole of sulfur trioxide from sulfur and oxygen.

Options:
A. $396$ kJ
B. $-296.75$ kJ
C. -396.5 kJ
D. -396.75 kJ
E. -196.75 kJ
F. -197.25
G. -196.25 kJ
H. -395.5 kJ
I. $-198$ kJ
J. $-199$ kJ
Answer: A
The rate constant for the decomposition of a hydrocarbon is $2.418\times 10^{-5}S^{-1}$ at $546$ K. If the energy of activation is $179.9$ kJ/mol. What will be the value of pre-exponential factor A?

Options:
A. $$[A]=4.812 \times 10^{12}s^{-1}$$
B. $$[A]=4.012 \times 10^{12}s^{-1}$$
C. $$[A]=6.245 \times 10^{12}s^{-1}$$
D. $$$[A]=7.83\times 10^{12}s^{-1}$.$$
E. $$[A]=5.412 \times 10^{12}s^{-1}$$
F. $$$[A]=9.245\times 10^{12}s^{-1}$.$$
G. $$None of these$$
H. $$[A]=4.912 \times 10^{12}s^{-1}$$
I. $$$[A]=3.912\times 10^{12}s^{-1}$.$$
J. $$[A]=3.812 \times 10^{12}s^{-1}$$
Answer: I
A string of length L, fixed at its both ends is vibrating in its first overtone mode. Consider two elements of the spring of the same small length at positions ${\ell _1} = 0.2\;{\text{and}}\;{\ell _2} = 0.45{\text{L}}$ from one end. If ${K_1}\;{\text{and}}\;{K_2}$ are their respective maximum kinetic energies then

Options:
A. $${K_1} = 1.25{K_2}$$
B. $${K_1} > {K_2}$$
C. $${K_1} < {K_2}$$
D. $${K_1} = 0.75{K_2}$$
E. $${K_1} = {K_2}$$
F. $${K_1} = 0.5{K_2}$$
G. $${K_1} = 1.5{K_2}$$
H. It is not possible to decide the relation.
I. $${K_1} = 2{K_2}$$
Answer: E
A swimmer is crossing the river of width d with velocity $v$ w.r.t river in such a way that drift of swimmer becomes minimum. The velocity of water current is u. If $\frac{u}{v}=n$  and $n \ge 1$ then

Options:
A. All the above are correct
B. minimum drift is $d \sqrt{n^2-1}$
C. Time taken to cross the river is $\frac{nd}{v \sqrt{n^2-1}}$
D. total displacement of swimmer is \(d \sqrt{n^2-2}\)
E. total displacement of swimmer is $nd$
F. total displacement of swimmer is \( d \sqrt{n^2 - 3} \)
G. total displacement of swimmer is $d \sqrt{n^2-1}$
H. total displacement of swimmer is \( d \sqrt{n^2 + 2} \)
I. time taken to cross the river is $\frac{nd}{v \sqrt{n^2+1}}$
J. total displacement of swimmer is $d \sqrt{n^2+1}$
Answer: E
Select the correct statement:

Options:
A. For a real gas following van der Waal's equation of state, the expansion of critical temperature is $\frac{8a}{27R.b}$
B. For a real gas following van der Waal's equation of state, the expansion of critical temperature is \(\frac{8a}{27R^2b}\)
C. The real gas behaves as an ideal gas at Boyle's temperature.
D. For a real gas following van der Waal's equation of state, the expansion of critical temperature is \(\frac{8a}{27R^2b^2}\)
E. For a real gas following van der Waal's equation of state, the expansion of critical temperature is $\frac{8a}{27R^2b}$
F. At low pressure, the compressibility factor$'Z'=1+\frac{P.b}{RT}$ for a van der Waal's gas.
G. For a real gas following van der Waal's equation of state, the expansion of critical temperature is $\frac{8a}{27R^2b^2}$
H. The value of compressibility factor $'Z'$ for $H_{2}$ gas is greater than one at room temperature and pressure.
I. For a real gas following van der Waal's equation of state, the expansion of critical temperature is $\frac{8a}{27Rb^2}$
J. For a real gas following van der Waal's equation of state, the expansion of critical temperature is \(\frac{8a}{27bR}\)
Answer: A
A particle of mass $m$ and positive charge $q$ is projected towards an infinitely long line of a charge (having linear density of charge $+\lambda$) from a distance $t_0$. The direction of initial velocity $v_0$ makes an angle $30^\circ$ with the normal to the line of charge as shown in figure. The minimum distance of approach of the charge particle with the line of charge will be (neglect gravity). Take $\lambda = \dfrac{\pi arepsilon_0 mv_0^2}{4q}$.

Options:
A. $$\dfrac{r_0}{e^3}$$
B. $$\dfrac{r_0}{e^7}$$
C. $$\dfrac{r_0}{e^5}$$
D. $$\dfrac{r_0}{e^2}$$
E. $$\dfrac{r_0}{e^4}$$
F. $$\dfrac{r_0}{2}$$
G. $$\dfrac{r_0}{e}$$
H. $$\dfrac{r_0}{e^6}$$
I. \(\dfrac{r_0}{e^4}\)
Answer: A
Calculate the heat effect produced when a solution of 1 mole of ethanol in 5 moles of water is mixed with a solution of 2 moles of ethanol in 20 moles of water. Given that, $C_2H_5OH + 5H_2O \rightarrow C_2H_5OH (5H_2O)$;  $\Delta H = -2140$ cal $C_2H_5OH + 10H_2O \rightarrow C_2H_5OH (10H_2O)$;  $\Delta H = -2730$ cal $C_2H_5OH + 25/3H_2O \rightarrow C_2H_5OH (25/3H_2O)$; $\Delta H = -2670$cal.

Options:
A. 103.33 cal is released
B. 510 cal is absorbed
C. 210 cal is released
D. 210 cal is absorbed
E. 410 cal is absorbed
F. 310 cal is released
G. 510 cal is released
H. 103.33 cal is absorbed
I. 410 cal is released
J. 310 cal is absorbed
Answer: I
The standard enthalpy of combustion of sucrose is-5645 kJ $mol^{-1}$.What is the advantage (in kJ $mol^{-1}$ of energy released as heat) of complete aerobic oxidation compared to anaerobic hydrolysis of sucrose to lactic acid? $\triangle H^{^{\circ}}_{\int }$  for lactic acid,CO_{2} and $H_{2}$O is -694,-395 and -286.0 respectively

Options:
A. $$advantage=5496 kJ.$mol^{-1}$$$
B. $$advantage=5536 kJ.$mol^{-1}$$$
C. $$advantage=5657 kJ.$mol^{-1}$$$
D. $$advantage=4756 kJ.$mol^{-1}$$$
E. $$advantage=5456 kJ.$mol^{-1}$$$
F. $$advantage=5256 kJ.$mol^{-1}$$$
G. $$advantage=5596 kJ.$mol^{-1}$$$
H. $$advantage=5396 kJ.$mol^{-1}$$$
I. $$advantage=5296 kJ.$mol^{-1}$$$
J. $$advantage=4356 kJ.$mol^{-1}$$$
Answer: H
Two moles of air,When heated through $10k$ expand by an amount of $1.66 \times {10^{ - 3}}{m^{ - 3}}$ under a constant pressure of ${10^5}N/{m^2}.$ If ${C_v} = 20.81J/mole\,K,$ then $C_p$ is

Options:
A. $$39.11J/mole\,K$$
B. $$49.11J/mole\,K$$
C. $$32.11J/mole\,K$$
D. $$56.11J/mole\,K$$
E. $$46.11J/mole\,K$$
F. $$42.11J/mole\,K$$
G. $$29.11J/mole\,K$$
H. $$19.11J/mole\,K$$
I. $$52.11J/mole\,K$$
J. $$59.11J/mole\,K$$
Answer: B
$100 g$ of $C_6H_12O_6$(aq.) solution has vapour pressure is equal to 40 torr at certain temperature. Vapour pressure of $H_2O(l)$ is 4.80 torr at same temperature. If this solution is cooled to $-0.93^{\circ}C$, what mass of ice will be separated out? ($K_f = 1.86 kg mol^{-1}$)

Options:
A. $$51.2 g$$
B. $$43.5 g$$
C. $$45.5 g$$
D. $$95.5 g$$
E. $$44.3 g$$
F. $$49.3 g$$
G. $$4.5 g$$
H. $$47.8 g$$
I. $$50.2 g$$
Answer: H
A container of a large uniform cross-sectionl area resting a horizontal surface holds two immiscible.non-viscous and incompressible liquids of densitites 'd' and 3'd' each of height (1/2) H as shown.The smaller density liquid is poen to atmosphere,A homogeneous solid Cylinder of length $ L \left( <\frac { 1 }{ 2 } H \right) $ cross-sectional area (1/5) A is immersed such that it floats with its axis vertical to the liquid-liquid interface with length (1/4) L in denser liquid.If D is the density of the solid cylinder then:

Options:
A. $$D = \frac{4d}{3}$$
B. $$D = \frac{7d}{4}$$
C. $$D = \frac{3d}{4}$$
D. $$ D = \frac {3d}{2} $$
E. $$ D = \frac {2d}{3} $$
F. $$ D = \frac {5d}{4} $$
G. $$D = \frac{9d}{8}$$
H. $$D = \frac{6d}{5}$$
I. $$D = \frac{7d}{8}$$
J. $$ D = \frac {d}{2} $$
Answer: F
One mole of an ideal monatomic gas undergoes thermodynamic cycle $ 1 \rightarrow 2 \rightarrow 3 \rightarrow 1 $ as shown in Fig. $ 2.161 . $ Initial temperature of gas is $ T_{0}=300 \mathrm{K} $
Process $ 1 \rightarrow 2: P=a V $
Process $ 2 \rightarrow 3: P V= $ Constant
Process $ 3 \rightarrow 1: P= $ Constant 
(Take $ \ln |3|=1.09) $
Find the net work done by the cycle.

Options:
A. $$6.12 R T_{0}$$
B. $$4.92 R T_{0}$$
C. $$ 5.81 R T_{0} $$
D. $$5.12 R T_{0}$$
E. $$5.32 R T_{0}$$
F. $$ 4.53 R T_{0} $$
G. $$5.92 R T_{0}$$
H. $$ 3.27 R T_{0} $$
I. $$5.61 R T_{0}$$
J. $$ 6.83 R T_{0} $$
Answer: C
Initially an object is kept at a distance of 10 cm from the convex lens and a sharp image is formed at 10 cm ahead of lens on the screen. Now a glass plate of $\mu =1.5$ cm and thickness 1.5 cm is placed between object and lens. The distance by which the screen be shifted to get sharp image on the screen will be

Options:
A. $$\dfrac { 5 }{ 9 } cm$$
B. $$\dfrac { 8 }{ 5 } cm$$
C. $$\dfrac { 9 }{ 5 } cm$$
D. $$\dfrac{11}{5} cm$$
E. 5 cm
F. $$1 cm$$
G. $$\dfrac{7}{9} cm$$
H. $$\dfrac{10}{5} cm$$
I. $$\dfrac { 6 }{ 5 } cm$$
J. $$\dfrac { 7 }{ 5 } cm$$
Answer: C
An infinite number of charges 'q' each are placed along the x - axis at x =1, x=4 ,x=8 and so on. If the distance are in meters calculate the electric potential at x=0

Options:
A. $$\frac{3q}{6\pi \epsilon_0}$$
B. $$\frac { q } { 2 \pi \epsilon _ { 0 } }$$
C. $$\frac{5q}{8\pi \epsilon_0}$$
D. $$\frac { 2 q } { \pi \epsilon _ { 0 } }$$
E. $$\frac { 4 q } { \pi \epsilon _ { 0 } }$$
F. $$\frac{3q}{12\pi \epsilon_0}$$
G. $$\frac{3q}{2\pi \epsilon_0}$$
H. $$\frac{3q}{4\pi \epsilon_0}$$
I. $$\frac{{3q}}{{8\pi {E_0}}}$$
J. $$\frac{3q}{10\pi \epsilon_0}$$
Answer: I
If for $2A_2B(g) \rightleftharpoons 2A_2(g) + B_2(g)$, $K_p$ = total pressure (at equilibrium ) and starting the dissociation from 4 mol of $A_2B$ then :

Options:
A. degree of dissociation of $A_2B$ will be (2/3).
B. at equilibrium the no. of moles of $A_2B$ are equal to the no. of moles of $A_2$
C. at equilibrium the no. of moles of $A_2B$ are equal to the no. of moles of $A_2$.
D. at equilibrium the no. of moles of \( A_2B \) are not equal to the no. of moles of \( B_2 \) and \( A_2 \)
E. at equilibrium the no. of moles of $A_2B$ are not equal to the no. of moles of $A_2$
F. total no. of moles at equilibrium will be (14/3).
G. at equilibrium the no. of moles of $A_2B$ are not equal to the no. of moles of $B_2$.
H. at equilibrium the no. of moles of $A_2B$ are equal to the no. of moles of $B_2$
I. at equilibrium the no. of moles of \( A_2B \) are equal to the no. of moles of \( B_2 \)
J. at equilibrium the no. of moles of \( A_2B \) are not equal to the no. of moles of \( A_2 \)
Answer: G
A car beings from rest at time t= 0, and then accelerates along a straight track during the interval $\displaystyle 0< t\leq 2s$ and thereafter with constant velocity as shown in the graph. A coin is initially at rest on the floor of the car. At t= 1 s, the coin begins to slip and its stops slipping at t= 3s. The coefficient of static friction between the floor and the coin is $\displaystyle \left ( g=10 m/s^{2} \right )$

Options:
A. 0.42
B. 0.45
C. 0.3
D. 0.5
E. 0.6
F. 0.38
G. 0.35
H. 0.4
I. 0.2
J. 0.7
Answer: H
The mass in grams of potassium permanganate (MM = 158) crystals required to oxidize $750\,\,c{m^3}$ of 0.1 M Mohr's salt solution in acidic medium is about?

Options:
A. 5.9
B. 5.7
C. 6.0
D. 6.1
E. 11.7
F. 11.9
G. 11.6
H. 11.8
I. 5.8
J. 2.4
Answer: H
An arbitrary compound $P_2Q$ decomposes according to reaction:$2P_2Q (g) \rightleftharpoons 2P_2 (g) + Q_2 (g)$If one starts the decomposition reaction with $4$ moles of $P_2Q$ and value of equilibrium constant $K_p$ is numerically equal to total pressure at equilibrium. Then which option is/are correct at equilibrium:

Options:
A. total number of moles of product $(P_2$ & $Q_2)$ at equilibrium is $4$
B. total number of moles of product $(P_2$ & $Q_2)$ at equilibrium is $5
C. total number of moles of product $(P_2$ & $Q_2)$ at equilibrium is $6
D. total number of moles of product $(P_2$ & $Q_2)$ at equilibrium is $8
E. degree of dissociation is $\alpha = (2/3)$
F. total number of moles of product $(P_2$ & $Q_2)$ at equilibrium is $7
G. moles of $n_{P_2Q} = n_{Q_2}$
H. total number of moles of product $(P_2$ & $Q_2)$ at equilibrium is $2
Answer: A
Small identical balls are attached at each end of a spring of force constant k = 200 N/m and relaxed length l = 20 cm. Unknown amount of charges are gradually transferred to the balls in unequal amounts until the spring length becomes twice of its relaxed length. What amount of work must be done by an external agency in slowly compressing the spring back to its relaxed length?

Options:
A. 14 J
B. 10 J
C. 13 J
D. 20 J
E. 18 J
F. 15 J
G. 16 J
H. 12 J
I. 8 J
J. 9 J
Answer: H
One $dm^{3}$ solution containing $10^{-5}$ moles each $Cl^{-}$ ions is treated with $10^{-4}$ moles of silver nitrate .Which one of the following observations is made
$[K_{sp}AgCrO_{4}=4\times10^{-12}]$$[K_{sp}AgCl=1\times10^{-10}$]

Options:
A. Silver chloride gets precipitated first
B. Precipitation does not occur
C. Silver chloride precipitates first, then silver chromate
D. Silver chloride and silver chromate precipitate simultaneously but incompletely
E. Both silver chromate and silver chloride precipitate immediately and completely
F. Silver chloride and silver chromate precipitate at the same rate but not completely
G. Silver chromate gets precipitated first
H. Silver chloride and silver chromate precipitate at different rates
I. Silver chloride gets precipitated first but incompletely
J. Both silver chromate and silver chloride start precipitation simultaneously
Answer: B
The charge through a conductor of resistance $R$ varies with time as $q=at-bt^2$ where a and b are constant. Total heat produced in R is

Options:
A. $$\(\displaystyle Ra^3/5b\)$$
B. $$\displaystyle Ra/6b$$
C. $$\displaystyle Ra^3/3b$$
D. \(\displaystyle Ra^3/9b \)
E. \(\displaystyle Ra^3/12b\)
F. \(\displaystyle Ra^3/2b\)
G. \(\displaystyle Ra^3/8b\)
H. \(\displaystyle Ra^3/4b \)
I. $$\displaystyle Ra^2/9b$$
J. $$\displaystyle Ra^3/6b$$
Answer: C
The gas A decomposes as $A\longrightarrow B+5C$. A partially decomposed gaseous mixture is allowed to effuse through a pin hole and the gas coming out initially was analysed. The mole of fraction C is effused gas was found to be 0.6. Determine ?

Options:
A. Molecular mass of B = 160 & C = 410
B. 5/3
C. 7/4
D. Given molecular mass of A = 360
E. 4/3
F. 3/2
G. 5/2
H. (b) the degree of dissociation of "A". Fill your answer as sum of digits (excluding decimal places)till you get single digit answer.
I. 5/4
J. (a) ratio of rate of effusion of "C" and "B". if your simplest ratio is x/y the fill the value of x+y in OMR sheet till you get single digit answer.
Answer: A
A seconds pendulum clock has a steel wire, The clock shows correct time at $25 ^ { \circ } \mathrm { C }$. How much time does the clock lose or gain, in one week, when the temperature is increased to $35 ^ { \circ } \mathrm { C }$ ? $\alpha_{\text{steel}} = 1.2 \times 10^{-5} \, /\, ^\circ \mathrm{C}$

Options:
A. $$311.5 \mathrm { s }$$
B. $$36.28 s$$
C. $$241.5 \mathrm { s}$$
D. $$3.828 s$$
E. $$82.35 s$$
F. $$261.5 \mathrm { s }$$
G. $$221.5 \mathrm { s}$$
H. $$281.5 \mathrm { s}$$
I. $$331.5 \mathrm { s }$$
J. $$321.5 \mathrm { s }$$
Answer: J
A block mass m is kept on an inclined plane of a lift moving down with acceleration of 2 $ms^{-2}$. What should be minimum coefficient of friction to let the block move down with constant velocity?

Options:
A. \(\mu = 0.6 \)
B. \(\mu = 0.75\)
C. \(\mu = 0.9\)
D. $\mu$ =not defined
E. $$\(\mu = 0.5\)$$
F. \(\mu = 0.3\)
G. $$\mu =\frac{1}{\sqrt3}$$
H. $$\mu = 0.4$$
I. $$\mu = 0.8$$
J. \(\mu = 0.7\)
Answer: I
A large rectangular box $ABCD$ falls vertically with an acceleration $a$. A toy gun fixed at $A$ and aimed towards $C$ fires a particle $P$.

Options:
A. $P$ will hit the wall $CD$ or the floor $AD$ if $a< g$.
B. P will hit the wall BC or the floor AD if a > g
C. May be options either $(A),(B)$ or $(C)$, depending on the speed of projection of $P$.
D. P will hit the wall $CD$ or the floor $AD$ if $a > g$
E. P will hit the wall BC or the floor AD if a > g.
F. $P$ will hit the roof $BC$ if $a> g$
G. $P$ will hit the wall $CD$ or the roof $AB$ if $a< g$
H. $P$ will hit the wall $CD$ or the floor $AD$ if $a= g$
I. $P$ will hit the wall $BC$ or the floor $AD$ if $a< g$.
J. $P$ will hit $C$ if $a=g$
Answer: A
To $20 ml$ of $0.1 M - NaOH$ solution, $3 ml$ of $1 M$ acetic acid solution is added. Is the solution now neutral, acidic or alkaline? How much more of the acetic acid solution we add to produce a change of $pH = 0.3$ unit? ($pK_{a}$ for $CH_{3}COOH = 4.74$, $log 2 = 0.3$)

Options:
A. alkaline, $1 ml$
B. acidic, $7 ml$
C. acidic, $8 ml$
D. acidic, $6 ml$
E. neutral, $2 ml$
F. acidic, $5 ml
G. acidic, $4 ml$
H. acidic, $9 ml$
I. acidic, $3 ml$
Answer: A
Two protons are kept at a separation of ${40} \ A^{o}$. ${F}_{n}$ is the nuclear force and ${F}_{e}$ is the electrostatic force between them. Then

Options:
A. $${F}_{n} = {F}_{e}$$
B. $${F}_{n} \ll {F}_{e}$$
C. ${F}_{n} \approx {F}_{e}$
D. $${F}_{n} \approx 2{F}_{e}$$
E. $${F}_{n} \ggg {F}_{e}$$
F. $${F}_{n} \gg {F}_{e}$$
G. $${F}_{n} \approx 3{F}_{e}$$
H. $$ {F} _ {n} \approx {F} _ {e}$$
Answer: F
$0.2\ moles$ of an ideal gas is taken round the cycle $ABC$ as shown in the figure. The path $B\rightarrow C$ is an adiabatic process, $A\rightarrow B$ is an isochoric process and $C\rightarrow A$ is an isobaric process. The temperature $T_{B} = 500\ K$ and pressure at $A$ is $1\ atm$ and volume at $A$ is $4.9\ L$. The volume at $C$ is (Given : $\gamma =\dfrac {C_{P}}{C_{V}} = \dfrac {5}{3}, R = 8.205\times 10^{-2} L\ atm\ mol^{-1} K^{-1}, \left (\dfrac {3}{2}\right )^{2/5} = 0.81)$.

Options:
A. $$5.9 L$$
B. $$5.8\ L$$
C. $$5.5\ L$$
D. $$6.4\ L$$
E. $$6.6\ L$$
F. $$6.1\ L$$
G. $$7.0\ L$$
H. $$6.8\ L$$
I. $$6.7\ L$$
J. $$6.3\ L$$
Answer: H
The temperature of the cold junction of thermo-couple is $0^{\circ}C$ and the temperature of hot junction is $T^{\circ}C$ . The e.m.f. is $E=16T-0.04T^{2}\mu $ volts. The temperature of inversion is

Options:
A. $$175^{\circ}C$$
B. $$200^{\circ}C$$
C. $$400^{\circ}C$$
D. $$300^{\circ}C$$
E. $$120^{\circ}C$$
F. $$250^{\circ}C$$
G. $$150^{\circ}C$$
Answer: C
A small block slides along a path, without friction, until the block reaches the section $L=3m$, which begins at height $h=3m$ on a flat incline of angle ${37}^{o}$, as shown. In that section, the coefficient of kinetic friction is $\mu_{k}$=$0.50$. The block passes through point $A$ with a speed of $\sqrt {136}m/s$. Find the speed (in m/s) of the block as it passes through the point $B$ where the friction ends. (Take $g=10m/{s}^{2}$)

Options:
A. $$7\ m/s$$
B. $$10\ m/s$$
C. $$7.5\ m/s$$
D. $$6\ m/s$$
E. $$7.2\ m/s$$
F. $$9\ m/s$$
G. $$4\ m/s$$
H. $$8\ m/s$$
I. $$5.5\ m/s$$
J. $$5\ m/s$$
Answer: D
Consider the following equation for a cell reaction:
$$A+B\longrightarrow C+D\quad ;\quad E^\circ=x\quad volt,\triangle G=\triangle G_{1}$$
$$2A+2B\longrightarrow 2C+2D\quad ;\quad E^\circ=y\quad volt,\triangle G=\triangle G_{2}$$

Options:
A. $$x=y,\quad \triangle G_{2}=\frac{3}{2}\triangle G_{1}$$
B. $x<y,\quad \triangle G_{2}=\frac{1}{2}\triangle G_{1}$
C. $$x=y,\quad \triangle G_{2}=3\triangle G_{1}$$
D. $$x=y,\quad \triangle G_{2}=4\triangle G_{1}$$
E. $$x=y,\quad \triangle G_{2}=\frac{1}{2}\triangle G_{1}$$
F. $$x=y,\quad \triangle G_{2}=2\triangle G_{1}$$
G. $$x=y,\quad \triangle G_{2}=\frac{1}{3}\triangle G_{1}$$
H. $$x<y,\quad 2\triangle G_{ 2 }=\triangle G_{ 1 }$$
I. $$x=y,\quad \triangle G_{1}=\triangle G_{2}$$
J. $$x>y,\quad \triangle G_{1}>\triangle G_{2}$$
Answer: I
For a gaseous reaction : A(g)$\rightarrow $ Products (order=n), the rate may be expressed as:(i)$-\dfrac {P_{A}}{dt}=K_{1}.P_{A}$(ii)$-\cfrac{1}{V}\cfrac{dn_{A}}{dt}=K_{2}.C_{A}$The rate constants,$K_{1}$ and $K_{2}$ are related as $(P_{A}$ and $C_{A})$ are the partial pressure and molar concentration of A at time 't', respectively:

Options:
A. $$K_{2}=K_{1}.(RT)^{2n}$$
B. $$K_{2}=K_{1}.(RT)^{n-1}$$
C. $$K_{2}=K_{1}.(RT)^{n+2}$$
D. $$K_{2}=K_{1}.(RT)^{n+1}$$
E. $$K_{1}=K_{2}$$
F. $$K_{2}=K_{1}.(RT)^{n-2}$$
G. $$K_{2}=K_{1}.(RT)^{n}$$
H. $$K_{2}=K_{1}.(RT)^{n-3}$$
I. $$K_{2}=K_{1}.(RT)^{1-n}$$
J. $$K_{2}=K_{1}.(RT)^{-n}$$
Answer: B
In a region, electric field depends on X-axis as $E=E_0x^2$. There is a cube of edge a as shown. Then find the charge enclosed in that cube.

Options:
A. $$2\epsilon_0a^4E_0$$
B. $$4\epsilon_0a^4E_0$$
C. $$3\epsilon_0a^4E_0$$
D. $$10\epsilon_0a^4E_0$$
E. $$6\epsilon_0a^4E_0$$
F. $$7\epsilon_0a^4E_0$$
G. Zero
H. $$8\epsilon_0a^4E_0$$
I. $$9\epsilon_0a^4E_0$$
J. $$5\epsilon_0a^4E_0$$
Answer: J
An ammeter and a voltmeter are connected in series to a battery with an emf $arepsilon=6.0 \ V$. When a certain resistance is connected in parallel with the voltmeter, the reading of the voltmeter decreases $2.0\ times$, whereas the reading of the ammeter increase the same number of times. Find the voltmeter reading after the connection of the resistance.

Options:
A. $$5V$$
B. $$4V$$
C. 3V
D. $$8V$$
E. $$6V$$
F. $$2V$$
G. $$12V$$
H. $$18V$$
Answer: B
The fraction of the volume of a glass flask must be filled with mercury so that the volume of the empty space may be the same at all temperature is (${\alpha  }_{ glass }=9\times { 10 }^{ -6 }/C,{ \gamma  }_{ Hg }=18.9\times { 10 }^{ -5 }/C$)

Options:
A. $$\dfrac { 1 }{ 8 }$$
B. $$\dfrac { 1 }{ 4 } $$
C. $$\dfrac { 1 }{ 10 }$$
D. $$\dfrac { 1 }{ 5 } $$
E. $$\dfrac { 1 }{ 11 }$$
F. $$\dfrac { 1 }{ 7 } $$
G. $$\dfrac { 1 }{ 6 }$$
H. $$\dfrac { 1 }{ 2 } $$
I. $$\dfrac { 1 }{ 3 }$$
J. $$\dfrac { 1 }{ 9 }$$
Answer: B
The pressure in a bulb dropped from 2000 to 1500 mm Hg in 47 min when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight 79 in the molar ratio of 1:1 at a total pressure of 4000 mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 74 min.

Options:
A. $$\displaystyle \frac{M_{gas}}{M_{O_2}}=0.666$$
B. $$\displaystyle \frac{M_{gas}}{M_{O_2}}=1.136$$
C. $$\displaystyle \frac{M_{gas}}{M_{O_2}}=1.236$$
D. $$\displaystyle \frac{M_{gas}}{M_{O_2}}=1.952$$
E. $$\frac{M_{gas}}{M_{O_2}}=1.336$$
F. $$\displaystyle \frac{M_{gas}}{M_{O_2}}=1.572$$
G. $$\displaystyle \frac{M_{gas}}{M_{O_2}}=5.236$$
H. $$\frac{M_{gas}}{M_{O_2}}=1.672$$
I. $$\frac{M_{gas}}{M_{O_2}}=1.472$$
J. $$\displaystyle \frac{M_{gas}}{M_{O_2}}=2.472$$
Answer: C