MikaStars39/nano-eval huggingface dataset:
A large slab of mass $5 kg$ lies on a smooth horizontal surface, with a block of mass $4 kg$ lying on the top of it, the coefficient of friction between the block and the slab is $0.25$. If the block is pulled horizontally by a force of $F=6 N$, the work done by the force of friction on the slab between the instant $t=2 s$ and $t=3 s$ is :

Options:
A. 5.65 J
B. 4.55 J
C. 6.66 J
D. 6.55 J
E. 5.44 J
F. $$10 J$$
G. 4.44 J
H. $$2.4 J$$
I. $$5.55 J$$
Answer: I
A metal (atomic mass = 50) has a body centered cubic crystal structure. The density of metal is 5.96 g $cm^{-3}$. Find the volume of the unit cell.

Options:
A. $$13.9 \times 10^{-24} \ cm^3$$
B. $$27.8 \times 10^{-24} \ cm^3$$
C. $$55.6 \times 10^{-24} \ cm^3$$
D. 13.8 \times 10^{-24} \ cm^3
E. 13.7 \times 10^{-24} \ cm^3
F. $$6.95 \times 10^{-24} \ cm^3$$
G. 17.9 \times 10^{-24} \ cm^3
H. 17.85 \times 10^{-24} \ cm^3
I. 13.85 \times 10^{-24} \ cm^3
J. 14.0 \times 10^{-24} \ cm^3
Answer: A
A metal plate of area $1\times { 10 }^{ -4 }{ m }^{ 2 }$ is  illuminated by a radiation of intensity 16 m $W/{ m }^{ 2 }.$ The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be :
$$\left[ { 1eV=1.6\times 10^{ 19 }J } \right] $$

Options:
A. $${ 10 }^{ 12 } and\ 5eV$$
B. $${ 10 }^{ 11 } and\ 2.5eV$$
C. $${ 10 }^{ 14 } and\ 5eV$$
D. $${ 10 }^{ 10 } and\ 5eV$$
E. { 10 }^{ 13 } and\ 5eV
F. { 10 }^{ 13 } and\ 4.5eV
G. { 10 }^{ 11 } and\ 4.5eV
H. ${ 10 }^{ 12 } and\ 2.5eV$
I. { 10 }^{ 13 } and\ 2.5eV
J. { 10 }^{ 11 } and\ 5eV
Answer: A
How long would it take a current of 10 amperes to put a layer of copper 0.1 mm thick on both sides of a rectangular plate of 10 cm x 1 x 65 cm (e.c.e. of copper = 0 x 00033 gm/coul, density of Copper = 9 gin/cc)?

Options:
A. 450 sec
B. 900 sec
C. 9 min
D. 45000 sec
E. 90 sec
F. 9 sec
G. 9000 sec
H. 900 min
I. 45 sec
J. 4500 sec
Answer: A
The magnetic field of a plane electromagnetic wave is given by:
$\overrightarrow { B } ={ B }_{ 0 }\hat { i } \left[ \cos { \left( kz-\omega t \right)  }  \right] +{ B }_{ 1 }\hat { j } \left[ \cos { \left( kz+\omega t \right)  }  \right] $ where ${ B }_{ 0 }=3\times { 10 }^{ -5 }T;{ B }_{ 1 }=2\times { 10 }^{ -6 }T\quad $. The rms value of the force experienced by a stationary charge $Q={ 10 }^{ -4 }C$ at $z=0$ is closest to:

Options:
A. $$0.9N$$
B. 0.4N
C. $$0.1N$$
D. 1.2N
E. $$0.6N$$
F. 0.5N
G. $$3\times{10}^{-2}N$$
H. 0.7N
I. $$0.8N$$
J. 0.3N
Answer: E
A parallel-plate air condenser of plate area A and separation d is charged to potential V and then the battery is removed. Now a slab of dielectric constant k is introduced between the plates. If Q, E and W denote respectively the magnitude of charge on each plate, the electric field between the plates (after introduction of dielectric slab) and work done on the system in the process of introducing the slab, then

Options:
A. \frac{V}{d} \left(1 - \frac{1}{2k}\right)
B. \(\displaystyle E= \frac{V}{d(1+k)} \)
C. \(\displaystyle E= \frac{V}{d(1-k)} \)
D. $$\displaystyle W=\frac{arepsilon _0 A Vh^2}{2d}(1-1/k)$$
E. \frac{V}{d}
F. $$\displaystyle Q=\frac{arepsilon _0 A V}{d}$$
G. $$\displaystyle E=\frac{V}{kd}$$
H. \frac{V}{d} \left(1 - \frac{1}{k}\right)
I. $$\displaystyle Q=\frac{arepsilon _0 K A V}{d}$$
J. \frac{V}{d(1+\frac{1}{k})}
Answer: G
When a beam of light goes from a denser medium $ \left(\mu_{d}\right)  $ to a rarer medium ( $  \mu_{r} ),  $ then it is generally observed that magnitude of angle of incidence is half that of angle of refraction. Then magnitude of incident angle will be - (here $  \mu=\mu_{d} / \mu_{r} )  $

Options:
A. 2$ \cos ^{-1} \mu $
B. 2 \sin ^{-1} \mu
C. 2 \sin ^{-1} \left( \frac{\mu}{4} \right)
D. $$ \cos ^{-1}\left(\frac{\mu}{2}\right) $$
E. 2$ \sin ^{-1}\left(\frac{\mu}{2}\right) $
F. 2 \sin ^{-1} \left( \frac{\mu}{3} \right)
G. 2 \tan ^{-1}\left(\frac{\mu}{2}\right)
H. 2 \cos ^{-1} \left( \frac{\mu}{3} \right)
I. 2$ \cos ^{-1}\left(\frac{\mu}{2}\right) $
Answer: E
Overall changes in volume and radii of a uniform cylindrical steel wire are $0.2\% $ and $0.002\%$ respectively when subjected to some suitable force. Longitudinal tensile stress acting on the wire is $\left( {Y = 2.0 \times {{10}^{11}}N{m^{ - 2}}} \right)$

Options:
A. 0.394 \times {10^9}N{m^{ - 2}}
B. 0.398 \times {10^9}N{m^{ - 2}}
C. $$0.392 \times {10^9}N{m^{ - 2}}$$
D. 0.382 \times {10^9}N{m^{ - 2}}
E. 0.390 \times {10^9}N{m^{ - 2}}
F. 3.12 \times 10^7 Nm^{-2}
G. $$3.2 \times {10^9}N{m^{ - 2}}$$
H. $$3.2 \times {10^7}N{m^{ - 2}}$$
I. $$3.6 \times {10^7}N{m^{ - 2}}$$
J. 0.402 \times {10^9}N{m^{ - 2}}
Answer: C
The principal axis of an optical device is along $y=-1$, image of a small body placed at ($-30, 3$) is formed at a point ($60, -3$). Then the optical device is:-

Options:
A. A concave lens of focal length $20\ cm$
B. A convex mirror of focal length $60\ cm$
C. A concave lens of focal length $30\ cm$
D. A convex lens of focal length $20\ cm$
E. A convex lens of focal length $18\ cm$
F. A convex mirror of focal length $15\ cm$
G. A concave mirror of focal length $60\ cm$
H. A concave lens of focal length $15\ cm$
I. A convex lens of focal length $15\ cm$
J. A convex lens of focal length $25\ cm$
Answer: D
Find the heat capacity for a non-atomic gas undergoing the following process

Options:
A. PV^{-2}=Const
B. $$ \dfrac{V}{T^3} =Const$$
C. PV^{-3}=Const
D. PV^{-4}=Const
E. $$ \dfrac{P}{T^2} =Const$$
F. $$PV^{-1}=Const$$
G. \dfrac{P}{T^4} = Const
H. PV^{-5}=Const
I. $$PV^3=Const$$
Answer: F
There is layer of medium of variable refractive index $\mu = \sqrt { 2 } - \frac { 1 } { \sqrt { 2 } } \mathrm { y } ($ where $0 \leq \mathrm { y } < 1 / 2 )$ sandwitched between the layer of glass and air. A beam of light travelling in air at an anlge $45 ^ { \circ }$ has a width $\Delta \omega$ . When the beam
enters the layer of glass its width becomes: ( $\mu _ { glass } = \sqrt { 2 } )$

Options:
A. \sqrt { \frac { 5 } { 3 } } \Delta \omega
B. $$\sqrt { \frac { 2 } { 3 } } \Delta \omega$$
C. \sqrt { \frac { 4 } { 3 } } \Delta \omega
D. \sqrt { \frac { 2 } { 5 } } \Delta \omega
E. $$\sqrt { 2 } \Delta \omega$$
F. $$\sqrt { \frac { 1 } { 2 } } \Delta \omega$$
G. \sqrt { \frac { 7 } { 4 } } \Delta \omega
H. \sqrt { \frac { 4 } { 5 } } \Delta \omega
I. $$\sqrt { \frac { 3 } { 2 } } \Delta \omega$$
J. \sqrt { \frac { 5 } { 2 } } \Delta \omega
Answer: I
Two charges $Q_1 = 18 \mu C$ and $Q_2 = - 2 \mu C$ are separated by a distance R, and $Q_1$ is on the left of $Q_2$. The distance of the point where the net electric field is zero is :

Options:
A. right of $Q_2$ at R/3
B. right of $Q_2$ at R/4
C. left of $Q_1$ at R/2
D. right of $Q_2$ at 3R/4
E. between $Q_1$ and $Q_2$
F. right of $Q_2$ at R/2
G. right of $Q_2$ at R/6
H. right of $Q_2$ at R/5
I. right of $Q_2$ at R
J. right of $Q_2$ at 2R/3
Answer: F
A hydrogen atom initially at rest in n = 3 state comes to the ground state by emission of a photon . The momentum of hydrogen atom will be ( in N-m):

Options:
A. $$1.2 \times 10^{-27} $$
B. 5.6 \times 10^{-25}
C. 5.9 \times 10^{-25}
D. 6.65 \times 10^{-25}
E. 6.4 \times 10^{-25}
F. 6.0 \times 10^{-25}
G. $$6.3 \times 10^{-25} $$
H. $$3.5 \times 10^{-24} $$
I. $$6.2 \times 10^{-25} $$
J. $$6.45 \times 10^{-27} $$
Answer: I
Four identical rods of mass $M=6 kg$ each are welded at their ends to form a square and then welded to a massive ring having mass $m = 4 k g$ having radius $R = 1 m$. If the system is allowed to roll down the incline of inclination $\theta = 30 ^ { \circ }$, assume there is no slipping. The acceleration of the system will be

Options:
A. $$\mathrm { g } / 2$$
B. \mathrm { 5 g } / 12
C. \mathrm { 7 g } / 12
D. $$\mathrm { g } / 4$$
E. \mathrm { 3 g } / 8
F. \mathrm { 3 g } / 4
G. \mathrm {5 g } / 24
H. $$\mathrm { g } / 8$$
I. \mathrm { 3 g } / 16
J. $$\mathrm {7 g } / 24$$
Answer: A
In the cell $Ag | Ag^+ || Cu^{2+} | Cu$ containing 1 M $Cu^{2+}$ and 1 M $Ag^+$ ions, 9.65 ampere current is passed for 1 hour. Select the correct statement.

Options:
A. $Cu^{2+}$ will reduce to Cu and new $[Cu^{2+}] = 1.18 M$
B. $Ag^+$ will reduce to Ag and new $[Ag^+] = 1.18 M$
C. $Cu^{2+}$ will reduce to Cu and new $[Cu^{2+}] = 0.92 M$
D. Cu will oxidise to $Cu^{2+}$ and new $[Cu^{2+}] = 0.82 M$
E. $Cu^{2+}$ will reduce to Cu and new $[Cu^{2+}] = 0.64 M$
F. $Ag^+$ will reduce to Ag and new $[Ag^+] = 0.64 M$
G. $Cu^{2+}$ will reduce to Cu and new $[Cu^{2+}] = 1.12 M$
H. $Cu^{2+}$ will reduce to Cu and new $[Cu^{2+}] = 0.82 M$
I. Ag will oxidise to $Ag^+$ and new $[Ag^+] = 1.36 M$
J. $Cu^{2+}$ will reduce to Cu and new $[Cu^{2+}] = 1.02 M$
Answer: H
The charge  $Q$  (in coulomb) flowing through a resistance  $R = 10\Omega$  varies with time  $t$  (in second) as  $Q = 2 t - t ^ { 2 } .$  The total heat produced in resistance  $R$  is

Options:
A. \dfrac { 100 } { 7 } \mathrm { J }
B. $$\dfrac { 40 } { 3 } \mathrm { J }$$
C. $$\dfrac { 80 } { 7 } \mathrm { J }$$
D. \dfrac { 100 } { 3 } \mathrm { J }
E. \dfrac { 120 } { 7 } \mathrm { J }
F. $$\dfrac { 80 } { 3 } \mathrm { J }$$
G. \dfrac { 60 } { 7 } \mathrm { J }
H. \dfrac { 20 } { 7 } \mathrm { J }
I. \dfrac { 60 } { 3 } \mathrm { J }
J. $$\dfrac { 20 } { 3 } \mathrm { J }$$
Answer: B
A $40.0 ml$ solution of weak base, $BOH$ is titrated with $0.1 N - HCl$ solution. The $pH$ of the solution is found to be $10.0$ and $9.0$ after adding $5.0 ml$ and $20.0 ml$ of the acid, respectively. The dissociation constant of the base is ($log 2 = 0.3$)

Options:
A. 3 \times 10^{-5}
B. $$5 \times 10^{-5}$$
C. $$1 \times 10^{-5}$$
D. 9 \times 10^{-5}
E. 8 \times 10^{-5}
F. $$4 \times 10^{-5}$$
G. $$2 \times 10^{-5}$$
H. 6 \times 10^{-5}
I. 7 \times 10^{-5}
Answer: G
The steel railway track is to be stopped from expanding at the joints due to temperature increases. If the area of cross-section of the track is $100\ cm^2$ then the elastic temperature increases $20^oC$ to $40^oC$, then the work done per unit length will be: (if a = 1.2 $\times$ $10^{-5}$ $^oC^{-1}$ and $Y=10^{11}$ N/$m^2$)

Options:
A. 144.0\ J
B. $$7.2\ J$$
C. $$28.8\ J$$
D. 43.2\ J
E. 22.8\ J
F. $$57.6\ J$$
G. 115.2\ J
H. $$14.4\ J$$
I. 34.56\ J
Answer: F
Starting from rest a particle moves in a straight line with acceleration $a={(25-{t}^{2})}^{1/2}m/{s}^{2}$ for $0\le t\le 5s$, $a=\cfrac{3\pi}{8}m/{s}^{2}$ for $t> 5s$. The velocity of particle at $t=7s$ is

Options:
A. $$22m/s$$
B. 18m/s
C. 12m/s
D. $$33m/s$$
E. $$11m/s$$
F. 16m/s
G. 21m/s
H. $$44m/s$$
I. 31m/s
Answer: E
A point moves in the plane so that its tangential acceleration $\omega_\tau=a$, and its normal acceleration $\omega_n=bt^4$, where $a$ and $b$ are positive constants, and $t$ is time. At the moment $t=0$ the point was at rest. Find how the curvature radius $R$ of the point's trajectory and the total acceleration $\omega$ depend on the distance covered $s$.

Options:
A. \omega=a\sqrt{1+{\left(\frac{3bs^2}{a^3}\right)}^2}
B. $$\displaystyle R=\frac{a^3}{2bs}$$
C. $$\displaystyle \omega=a\sqrt{1+{\left(\frac{4bs^2}{a^3}\right)}^2}$$
D. $$\displaystyle \omega=a\sqrt{1+{\left(\frac{2bs^2}{a^3}\right)}^2}$$
E. \omega=a\sqrt{1+{\left(\frac{5bs^2}{a^3}\right)}^2}
F. \omega=a\sqrt{1+{\left(\frac{6bs^2}{a^3}\right)}^2}
G. \displaystyle \omega=a\sqrt{1+{\left(\frac{7bs^2}{a^3}\right)}^2}
H. \omega=a\sqrt{1+{\left(\frac{8bs^2}{a^3}\right)}^2}
Answer: C
The uncertainity in the position and velocity of a particle are respectively $10^{15} m$ and $5.25\times10^{-24} ms^{-1}$. The mass of the particle is: $[h=6.6\times10^{-34} Js]$

Options:
A. 20g
B. 200g
C. 10g
D. $$100g$$
E. 1000g
F. 50g
G. 500g
H. $$0.1g$$
Answer: D
A bomber plane moving at a horizontal speed of 20 m/s releases a bomb at a height of 80 m above ground as shown. At the same instant a Hunter of negligible height starts running from a point below it, to catch the bomb with speed 10 m/s. After two seconds he realized that he cannot make it, he stops running and immediately holds his gun and fires in such direction so that just before bomb hits the ground, bullet will hit it. What should be the firing speed of bullet. (Take $g=10m/{ s }^{ 2 }$

Options:
A. 65 m/s
B. $$70 m/s$$
C. 50 m/s
D. 60 m/s
E. None of these
F. $$20\sqrt { 10 } m/s$$
G. $$10\sqrt { 10 } m/s$$
H. 80 m/s
I. 55 m/s
Answer: B
A solenoid is placed inside another solenoid, the length of both being equal carrying same magnitude of current. The other parameters like radius and number of turns are in the ratio 1 : 2 for the two solenoids. The mutual inductance on each other would be

Options:
A. M_{12} = \frac{1}{4}M_{21}
B. $M_12$ = $M_{21}$
C. $M_{12} = \frac{3}{4}M_{21}$
D. $M_12$ = $2 M_{21}$
E. M_{12} = \frac{1}{3}M_{21}
F. $2M_{12}$ = $M_{21}$
G. $M_{12} = \frac{1}{2}M_{21}$
H. $M_{12} = 3M_{21}$
I. $M_{12}$ = $4M_{21}$
J. M_{12} = \frac{2}{3}M_{21}
Answer: B
A smooth sphere $A$ of mass $0.1kg$ is moving with speed $5m/s$ when it collides head on with another smooth stationary sphere of same radius. If $A$ is brought to rest by the impact and $e=\cfrac { 1 }{ 2 } $, then,

Options:
A. the mass of $B$ is $0.4kg$
B. magnitude of impulse during collision is $0.75Ns$
C. magnitude of impulse during collision is $0.6Ns$
D. the mass of $B$ is $0.2kg$
E. magnitude of impulse during collision is $1.25Ns$
F. magnitude of impulse during collision is $0.5Ns$
G. magnitude of impulse during collision is $0.3Ns$
H. magnitude of impulse during collision is $1.5Ns$
I. magnitude of impulse during collision is $1.0Ns$
Answer: F
The Young's moduli of Steel and Brass are $Y_S$ and $Y_B$ respectively. Both the wires are geometrically identical. Mass (m) is hung with the support of a light rod of length 2m. Find the value of x so that the strain produced in both the wires be same. Take $\frac{Y_S}{Y_B} = 2$

Options:
A. \frac{1}{4} m
B. \(\frac{1}{6} m\)
C. \frac{3}{4} m
D. $$\frac{1}{3} m$$
E. \(\frac{1}{2} m \)
F. $$\frac{4}{3} m$$
G. \(\frac{2}{5} m\)
H. \frac{3}{5} m
I. $$\frac{2}{3} m$$
J. 1 m
Answer: D
Which of the following statement is correct regarding induced electric field (symbols have their usual meanings)?

Options:
A. Induced e.m.f. in the loop in $\displaystyle arepsilon =\oint \overrightarrow{E}.\overline{d\iota }=frac{d\phi }{dt^2}$
B. Induced e.m.f. in the loop in $\displaystyle \varepsilon =\oint \overrightarrow{E}.\overline{d\iota }=-\frac{d\phi^2}{dt^2}$
C. Induced electric field is non-conservative is nature
D. Induced e.m.f. in the loop in \(\displaystyle \arepsilon =\oint \overrightarrow{E}.\overline{d\iota }=-\frac{d\phi^2}{dt}\)
E. Induced e.m.f. in the loop in $\displaystyle \varepsilon =\oint \overrightarrow{E}.\overline{d\iota }=-\frac{2d\phi }{dt}$
F. Induced e.m.f. in the loop in $\displaystyle arepsilon =\oint \overrightarrow{E}.\overline{d\iota }=-\frac{d\phi }{dt}$
G. Induced e.m.f. in the loop in $\displaystyle arepsilon =\oint \overrightarrow{E}.\overline{d\iota }=\frac{d\phi }{dt}$
H. Induced electric lines of force form closed loops
I. Work done in moving a test charge in an induced electric field can be zero
J. Induced e.m.f. in the loop in $\displaystyle arepsilon =\oint \overrightarrow{E}.\overline{d\iota }=-\frac{d\phi }{2dt}$
Answer: F
You were designing a project in which you required triangular wave. The function generator available to you gives only sine and square wave. You had listened to your physics teacher very carefully in the previous lecture. He stressed how series $RC$ circuit can be used as an integrater and differentiator. You wish to apply the same idea to convert square wave into triangular wave. Your project demands amplitude of triangular wave $3.6 V$ at $2 kHz$. What should be the value of $R$ and $C$?

Options:
A. $$\displaystyle X_C = 3/4 R$$
B. \displaystyle X_C = 11/8 R
C. \displaystyle X_C = 5/3 R
D. \displaystyle X_C = 2/3 R
E. \displaystyle X_C = 7/4 R
F. \displaystyle X_C = 3/2 R
G. $$\displaystyle R = X_C$$
H. none of these
I. \displaystyle X_C = 5/4 R
J. $$\displaystyle X_C = 4/3 R$$
Answer: J
The molar heat capacity of rock salt at low temperatures varies with temperature according to Dobey's $T^3$ law. Thus $C = k\frac{{{T^3}}}{{{\theta ^3}}}\,\,\,k = 1940Jmo{l^{ - 1}}{K^{ - 1}}\,\,\theta  = 281k$.How much heat is required to raise the temperature of $2$ moles of rock salt from $10 k$ to $50K$ ?

Options:
A. $$800J$$
B. $$273J$$
C. 373J
D. $$274J$$
E. 270J
F. 272J
G. 275J
H. None of these
I. 271J
Answer: B
Which of the following options will be correct for the stage of half completion of the reaction $A \rightleftharpoons B$?

Options:
A. \(\triangle G^0 = -RT \ln 4\)
B. $$\triangle G^0 = -RT ln 2$$
C. \triangle G^0 = -RT \ln 2.5
D. \(\triangle G^0 = -2RT \ln 2\)
E. \(\triangle G^0 = -RT \ln 1.5\)
F. \(\triangle G^0 = -RT \ln 5\)
G. $$\triangle G^0 < 0$$
H. \(\triangle G^0 = -RT \ln 3\)
I. $$\triangle G^0 > 0$$
J. $$\triangle G^0 = 0$$
Answer: J
An inductor $(L=100\ mH)$, a resistor $(R=100\ Omega)$ and a battery $(E=100\ V)$ are initially connected in series as shown in figure. After a long time the battery is disconnected after short circuiting the points $A$ and $B$. The current in the circuit after $1\ m$ sec is

Options:
A. $$1\ A$$
B. $$10e^{-0.1} A$$
C. $$\dfrac{1}{e}\ A$$
D. 10e\ A
E. \dfrac{1}{10}e\ A
F. 10e^{0.1} A
G. $$0.1\ A$$
H. 10e^{-1} A
I. $$e\ A$$
Answer: I
Find the work requires to compress adiabatically $ 1\ g$ of air initially at $NTP$ to half its volume. Density of air at $NTP = 0.001129\ gcm^{-3}$ and $\dfrac{C_p}{C_v} = 1.4$

Options:
A. 64.64 J
B. -64.64 J
C. $$62.64 J$$
D. $$-62.64 J$$
E. -61.64 J
F. $$32.64 J$$
G. -31.32 J
H. -63.64 J
I. $$-32.64 J$$
J. -61.32 J
Answer: D
As per Heisenberg uncertainty principle if electron is found within $2\mathring { A } $, then what is the minimum uncertainty of electron in $m{ s }^{ -1 }$?

Options:
A. \triangle V=\dfrac{h \times 10^{10}}{8\pi m_e}
B. \triangle V=\dfrac{h}{4\pi m_e}
C. $$\triangle V=\frac { h }{ 8\pi { m }_{ e } } $$
D. $$\triangle V=\frac { { 10 }^{ 10 }h }{ 8\pi } $$
E. $$\triangle V=\frac { h\times { 10 }^{ 10 } }{ 4\pi { m }_{ e } } $$
F. \triangle V=\dfrac{h \times 10^{10}}{14\pi m_e}
G. \(\triangle V=\dfrac{h}{10\pi m_e} \)
H. $$\triangle V=\frac { { 10 }^{ 10 }h }{ 8\pi { m }_{ e } } $$
I. \(\triangle V=\dfrac{h \times 10^{10}}{16\pi m_e} \)
J. \triangle V=\dfrac{h \times 10^{10}}{12\pi m_e}
Answer: C
Two spherical planets P and Q have the same uniform density $\rho$, masses $M_P$ and $M_Q$, and surface areas $A$ and $4A$, respectively. A spherical planet R also has uniform density $\rho$ and its mass is $(M_P+M_Q)$. The escape velocities from the planets P, Q and R, are $V_P, V_Q$ and $V_R$, respectively. Then.

Options:
A. V_R/V_P=2
B. $$V_R/V_P=3$$
C. $$V_Q > V_R > V_P$$
D. V_R/V_P=1.5
E. V_R/V_P=2.25
F. $$V_R > V_Q > V_P$$
G. V_R/V_P=1.75
H. V_R/V_P=2.5
I. $$V_P/V_Q=1/2$$
J. V_P/V_Q=1/4
Answer: I
A circular race track has radius = 20m and bank angle = $37^{o}$ with horizontal. If $ mu _{8} = 1/4 $ , find maximum safe speed of car . [ $ \sin {37^{o}} = 0.6 $ ] g = 10 g/$s^{2}$

Options:
A. none of these
B. 10 \sqrt{ \frac{13}{32} } m/s
C. 15 \sqrt { \frac{13}{16} } m/s
D. 25 \sqrt { \frac{13}{32} } m/s
E. 20 \sqrt { \frac{13}{32} } m/s
F. $$30 \sqrt {\frac {13}{32}} m/s $$
G. 20 \sqrt { \frac{13}{16} } m/s
H. 10 \sqrt { \frac{13}{64} } m/s
I. $$15 \sqrt {\frac {13}{32}} m/s $$
J. $$10 \sqrt {\frac {32}{13}} m/s $$
Answer: F
Not considering electronic spin, the degeneracy of second excited state ($n=3$) of $H$ atom is $9$, while the degeneracy of fourth excited state of hydride ion (${H}^{-}$) is:

Options:
A. five
B. nine
C. seven
D. three
E. one
F. six
G. eight
H. ten
I. two
J. four
Answer: D
In the combustion of solid naphthalene ($ C_{10}H_{8} $) at 298 K and atmospheric pressure, 5157 kJ/mol of heat is evolved. Calculate the resonance energy of naphthalene. Given that:Latent heat of sublimation of naphthalene = 72.0 kJ/mol Enthalpy of formation of $ H_{2}O $ = 286.0 kJ/molEnthalpy of formation of $ CO_{2} $ = 393.5 kJ/molEnthalpy of atomization of C = 715.5 kJ/mol B.E. for: $C-C$ bond = 345.5 kJ/mol $C=C$ bond = 617.0 kJ/mol  $C-H$ bond = 413.0 kJ/mol $H-H$ bond = 436 kJ/mol

Options:
A. 277 kJ mol^{-1}
B. 237 kJ mol^{-1}
C. 287 kJ mol$^{-1} $
D. 33 kJ mol$^{-1} $
E. 286 kJ mol^{-1}
F. 287.0 kJ mol$^{-1} $
G. 288 kJ mol^{-1}
H. 330 kJ mol^{-1}
I. 120 kJ mol$^{-1} $
Answer: C
If the mass of neutron $=1.7\times{10}^{27}kg$, then the De broglie wavelength of neutron of energy $3eV$ is:

Options:
A. 1.8 \times{10}^{-10}m
B. 1.8 \times{10}^{-9}m
C. 1.4 \times{10}^{-10}m
D. $$1.6\times{10}^{-10}m$$
E. $$1.6\times{10}^{-11}m$$
F. 1.6 \times{10}^{-12}m
G. 1.4 \times{10}^{-12}m
H. 1.6 \times{10}^{-9}m
I. $$1.4\times{10}^{-11}m$$
Answer: D
A star appears red if the wavelength of maximum emission is in the range $620$nm to $780$nm. The corresponding temperatures of the star for maximum emission are $T_{1} $ and $ T_{2}$ respectively. If Wien's constant is $b=2.9\times 10$$^{-3}$mK, then

Options:
A. $$T_{2}=4676K$$
B. $$T_{2}=2318K$$
C. T_{1}=3690K
D. T_{2}=4515K
E. $$T_{1}=3718K$$
F. $$T_{1}=2690K$$
G. T_{2}=4777K
H. $$T_{2}=4677K$$
I. T_{2}=3918K
J. $$T_{2}=4967K$$
Answer: H
An electron revolves round a nucleus of charge Ze. In order to excite the electron from the state $n=3$ to $n = 4$ the energy required is $66.0$ eV. Z will be

Options:
A. 36
B. 20
C. $$25$$
D. 18
E. $$30$$
F. $$4$$
G. 27
H. $$10$$
I. $$5$$
J. 35
Answer: C
The rate of reaction is expressed in different ways as follows:
$$+\dfrac{1}{2} \dfrac{d[C]}{dt} = -\dfrac{1}{3} \dfrac{d[D]}{dt} = +\dfrac{1}{4} \dfrac{d[A]}{dt} = -\dfrac{d[B]}{dt}$$
The reaction is:

Options:
A. B + 3D \longrightarrow 3A + 2C
B. $$B + D \longrightarrow A + C$$
C. B + 3D \longrightarrow 3A + 3C
D. B + 2D \longrightarrow 4A + 3C
E. B + 2D \longrightarrow 3A + 2C
F. $$A + B \longrightarrow C + D$$
G. $$4A + B \longrightarrow 2C + 3D$$
H. $$B + 2D \longrightarrow 4A + 2C$$
I. $$B + 3D \longrightarrow 4A + 2C$$
J. B + 3D \longrightarrow 4A + 3C
Answer: I
If $\mathrm { E } _ { 1 }$ is the energy required to raise a satellite to a height h = R (radius of the earth) above the surface of the earth and $\mathrm { E } _ { 2 }$  is the energy required to put it into a circular orbit at that height then the ratio $E _ { 1 / E 2 }$

Options:
A. 1
B. \frac{5}{4}
C. \frac{5}{2}
D. \frac{4}{3}
E. \frac{7}{4}
F. \frac{3}{2}
G. $$\frac { 2 } { 3 }$$
H. \frac{3}{4}
I. $$\frac { 1 } { 3 }$$
J. $$\frac { 1 } { 2 }$$
Answer: A
The radii of a spherical capacitor are equal to $a$ and $b (b > a$) The space between them is filled with a dielectric constant $K$ and resistivity $\displaystyle \rho $ At $t = 0$ the inner electrods is given a charge $\displaystyle q_{0} $ Choose the correct options

Options:
A. \displaystyle H=\left ( \frac{1}{2a}-\frac{1}{b} \right )\frac{q_{0}^{2}}{8\pi arepsilon _{0}K}
B. H=\left ( \frac{1}{2a} - \frac{1}{2b} \right )\frac{q_{0}^{2}}{8\pi \varepsilon_{0}K}
C. The total amount of heat generated during the spreading of charge will be given by $\displaystyle H=\left ( \frac{1}{2a}-\frac{1}{b} \right )\frac{q_{0}^{2}}{8\pi arepsilon _{0}K}$
D. \displaystyle H=\left ( \frac{1}{a}-\frac{1}{2b} \right )\frac{q_{0}^{2}}{8\pi \varepsilon _{0}K}
E. In short time the charge on the inner electrode will become zero
F. The total amount of heat generated during the spreading of charge will be given by $\displaystyle H=\left ( \frac{1}{b}-\frac{1}{a} \right )\frac{q_{0}^{2}}{8\pi arepsilon _{0}K}$
G. The total amount of heat generated during the spreading of charge will be given by $\displaystyle H=\left ( \frac{1}{a}-\frac{1}{b} \right )\frac{q_{0}^{2}}{8\pi arepsilon _{0}K}$
H. Charge q on the electrode as a function of time is given by $\displaystyle q=q_{0}e^{-\dfrac{1}{\rho Karepsilon _{0}}} $
I. H=\left( \frac{1}{a} - \frac{1}{2b} \right) \frac{q_{0}^{2}}{8\pi \varepsilon_0 K}
Answer: G
Find the quantum number of the excited state of electron in ${ He }^{ + }$ ion which on transition to first excited state emit photon of wavelength 108.5 nm (${ R }_{ H }=\quad 1.09678*{ 10 }^{ 7 }\quad { m }^{ -1 }$)

Options:
A. 12
B. 9
C. 8
D. 10
E. 6
F. 7
G. 5
H. 2
I. 3
J. 4
Answer: E
Force acting on a particle moving in a straight line varies with the velocity of the particle as F = KV, Where K is constant. The work done by this force in time t is (initial  velocity is 1 m/s)

Options:
A. \dfrac { m }{ 7k } \left[ e^ \{ \dfrac { k }{ 2m } t\} -1 \right]
B. \dfrac { m }{ 4k } \left[ e^ \{ \dfrac { k }{ 2m } t\} -1 \right]
C. $$ m \left[ e^ \{ \dfrac { k }{ 2m } t\} -1 \right] $$
D. $$ \dfrac { m }{ k } \left[ e^ \{ \dfrac { k }{ 2m } t\} -1 \right] $$
E. $$ \dfrac { m }{ 2k } \left[ e^ \{ \dfrac { k }{ 2m } t\} -1 \right] $$
F. \dfrac { m }{ 5k } \left[ e^ \{ \dfrac { k }{ 2m } t\} -1 \right]
G. $$ \dfrac { m }{ 2 } \left[ e^ \{ \dfrac { k }{ 2m } t\} -1 \right] $$
H. \dfrac { m }{ 6k } \left[ e^ \{ \dfrac { k }{ 2m } t\} -1 \right]
I. \dfrac { m }{ 3k } \left[ e^ \{ \dfrac { k }{ 2m } t\} -1 \right]
J. \dfrac { m }{ 2k } \left[ e^ \{ \dfrac { 3k }{ 2m } t\} -1 \right]
Answer: E
Dichromate ion in acidic medium oxidizes stannous ion as: $xSn^{2+}+yCr_{2}O_{7}^{2-}+zH^{+} \rightarrow aSn^{4+}+bCr^{3+}+cH_{2}O$

Options:
A. the value of \( z-c=9 \)
B. the value of $z-c=5$
C. the value of \( z-c=10 \)
D. the value of \( z-c=8 \)
E. $$a:b=3:2$$
F. the value of $z-c=7$
G. the value of \( z-c=6 \)
H. the value of \( z-c=11 \)
Answer: F
If velocity of particle moving along x-axis is given as $ v = k \sqrt {x} $ Then (a is acceleration ) :-

Options:
A. x \quad \propto \quad t^{3/2}
B. $$ a \propto \quad x $$
C. x \quad \propto \quad t^{1/4}
D. x \quad \propto \quad t^{1/3}
E. x \quad \propto \quad t^{1/5}
F. $$ a = constant $$
G. $$ x\quad \propto \quad \sqrt { t } $$
H. x \quad \propto \quad t^2
I. x \quad \propto \quad t^{1/6}
J. $$ x\quad \propto \quad t $$
Answer: G
An astronomical telescope has $f_{0}=200\ cm$ and $f_{E}=4\ cm$. Telescope is focused to see an object $10\ km$ from the objective. Final image is formed at infinity. The length $L$ of the tube and the angular magnification of telescope are respectively.

Options:
A. Length $=196\ cm$, angular magnification $=50$
B. Length $=204\ cm$, angular magnification $=50$
C. Length $=196\ cm$, angular magnification $=45$
D. Length $=200\ cm$, angular magnification $=50$
E. Length $=196\ cm$, angular magnification $=40$
F. Length $=200\ cm$, angular magnification $=45$
G. Length $=204\ cm$, angular magnification $=25$
H. Length $=204\ cm$, angular magnification $=45$
I. Length $=200\ cm$, angular magnification $=40$
J. Length $=204\ cm$, angular magnification $=40$
Answer: D
The electrons represented by following set of quantum numbers in decreasing order of energy is:a) $n=4,l=0,m=0,s=+\frac{1}{2}$b) $n=3,l=1,m=1,s=-\frac{1}{2}$c) $n=3,l=2,m=0,s=+\frac{1}{2}$d) $n=3,l=0,m=0,s=-\frac{1}{2}$

Options:
A. c > a > d > b
B. b > a > c > d
C. c > a > b >d
D. a > c > b > d
E. a > b > c > d
F. b > c > a > d
G. c > b > a > d
H. c > b > d > a
I. a > c > d > b
J. c > d > b > a
Answer: C
The speed of proton is one-hundredth of the speed of light in vacuum. What is its de-Broglie wavelength?Assume that one mole of protons has a mass equal to one gram.[ $h= 6.626\times10^{-34}$Js]

Options:
A. 1.33 \times 10^{-14}\mathring{A}
B. 1.33 \times 10^{-10}\mathring{A}
C. 1.33 \times 10^{-11}\mathring{A}
D. 1.33 \times 10^{-12}\mathring{A}
E. $$1.33\times10^{-3}\mathring{A}$$
F. $$13.13\times10^{-13}\mathring{A}$$
G. 1.33 \times 10^{-15}\mathring{A}
H. $$1.33\times10^{-13}\mathring{A}$$
I. $$13.33\times10^{-3}\mathring{A}$$
J. 13.33 \times 10^{-14}\mathring{A}
Answer: H
Current $i = 2.5$ A flows along the circle $x^{2} + y^{2} = 9\ cm^{2}$ (here $x$ and $y$ are in cm) as shown. Magnetic field (in Tesla) at point $(0, 0, 4\ cm)$ is

Options:
A. \left (\dfrac {9\pi}{5} \times 10^{-7} \right )\hat {i}
B. \left (\dfrac {9\pi}{5} \times 10^{-7} \right )(-\hat {k})
C. $$\left (\dfrac {9\pi}{5}\times 10^{-7}\right )\hat {k}$$
D. \left (\dfrac {9\pi}{5} \times 10^{-7} \right )\hat {j} + \left (\dfrac {9\pi}{5} \times 10^{-7} \right )(-\hat {k})
E. (36\pi \times 10^{-7})\hat {i}
F. $$(36\pi \times 10^{-7})\hat {k}$$
G. $$(36\pi \times 10^{-7})(-\hat {k})$$
H. \left (\dfrac {9\pi}{5} \times 10^{-7} \right )\hat {i} + \left (\dfrac {9\pi}{5} \times 10^{-7} \right )\hat {k}
I. \left (\dfrac {9\pi}{5} \times 10^{-7} \right )\hat {j}
Answer: F
From which height a block of ice must be dropped in order that it melts completely. Assume that all the energy is retained by the ice .($g=10ms^{-2}$, $L=80Jgm^{-1}$and J $=$ 4.2 J/cal)

Options:
A. 1000Km
B. 33.6Km
C. 16.7Km
D. 33.8Km
E. 67.2Km
F. 16.8Km
G. 100Km
H. 16.9Km
I. 1Km
J. 33.5Km
Answer: B
Length of the latus rectum of the hyperbola $xy-3x-4y+8=0$

Options:
A. $$2\sqrt{10}$$
B. $$2\sqrt{2}$$
C. $$8$$
D. $$4$$
E. 4\sqrt{6}
F. 4\sqrt{5}
G. $$4\sqrt{2}$$
H. 4\sqrt{3}
I. none of these
J. 4\sqrt{7}
Answer: G
One mole of an ideal monatomic gas is taken from temperature $T_{0}$ to $2T_{0}$ by the process $T^{4}P^{-1}= C$. Then,

Options:
A. work done by the gas is $-3RT_{0}$
B. molar specific heat of the gas is $\dfrac{3R}{2}$
C. molar specific heat of the gas is $-\dfrac{3R}{2}$
D. heat released by the gas is $\dfrac{3}{2} RT_{0}$
E. heat released by the gas is \dfrac{7}{4} RT_{0}
F. heat released by the gas is \dfrac{9}{4} RT_{0}
G. heat released by the gas is $\dfrac{5}{4} RT_{0}$
H. heat released by the gas is \dfrac{5}{2} RT_{0}
I. heat released by the gas is $\dfrac{3}{4} RT_{0}$
J. heat released by the gas is $\dfrac{1}{2} RT_{0}$
Answer: D
The increase in entropy of 1 kg of ice at 200 K which is heated to 400 K (super heated steam) at constant atmospheric pressure, will be -
Given that ${C_P}(ice) = 2.09 \times {10^3}J/Kg$degree; ${C_P}(water) = 4.18 \times {10^3}J/Kg$degree; ${C_P}(steam) = 2.09 \times {10^3}J/Kg$degree       Latent Heat (Ice, 273 K) = $3.34 \times {10^5} J/kg$;           Latent Heat (water, 373 K) = $22.6 \times {10^5} J/kg$

Options:
A. $$9750\,J/K{g^{ - 1}}$$
B. 9190\,J/K{g^{ - 1}}
C. 9250\,J/K{g^{ - 1}}
D. 8930\,J/K{g^{ - 1}}
E. $$8750\,J/K{g^{ - 1}}$$
F. 9130\,J/K{g^{ - 1}}
G. $$9390\,J/K{g^{ - 1}}$$
H. 9210\,J/K{g^{ - 1}}
I. $$8390\,J/K{g^{ - 1}}$$
J. 8990\,J/K{g^{ - 1}}
Answer: G
A non-conducting insulating ring of mass m and radius $R$, having charge $Q$ uniformly distributed over its circumference is hanging by a insulated thread with the help of a small smooth ring (not rigidly fixed with bigger ring). A time varying magnetic field $B = B_0 sin\omega t$ is switched on at $t = 0$ and the ring is released at the same time. Work done by all forces on ring in $0$ to $(\dfrac{\pi}{\omega})$ time interval is :

Options:
A. $$\dfrac{1}{2}\dfrac{(B_0QR)^2}{m}$$
B. \dfrac{1}{12}\dfrac{(B_0QR)^2}{m}
C. $$0$$
D. \dfrac{1}{14}\dfrac{(B_0QR)^2}{m}
E. \dfrac{1}{10}\dfrac{(B_0QR)^2}{m}
F. $$\dfrac{1}{4}\dfrac{(B_0QR)^2}{m}$$
G. \dfrac{1}{6}\dfrac{(B_0QR)^2}{m}
H. $$\dfrac{1}{8}\dfrac{(B_0QR)^2}{m}$$
I. \dfrac{1}{3}\dfrac{(B_0QR)^2}{m}
J. \dfrac{1}{9}\dfrac{(B_0QR)^2}{m}
Answer: C
The wavelength of limiting line of Lyman series in ${He}^{\oplus }$ ion would be:  ${R}_{H}=109677{cm}^{-1}$

Options:
A. 910.5\mathring{A}
B. 911.2\mathring{A}
C. 911.4\mathring{A}
D. $$303.8\mathring{A}$$
E. $$227.9\mathring{A}$$
F. 912.7\mathring{A}
G. 911.9\mathring{A}
H. 102.6\mathring{A}
I. $$911.7\mathring{A}$$
J. $$1215.1\mathring{A}$$
Answer: I
A metal tape is calibrated at $30 ^ { \circ } \mathrm { C }$ . It gave the measurement1000$\mathrm { m }$ at a lesser temperature. It is found that the error in measurement is 2.5$\cdot \mathrm { cm } .$ The temperature at which themeasurement is made is: $\left( lpha = 10 ^ { - 6 } / ^ { \circ } \mathrm { C } \right)$

Options:
A. 17 ^ { \circ } \mathrm { C }
B. $$10 ^ { \circ } \mathrm { C }$$
C. 25 ^ { \circ } \mathrm { C }
D. $$5 ^ { \circ } \mathrm { C }$$
E. 18 ^ { \circ } \mathrm { C }
F. 22 ^ { \circ } \mathrm { C }
G. 12 ^ { \circ } \mathrm { C }
H. $$20 ^ { \circ } \mathrm { C }$$
I. $$15 ^ { \circ } \mathrm { C }$$
J. 19 ^ { \circ } \mathrm { C }
Answer: H
Two sources of sound placed close to each other, are emitting progressive waves given by  $y _ { 1 } = 4 \sin 600 \pi t$  and  $y _ { 2 } = 5 \sin 608 \pi t .$  An observer located near these two sources of sound will hear
(1)  $4$ beats per second with intensity ratio  $25 : 16$  between waxing and waning
(2)  $8$  beats per second with intensity ratio  $25 : 16$  between waxing and waning
(3)  $8$  beats per second with intensity ratio  $81 : 1$  between waxing and waning
(4)  $4$  beats per second with intensity ratio  $81 : 1$  between waxing and waning.
Maximum number of beats frequency heard by a human being is

Options:
A. $$4$$
B. $$20$$
C. 16
D. 6
E. 12
F. 14
G. $$10$$
H. 8
Answer: G
Given the ionic equivalent conductivities for the following ions : $\lambda^{\circ}_{eq}K^{\oplus}\, =\, 73.5\, cm^{2}\, ohm^{-1}\, eq^{-1}$ $\lambda^{\circ}_{eq}Al^{3+}\, =\, 189\, cm^{2}\, ohm^{-1}\, eq^{-1}$$\lambda^{\circ}_{eq}SO_{4}^{2-}\, =\, 85.8\, cm^{2}\, ohm^{-1}\, eq^{-1}$The $\Lambda^{\circ}_{eq}$ for potash alum $(K_2SO_4.Al_2(SO_4)_3. 24H_2O)$ is :

Options:
A. 245.12
B. $$108.52$$
C. $$245.92$$
D. 347.3
E. 245.62
F. $$348.3$$
G. 246.12
H. 346.3
I. 245.32
J. $$368.2$$
Answer: C
Electric field in the region is given as $E=(10-5x)\hat i$. A charge of the mass 5 kg and charge Q (=1 C) is situated at origin and free to move in given electric field, then choose the correct option (neglect gravity).

Options:
A. motion of charge particle is oscillatory
B. the position of charge particle, when velocity gained by particle is maximum is 1.8 SI units
C. the position of charge particle, when velocity gained by particle is maximum is 2.1 SI units
D. maximum displacement of charge particle from origin is 4 SI units
E. maximum velocity gain by charge particle is 2 SI units
F. the position of charge particle, when velocity gained by particle is maximum is 3 SI units
G. the position of charge particle, when velocity gained by particle is maximum is 1.75 SI units
H. the position of charge particle, when velocity gained by particle is maximum is 1.5 SI units
I. the position of charge particle, when velocity gained by particle is maximum is 2 SI units
J. the position of charge particle, when velocity gained by particle is maximum is 1 SI units
Answer: I
When $0.1\ mol$ arsenic acid, $H_{2}AsO_{4}$ is dissolved in $1L$ buffer solution of $pH = 4$, which of the following hold good? $K_{1} = 2.5\times 10^{-4}, K_{2} = 5\times 10^{4}, K_{3} = 2\times 10^{-23}$ for arsenic acid $['< <'$ sign denotes that the higher concentration is at least $100$ times more than the lower one].

Options:
A. $$[H_{2}AsO_{4}] < < [HAsO_{4}^{2-}]$$
B. [H_{2}AsO_{4}] < < [AsO_{4}^{2-}]
C. [H_{2}AsO_{4}^{-}] < < [HAsO_{4}^{2-}]
D. $$[HAsO_{4}^{2-}] < < [HAsO_{4}^{-}]$$
E. [H_{2}AsO_{4}^{-}] < < [AsO_{4}^{3-}]
F. $$[AsO_{4}^{2-}] < < [HAsO_{4}^{2-}]$$
G. $$[H_{2}AsO_{4}] < < [H_{2}AsO_{4}^{-}]$$
H. [H_{2}AsO_{4}^{-}] < < [H_{2}AsO_{4}]
Answer: F
In the concentration cell
$$Pt\left( { H }_{ 2 } \right) \begin{vmatrix} HA(0.1M) \ NaA(1M) \end{vmatrix}\begin{vmatrix} HA(1M) \ NaA(1M) \end{vmatrix}\left( { H }_{ 2 } \right) Pt$$
($p{ K }_{ a }$ of $HA=4$)
Cell potential will be:

Options:
A. 0.006V
B. $$-0.03V$$
C. 0.003V
D. $$-0.06V$$
E. $$0.03V$$
F. $$-0.006V$$
G. -0.003V
H. $$0.06V$$
Answer: E
The current voltage relation of diode is given by $I=\left( { e }^{ 1000V/T }-1 \right)mA$, where the applied voltage $V$ is in kelvin. If a student makes an error measuring $\pm 0.01V$ while measuring the current of $5mA$ at $300k$, what will be the error in the value of current in mA?

Options:
A. 0.03mA
B. 0.25mA
C. 0.01mA
D. $$0.02mA$$
E. $$0.2mA$$
F. $$0.5mA$$
G. $$0.05mA$$
H. 0.04mA
I. 0.06mA
J. 0.08mA
Answer: D
A particle is moving with a velocity of $ec{v}=(3\hat{i}+4t\hat{j})$m/s. Find the ratio of tangential acceleration to that of normal acceleration at $t=1$ sec.

Options:
A. \dfrac{5}{3}
B. \dfrac{5}{6}
C. $$\dfrac{4}{3}$$
D. $$\dfrac{4}{5}$$
E. \dfrac{3}{5}
F. \dfrac{3}{2}
G. \dfrac{5}{4}
H. $$\dfrac{3}{4}$$
Answer: D
A nugget of gold and quartz was found to contain x g of gold and y g of quartz and has density d if the densities of gold and quartz are $d_{1}$ and $d_{2}$ respectively then the correct relation is :

Options:
A. $$\dfrac{x + y}{d} + \dfrac{x}{d_{1}} + \dfrac{x}{d_{2}} = 0$$
B. $$xd_{1} + yd_{2} = (x + y) d$$
C. \dfrac{x}{d_{1}} + \dfrac{y}{d_{2}} = \dfrac{x + y}{d_{1}} + \dfrac{x + y}{d_{2}}
D. \dfrac{x}{d_{1}} + \dfrac{y}{d_{2}} = \dfrac{x + y}{d_{1}}\dfrac{x}{d_{2}} + \dfrac{y}{d} = \dfrac{x + y}{d}
E. $$\dfrac{x}{d_{2}} + \dfrac{y}{d_{1}} = \dfrac{x + y}{d}$$
F. \dfrac{x}{d_{1}} + \dfrac{y}{d} = \dfrac{x + y}{d_{2}}
G. $$\dfrac{x}{d_{1}} + \dfrac{y}{d_{2}} = \dfrac{x + y}{d}$$
H. \dfrac{x}{d_{1}} + \dfrac{y}{d} = \dfrac{x + y}{d_{1}}\dfrac{x}{d_{2}} + \dfrac{y}{d_{1}} = \dfrac{x + y}{d}
I. \dfrac{x}{d} + \dfrac{y}{d_{1}} = \dfrac{x + y}{d_{2}}
J. \dfrac{x}{d_{2}} + \dfrac{y}{d_{1}} = \dfrac{x + y}{d_{1}}
Answer: E
A stationary fish in the water notices that a bird in the air approaches the fish with a velocity $12ms^{-1}$. The true velocity of bird is $\left ( \mu _{water}=\dfrac{4}{3} \right )$

Options:
A. 8\ ms^{-1}
B. 13\ ms^{-1}
C. 11\ ms^{-1}
D. $$9\ ms^{-1}$$
E. 14\ ms^{-1}
F. $$10.5 \ ms^{-1}$$
G. $$16\ ms^{-1}$$
H. $$15\ ms^{-1}$$
I. 12\ ms^{-1}
J. 10\ ms^{-1}
Answer: D
A thin half ring of radius $R = 20$ cm is uniformly charged with a total charge $q = 0.70$nC. The magnitude of electric field strength at the curvature centre of this half ring is

Options:
A. 50 V / m
B. $$4 V / m$$
C. $$100 V /m$$
D. 150 V / m
E. 125 V / m
F. $$225 V / m$$
G. $$20 V / m$$
H. 120 V / m
I. 80 V / m
Answer: C
A hydrogen electrode $\left( { p }_{ { H }_{ 2 } }=1atm;T=298K \right) $ is placed in buffer solution of ${ CH }_{ 3 }COONa$ and ${ CH }_{ 3 }COOH$ in the molar ratio $x:y$ and then in $y:x$ has reduction potential values as ${E}_{1}$ and ${E}_{2}$ volts respectively. Then $p{K}_{e}$ of ${ CH }_{ 3 }COOH$ will be:

Options:
A. $$\cfrac { { E }_{ 1 }+{ E }_{ 2 } }{ 0.118 } $$
B. $$\cfrac { { E }_{ 2 }-{ E }_{ 1 } }{ 0.118 } $$
C. $$-\cfrac { { E }_{ 1 }+{ E }_{ 2 } }{ 0.118 } $$
D. \cfrac { { E }_{ 2 }+{ E }_{ 1 } }{ 0.059 }
E. \cfrac { { E }_{ 1 }-{ E }_{ 2 } }{ 0.118 }
F. \cfrac { { E }_{ 1 }+{ E }_{ 2 } }{ 0.236 }
G. \cfrac { { E }_{ 1 }-{ E }_{ 2 } }{ 0.059 }
H. \cfrac { { E }_{ 2 }-{ E }_{ 1 } }{ 0.236 }
I. $$\cfrac { { E }_{ 1 }+{ E }_{ 2 } }{ 0.059 } $$
J. \cfrac { { E }_{ 2 }-{ E }_{ 1 } }{ 0.059 }
Answer: C
The average human eye sees colors with wavelength between $430nm$ to $680nm$. For what visible wavelength(s) will a $350nm$ thick ($n=1.35$) soap film produce maximum destructive interference?

Options:
A. 475nm
B. 471nm
C. $$473nm$$
D. None of these
E. 474nm
F. 476nm
G. 470nm
H. $$945nm$$
I. 472nm
J. $$315nm$$
Answer: C
If an alternating current $i=i_{m} \sin \omega t$ is flowing through a capacitor then voltage drop $\Delta V_{C}$ across capacitor $C$ will be?

Options:
A. -\dfrac{i_{m}}{\omega C} \cos \left(\omega t - \dfrac{\pi}{2}\right)
B. -\dfrac{i_{m}}{\omega C} \sin \left(\omega t + \dfrac{\pi}{2}\right)
C. -\dfrac{i_{m}}{\omega C} \sin \left(\omega t - \dfrac{\pi}{4}\right)
D. $$-\dfrac{i_{m}}{\omega C}\left(\sin \omega t +\dfrac{\pi}{4} \right)$$
E. $$-\dfrac{i_{m}}{\omega C} \cos \omega t$$
F. -\dfrac{i_{m}}{\omega C} \left(\sin \omega t + \dfrac{\pi}{2}\right)
G. -\dfrac{i_{m}}{\omega C} \sin \left(\omega t - \dfrac{\pi}{2}\right)
H. $$\dfrac{i_{m}}{\omega C}\left(\sin \omega t -\dfrac{\pi}{4} \right)$$
I. -\dfrac{i_{m}}{\omega C} \sin \left(\omega t + \dfrac{\pi}{4}\right)
J. $$-\dfrac{i_{m}}{\omega C} \sin \omega t$$
Answer: J
A monochromatic ray of light is incident on the surface of a glass slab of thickness $8\ cm$, making and angle $30^{\circ}$ with the surface. The lateral displacement of the ray of light will be $(\mu_{g} = \sqrt {3})$.

Options:
A. $$8\ cm$$
B. 5.62\ cm
C. 6.62\ cm
D. 5.82\ cm
E. 5.42\ cm
F. 7.62\ cm
G. $$4.62\ cm$$
H. 5.22\ cm
I. $$3.62\ cm$$
Answer: G
A man of height 1.47 m stands on a straight road on a hot say. The vertical temperature in the air results in a variation of refractive index with height y as $\mu = \mu_o (1+ay) $ where $\mu_o$ is the refractive index of air near the road and $lpha=1.5\times 10^{-6}$/m. What is the apparent length of the road man is able to see

Options:
A. infinite distance
B. 700\sqrt{3}
C. 350\sqrt{3}
D. 390 m
E. 390\sqrt{3}
F. 700\sqrt{2}
G. 350 m
H. 700 m
I. 390\sqrt{2}
Answer: H
Suppose, the electron in a hydrogen atom makes transition from n= $3$ to n= $2$ in  $10^{-8}s.$ The order of the torque acting on the electron in this period, using the relation between torque and angular momentum as discussed in the chapter on rotational mechanics is

Options:
A. 10^{-36}N-m
B. 10^{-44}N-m
C. 10^{-40}N-m
D. $$10^{-24}N-m$$
E. 10^{-46}N-m
F. 10^{-32}N-m
G. 10^{-38}N-m
H. $$10^{-42}N-m$$
I. $$10^{-8}N-m$$
J. $$10^{-34}N-m$$
Answer: H
Product of the perpendiculars from the foci on any tangent o the hyperbola $\frac{x^{2}}{12}-\frac{y^{2}}{9}=1$ is

Options:
A. 10
B. 14
C. 8
D. 24
E. 21
F. 15
G. 12
H. 18
I. 16
J. 9
Answer: G
Ravi can throw a ball at a speed on the earth which can cross a river of width $10$m. Ravi reaches on an imaginary planet whose mean density is twice that of the earth. Find out the maximum possible radius of the planet so that if Ravi throws the ball at the same speed it may escape from the planet. Given radius of the earth $=6.4\times 10^6$m.

Options:
A. 48 km
B. 16 km
C. 8 km
D. $9$km.
E. $8$km.
F. $44$km.
G. $41$km.
H. $48$km.
I. 44 km
J. $4$km.
Answer: J
$$\lim_{x\rightarrow 0}\left ( \frac{1+tan x}{1+sinx} \right )^{cosecx}=$$

Options:
A. 0
B. 2
C. \frac{1}{e^3}
D. -1
E. does not exist
F. \frac{1}{2}
G. \frac{1}{e^2}
H. 1
I. -2
J. \frac{1}{e}
Answer: A
The maximum range of a bullet fired from a toy pistol mounted on a car at rest $R_{0} = 40\ m$. What will the acute angle of inclination of the pistol for maximum range when the car is moving in the direction of firing with uniform velocity $v = 20\ m/s$, on a horizontal surface? $(g = 10\ m/s^{2})$.

Options:
A. 65^{o}
B. $$45^{o}$$
C. $$55^{o}$$
D. 40^{o}
E. $$60^{o}$$
F. $$75^{o}$$
G. $$50^{o}$$
H. $$30^{o}$$
I. 35^{o}
Answer: E
The standard reduction potential data at $25^{\circ}C$ is given below:
$E^{\circ}_{Fe^{3+}/ Fe^{2+}} = +0.77\ V; E^{\circ}_{Fe^{2+}/ Fe} = -0.44\ V$;
$E^{\circ}_{Cu^{2+}/ Cu} = +0.34\ V; E^{\circ}_{Cu^{+}/ Cu} = +0.52\ V$;
$$E^{\circ}_{[O_{2}(g) + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O]} = +1.23\ V$$
$E^{\circ}_{[O_{2}(g) + 2H_{2}O + 4e^{-}\rightarrow 4OH]} = +0.40\ V$;
$$E^{\circ}_{(Cr^{3+} / Cr)} = -0.74\ V; E^{\circ}_{(Cr^{2+}/Cr)} = -0.91\ V$$
Match $E^{\circ}$ of the redox pair in List-I with the values given in List - II and select the correct answer using the code given below the lists:List - IList - II(P)$E^{\circ}_{Fe^{3+}/ Fe}$
(1)$-0.18\ V$
(Q)$E^{\circ}_{(4H_{2}O \rightleftharpoons 4H^{+} + 4OH^{-})}$
(2)$-0.4\ V$
(R)$E^{\circ}_{(Cu^{2+} + Cu \rightarrow 2Cu^{+})}$(3)$-0.04\ V$(S)$E^{\circ}_{(Cr^{3+}/ Cr^{2+})}$(4)$-0.83\ V$

Options:
A. P = 1, Q = 4, R = 3, S = 2
B. $$P = 1, Q = 3, R = 4, S = 2$$
C. $$P = 2, Q = 3, R = 4, S = 1$$
D. P = 1, Q = 4, R = 2, S = 3
E. $$P = 4, Q = 2, R = 3, S = 1$$
F. P = 4, Q = 3, R = 1, S = 2
G. $$P = 1, Q = 2, R = 3 S = 4$$
H. $$P = 4, Q = 1, R = 2, S = 3$$
I. $$P = 3, Q = 4, R = 1, S = 2$$
J. P = 4, Q = 3, R = 2, S = 1
Answer: I
The distance between the genes $ a, b, c ,d $ in mapping units are$ a-d=3.5;b-c=1;a-b=6;c-d=1.5;a-c=5$find out the sequence of arrangement of these genes.

Options:
A. $$adcb$$
B. acbd
C. adbc
D. $$acdb$$
E. $$abdc$$
F. $$dbca$$
G. abcd
Answer: D
Maximum emitted wavelength by photon during a transition in $H-$atom that produce a visible light $\left(\dfrac{1}{R}\right)=912 \mathring { A } ,R \Rightarrow $ Rydberg constant ) is(approx):

Options:
A. 1217\ \mathring { A }
B. 1214\ \mathring { A }
C. 1215\ \mathring { A }
D. $$7172.3\ \mathring { A } $$
E. $$6566.4\ \mathring { A } $$
F. 1212\ \mathring { A }
G. $$3536\ \mathring { A } $$
H. 1216\ \mathring { A }
I. $$1213\ \mathring { A } $$
J. 3537\ \mathring { A }
Answer: I
Assertion (A): The shadow of a tower on a level plane is found to be 60 meters longer when sun's altitude is $30^{0}$ than that when it is $45^{0}$ Then the height of the tower is $30$ $(\sqrt{3}+1)$ {meters}Reason (R): The angle of elevation of a top of a tower standing an horizontal plane from a Point $A$ is $lpha$. After wlaking a distance $d$ meters towards the foot of the tower, the angle elveation is found to be $\beta$ then the height of the tower is $h=\displaystyle \frac{d}{\cotlpha-\cot\beta}$

Options:
A. Both $A$ and $R$ are true and  $R$ is not correct explanation of $A$
B. Both $A$ and $R$ are ture and $R$ is the correct explanation of $A$
C. Both $A$ and $R$ are true and $R$ is the correct explanation of $A$ but with a minor calculation error
D. Both $A$ and $R$ are true, but $R$ is not the correct explanation of $A$ because the distances used in the formula are interchanged.
E. Both $A$ and $R$ are true, but $R$ is not the correct explanation of $A$ because the angles used in the formula are swapped.
F. $A$ is false but $R$ is true.
G. Both $A$ and $R$ are true, but $R$ is not the correct explanation of $A$ due to a different approach being used in $A$.
H. Both $A$ and $R$ are true, but $R$ is not the correct explanation of $A$ because the angles used in the formula are interchanged.
I. $A$ is true but $R$ is false
J. Both $A$ and $R$ are true and $R$ is not the correct explanation of $A$
Answer: B
$10\space mL$ of $0.1\space M$ tribasic acid $H_3A$ is titrated with $0.1\space M\space NaOH$ solution. What is the ratio of $\displaystyle\frac{[H_3A]}{[A^{3-}]}$ at $2^{nd}$ equivalence point? [Given that: $K_{a_1} = 10^{-3},\space K_{a_2} = 10^{-8}, \space K_{a_3} = 10^{-12}$]

Options:
A. \simeq 10^{-5}
B. \simeq 10^{-10}
C. \simeq 10^{-11}
D. \simeq 10^{-9}
E. $$\simeq 10^{-4}$$
F. $$\simeq 10^{+6}$$
G. $$\simeq 10^{-7}$$
H. \simeq 10^{-8}
I. \simeq 10^{-6}
J. $$\simeq 10^{+4}$$
Answer: G
The decomposition of a compound $P$, at temperature $T$ according to the equation $\displaystyle2P_{(g)}\rightarrow4Q_{(g)}+R_{(g)}+S_{(l)}$ is the first order reaction. After $30$ minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be $317$ mm $Hg$ and after a long period of time the total pressure observed to be $617$ mm $Hg$. The total pressure of the vessel after $75$ minute, if the volume of liquid $S$ is supposed to be negligible is:(Given : Vapour pressure of $S (l)$ at temperature $T=32.5 $mm $Hg$)

Options:
A. $P_{t}=489.44$ mm $Hg$
B. $P_{t}=379.55$ mm $Hg$
C. P_{t}=391.73 \text{ mm Hg}
D. $P_{t}=389.55$ mm $Hg$
E. $P_{t}=384.55$ mm $Hg$
F. P_{t}=374.55 \text{ mm Hg}
G. $P_{t}=387.64$ mm $Hg$
H. $P_{t}=382.45$ mm $Hg$
I. $P_{t}=468.23$ mm $Hg$
J. P_{t}=479.44 \text{ mm Hg}
Answer: B
A ray of light travelling along the line $x+y=1$ is incident on the x-axis and after refraction it enters the other side of the x-axis by turning $\dfrac{\pi}{6}$ away from positive direction of x-axis. The equation of the line along which the refracted ray travels is

Options:
A. y+(2-\sqrt 3)x=2+\sqrt 3
B. $$(2-\sqrt 3)x+y=1$$
C. y+(2+\sqrt 3)x=1
D. $$y+(2+\sqrt 3)x=2+\sqrt 3$$
E. $$x+(2-\sqrt 3)y=1$$
F. y+(2+\sqrt 3)x=2-\sqrt 3
G. y+(2-\sqrt 3)x=2-\sqrt 3
H. none of these
I. y-(2+\sqrt 3)x=2+\sqrt 3
J. y+(2-\sqrt 3)x=1
Answer: D
The series limit for Balmer series of $H$ spectrum occurs at $3664\mathring{A}$. Calculate (a) ionization energy of $H$-atom.(b) the wavelength of the photon that would remove the electron in the ground state of the $H$-atom.

Options:
A. (a)$E=4.12eV$
B. (a)$E=4.25eV$
C. (a)$E=3.38eV$
D. (a)$E=1.42\times 10^{-19}J$(b)$\lambda=16\mathring{A}$
E. (a)$E=7.25eV$
F. (a)$E=5.42\times 10^{-19}J$(b)$\lambda=916\mathring{A}$
G. (a)$E=2.21eV$
H. (a)$E=13.60eV$
I. (a)$E=6.80eV$
J. (a)$E=5.42\times 10^{-19}J$
(b)$wave\,number=27\times10^5m^{-1}$
Answer: C
An uranium mineral containing $^{238}\textrm{U}$ and $^{206}\textrm{Pb}$ in the ratio 4 : 1 by mass.$t_{1/2} \: ^{238}\textrm{U}=4.5\times 10^9 \: year$. Assume all the lead present in mineral is formed from disintegration of $^{238}\textrm{U}$. Age of mineral is:

Options:
A. $$[4.243\times 10^9\: year]$$
B. [2.143 \times 10^9\: year]
C. $$[1.449\times 10^9\: year]$$
D. [1.513 \times 10^9\: year]
E. $$[1.648\times 10^9\: year]$$
F. [1.814 \times 10^9\: year]
G. $$[3.656\times 10^9\: year]$$
H. [1.342 \times 10^9\: year]
I. 1.746 \times 10^9\: year
J. [1.847 \times 10^9\: year]
Answer: C