MikaStars39/nano-eval huggingface dataset:
A block is projected with velocity$ v_0 $ up the inclined plane from its bottom at t = 0. The plane makes an angle $ \theta $ with the horizontal. If the coefficient of friction between the block and the incline is $\mu = \tan \alpha \quad (\alpha > \theta)$ then frictional force applied by the plane on the block for $t > \dfrac{v_0}{g [ \sin \theta + \tan \alpha \cos \theta]}$ will be :

Options:
A. $$ \tan \alpha mg \sin \theta $$
B. $$ \tan \alpha \cdot mg \cdot \cos \theta $$
C. $$ mg \tan \alpha \cos \theta $$
D. $$mg \cos \alpha$$
E. $$\tan \alpha \, mg \cos \alpha$$
F. $$ mg \sin \theta $$
G. $$tan \alpha mg sin \alpha$$
H. Zero
I. $$mg tan \alpha sin \theta$$
J. $$ mg sin \alpha $$
Answer: J
The equilibrium constant $(K_p)$ for the decomposition of gaseous $H_2O$
$${H_2}O(g)\leftrightharpoons {H_2}(g) + \frac{1}{2}{O_2}(g)$$
is related to degree of dissociated $\alpha $ at a total pressure $p$ is given by

Options:
A. $$K_p = \frac{a^{3/2} p^2}{(1 + \alpha)(2 + \alpha)^{1/2}}$$
B. $$K_p = \frac{a^3 p^{3/2}}{(1 + \alpha)(2 + \alpha)^{1/2}}$$
C. $${K_p} = \frac{{{a^{3/2}}{p^{1/2}}}}{{(1 + 2\alpha ){{(2 + \alpha )}^{1/2}}}}$$
D. $${K_p} = \frac{{{a^{5/2}}{p^{1/2}}}}{{(1 + \alpha ){{(2 + \alpha )}^{1/2}}}}$$
E. $$K_p = \frac{a^{3/2} p^{1/2}}{(1 + \alpha)(2 + \alpha)^{1/2}}$$
F. $${K_p} = \frac{{{a^{3/2}}{p^{1/3}}}}{{(1 + \alpha ){{(2 + \alpha )}^{1/2}}}}$$
G. $${K_p} = \frac{{{a^{3/2}}{p^{1/2}}}}{{(1 - \alpha ){{(2 + \alpha )}^{1/2}}}}$$
H. $$K_p = \frac{a^3 p^{1/2}}{(1 + \alpha)(2 + \alpha)^{1/2}}$$
I. $${K_p} = \frac{{{a^{3/2}}{p^{3/2}}}}{{(1 + \alpha ){{(2 + \alpha )}^{1/2}}}}$$
J. $${K_p} = \frac{{{a^{3/2}}{p^{1/2}}}}{{(1 + \alpha ){{(2 + 2\alpha )}^{1/2}}}}$$
Answer: G
A quantity of $12\ g$ of magnesium is burnt completely in air $(O_{2}=20\%$ and $N_{2}=80\%$ by volume). Which of the following is/are correct statement(s) regarding this combustion?

Options:
A. If air is consumed completely, then the total mass of products formed is $17.16\ g$
B. If air is consumed completely, then the total mass of products formed is $17.24\ g$
C. If air is consumed completely, then the total mass of products formed is $17.12\ g$
D. A minimum of $40\ g$ air is needed if all $Mg$ is converted into $MgO$ only
E. If air is consumed completely, then the total mass of products formed is $17.34\ g$
F. If air is consumed completely, then the total mass of products formed is $16.24\ g$
G. A minimum of $4.67\ g$ air is needed if all $Mg$ is converted into $Mg_{3}N_{2}$ only.
H. If air is consumed completely, then the total mass of products formed is $16.14\ g$
I. If air is consumed completely, then the total mass of products formed is $17.14\ g$
Answer: I
Two particles of masses $m_{1}$ and $m_{2}$ are released from rest on a smooth wedge of mass $M$, then

Options:
A. if $(m_{1}>m_{2}$, the wedge will accelerate towards right
B. if $(m_{1}<m_{2}$, the wedge will remain stationary
C. if $(m_{1}>m_{2}$, the wedge will accelerate towards left
D. The wedge remains stationary if $\dfrac{(m_{1}}{m_{2}}=\tan \theta$
E. if $(m_{1}<m_{2}$, the wedge will accelerate towards right
F. if $(m_{1}=m_{2}$, the wedge will accelerate towards left
G. The wedge remains at rest if $m_{1}=m_{2}$
H. if $(m_{1}<m_{2}$, the wedge will accelerate towards left
I. if $(m_{1}>m_{2}$, the wedge will remain stationary
Answer: A
A rectangular loop consists of $N$ closed wrapped turns and has dimensions $a\times b$. The loop is hinged along the Y-axis. What is the magnitude of the torque exerted on the loop by a uniform magnetic field $B={B}_{0}$ directed along the X-axis when current $i={i}_{0}$ in the direction shown. The torque acting on the loop is

Options:
A. $$-\cfrac { 4{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 5 } $$
B. $$-\cfrac { 3{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 5 }$$
C. $$\cfrac { 2{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 5 }$$
D. $$-\cfrac { 4{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 3 } $$
E. $$\cfrac { 4{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 4 }$$
F. $$\cfrac { 4{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 5 } $$
G. $$\cfrac { 4{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 3 }$$
H. $$\cfrac { 2{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 3 }$$
I. $$-\cfrac { 3{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 4 }$$
J. $$\cfrac { -2{ Ni }_{ 0 }ab{ B }_{ 0 }\hat { j } }{ 5 } $$
Answer: A
Buffer solution of pH = 9.56 was obtained by mixing 2 mole of $NH_3$ and 1 mole of $NH_4Cl$ to form 1 L solution. To 200 ml of this soln, 10 ml of 10 M NaOH is added. Calculate pH of resulting soln

Options:
A. 9.56
B. 9.96
C. 9.62
D. 9.26
E. 9.34
F. 9.68
G. 4.44
H. 4.78
I. 9.78
J. 9.54
Answer: D
A pendulum clock loses 12 s a day if the temperature is  ${40^ \circ }C$ and gains 4 s a day if the temperature is ${20^ \circ }C$. The temperature at which the clock will show correct time, and the co-efficient of linear expansion ()of the metal of the pendulum shaft are respectively.

Options:
A. $${40^ \circ }C; \alpha = 1.85 \times {10^{ - 3}}/{}^oC$$
B. $${55^ \circ }C; \alpha = 1.85 \times {10^{ - 3}}/{}^oC$$
C. $$30^\circ \text{C}; \, \alpha = 1.85 \times 10^{-3} \, /^\circ \text{C}$$
D. {35^ \circ }C; \alpha = 1.85 \times {10^{ - 3}}/{}^oC
E. $${40^ \circ }C; \alpha = 1.85 \times {10^{ - 4}}/{}^oC$$
F. $${45^ \circ }C; \alpha = 1.85 \times {10^{ - 4}}/{}^oC$$
G. $$55^\circ \text{C}; \, \alpha = 1.85 \times 10^{-2} \, /^\circ \text{C}$$
H. $$25^\circ \text{C}; \, \alpha = 1.85 \times 10^{-5} \, /^\circ \text{C}$$
I. $${35^ \circ }C; \alpha = 1.85 \times {10^{ - 4}}/{}^oC$$
J. $$60^\circ \text{C}; \, \alpha = 1.85 \times 10^{-4} \, /^\circ \text{C}$$
Answer: C
If a rubber ball is taken down to a 100 m deep lake, its volume decreases by 0.1%. If $g=10\quad m/{ s }^{ 2 }$ then the bulk modulus of elasticity for rubber, in N/${ m }^{ 2 }$, is

Options:
A. $${ 10 }^{ 8 }$$
B. $${ 10 }^{ 12 }$$
C. $${ 10 }^{ 10 }$$
D. $${ 10 }^{ 7 }$$
E. $${ 10 }^{ 9 }$$
F. $${ 10 }^{ 11 }$$
G. $${ 10 }^{ 6 }$$
H. $${ 10 }^{ 9.5 }$$
Answer: C
Calculate $\Delta H^o$ for the reaction,
$Na_2O(s) + SO_3(g) \rightarrow Na_2SO_4(g)$ using the following :
$$Na(s) + H_2O(l) \rightarrow NaOH(s) + \dfrac{1}{2}H_2(g); \Delta H^o = -146 kJ$$
$$Na_2SO_4(s) + H_2O(l) \rightarrow 2NaOH(s) + SO_3(g); \Delta H^o = +418 kJ$$
$$2NaO(s) + 2H_2(g) \rightarrow 4Na(s) + 2H_2O(l);\Delta H^o = + 259 kJ$$

Options:
A. + 435 kJ
B. + 531 kJ
C. + 823 kJ
D. - 435 kJ
E. +541 kJ
F. -541 kJ
G. + 581 kJ
H. - 581 kJ
I. -531 kJ
Answer: H
$40\ g\ NaOH (i=2)$ is dissolved in $1L$ volatile solvent. Vapour pressure of this solution becomes equal to its solid phase at $300\ K$. What is the freezing point of pure solvent. (Density of solvent $=1g/mL$) ($K_{f}=1.8\ K\ kg\ mol^{-}$)

Options:
A. $$298.6\ K$$
B. $$296.4\ K$$
C. $$297.5\ K$$
D. $$300.0\ K$$
E. $$295.4\ K$$
F. $$295.9\ K$$
G. $$298.4\ K$$
H. $$303.6\ K$$
I. $$302.4\ K$$
J. $$297.2\ K$$
Answer: B
Two block of mass m and 2m are placed one over the other as shown in fig. The coefficient of friction between m and 2m is $\mu $ and between 2m and ground is $\dfrac { \mu  }{ 3 } $. If a horizontal force F is applied on upper block and T is tension developed in string then choose the incorrect alternative.

Options:
A. IF $F=\dfrac { \mu }{ 3 } mg,T=\dfrac { \mu mg }{ 9 }$
B. IF $F=\dfrac { \mu }{ 3 } mg,T=0$
C. IF $F=\dfrac { 2\mu }{ 3 } mg,T=\dfrac { \mu mg }{ 2 }$
D. IF $F=2\mu mg,T=\dfrac { \mu mg }{ 3 } $
E. IF $F=\dfrac { 2\mu mg }{ 3 },T=\dfrac { \mu mg }{ 3 }$
F. IF $F=\mu mg,T=0$
G. IF $F=\dfrac { \mu }{ 3 } mg,T=\dfrac { \mu mg }{ 4 }$
H. IF $F=\dfrac { 2\mu mg }{ 3 },T=0$
I. IF $F=\dfrac { \mu }{ 3 } mg,T=\dfrac { \mu mg }{ 6 } $
J. IF $F=3\mu mg,T=0$
Answer: B
What is the Compressibility factor (Z) for 0.02 mole of a van des Waal's gas at pressure of 0.01 atm. Assume the size of gas molecules is negligible.
Given: RT=20 L atm ${ mole }^{ -1 }$ and  a=1000 atm ${ L }^{ 2 }{ mol }^{ -2 }$

Options:
A. 1.5
B. 0.2
C. 0.1
D. 1
E. 0.05
F. 0.45
G. 0.75
H. 0.01
I. 0.5
J. 2
Answer: I
The correct statement (s) regarding, $\left( i \right)\,\,HClO\,\,\left( {ii} \right)\,\,\,HCl{O_2}\,\,\,\,\left( {iii} \right)\,\,\,HCl{O_3}$ and $\left( {iv} \right)\,\,\,HCl{O_4}$, is:

Options:
A. Amongst $(i)$ to $(iv)$, the strongest acid is $(i)$.
B. The number of $Cl=O$ bond in $(ii)$ and $(iii)$ together is two.
C. The number of $Cl=O$ bonds in $(ii)$ and $(iii)$ together is four.
D. The numbers of lone pairs of electrons on $Cl$ in $(ii)$ and $(iii)$ together is three.
E. The number of $Cl=O$ bonds in $(ii)$ and $(iii)$ together is five.
F. The number of $Cl=O$ bond in $(ii)$ and $(iii)$ together is one.
Answer: D
The internal energy of an ideal gas is related as $U=a+b(PV)$ (where $P$ is the pressure of the gas, $V$ is the volume of gas '$a$' and '$b$' are the constant). For an adiabatic process, $\gamma$ will be

Options:
A. $$1 - \frac{a}{b}$$
B. $$1+ \frac {1}{b}$$
C. $$1+ \frac {b}{a}$$
D. $$1 - \frac{b}{a}$$
E. $$1 - \frac{1}{a}$$
F. $$\frac {a}{b}$$
G. $$1 + \frac{a}{b}$$
H. $$1+ \frac {1}{a}$$
Answer: B
The heat of hydration of $Cr^{2+}$ ion is $460\;Kcal/mole$. For $[Cr(H_2O)_6]^{2+}\;\Delta_0=13900\;cm^{-1}$. What heat of hydration would be, if there were no crystal field stabilization energy?

Options:
A. -434\;Kcal/mole
B. $$-420\;Kcal/mole$$
C. -438\;Kcal/mole
D. $$-245\;Kcal/mole$$
E. $$-436\;Kcal/mole$$
F. -430\;Kcal/mole
G. $$+436\ Kcal/mole$$
H. -442\;Kcal/mole
I. +434\ Kcal/mole
J. -450\;Kcal/mole
Answer: G
The indicator Constant for an acidic indicator, $HIn$, is $5 \times 10^{-6} M$. This indicator appears only in the color of acidic form when $\frac{[IN^{-}]}{[HIn]} \leq \frac{1}{20}$ and it appears only in the color of the basic form when $\frac{[HIn]}{[In^{-}]} \leq 40$. The $pH$ range of the indicator is ($log 2 = 0.3$)

Options:
A. $$4.0-6.9$$
B. $$4.0-6.6$$
C. $$4.3-6.9$$
D. $$4.7-6.9$$
E. $$4.0-6.3$$
F. $$4.7-6.3$$
G. $$3.7-6.6$$
H. $$4.3-6.6$$
I. $$4.3-6.3$$
Answer: I
Which of the following relationships is not true?

Options:
A. average kinetic energy of gas $= \dfrac {3}{4}kT$
B. average kinetic energy of gas $= \dfrac {5}{6}kT$
C. most probable velocity $= \sqrt {\dfrac {2RT}{M}}$
D. compressibility factor $Z = \dfrac {PV}{nRT}$
E. average kinetic energy of gas $= \dfrac {3}{2}kT$
F. average kinetic energy of gas $= \dfrac {2}{3}kT$
G. average kinetic energy of gas $= \dfrac {3}{5}kT$
H. average kinetic energy of gas $= \dfrac {1}{3}kT$
I. average kinetic energy of gas $= \dfrac {1}{2}kT$
J. $$PV = \dfrac {3}{2}kT$$
Answer: I
Degree of Hydrolysis of $\dfrac { N }{ 100 } $ solution of KCN is $\quad (Given\quad Ka\quad =1.4\times { 10 }^{ -9 })$

Options:
A. $$2.7\times { 10 }^{ -2 }$$
B. $$2.7 \times { 10 }^{ -5 }$$
C. $$2.7\times { 10 }^{ -4 }$$
D. $$2.7 \times { 10 }^{ -7 }$$
E. $$2.7 \times { 10 }^{ -6 }$$
F. $$2.7\times { 10 }^{ -3 }$$
G. $$2.7 \times { 10 }^{ -1 }$$
Answer: F
For an infinity long metal cylinder, the radius is 3 mm, K=6.28 and charge density =$4\mu C/{ m }^{ 2 }$. What is the electric intensity (E) at a distance of 1.5 m from the axis?
$$[\frac { 1 }{ 4\pi { arepsilon  }_{ 0 } } =9\times { 10 }^{ 9 }]$$

Options:
A. 156 N/C
B. $$144 N/C$$
C. $$160 N/C$$
D. $$72 N/C$$
E. 80 N/C
F. $$288 N/C$$
G. 192 N/C
H. 168 N/C
I. 120 N/C
J. 180 N/C
Answer: B
The minimum speed of a bucket full of water whirled in a vertical circle of radius  $10\mathrm { m }$  at the height point so that the water may not fall is  $\left( g = 10 \mathrm { ms } ^ { - 2 } \right) .$

Options:
A. $$6 m s ^ { - 1 }$$
B. $$4 m s ^ { - 1 }$$
C. $$1 m s ^ { - 1 }$$
D. $$8 m s ^ { - 1 }$$
E. $$2 m s ^ { - 1 }$$
F. $$10 m s ^ { - 1 }$$
G. $$5 m s ^ { - 1 }$$
H. $$7 m s ^ { - 1 }$$
I. $$3 m s ^ { - 1 }$$
Answer: C
Among the solubility rules is the statement that all chlorides are soluble except $Hg_2Cl_2,AgCl,PbCl_2$ and $CuCl$.
Write the expression for the equilibrium constant for the reaction represented by the equation.
$$AgCl(s)\rightleftharpoons Ag^{+}(aq)+Cl^{-}(aq)$$
Is K greater than 1, less than 1, or about equal to 1? Explain your answerWrite the expression for the equilibrium constant for the reaction represented by the equation
$$Pb^{2+}(aq)+2Cl^{-}(aq)\rightleftharpoons PbCl_2(s)$$
Is K greater than 1, less than 1, or about equal to 1? Explain you answer.

Options:
A. 1.$K=[Ag^{+}][Cl^{-}]$<1, 2. $K= 1/[Pb^{2+}][Cl^-]^2$ >1
B. 1.$K=[Ag^{+}][Cl^{-}]$=1.5, 2. $K= 1/[Pb^{2+}][Cl^-]^2$ <1
C. 1.$K=[Ag^{+}][Cl^{-}]$=0.5, 2. $K= 1/[Pb^{2+}][Cl^-]^2$ =0.5
D. 1.$K=[Ag^{+}][Cl^{-}]$=1
E. 1.$K=[Ag^{+}][Cl^{-}]$=0.75, 2. $K= 1/[Pb^{2+}][Cl^-]^2$ =0.75
F. 1.$K=[Ag^{+}][Cl^{-}]$<1, 2. $K= 1/[Pb^{2+}][Cl^-]^2$ <1
G. 1.$K=[Ag^{+}][Cl^{-}]$=1, 2. $K= 1/[Pb^{2+}][Cl^-]^2$ >1
H. 1.$K=[Ag^{+}][Cl^{-}]$=1.25, 2. $K= 1/[Pb^{2+}][Cl^-]^2$ <1
I. 1.$K=[Ag^{+}][Cl^{-}]$=1, 2. $K= 1/[Pb^{2+}][Cl^-]^2$ =1
J. 1.$K=[Ag^{+}][Cl^{-}]$>1, 2. $K= 1/[Pb^{2+}][Cl^-]^2$ <1
Answer: A
A gaseous compound $A$ reacts by three independent first order processes (as shown in figure) with rate constant $2\times { 10 }^{ -3 },3\times { 10 }^{ -3 }$ and $1.93\times { 10 }^{ -3 }$ $sec^{ -1 }$ for products $B,C$ and $D$ respectively. If initially pure $A$ was taken in a closed container with $P=8$ atm, then the partial pressure of $B$ ( in atm) after $100$ sec from start of experiment.

Options:
A. $$0.288$$
B. $$0.654$$
C. $$0.866$$
D. $$0.577$$
E. $$1.732$$
F. $$1.154$$
G. none of these
H. $$0.435$$
I. $$0.722$$
J. $$1.454$$
Answer: G
$9.0 g$ ice at $0^{\circ}\! C$ is mixed with $36 g$ of water at $50^{\circ}\! C$ in a thermally insulated container. Using the following data, answer the question that follow ?$C_P(H_2O)=4.18 \:Jg^{-1}K^{-1}; \:\Delta H_{fusion}(ice) =335\:J\:g^{-1}$ $\Delta S_{ice}$ is :

Options:
A. $$12.8 \:JK^{-1}$$
B. $$14.6 \:JK^{-1}$$
C. $$3.16 \:JK^{-1}$$
D. $$12.5 \:JK^{-1}$$
E. $$15.2 \:JK^{-1}$$
F. $$13.2 \:JK^{-1}$$
G. $$14.2 \:JK^{-1}$$
H. $$13.8 \:JK^{-1}$$
I. $$11.04 \:JK^{-1}$$
J. $$7.84 \:JK^{-1}$$
Answer: G
The spring block system lies on a smooth horizontal surface. the free end of the spring is being pulled towards right with constant speed $v_o = 2 $ m/s. At $t = 0$ sec, the spring of constant $T=400$ N/cm is unstretched and the block has a speed $1$ m/s to the left The maximum extension of the spring is

Options:
A. $8 cm$
B. $1.75 cm$
C. $1.5 cm$
D. $4 cm$
E. $3 cm$
F. $1 cm$
G. $5 cm$
H. $2 cm$
I. $6cm$
J. $2.5 cm$
Answer: H
At Boyle temperature, the value of compression factor $Z(={ PV }_{ M }/RT={ V }_{ real/ }{ V }_{ idea })$ has a value of one over a wide range of pressure. This is due to the fact that in the van der Waals equation?

Options:
A. the effect products due to the molecular attraction compensates the effects produced to the molecular volume
B. the constant b is negligible and not a
C. the constant a is negligible and b is not
D. the constant a is negligible and b is significant
E. the constant a is negligible and b is not negligible
F. the constant b is negligible and a is not
G. the constant b is negligible and a is not negligible
H. the constant a is negligible and not b
I. the effect products due to the molecular attraction compensates the effects produced by the molecular volume
J. both the constants a and b are negligible
Answer: J
Two electric dipoles, $A, B$ with respective dipole moments $ec {d}_{A} = -4qa\hat {i}$ and $ec {d}_{B} = -2qa\hat {i}$ placed on the x-axis with a separation $R$, as shown in the figure.
The distance from $A$ at which both of them produce the same potential is

Options:
A. $$\dfrac {\sqrt {2}R}{\sqrt {2} + 2}$$
B. $$\dfrac {R}{\sqrt {2} - 1}$$
C. $$\dfrac {\sqrt {2}R}{\sqrt {2} + 1}$$
D. $$\dfrac {R}{\sqrt {2} + 2}$$
E. $$\dfrac {R}{\sqrt {2} + 1}$$
F. $$\dfrac {R}{2\sqrt {2} - 1}$$
G. $$\dfrac {\sqrt {2}R}{\sqrt {2} - 1}$$
H. $$\dfrac {R}{2\sqrt {2} + 1}$$
Answer: G
$H^{+}(aq) + NaOH (aq) \to Na^{+} + H_{2}O(l)\ \triangle H_{1} = -1390\ cals$$HCN(aq) + NaOH(aq) \rightarrow Na^{+} + CN^- + H_{2}O(l)\ \triangle H_{2} = -2900\ cals$.What is $\triangle H$ value for $HCN (aq) \to H^{+}(aq) + CN^{-}(aq)$?

Options:
A. $$10790\ cals$$
B. $$11390\ cals$$
C. $$11400\ cals$$
D. $$9800\ cals$$
E. $$10990\ cals$$
F. $$10690\ cals$$
G. $$10590\ cals$$
H. $$12500\ cals$$
I. $$10490\ cals$$
J. $$10890\ cals$$
Answer: A
The pH of solution on mixing $100ml$ of $0.1M$ $HCOONa$ with $100ml$ of $0.1M$ $HCl$ ($Ka$ of $HCOOH={10}^{-4}$) is

Options:
A. $$2.05$$
B. $$1.5$$
C. $$2.35$$
D. $$2.45$$
E. $$2.25$$
F. $$2.15$$
G. $$1.75$$
H. $$2.55$$
I. $$1.35$$
J. $$2.65$$
Answer: A
p represents radiation pressure, c represents  speed of light and S represents radiation energy striking unit area per sec. The non zero integers x, y, z such that $P ^ { x } S ^ { y } c ^ { z }$  is dimensionless are

Options:
A. $$x = 0 , y = 0 , z = 0$$
B. $$x = 1 , y = - 1 , z = - 1$$
C. $$x = - 1 , y = 1 , z = 1$$
D. $$x = -1 , y = 1 , z = -2$$
E. $$x = -1 , y = 1 , z = -1$$
F. $$x = 1 , y ^ { - 1 } , z = - 1$$
G. $$x = 1 , y = 1 , z = 1$$
H. $$x = -1 , y = -1 , z = 1$$
I. $$x = 1 , y = 1 , z = -1$$
J. $$x = 1 , y = - 1 , z = 1$$
Answer: J
A solid homogeneous cylinder of height h and base radius r is kept vertically on a conveyer  belt moving horizontally with an increasing velocity $ v= a + bt^2 $ . if the cylinder is not allowed to slip then the time when the cylinder is about to topple, will be equal to

Options:
A. $$\dfrac {4rg}{3bh}$$
B. $$\dfrac {3rg}{5bh}$$
C. $$ \dfrac {2rg}{bh} $$
D. $$\dfrac {3rg}{2bh}$$
E. $$\dfrac {2rg}{3bh}$$
F. $$\dfrac {3rg}{bh}$$
G. $$\dfrac {3rg}{4bh}$$
H. $$ \dfrac {rg}{bh} $$
I. $$ \dfrac {2bg}{rh} $$
J. $$ \dfrac {rg}{2 bh} $$
Answer: H
The electric potential $V$ (n volts) varies with $x$ (in meter) according to the relation $V=5+4{x}^{2}$
The force experienced by a negative charge of $\mu C$ located at $x=0.5m$ is

Options:
A. $$4\times { 10 }^{ -6 }N$$
B. 18 \times { 10 }^{ -6 }N
C. 16 \times { 10 }^{ -6 }N
D. $$2\times { 10 }^{ -6 }N$$
E. 14 \times { 10 }^{ -6 }N
F. $$8\times { 10 }^{ -6 }N$$
G. 12 \times { 10 }^{ -6 }N
H. 20 \times { 10 }^{ -6 }N
I. 10 \times { 10 }^{ -6 }N
J. $$6\times { 10 }^{ -6 }N$$
Answer: F
A particle moves along the X-axis according to the equation $x = 10{\sin ^3}(\pi t)$. the amplitudes and frequencies of component SHMs are

Options:
A. $$amplitude 10 , 10 ; frequencies 1/2 , 2/2$$
B. $$amplitude 10 , 10 ; frequencies 3/2 , 3/2$$
C. $$amplitude 30/4, 10/4; frequencies 3/2,1/2$$
D. $$amplitude 10 , 10 ; frequencies 1/2 , 1/2$$
E. $$amplitude 10, 10; frequencies 3/2, 1/2$$
F. $$amplitude 30/4, 10/4; frequencies 1/2, 1/2$$
G. $$amplitude 30/4 ,10 ; frequencies 3/2,2$$
H. $$amplitude 10 , 10 ; frequencies 1/2 , 3/2$$
I. $$amplitude 30/4 , 10/4; frequencies 1/2 , 3/2$$
Answer: D
In the circuit shown the battery is ideal with emf $V$. The capacitor is initially uncharged. The switch $S$ is closed at $t = 0$. The charge on the capacitor at time $t$ is :

Options:
A. $$\displaystyle \frac{VC}{2}\left ( 1-e^{-2t/3RC} \right )$$
B. $$\displaystyle \frac{VC}{3}\left ( 1-e^{-2t/3RC} \right )$$
C. $$\displaystyle \frac{VC}{2}\left ( 1-e^{-t/2RC} \right )$$
D. $$\frac{VC}{3}\left ( 1-e^{-2t/3RC} \right )$$
E. $$\frac{VC}{3}\left ( 1-e^{-2t/5RC} \right )$$
F. $$\frac{VC}{3}\left ( 1-e^{-t/2RC} \right )$$
G. $$\displaystyle \frac{VC}{3}\left ( 1-e^{-t/3RC} \right )$$
H. $$\displaystyle \frac{VC}{3}\left ( 1-e^{-t/RC} \right )$$
I. $$\displaystyle \frac{VC}{3}\left ( 1-e^{-3t/2RC} \right )$$
J. $$\displaystyle \frac{VC}{2}\left ( 1-e^{-t/3RC} \right )$$
Answer: A
$1\ mole$ of an ideal gas is taken through a cyclic process. The minimum temperature during the cycle is $300 K$. Then net exchange of heat for complete cycle is:Note :$ ({ T }_{ A }=600K;\quad { T }_{ B }=1200K;\quad { T }_{ c }=600K$)

Options:
A. $$400 R \ln 2$$
B. $$- 300 R \ln 2$$
C. $$450 R \ln 2$$
D. $$600 R \ln 2$$
E. $$300 R \ln 3$$
F. $$900 R \ln 2$$
G. $$-600 R \ln 2$$
H. $$300 R \ln 2$$
I. $$150 R \ln 2$$
Answer: H
Which one of the following correctly represents the physical significance of Gibb's energy change?

Options:
A. $$\displaystyle \Delta G={ -W }_{ \text{expansion} }={ W }_{ \text{non-expansion} }$$
B. $$\displaystyle -\Delta G={ W }_{ \text{expansion} }$$
C. $$\displaystyle \Delta G={ W }_{ \text{non-expansion} }$$
D. $$\displaystyle -\Delta G={ W }_{ \text{non-expansion} }$$
E. $$\displaystyle \Delta G={ -W }_{ \text{compression} }$$
F. $$\Delta G={ W }_{ \text{compression} }$$
G. $$\displaystyle -\Delta G={ W }_{ \text{compression} }$$
H. $$\displaystyle \Delta G={ W }_{ \text{expansion} }$$
Answer: B
Mutual inductance of a system of two thin coaxial conducting loops of radius 0.1 m, and then center separated by distance 10 m is (Take $\mu^{2} = 10$)

Options:
A. $2 \times 10^{-18}$ H
B. $2 \times 10^{-14}$ H
C. $2 \times 10^{-15}$ H
D. $2 \times 10^{-20}$ H
E. $2 \times 10^{-13}$ H
F. $$2 \times 10^{-19} H$$
G. $$2 \times 10^{-16} H$$
H. $$2 \times 10^{-17} H$$
I. $2 \times 10^{-22}$ H
J. $$2 \times 10^{-21} H$$
Answer: D
A vertical narrow glass tube of uniform cross-section, closed at one and has air filled in it upto 40 cm of height, with a column of mercury of height 10 cm above it. If the tube is gradually inclined in the vertical plane and brought to a position making an angle of $60^{0}$ with the vertical, then new length of air column entrapped within the tube is(atmospheric pressure=76 cm of Hg)

Options:
A. 45.97 cm
B. 47.47 cm
C. 44.47 cm
D. 41.47 cm
E. 42.47 cm
F. 45.47 cm
G. 45.77 cm
H. 46.47 cm
I. 45.57 cm
J. 43.47 cm
Answer: F
A block starts from rest at the top frictionless slide at a height,$h_{1}$ above the ground. The block leaves the side moving perfectly horizontally at a height $h_{2}$ above the ground. The block eventually this the ground travelling an angle$ \theta ={ 30 }^{ o }$ below the horizontal. The ratio of $h_{1}$ and $h_{2}$ is

Options:
A. $$\dfrac { 2 }{ 1 }$$
B. $$\dfrac { 5 }{ 3 }$$
C. $$\dfrac { 5 }{ 2 }$$
D. $$ \dfrac { 4 }{ 1 }$$
E. $$\dfrac { 3 }{ 2 }$$
F. $$ \dfrac { 3 }{ 1 }$$
G. $$ \dfrac { 4 }{ 3 }$$
H. $$ \dfrac { 2 }{ 5 }$$
I. $$\dfrac { 5 }{ 1 }$$
Answer: F
An equilateral prism (side length $a=15\ m$) is arranged with its shown cross-section parallel to $xy$ plane. the refractive index of the material only varies along $x$ is given by $n(x)=\sqrt{1+x^{2/3}}$. A light ray is incident at origin (near-grazing incidence ). Find the location at which the ray touches the edge $BC$ for the first time.

Options:
A. $$\left(\sqrt{3},\dfrac{3^{3/2}}{2} \right)$$
B. $$\left(\sqrt{3},\dfrac{3^{3/5}}{2} \right)$$
C. $$\left(3\sqrt{3},\dfrac{9}{2}\right)$$
D. $$\left(\sqrt{3},\dfrac{3^{5/6}}{2}\right)$$
E. $$\left(\sqrt{3},\dfrac{3^{3/4}}{2} \right)$$
F. $$\left(\sqrt{3},\dfrac{3^{5/7}}{2} \right)$$
G. $$\left(\sqrt{3},3^{4/3}\right)$$
H. $$\left(3\sqrt{3},9\right)$$
I. $$\left(\sqrt{3},\dfrac{3^{2/3}}{2} \right)$$
J. $$\left(\sqrt{3},\dfrac{3^{4/5}}{2}\right)$$
Answer: J
If the enthalpy change for the reaction,${CH}_{4}(g)+{Cl}_{2}(g)\rightarrow {CH}_{3}Cl(g)+HCl(g); \Delta H=-25kcal$,If bond energy of $C-Cl$ and bond energies of $H-H$ and $H-Cl$ are the same in magnitude and the bond energy of $C-H$ is $20\ kcal\  mol$ greater than the bond energy of $C-Cl$, then find the enthalpy of the given reaction:${H}_{2}(g)+{Cl}_{2}(g)\rightarrow 2HCl(g)$

Options:
A. $$-22.5kcal/mol$$
B. $$-21.0kcal/mol$$
C. $$-24.5kcal/mol$$
D. $$-32.5kcal/mol$$
E. $$-20.5kcal/mol$$
F. $$-23.5kcal/mol$$
G. $$-25.5kcal/mol$$
H. $$-21.5kcal/mol$$
I. $$-12.5kcal/mol$$
Answer: A
A uniform electric field of strength exists in a region. An electron (charge $-e$, mass $m$) enters a point $A$ with velocity $V \hat{j}$. It moves through the electric field & exits at point $B$. Then:

Options:
A. Velocity at \( B \) is \(\dfrac{2av}{d}\hat{i} - \dfrac{v}{2}\hat{j} \)
B. Rate of work done by the electric field at $B$ is $\dfrac{4ma^2 v^3}{d^3}$
C. Velocity at \( B \) is \(\dfrac{2av}{d}\hat{i} + \dfrac{v}{2}\hat{j} \)
D. Velocity at $B$ is $\dfrac{2av}{d}\hat{i} - v\hat{j}$
E. Velocity at \( B \) is \(\dfrac{2av}{d}\hat{i} + 2v\hat{j} \)
F. $$\overrightarrow{E} = - \dfrac{2amv^2}{ed^2}\hat{i}$$
G. Velocity at $B$ is $\dfrac{2av}{d}\hat{i} + v\hat{j}$
H. Velocity at \( B \) is \(\dfrac{2av}{d}\hat{i} - 3v\hat{j} \)
I. Velocity at $B$ is $\dfrac{2av}{d}\hat{i} - 2v\hat{j}$
J. Rate of work by the electric field at $A$ is zero
Answer: G
An electric dipole is placed at the origin O such that its equator is y-axis. At a point P far away from dipole, the electric field direction is along y-direction. OP makes an angle $\alpha $ with the x-axis such that :

Options:
A. $$ tan \alpha =1$$
B. $$ tan \alpha =\sqrt{2}$$
C. $$tan \alpha =\sqrt{7}$$
D. $$ tan \alpha =\frac{1}{\sqrt{2}}$$
E. $$tan \alpha = \frac{1}{\sqrt{3}}$$
F. $$tan \alpha =\sqrt{5}$$
G. $$tan \alpha = \sqrt{4}$$
H. $$ tan \alpha =\sqrt{3}$$
I. $$tan \alpha =\sqrt{6}$$
Answer: H
Two moles of a gas at $1$ bar and $289 K$ are compressed at constant temperature by use of a constant pressure of $5$ bar. If the compression is driven by a $10-kg$ mass, how far will it fall in the earth's gravitational field? ($g=9.8{ms}^{-2}$)

Options:
A. $$18.82m$$
B. $$21.44m$$
C. $$21.22m$$
D. $$38.61m$$
E. $$20.22m$$
F. $$40.44m$$
G. $$20.02m$$
H. $$19.22m$$
I. $$19.61m$$
J. $$19.82m$$
Answer: E
The potential energy $U$ for a force field $\overrightarrow { F } $ is such that $U=-kxy$ where $K$ is a constant. Then

Options:
A. the force $\overrightarrow { F } = -kx\hat { i } + ky\hat { j }$ is a non-conservative force
B. the force $\overrightarrow { F } $ is a non-conservative force
C. the force $\overrightarrow { F } = -kx\hat { i } - ky\hat { j }$ is a non-conservative force
D. the force $\overrightarrow { F } = kx\hat { i } - ky\hat { j }$ is a conservative force
E. the force $V$ is a non-conservative force
F. $$\overrightarrow { F } =kx\hat { i } +ky\hat { j } $$
G. $$\overrightarrow { F } =ky\hat { i } +kx\hat { j } $$
H. the force $\overrightarrow { F } = -ky\hat { i } - kx\hat { j }$ is a conservative force
I. the force $\overrightarrow { F } = kx\hat { i } + ky\hat { j }$ is a non-conservative force
J. the force $\overrightarrow { F } $ is a conservative force
Answer: J
A pendulum clock keeps correct time at $20^0$C. The correction to be made during summer per day where the average temperature is $40^0$C, ($lpha = 10^{-5}/^0$C) will be :-

Options:
A. 5.64 sec
B. 8.64 sec
C. 6.44 sec
D. 4.54 sec
E. 5.84 sec
F. 5.44 sec
G. 6.54 sec
H. 4.64 sec
I. 7.64 sec
J. 6.64 sec
Answer: A
One end of a light spring of natural length $d$ and spring constant $k(=mg/d)$ is fixed on a rigid support and the other end is fixed to a smooth ring of mass $m$ which can slide without frcition on a vertical rod fixed at a distance $d$ from the support. Initally, the spring makes an angle of $37^{o}$ with the horizontal as shown in the figure, The system is released from rest. The speed of the ring at the same angle substended downward will be

Options:
A. $$\dfrac{\sqrt {gb}}{2}$$
B. $$\dfrac{\sqrt {2gb}}{3}$$
C. $$\sqrt {gb/2}$$
D. $$\dfrac{\sqrt {2gb}}{4}$$
E. $$\sqrt {2gb}$$
F. $$\dfrac{\sqrt {gb}}{4}$$
G. $$\dfrac{\sqrt {gb}}{6}$$
H. $$\sqrt {4gb}$$
I. $$\sqrt {3gb}$$
J. $$\dfrac{\sqrt {gb}}{5}$$
Answer: F
A solution contains a mixture of $\displaystyle Na_{2}CO_{3}$ and $NaOH$. Using phenolpthalein as indicator, $25$ mL of mixture required $19.5$ mL of $0.995\ N\ HCl$ for the end point. With methyl orange, $25$ mL of the solution required $25$ mL of the same $HCI$ for the end point. Calculate g/L of each substance in the mixture.

Options:
A. $Na_{2}CO_{3}=23.2$ g/L, $NaOH= 22.3$ g/L
B. $Na_{2}CO_{3}=23.2$ g/L, $NaOH= 21.3$ g/L
C. Na_{2}CO_{3}=22.2 g/L, NaOH= 23.2 g/L
D. Na_{2}CO_{3}=24.2 g/L, NaOH= 23.2 g/L
E. $Na_{2}CO_{3}=22.3$ g/L, $NaOH= 23.2$ g/L
F. Na_{2}CO_{3}=22.3 g/L, NaOH= 21.2 g/L
G. None of the above
H. $Na_{2}CO_{3}=24.2$ g/L, $NaOH= 21.2$ g/L
I. $Na_{2}CO_{3}=20$ g/L, $NaOH= 25$ g/L
J. Na_{2}CO_{3}=23.2 g/L, NaOH= 21.2 g/L
Answer: A
Two waves are represented by ${ y }_{ 1 }$ = $ 5sin2\pi(75t - 0.25x)$, ${ y }_{ 2 }$= $10sin2\pi(150t - 0.50x)$. The intensity ratio $\dfrac { { I }_{ 1 } }{ { I }_{ 2 } } $ of the two waves is:-

Options:
A. $$1:2$$
B. $$1:8$$
C. $$1:256$$
D. $$1:64$$
E. $$1:16$$
F. $$1:4$$
G. $$1:32$$
H. $$1:128$$
Answer: E
The correct order of True/False of the following statement is: $S_1$: Rate of diffusion and effusion of a gas increases by increasing the pressure and decreasing the temperature.$S_2$: An ideal gas has van der Waal's constant a and b equal to zero.$S_3$: If an ideal gas in expanded adiabatically against a constant pressure, then $TV^{\gamma -1}=$constant.

Options:
A. FFF
B. FTT
C. TTF
D. TFT
E. TFC
F. FTF
G. TCT
H. TFF
I. FFT
J. TTT
Answer: C
A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $F \ sin \omega.$ If the amplitude of the particle is maximum for $\omega = \omega_1$ and the energy of the particle is maximum for $\omega = \omega_2$ then (where $\omega_0$ natural frequency of oscillation of particle)

Options:
A. \omega_1 \neq \omega_0 \ and \ \omega_2 = \frac{\omega_0}{2}
B. $$\omega _1
eq \omega_0 \ and \ \omega_2 =\omega_0$$
C. $$\omega _1 = \omega_0 \ and \ \omega_2 = \omega_0$$
D. $\omega _1 = \omega_0 \ and \ \omega_2
eq \omega_0$
E. $$\omega _1 = \omega_0 \ and \ \omega_2 = \frac{\omega_0}{2}$$
F. $$\omega _1 = \omega_0 \ and \ \omega_2 = \frac{3\omega_0}{2}$$
G. $$\omega _1 = \frac{\omega_0}{2} \ and \ \omega_2 = \omega_0$$
H. $$\omega _1 = \omega_0 \ and \ \omega_2 = 2\omega_0$$
I. $$\omega_1 = \frac{\omega_0}{2} \ and \ \omega_2 = 2\omega_0$$
Answer: C
Two particles of equal mass m are projected from the ground with speeds $v_1$ and $v_2$ at angles $\theta_1$ and $\theta_2$ as shown in the figure. Given $\theta_2 > \theta_1$ and $v_1   cos \theta_1 = v_2 cos  \theta_2$. Which statement/s is/are correct?

Options:
A. Centre of mass of particles will move along a line inclined at some angle with vertical.
B. Particle '2' will be at the same level as the centre of mass when both particles are in air.
C. Centre of mass of particles will move along a vertical line.
D. Particle '2' will be above centre of mass level when both particles are in air.
E. Particle '1' will be above centre of mass level when both particles are in air.
F. Centre of mass of particles will move along a horizontal line.
Answer: D
A disc of mass 3$\mathrm { m }$ and a disc of mass $\mathrm { m }$ are connected by a massless spring of stiffness $\mathrm { k }$ . The heavier disc is placed on the ground with the spring vertical and lighter disc on top. From its equilibrium position, the upper disc is pushed down by a $\delta$ distance and released. Then:

Options:
A. If $\delta = 2 m g / k ,$ maximum normal reaction from ground on lower disc $= 4 \mathrm { mg }$
B. If $\delta = 2 m g / k ,$ maximum normal reaction from ground on lower disc $= 6 \mathrm { mg }$
C. if $\delta > 2.75 m g / k ,$ the lower disc will bounce up
D. if $\delta > 3.5 m g / k ,$ the lower disc will bounce up
E. if $\delta > 2.5 m g / k ,$ the lower disc will bounce up
F. if $\delta > 2 m g / k ,$ the lower disc will bounce up
G. if $\delta > 3.25 m g / k ,$ the lower disc will bounce up
H. if $\delta > 3 m g / k ,$ the lower disc will bounce up
I. if $\delta > 4 m g / k ,$ the lower disc will bounce up
J. if $\delta > 5 m g / k ,$ the lower disc will bounce up
Answer: H
The energy stored in the capacitors is U when S is open. Now S is closed, the charge passed through S is Q, then?

Options:
A. $$U=0, Q=\dfrac{3}{4}CE$$
B. $$U=0, Q=2CE$$
C. $$U=CE^2, Q=0$$
D. $$U=0, Q=\dfrac{5}{4}CE$$
E. $$U=0, Q=\dfrac{1}{3}CE$$
F. $$U=0, Q=\dfrac{4}{3}CE$$
G. $$U=0, Q=\dfrac{3}{2}CE$$
H. $$U=0, Q=\dfrac{7}{6}CE$$
I. $$U=0, Q=\dfrac{2}{3}CE$$
J. $$U=0, Q=\dfrac{1}{2}CE$$
Answer: I
A point charge $q$ is placed at distance $a$ from the centre of an uncharged this spherical conducting shell of radius $R=2a$. A point $P$ is located at a distance $4a$ from the centre of the conducting shell as shown. The electric potential due to induced charge on the inner surface of the conducting shell at point $P$ is

Options:
A. $$-\cfrac{kq}{5a}$$
B. $$-\cfrac{kq}{7a}$$
C. $$\cfrac{kq}{7a}$$
D. $$\cfrac{kq}{5a}$$
E. $$-\cfrac{kq}{6a}$$
F. $$-\cfrac{kq}{4a}$$
G. $$-\cfrac{kq}{8a}$$
H. $$\cfrac{kq}{4a}$$
I. $$\cfrac{kq}{6a}$$
Answer: D
In a certain gas, the ratio of the speed of sound and root mean square speed is $\sqrt { \dfrac { 5 } { 9 } }$ The molar heat capacity of the gas in a process given by PT = constant is (Take R = 2 cal /mole K )Treat the gas is ideal

Options:
A. $$\dfrac { 9 R } { 2 }$$
B. $$\dfrac { 5 R } { 2 }$$
C. $$\dfrac { 9 R } { 4 }$$
D. $$\dfrac { R } { 2 }$$
E. $$\dfrac { 3R } { 2 }$$
F. $$\dfrac { 5 R } { 3 }$$
G. $$\dfrac { 3 R } { 4 }$$
H. $$\dfrac { 11 R } { 2 }$$
I. $$\dfrac { 7 R } { 2 }$$
Answer: I
In the given circuit $R_1 R_4 = R_2 R_3.$ Then choose the correct statement:

Options:
A. If position of battery and galvanometer are interchanged then galvanometer will still show zero deflection.
B. If the emf of the battery is quadrupled, the galvanometer will still show zero deflection.
C. If galvanometer is not ideal it will show deflection if e.m.f. is quadrupled
D. If the emf of the battery is increased by a factor of 3, the galvanometer will still show zero deflection.
E. If galvanometer is not ideal it will show deflection if e.m.f. is halved
F. If galvanometer is not ideal it will show will show deflection if e.m.f. is doubled
G. If the emf of the battery is halved, the galvanometer will still show zero deflection.
H. If e.m.f. $\varepsilon$ is doubled still no direction is shown by galvanometer
I. Rate of heat dissipation through $R_1$ will change if position of battery and galvanometer are inter changed
J. If the emf of the battery is doubled, the galvanometer will still show zero deflection.
Answer: F
A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at $t=0$. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at $t=0$ along the horizonal string AB, with the speed v. Friction between the bead and the string may be neglected. Let $t_P$ and $t_Q$ be the respective times taken by P and Q to reach the point B. Then?

Options:
A. $$\dfrac{t_P}{t_Q}=\dfrac{length \quad of \quad arc \quad ACB}{length \quad of \quad chord \quad AB}$$
B. \( t_P = \frac{length \quad of \quad chord \quad AB}{length \quad of \quad arc \quad ACB} \cdot t_Q \)
C. $$t_P = t_Q$$
D. $$t_P > t_Q$$
E. t_P = \frac{1}{2} t_Q
F. t_P = \frac{3}{4} t_Q
G. t_P = \frac{length \quad of \quad chord \quad AB}{3 \cdot length \quad of \quad arc \quad ACB} \cdot t_Q
H. $$t_P < t_Q$$
I. t_P = \frac{length \quad of \quad chord \quad AB}{2 \cdot length \quad of \quad arc \quad ACB} \cdot t_Q
J. t_P = 2t_Q
Answer: H
In an interference arrangement similar to young's double sit experiment , slits ${ s }_{ 1 }$ and ${ s }_{ 2 }$ are illuminated with coherent microwave sources each of frequency 1 MHz. The sources are synchronized to have zero phase difference. This slits are separated by distance d=1501 m. The intensity of light on the screen at an angle ${ 30 }^{ \circ  }$ with the center of the sits is

Options:
A. $$\dfrac { { I }_{ 0 } }{ 8 }$$
B. $$2{ I }_{ 0 }$$
C. $$\dfrac { { I }_{ 0 } }{ 2 } $$
D. $$\dfrac { { I }_{ 0 } }{ 9 }$$
E. $${ I }_{ 0 }$$
F. $$\dfrac { { I }_{ 0 } }{ 4 } $$
G. $$\dfrac { { I }_{ 0 } }{ 7 }$$
H. $$\dfrac { { I }_{ 0 } }{ 6 }$$
I. $$\dfrac { { I }_{ 0 } }{ 5 }$$
J. $$\dfrac { { I }_{ 0 } }{ 3 }$$
Answer: C
Two adiabatic processes bc and ad for the same gas are given to intersect two intersect two isotherms at ${ T }_{ 1 }$ and ${ T }_{ 2 }$ then

Options:
A. $$\dfrac { { V }_{ a } }{ { V }_{ b } } ={ \left( \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ \dfrac { 1 }{ \gamma -1 } }$$
B. $$\dfrac { { V }_{ b } }{ { V }_{ c } } ={ \left( \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }$$
C. $$\dfrac { { V }_{ a } }{ { V }_{ d } } ={ \left( \dfrac { { T }_{ 1 } }{ { T }_{ 2 } } \right) }^{ \dfrac { 1 }{ \gamma -1 } }$$
D. $${ V }_{ a }{ V }_{ d } = { V }_{ b }{ V }_{ c }$$
E. $${ V }_{ a }{ V }_{ c } = { V }_{ b }{ V }_{ d }$$
F. $$\dfrac { { V }_{ a } }{ { V }_{ d } } ={ \left( \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ \dfrac { 1 }{ \gamma -1 } }$$
G. $$\dfrac { { V }_{ a } }{ { V }_{ d } } ={ \left( \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } \right) }^{ \dfrac { 1 }{ \gamma +1 } }$$
H. $${ V }_{ a }{ V }_{ b } = { V }_{ c }{ V }_{ d }$$
I. $${ V }_{ a }{ V }_{ d } = { V }_{ c }{ V }_{ b }$$
Answer: E
A particle having a mass $ m$ and velocity $V_m$ in the y-direction is projected on to a horizontal belt that is moving with uniform velocity $V_b$ in the $x-direction$ as shown in figure. $\mu$ is the coefficient of friction between particle and belt. Assuming that the particle first touches the belt at the origin at the fixed $x-y$ coordinate system and remains on the belt, find the coordinates $(x,y)$ of the point where the sliding stops.

Options:
A. $$\displaystyle x= \frac{V_m}{\mu g}\sqrt{V_m^2+V_b^2}, \ y= \frac{V_b}{2\mu g}\sqrt{V^2_m+V^2_b}$$
B. $$\displaystyle x= \frac{V_m}{\mu g}\sqrt{V_m^2+V_b^2}, \ y= \frac{V_b}{\mu g}\sqrt{V^2_m+V^2_b}$$
C. $$\displaystyle x=\frac{V_b}{\mu g}\sqrt{V_m^2+V_b^2}, \ y=\frac{V_m}{\mu g}\sqrt{V^2_m+V^2_b}$$
D. $$\displaystyle x=\frac{V_m}{2\mu g}\sqrt{V_m^2+V_b^2}, \ y=\frac{V_b}{\mu g}\sqrt{V^2_m+V^2_b}$$
E. $$\displaystyle x= \frac{V_m}{2\mu g}\sqrt{V_b^2+V_m^2}, \ y= \frac{V_b}{\mu g}\sqrt{V^2_b+V^2_m}$$
F. $$\displaystyle x= \frac{V_m}{2\mu g}\sqrt{V_b^2+V_m^2}, \ y= \frac{V_b}{2\mu g}\sqrt{V^2_b+V^2_m}$$
G. $$\displaystyle x=\frac{V_m}{2\mu g}\sqrt{V_m^2+V_b^2}, \ y=\frac{V_b}{2\mu g}\sqrt{V^2_m+V^2_b}$$
H. $$\displaystyle x=\frac{V_b}{2\mu g}\sqrt{V_m^2+V_b^2}, \ y=\frac{V_m}{2\mu g}\sqrt{V^2_m+V^2_b}$$
I. $$\displaystyle x=\frac{V_b}{\mu g}\sqrt{V_m^2+V_b^2}, \ y=\frac{V_m}{2\mu g}\sqrt{V^2_m+V^2_b}$$
Answer: H
The amount of urea to be dissolved in $500 \,ml$ of water $(K = 18.6 \,K \,mole^{-1}$ in $100 \,g \,solvent)$ to produce a depression of $0.186^\circ C$ in freezing point is

Options:
A. $$2.7 \,g$$
B. $$0.3 \,g$$
C. $$9 \,g$$
D. $$3.5 \,g$$
E. $$2.5 \,g$$
F. $$1.5 \,g$$
G. $$3 \,g$$
H. $$4.5 \,g$$
I. $$4 \,g$$
J. $$6 \,g$$
Answer: G
White light reflected from a soap film(Refractive Index$=1.5$) has a maxima at $600$nm and a minima at $450$nm with no minimum in between. Then the thickness of the film is _____$\times 10^{-7}$m.

Options:
A. $$5$$
B. $$6$$
C. $$2.5$$
D. $$1$$
E. $$2$$
F. $$4$$
G. $$3$$
Answer: G
Given below are the half cell reactions${ Mn }^{ 2+ }+{ 2e }^{ - }\longrightarrow { Mn }  E=-1.81V$$2\left( { Mn }^{ 3+ }+{ e }^{ - }\longrightarrow { Mn }^{ 2+ } \right) E=+1.51V$, $E_{Cell}$ of the reaction will be

Options:
A. 0.33V ; the reaction will not occur .
B. 2.33V ; the reaction will not occur
C. 2.33V ; the reaction will occur
D. 0.33V ; the reaction will occur.
E. 1.33V ; the reaction will not occur.
F. 2.33V ; the reaction will not occur.
G. 2.69V ; the reaction will not occur.
H. 2.69V ; the reaction will occur
I. 1.69V ; the reaction will not occur
Answer: G
The active ingredient in aspirin is acetyl salicylic acid with $K_{a}=4.0 \times 10^{-9}$. The $pH$ of the solution obtained by dissolving two aspirin tablets (containing $0.36 g$ of acetyl salicylic acid in each tablet) in $250 ml$ of water is ($log 2 = 0.3$)

Options:
A. $$5.1$$
B. $$5.05$$
C. $$5.12$$
D. $$5.15$$
E. $$5.20$$
F. $$8.9$$
G. $$5.35$$
H. $$5.25$$
I. $$10.2$$
J. $$5.40$$
Answer: H
A charged particle is deflected by two mutually perpendicular oscillating electric fields such that the displacement of the particle due to each one of them is given by $x=lpha cos \left ( \omega t \right ) and y= lpha cos\left ( \omega t+\frac{\pi }{6} \right )$ respectively. The trajectory followed by the charged particle is:

Options:
A. An ellipse with equation $x^{2}+y^{2}-xy=\frac{1}{3}a^{2}$
B. An ellipse with equation $x^{2}+y^{2}-xy=\frac{3}{2}a^{2}$
C. An ellipse with equation $x^{2}+y^{2}-xy=\frac{2}{3}a^{2}$
D. An ellipse with equation \( x^{2} + y^{2} - xy = \frac{4}{3}a^{2} \)
E. A circle with equation $x^{2}+y^{2}=a^{2}$
F. A straight line with equation $y=\frac{x}{\sqrt{3}}$
G. An ellipse with equation $x^{2}+y^{2}-xy=\frac{1}{2}a^{2}$
H. A straight line with equation $y=\sqrt{3}x$
I. An ellipse with equation $x^{2}+y^{2}-\sqrt{3}xy=\frac{1}{4}a^{2}$
J. An ellipse with equation $x^{2}+y^{2}-xy=\frac{3}{4}a^{2}$
Answer: E
Carbonic acid, $H_2CO_3$, is a diprotic acid for which $K_1=10^{-7}$ and $K_2=10^{-11}$. Which solution will produce a pH closet to $9$?

Options:
A. $0.1$M $Na_2CO_3$
B. $0.1$M $NaHCO_3$ and $0.1$M $Na_2CO_3$
C. 0.05M $Na_2CO_3$ and 0.05M $NaHCO_3$
D. $0.1$M $H_2CO_3$
E. 0.1M Na_2CO_3 and 0.05M H_2CO_3
F. 0.01M $Na_2CO_3$
G. $0.1$M $NaHCO_3$
H. 0.05M $H_2CO_3$ and 0.05M $NaHCO_3$
I. 0.01M H_2CO_3
Answer: D
In the given LC circuit if initially capacitor $C$ has charge $Q$ on it and $2C$ has charge $2Q$. The polarities are as shown in the figure. Then after closing switch $S$ at $t=0$

Options:
A. initial rate of growth of current in inductor will be \( \frac{Q}{2CL} \)
B. initial rate of growth of current in inductor will be \( \frac{3Q}{2CL} \)
C. initial rate of growth of current in inductor will be \( \frac{2Q}{4CL} \)
D. energy will get equally distributed in both the capacitor just after closing the switch
E. maximum energy in the inductor will be $3{Q}^{2}/2C$
F. initial rate of growth of current in inductor will be \( \frac{3Q}{4CL} \)
G. none of these
H. initial rate of growth of current in inductor will be \( \frac{5Q}{3CL} \)
I. initial rate of growth of current in inductor will be \( \frac{4Q}{3CL} \)
J. initial rate of growth of current in inductor will be $2Q/3CL$
Answer: D
Rate of a chemical reaction 2 A (g) $\rightarrow$ B (g) is defined as:
$$\displaystyle 'r_B' = \frac{1}{V} \frac{dn_B}{dt}$$
Where $n_B$ is the number of moles of B formed and $C_B$ is concentration of B then which of the following relations is/are correct?

Options:
A. r_B = \displaystyle \frac{dC_B}{dt} + \frac{2C_B}{V} \frac{dV}{dt}
B. r_B = \displaystyle \dfrac{dC_B}{dt} - \dfrac{C_B}{V} \dfrac{dV}{dt} (If volume V is constant)
C. $r_B = -\dfrac{dC_B}{dt}$ (If volume V is constant)
D. r_B = \displaystyle \dfrac{dC_B}{dt} - \dfrac{2C_B}{V} \dfrac{dV}{dt}
E. $\displaystyle r_B = \frac{dC_B}{dt} - \frac{C_B}{V} \frac{dV}{dt}$ (If volume V is not constant)
F. r_B = - \dfrac{dC_B}{dt} - \dfrac{C_B}{V} \dfrac{dV}{dt}
G. $r_B = - \dfrac{dC_B}{dt} + \dfrac{C_B}{V} \dfrac{dV}{dt}$ (If volume V is not constant)
H. $r_B = \displaystyle \frac{dC_B}{dt}$ (If volume V is constant)
Answer: H
Two moles of an ideal gas at a temperature of $T=273\ K$ was isothermally expanded $4$ times the initial volume and then heated isochorically, so that the final pressure becomes equal to the initial pressure. The ratio of molar specific heat capacities if total amount of heat imparted to the gas equals $Q=27.7\ kJ$, is

Options:
A. $$1.49$$
B. $$1.50$$
C. $$1.63$$
D. $$1.54$$
E. $$1.51$$
F. $$1.47$$
G. $$1.66$$
H. $$1.58$$
I. $$2.63$$
J. $$1.52$$
Answer: A
Nodal planes of $\pi$-bonds (s) in $CH_2\, =\, C\, =\, C\, =\, CH_2$ are located in:

Options:
A. one in molecular plane and one in a plane perpendicular to molecular plane which bisects $C -C$ $\sigma$-bonds at right angle
B. two in molecular plane and one in a plane perpendicular to molecular plane which contains $C - C$ $\sigma$-bonds
C. two in molecular plane and two in a plane perpendicular to molecular plane which contains $C -C$ $\sigma$-bonds
D. one in molecular plane and one in a plane perpendicular to molecular plane which contains $C - C$ $\sigma$-bonds
E. perpendicular plane which contains $C-  C$ $\sigma$-bond
F. two in molecular plane and one in a plane perpendicular to molecular plane which bisects $C-  C$ $\sigma$-bonds at right angle
G. one in molecular plane and two in a plane perpendicular to molecular plane which contains $C -C$ $\sigma$-bonds
H. one in molecular plane and two in a plane perpendicular to molecular plane which bisects $C - C$ $\sigma$-bonds at right angle
I. all are in molecular plane
J. one in molecular plane and three in a plane perpendicular to molecular plane which contains $C -C$ $\sigma$-bonds
Answer: E
The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T=298 J are
$$\Delta f{ G }^{ 0 }[C(graphite)]=0\quad KJ{ mol }^{ -1 }$$
$$\Delta f{ G }^{ 0 }[C(diamond)]=2.9\quad KJ{ mol }^{ -1 }$$
The standard state means that the pressure should be 1 bar. and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $2\times { 10 }^{ 6 }{ m }^{ 3 }{ mol }^{ -1 }$. If C(graphite) is converted to C(diamond) isothermally at T=298K, the pressure at which C(graphite) is in equilibrium with C(diamond),is
[Useful information : 1J=1 kg ${ m }^{ 2 }{ s }^{ -2 }$, 1 pa=1kg${ m }^{ -1 }{ s }^{ -2 }$; 1 bar = ${ 10 }^{ 5 }$]

Options:
A. 58001 bar
B. 29000 bar
C. 58501 bar
D. 14501 bar
E. 58002 bar
F. 1450 bar
G. 29001 bar
H. 58000 bar
I. 58500 bar
J. 14500 bar
Answer: J
A body is riding on a flat car of a train moving with velocity 10 m/s. The body throws a ball which according to him moves an angle of $ 60^o $ with the horizontal and in line with thr track. An observer on the ground observed the ball to rise vertically. how high does he see the ball rise? $ (g = 10 m/s^2) $

Options:
A. 20 m
B. 25 m
C. 20 \, m
D. 35 m
E. 10 \sqrt{3} \, m
F. 10 m
G. 30 \sqrt{3} \, m
H. 15 m
I. 20 \sqrt{3} \, m
J. 30 m
Answer: B
A radio transmitting station operating at a frequency of $120$MHz has two identical antennas that radiate in phase. Antenna $B$ is $9m$ to the right of antenna $A$. Consider point $P$ at a horizontal distance $x$ to the right of antenna $A$. The value of $x$ and order for which the constructive interference will occur at point $P$ are ____.

Options:
A. $$x=5.6m,n=2$$
B. $$x=1.65m,n=3$$
C. $$x=14.95m,n=1$$
D. $$x=1.95m, n=3$$
E. $$x=0,n=3.6$$
F. $$x=1.75m,n=3$$
G. $$x=1.45m, n=3$$
H. $$x=1.55m,n=3$$
I. $$x=1.55m,n=4$$
J. $$x=2.25m, n=3$$
Answer: B
Two identical coherent sources are placed on a diameter of a circle of radius $R$ at separation $x(<<R)$ symmetrical about the center of the circle. The sources emit identical wavelength $\lambda$ each. The number of points on the circle of maximum intensity is $(x=5\lambda$):

Options:
A. $$24$$
B. 21
C. $$20$$
D. $$22$$
E. $$26$$
F. 28
G. 19
H. 23
I. 25
Answer: C
A small uncharged sphere of mass $1.5 gm$ is suspended from a point by a 1m long thread. The sphere is illuminated for some time by light of a frequency greater than its photoelectric threshold frequency. An electric field of magnitude $1.5\times 10^4N/C$ is now applied, as shown. Consequently, the sphere shifts its position to a new equilibrium position, such that the thread is now inclined to the vertical at $45^{\circ}$. Number of electrons removed from the sphere due to its illuminations is ?  $(g=10m/s^2)$

Options:
A. $$6 \times 10^{12}$$
B. $$6.25\times 10^{12}$$
C. $$5 \times 10^{12}$$
D. $$5\times 10^{11}$$
E. $$7.2\times 10^{13}$$
F. $$9 \times 10^{12}$$
G. $$6.5 \times 10^{12}$$
H. $$8\times 10^{12}$$
I. $$7.5 \times 10^{13}$$
J. $$7.5 \times 10^{12}$$
Answer: H
Two protons are kept at a separation of 10 nm Let Fn and Fe be the nuclear force and electromagnetic force between them.

Options:
A. $F _ { e } \text { and } F _ { n }$ differ only slightly
B. $$F _ { e } \approx 2F _ { n }$$
C. $$F _ { e } < < F _ { n }$$
D. $$F _ { e } \approx F _ { n }$$
E. $$F _ { e } = F _ { n }$$
F. $$F _ { e } > > F _ { n }$$
G. $$F _ { e } \approx 0.5F _ { n }$$
H. $$F _ { e } \approx 0.1F _ { n }$$
I. $$F _ { e } \approx 10F _ { n }$$
Answer: C
Consider the Hofmann ammonolysis reaction:If $R = {C}_{2}{H}_{5}$, the correct order of basic character of the above amines in aqueous medium is :

Options:
A. $$ 3 > 2 > 1 > N{H}_{3}$$
B. $$ 2 > 3 > 1 > N{H}_{3}$$
C. $$3 > 2 > N{H}_{3} > 1$$
D. $$ N{H}_{3} > 1 > 2 > 3$$
E. $$2 > 1 > N{H}_{3} > 3$$
F. $$ 3 > 1 > 2 > N{H}_{3}$$
G. $$1 > 2 > 3 > N{H}_{3}$$
H. $$ 2 > 1 > 3 > N{H}_{3}$$
Answer: B
A 15 kg mass is accelerated from rest with a force of 100 N. As it moves faster, friction and air resistance create an oppositively directed retrading force given by $ F_R = A + BV, $ Where A = 25 N and B = $0.5 \frac {N}{m/s} $. At what velocity does the acceleration equal to one half of the initial acceleration?

Options:
A. 20 m/s
B. $90 ms^{-1}$
C. 100 m/s
D. 75 m/s
E. $ 25 ms^{-1} $
F. 105 m/s
G. 50 m/s
H. $85 ms^{-1}$
I. 65 m/s
J. 125 m/s
Answer: D