MikaStars39/nano-eval huggingface dataset:
$v_1$ and $v_2$ are velocities of a particle under the resultant motion of two parallel SHMs of amplitudes $r_1$ and $r_2$ and same period with zero phase difference. At the displacement $x_1 = r_1$ and $x_2 = r_2$. The ratio ($v_1/v_2$)

Options:
A. $$\sqrt{r_1(r_2 + 2r_2)/r_2(r_2+2r_1)}$$
B. $$\sqrt{r_2(r_2 + 2r_1)/r_1(r_1+2r_2)}$$
C. $$\sqrt{(r_1^2 - r_2^2)/(r_1^2 + r_2^2)}$$
D. $$\sqrt{r_2(r_1 + 2r_2)/r_1(r_2+2r_1)}$$
E. $$\sqrt{r_1(r_2 + 2r_1)/r_2(r_2 + 2r_1)}$$
F. $$\sqrt{r_1(r_2 + 3r_1)/r_2(r_2 + 3r_1)}$$
G. $$\sqrt{r_2(r_1 + 3r_2)/r_1(r_2+3r_1)}$$
H. $$\sqrt{r_1(r_2 + r_1)/r_2(r_2 + r_1)}$$
I. $$\sqrt{r_1(r_1 + 2r_2)/r_2(r_2+2r_1)}$$
J. $$\sqrt{(r_1^2 + r_2^2)/(r_1^2 - r_2^2)}$$
Answer: B
A thin uniform rigid rod of length $l$ is hinged at one end so that it can move in a vertical plane by rotating about a horizontal axis through upper end. The lower end is given a sharp blow and made to acquire a linear velocity $v_{0}$.
Maximum height attained by lower end of the rod is:

Options:
A. $\dfrac{v_{0}^{2}}{3g}$ for $v_{0}<\sqrt{6gl}$
B. $l$ for $v_{0}=\sqrt{4.5gl}$
C. $2l$ for $v_{0}\ge \sqrt{6gl}$
D. $l$ for $v_{0}=\sqrt{5gl}$
E. $l$ for $v_{0}=\sqrt{4gl}$
F. $\dfrac{3v_{0}^{2}}{g}$ for $v_{0}<\sqrt{6gl}$
G. $2l$ for $v_{0}=\sqrt{3gl}$
H. $l$ for $v_{0}=\sqrt{7gl}$
I. $l$ for $v_{0}=\sqrt{2gl}$
J. $l$ for $v_{0}=\sqrt{3gl}$
Answer: J
A block of mass $m=0.1kg$ is released from a height of $4\ m$ on a curved smooth surface. On the horizontal surface, path $AB$ is smooth and path $BC$ offers coefficient of friction $\mu =0.1$. If the impact of block with the vertical wall at $C$ be perfectly elastic, the total distance covered by the block on the horizontal surfaces before coming to rest will be (Take $g=10\ m/{ s }^{ -2 }$)

Options:
A. $$79\ m$$
B. $$44\ m$$
C. $$69\ m$$
D. $$49\ m$$
E. $$109\ m$$
F. $$39\ m$$
G. $$64\ m$$
H. $$29\ m$$
I. $$59\ m$$
Answer: I
Two soap bubbles combine to form a single bubble. In this process, the change in volume and surface area are $V$ and $A$ respectively. If $P$ is the atmospheric pressure, and $T$ is the surface tension of the soap solution, then which of the following relation is true?

Options:
A. 4 PV - 5 TA $=$ 0
B. 4 PV - 4 TA $=$ 0
C. 4 PV + 3 TA = 0
D. 3 PV - 4 TA $=$ 0
E. 3 PV + 4 TA $=$ 0
F. 4 PV + 4 TA $=$ 0
G. 4 PV  - 3 TA $=$ 0
Answer: E
Write balanced equations for the half reactions and calculate the reduction potentials at $25^{\circ}C$ for the following half cells:
(a) $Cl^{-}(1.2M) | Cl_{1}(g, 3.6\ atm)\ E^{\circ} = 1.36\ V$
(b) $Fe^{2+} (2M) | Fe(s)\ E^{\circ} = -0.44\ V$.

Options:
A. -2.372 V
B. 1.472 V
C. 4.272 V
D. 1.272 V
E. -1.372 V
F. 2.372 V
G. 2.272 V
H. 3.272 V
I. 1.372 V
J. -1.272 V
Answer: E
Two identical conducting very large plates $P_1$ and $P_2$ having charges $+ 4Q$ and $+ 6Q$ are placed very closed to each other at separation $d$. The plate area of either face of the plate is $A$, The potential difference between plates $P_1$ and $P_2$ is:

Options:
A. $$V_{P1} - V_{P2} = \dfrac{9Qd}{A \varepsilon_0}$$
B. $$V_{P1} - V_{P2} = \dfrac{- 5 Qd}{A arepsilon_0}$$
C. $$V_{P1} - V_{P2} = \dfrac{3Qd}{A arepsilon_0}$$
D. $$V_{P1} - V_{P2} = \dfrac{4Qd}{A arepsilon_0}$$
E. $$V_{P1} - V_{P2} = \dfrac{- Qd}{A arepsilon_0}$$
F. $$V_{P1} - V_{P2} = \dfrac{7Qd}{A arepsilon_0}$$
G. $$V_{P1} - V_{P2} = \dfrac{2Qd}{A INCLUDING>"._0}$$
H. $$V_{P1} - V_{P2} = \dfrac{6Qd}{A \varepsilon_0}$$
I. $$V_{P1} - V_{P2} = \dfrac{Qd}{A arepsilon_0}$$
J. $$V_{P1} - V_{P2} = \dfrac{ 5 Qd}{A arepsilon_0}$$
Answer: I
the particle of mass m is moving on a straight line such that its speed changes with distance traveled as $v=\beta \sqrt { x }$. The work done by net force on the particle during first t second is

Options:
A. $$\frac { m{ \beta }^{ 2 }{ t }^{ 2 } }{ 6 }$$
B. $$\frac { m{ \beta }^{ 2 }{ t }^{ 2 } }{ 16 }$$
C. $$\frac { m{ \beta }^{ 4 }{ { t }^{ 2 } } }{ 8 }$$
D. $$\frac { m{ \beta }^{ 2 }{ t }^{ 2 } }{ 8 }$$
E. $$\frac { m{ \beta }^{ 2 }{ t }^{ 2 } }{ 2 }$$
F. $$\frac { m{ \beta }^{ 2 }{ t }^{ 2 } }{ 4 }$$
G. $$\frac { m{ \beta }^{ 4 }{ { t } } }{ 4 }$$
H. $$\frac { m{ \beta }^{ 2 }{ t } }{ 4 }$$
I. $$\frac { m{ \beta }^{ 2 }{ t }^{ 3 } }{ 8 }$$
J. $$\frac { m{ \beta }^{ 4 }{ t } }{ 8 }$$
Answer: F
Sodium light has frequency of $5.093\times 10^{14}Hz$. The number of waves of sodium light in $5m$ of water $(^{a}{\mu}_{w}=4/3)$

Options:
A. $$1.132\times10^{8}$$
B. $$2.45\times10^{6}$$
C. $$2.45 \times 10^{5}$$
D. $$1.132\times10^{7}$$
E. $$2.45 \times 10^{10}$$
F. $$2.45 \times 10^{7}$$
G. $$2.45\times10^{8}$$
H. $$2.45 \times 10^{9}$$
I. $$2.45 \times 10^{11}$$
J. $$2.45 \times 10^{4}$$
Answer: B
A uniform solid cylinder of mass m and radius R is free to rotate about its fixed axis which is smooth and vertical. A light inextensible string is wound round the cylinder and its free end is pulled horizontally with a constant force $F=2$mg. The angular velocity of the cylinder when the free end of string has moved through a distance $4$R is?

Options:
A. $$9\sqrt{\dfrac{2g}{R}}$$
B. $$7\sqrt{\dfrac{2g}{R}}$$
C. $$8\sqrt{\dfrac{2g}{R}}$$
D. $$\sqrt{\dfrac{2g}{R}}$$
E. $$2\sqrt{\dfrac{2g}{R}}$$
F. $$4\sqrt{\dfrac{2g}{R}}$$
G. $$6\sqrt{\dfrac{2g}{R}}$$
H. $$5\sqrt{\dfrac{2g}{R}}$$
I. $$10\sqrt{\dfrac{2g}{R}}$$
J. $$3\sqrt{\dfrac{2g}{R}}$$
Answer: D
The velocity of centre of mass $\left({v}_{cm}\right)$of the plank-cylinder system when the cylinder is rolling over the plank is

Options:
A. Constant and equals to \(\dfrac{1}{4}\sqrt{2gh}\)
B. Variable with $\left({{v}_{cm}}_{max}\right)=\dfrac{1}{2}\sqrt{2gh}$
C. Constant and equals to $\sqrt{2gh}$
D. Constant and equals to \(\dfrac{5}{4}\sqrt{2gh}\)
E. Variable with \(\left({{v}_{cm}}_{max}\right)=\dfrac{3}{2}\sqrt{2gh} \)
F. Constant and equals to \(\dfrac{7}{4}\sqrt{2gh}\)
G. Constant and equals to $\dfrac{3}{4}\sqrt{2gh}$
H. Variable with $\left({{v}_{cm}}_{max}\right)=\sqrt{2gh}$
I. Constant and equals to $\dfrac{1}{2}\sqrt{2gh}$
J. Variable with \(\left({{v}_{cm}}_{max}\right)=\sqrt{\dfrac{gh}{2}} \)
Answer: C
A particle starts from a point $P$ at a distance of $A/2$ from the mean position $O$ & travels towards left as shown in the figure. If the time period of SHM, executed about $O$ is $T$ and amplitude $A$ then the equation of motion of particle is :

Options:
A. $$x = A \sin\left ( \dfrac{2\pi }{T} t+\dfrac{7\pi }{6} \right )$$
B. $$x = A \sin\left ( \dfrac{2\pi }{T} t+\dfrac{11\pi }{6} \right )$$
C. $$x = A \sin\left ( \dfrac{2\pi }{T} t+\dfrac{5\pi }{6}\right )$$
D. $$x = A \sin\left ( \dfrac{2\pi }{T} t+\dfrac{\pi }{3} \right )$$
E. $$x = A \cos\left ( \dfrac{2\pi }{T} t+\dfrac{\pi }{6}\right )$$
F. $$x = A \sin\left ( \dfrac{2\pi }{T} t+\dfrac{\pi }{6}\right )$$
G. $$x = A \cos\left ( \dfrac{2\pi }{T} t+\dfrac{5\pi }{6} \right )$$
H. $$x = A \cos\left ( \dfrac{2\pi }{T} t+\dfrac{7\pi }{6} \right )$$
Answer: C
A perfect gas contained in a cylinder is kept in vacuum. If the cylinder suddenly bursts, then the temperature of the gas

Options:
A. $$$W = 0, Q > 0, $ and $\bigtriangleup E_{int} = Q$$$
B. $$$W > 0, Q >0$ and $\bigtriangleup E_{int} = 0$$$
C. W > 0, Q = 0 and \bigtriangleup E_{int} = W
D. $$$Q = 0, W > 0$ and $ \bigtriangleup E_{int} = - W$$$
E. $$$W > 0, Q > 0$ and $\bigtriangleup E_{int} = W$$$
F. $$$W > 0, Q = 0$ and $\bigtriangleup E_{int} = 0$$$
G. $$$Q = 0, W > 0$ and $ \bigtriangleup E_{int} = 0$$$
H. $$$W = 0, Q > 0$ and $\bigtriangleup E_{int} = 0$$$
I. $$$W > 0, Q = 0$ and $\bigtriangleup E_{int} = -W$$$
J. $$$W > 0, Q > 0$ and $\bigtriangleup E_{int} = Q$$$
Answer: B
The reduction of nitric oxide gas by hydrogen gas as represented below:$2NO+2H_{2}\rightarrow N_{2}+2H_{2}O$takes place in the sequence of following steps:$I.NO+NO \rightleftharpoons N_{2}O_{2}\qquad (fast\  and\ reversible)$$II.N_{2}O_{2}+H_{2} \rightleftharpoons N_{2}O+H_{2}O\qquad (slow)$$I.NO+NO \rightleftharpoons N_{2}O_{2}\qquad (fast\  and\ reversible)$$III.N_{2}O+H_{2} \rightleftharpoons N_{2}+H_{2}-O\qquad (fast)$The reaction follows the rate law$+\dfrac {d[N_{2}]}{dt}=k[NO]^{2}[H_{2}]_{p}$If $k=6.0\ \times 10^{-4}\ mole^{-2}L^{2}min^{-1}$ and $k_{2}$ the rate constant of Step $II$ be be $3.0\ \times 10^{-2} mole^{-1}L min^{-1}$, the numerical value of the equilibrium constant$(K_{c})$ for the step $I$ is?

Options:
A. $$3 \times 10^{-6}$$
B. $$2\ \times 10^{-2}$$
C. $$7 \times 10^{-7}$$
D. $$1.8\ \times 10^{-8}$$
E. $$None\ of\ these$$
F. $$4 \times 10^{-7}$$
G. $$3 \times 10^{-7}$$
H. $$4 \times 10^{-6}$$
I. $$5\ \times 10^{-7}$$
J. $$8 \times 10^{-7}$$
Answer: I
A system shown in figure, consists of a massless pulley, a spring of force constant $k$ and a block of mass $m$. If block is just slightly displaced vertically down from its equilibrium position and released then the period of vertical oscillations is:

Options:
A. T=2\pi\sqrt{\left(\dfrac{m}{7k} \right)}
B. $$T=\pi \sqrt{\left(\dfrac{m}{4k}\right)}$$
C. $$T=2\pi\sqrt{\left(\dfrac{m}{2k}\right)}$$
D. $$T=2\pi\sqrt{\left(\dfrac{m}{10k} \right)}$$
E. $$T=2\pi\sqrt{\left(\dfrac{m}{9k} \right)}$$
F. $$T=2\pi\sqrt{\left(\dfrac{m}{8k} \right)}$$
G. $$T=2\pi\sqrt{\left(\dfrac{m}{3k}\right)}$$
H. $$T=2\pi\sqrt{\left(\dfrac{m}{4k}\right)}$$
I. $$T=2\pi\sqrt{\left(\dfrac{m}{5k} \right)}$$
J. $$T=2\pi\sqrt{\left(\dfrac{m}{6k} \right)}$$
Answer: H
Assume that the terrestrial magnetism is due to a short magnet at the centre of the earth. Refer to a point on the surface of the earth. Given : magnetic latitude of the reference point = $\lambda, \delta $ = angle of dip and R = radius of earth. Then, the value of vertical component V of the earth's field is

Options:
A. $$2 sin \lambda / R^3$$
B. $$2 tan \delta$$
C. $$tan \delta$$
D. $$\dfrac{2 M cos \lambda}{R^2}$$
E. $$\dfrac{2 M sin \lambda}{R^2}$$
F. $$2 cos \lambda / R^3$$
G. $$\dfrac{2 M sin \lambda}{R^3}$$
H. $$tan \lambda$$
I. $$2 M cos \lambda / R^3$$
Answer: H
Which is not the correct relation between different current gain $( lpha , \beta , \gamma )$  of transistor amplifier :-

Options:
A. $$\alpha = \frac { \beta } { \beta + 1 }$$
B. \beta = lpha + 1
C. \beta = \frac{\alpha + 1}{\alpha - 1}
D. \beta = \frac{\alpha}{1 + \alpha}
E. \beta = \frac{\alpha + 1}{\alpha}
F. $$\alpha = \frac { \beta } { \gamma }$$
G. $$\beta = \frac {\alpha } { 1 - \alpha }$$
H. \beta = \frac{\alpha}{\alpha - 1}
I. $$\beta = \gamma + 1$$
J. \beta = \frac{\alpha}{1 - \alpha - 1}
Answer: G
in a certain region of free space a non-uniform electric field which depends on x-coordinates is given by $ec E = E_0 x \hat i$. What is the total amount of electric potential energy contained within the cube of sides of length L as shown below?

Options:
A. $$ \dfrac{1}{2} \varepsilon_0 E_0^2 L^5 $$
B. $$\dfrac{1}{7} \epsilon_0 E_0^2 L^6$$
C. $$\dfrac{1}{4} \epsilon_0 E_0^2 L^5$$
D. $$\dfrac{1}{4} \epsilon_0 E_0^2 L^6$$
E. $$ \dfrac{1}{5} \varepsilon_0 E_0^2 L^5 $$
F. $$\dfrac{1}{3} \epsilon_0 E_0^2 L^6$$
G. $$\dfrac{1}{8} \epsilon_0 E_0^2 L^5$$
H. $$ \dfrac{1}{6} \varepsilon_0 E_0^2 L^6 $$
I. $$ \dfrac{1}{5} \varepsilon_0 E_0^2 L^6 $$
J. $$\dfrac{1}{3} \epsilon_0 E_0^2 L^5$$
Answer: A
When the air temperature is $-\theta^{\circ}C$, the ice on a large shallow lake increases by $x\ m$ in $t$ seconds. The thermal conductivity of ice is:
(Density of ice $= \rho kg\ m^{-3}$ and $L =$ Latent heat in $J\ k$).

Options:
A. $$\dfrac{5\rho Lx^{2}}{6\theta t}$$
B. \dfrac{\rho Lx^{2}}{3\theta t}
C. $$\dfrac{7\rho Lx^{2}}{8\theta t}$$
D. $$\dfrac{3\rho Lx^{2}}{2\theta t}$$
E. $$\dfrac {\rho Lx^{2}}{\theta t}$$
F. $$\dfrac {\rho Lx^{2}}{2\theta t}$$
G. $$\dfrac {2\rho Lx^{2}}{3\theta t}$$
H. $$\dfrac {3\rho Lx^{2}}{4\theta t}$$
I. $$\dfrac {\rho Lx^{2}}{4\theta t}$$
J. $$\dfrac {2\rho Lx^{2}}{\theta t}$$
Answer: F
The maximum speed with which a car can cross a convex bridge over a river with a radius of curvature 9.2 m is (centre of gravity of the car is 1 m above the bridge and g = 9.8 $m/s^2$)

Options:
A. 25 m/s
B. 30 m/s
C. 10 m/s
D. 48 m/s
E. 40 m/s
F. 45 m/s
G. 47 m/s
H. 20 m/s
I. 55 m/s
J. 35 m/s
Answer: A
How many turns of a nichrome wire 0.5 mm in diameter should be wound around a porcelain cylinder with radius 2.5 cm to obtain a furnace with a resistance of 20 ohm?
(p for nichrome $[=10^{-6}]$ cm)

Options:
A. 110
B. 120
C. 130
D. 200
E. 90
F. 140
G. 150
H. 50
I. 160
J. 100
Answer: J
A system is taken from state A to state B along two different paths $1$ and $2$ $.$ The work done on the system along these two paths are ${W_1}$ and ${W_2}$ respectively$.$ The heat absorbed by the system along these two paths are ${Q_1}$ and ${Q_2}$ respectively. The internal energies at A and B are ${U_A}$ and ${U_B}$ respectively. Then,

Options:
A. $${W_1} = {W_2} = {U_B} - {U_A}$$
B. $${Q_1} + ( - {W_1}) = {Q_2} + ({W_2}) = {U_B} - {U_A}$$
C. $${Q_1} + {W_1} = {Q_2} + {W_2} = {U_B} - {U_A}$$
D. $$${W_1} = {W_2} = {U_A} - {U_B}$$$
E. $${W_1} = {W_2} = {U_A} + {U_B}$$
F. $${W_1} = {W_2} = {U_B} + {U_A}$$
G. $${Q_1} = {Q_2} = {U_A} - {U_B}$$
H. $${Q_1} + {W_1} = {Q_2} + {W_2} = {U_A} + {U_B}$$
Answer: A
At 298 K the standard free energy of formation of H$_2$O$(l) $ is -237.20 kJ/mole while that of its ionisation into H$^+$ ion and hydroxyl ions is 80 kJ/mole, then the emf of the following cell at 298 K will be: (take F = 96500 C)$H_2 (g , 1 bar) | H^+ (1 M) || OH^- (1M) | O_2 (g, 1 bar)$

Options:
A. $$0.41 V$$
B. $$1.21 V$$
C. $$0.42 V$$
D. $$0.40 V$$
E. $$0.81 V$$
F. $$0.82 V$$
G. $$1.23 V$$
H. $$1.20 V$$
I. $$-0.40 V$$
Answer: D
If the velocity of a moving particle is $v\propto x^n$ where x is the displacement, then:

Options:
A. $$n = \dfrac{2}{5}$$
B. $$n = -\dfrac{1}{2}$$
C. $$n < \dfrac{1}{2}$$
D. $$n > \dfrac{1}{2}$$
E. $$n = \dfrac{1}{4}$$
F. $$n = \dfrac{3}{4}$$
G. (a) and (b) are correct
H. $$n = \dfrac{1}{3}$$
I. $$n = \dfrac{1}{2}$$
J. When $x=0$, the velocity and acceleration are zero
Answer: C
White light may be considered to be a mixture of waves with $\lambda$ ranging between $3900$ $\overset{o}{A}$ and $7800$ $\overset{o}{A}$. An oil film thickness $10,000$ $\overset{o}{A}$ is examined normally by the reflected light. If $\mu =1.4$, then the film appears bright for.

Options:
A. $4000$ $\overset{o}{A}$, $5091$ $\overset{o}{A}$, $5600$ $\overset{o}{A}$
B. $4000$ $\overset{o}{A}$, $4667$ $\overset{o}{A}$, $5600$ $\overset{o}{A}$
C. $$4000 \overset{o}{A}, 5600 \overset{o}{A}, 7000 \overset{o}{A}$$
D. $$4308 \overset{o}{A}, 5600 \overset{o}{A}, 6222 \overset{o}{A}$$
E. $4308$ $\overset{o}{A}$, $5091$ $\overset{o}{A}$, $6222\overset{o}{A}$
F. $4667$ $\overset{o}{A}$, $6222$ $\overset{o}{A}$, $7000$ $\overset{o}{A}$
G. $$4667 \overset{o}{A}, 5385 \overset{o}{A}, 6222 \overset{o}{A}$$
H. $$4000 \overset{o}{A}, 5091 \overset{o}{A}, 6000 \overset{o}{A}$$
I. $$4308 \overset{o}{A}, 5385 \overset{o}{A}, 6000 \overset{o}{A}$$
J. $$4667 \overset{o}{A}, 5600 \overset{o}{A}, 6222 \overset{o}{A}$$
Answer: E
An engine of mass 20 tons pulls a train of 20 wagons, each of mass 20 tons with constant velocity of $ 72 km/hr $ on a level track. If the coefficient of kinetic friction is $  \mu k = 0.01 $, find the power developed by the engine.

Options:
A. 440 KW
B. $$ 1000 KW $$
C. 500 KW
D. 800 KW
E. 400 KW
F. 480 KW
G. $$ 100 KW $$
H. $$ 840 KW $$
I. $$ 420 KW $$
Answer: I
Two different coils have self inductance $L_{1}=8\ mH, L_{2}=2\ mH$. The current in the second coil is also increased at the same constant rate. The current in the second coil is also increased at the same constant rate. At a certain instant of time, the power given to the two coil is the same. At that time, the current, the induced voltage and the energy stored in the first coil are $i_{1}, V_{1}$ and $W_{1}$ respectively. Corresponding values for the second coil at the same instant are $i_{2}, V_{2}$ and $W_{2}$ respectively. Then

Options:
A. \dfrac{V_{2}}{V_{1}}=\dfrac{1}{3}
B. $$\dfrac{i_{1}}{i_{2}}=4$$
C. \dfrac{W_{2}}{W_{1}}=16
D. \dfrac{V_{2}}{V_{1}}=\dfrac{1}{2}
E. \dfrac{V_{2}}{V_{1}}=8
F. $$\dfrac{W_{2}}{W_{1}}=8$$
G. $$\dfrac{i_{1}}{i_{2}}=\dfrac{1}{4}$$
H. \dfrac{V_{2}}{V_{1}}=2
I. \dfrac{V_{2}}{V_{1}}=4
J. $$\dfrac{W_{2}}{W_{1}}=4$$
Answer: F
Using the Gibbs energy change, $\triangle G^o=+63.3kJ$, for the following reaction, $Ag_2CO_2(s)\rightleftharpoons 2Ag^+(aq)+CO_3^{2-}(aq)$
$K_{sp}$ of the $Ag_2CO_3(s)$ in water at $25^oC$ is 
$$(R=8.314JK^{-1}mol^{-1})$$

Options:
A. $$8.0\times 10^{-12}$$
B. $$3.2 \times 10^{-24}$$
C. $$3.2 \times 10^{-23}$$
D. $$2.9\times 10^{-3}$$
E. $$3.2 \times 10^{-29}$$
F. $$7.9\times 10^{-2}$$
G. $$3.2 \times 10^{-25}$$
H. $$3.2 \times 10^{-27}$$
I. $$3.2 \times 10^{-28}$$
J. $$3.2\times 10^{-26}$$
Answer: J
In qualitative analysis, cations of group II, as well as group IV, precipitated in the form of sulphides. Due to the low value of $K_{sp}$ of group II sulphides, the group reagent is $H_2S$ in presence of dil. $HCl$ and due to high value of $K_{sp}$ of group IV sulphides, group reagent is $H_2S$ in presence of $NH_4OH$ and $NH_4Cl$. In 0.1 M $H_2S$ solution, $Sn^{2+}, Cd^{2+}$ and $Ni^{2+}$ ions are present in equimolar concentration 0.1 M. At what $pH$ precipitate of $NiS$ will form?Given: $Ka_1(H_2S) = 10^{-7},$ $Ka_2(H_2S) = 10^{-14},$ $K_{sp}(SnS) = 8 \times 10^{-29}$, $K_(sp) (CdS) = 10^{-28},$ $K_{sp}(NiS) = 3 \times 10^{-21}$

Options:
A. $$1.42$$
B. $$1.24$$
C. $$0.74$$
D. $$10.76$$
E. $$8.76$$
F. $$5.24$$
G. $$2.76$$
H. $$4.24$$
I. $$7$$
J. $$4$$
Answer: B
A fish is vertically below a flying bird moving vertically down toward water surface. The bird will appear to the fish to be

Options:
A. Moving slower than its real speed and farther than its real distance
B. Moving slower than its real speed and away from the real distance
C. Moving faster than its real speed and at the same distance as its real distance
D. Moving faster than its real speed and at a different distance than its real distance
E. Moving faster than its real speed and closer than its real distance
F. Moving faster than its speed and at the same distance as its real distance
G. Moving slower than its real speed and also nearer than its real distance
H. Moving faster than its real speed and nearer than its real distance
I. Moving faster than its speed and also away from the real distance
Answer: I
For the electrochemical cell, $(M | {M}^{+})$ || $({X}^{-} | X)$, ${E}^{o} ({M}^{+} | M)$ = 0.44 V and ${E}^{o} ({X} | {X}^{-})$ = 0.33V. From this data one can deduce:

Options:
A. ${E}_{cell}$ = 0.34 V
B. ${E}_{cell}$ = -0.77 V
C. ${E}_{cell}$ = 0.33 V
D. ${M}^{+} + {X}^{-} \rightarrow M + X$ is a spontaneous reaction
E. ${E}_{cell}$ = -0.34 V
F. ${E}_{cell}$ = 0.11 V
G. ${E}_{cell}$ = 0.77 V
H. ${E}_{cell}$ = -0.11 V
I. $M + X \rightarrow {M}^{+} + {X}^{-}$ is a spontaneous reaction
Answer: G
An inclined plane making an angle ${ 30 }^{ o }$ with the horizontal is placed in a uniform horizontal electric field of $100 V { m }^{ -1 }$. A particle of mass $1 kg$ and charge $0.01 C$ is allowed to slide down from rest from the top of the inclined plane. If the coefficient of friction is $0.2$, the particle reaches the bottom of the inclined plane in $1$ second. Then the length of the inclined plane is (Acceleration due to gravity $= 10 { ms }^{ -2 }$)

Options:
A. $$1.150 m$$
B. 1.322 m
C. $$1.151 m$$
D. $$1.323 m$$
E. 1.175 m
F. 1.160 m
G. 1.171 m
H. $$1.172 m$$
I. 1.152 m
J. 1.155 m
Answer: A
A particle of mass $\dfrac{2}{3}kg$ with velocity $c = -15m/s$ at $t = 2s$ is acted up on by a force $f = k - \beta t^2$. Here $k = 8N$ and $\beta = 2 N/s^2$. The motion is one dimensional. Then the speed at which the particle acceleration is zero again, is :

Options:
A. $$20 m/s$$
B. $$32 m/s$$
C. $$18 m/s$$
D. $$14 m/s$$
E. $$1 m/s$$
F. $$15 m/s$$
G. $$16 m/s$$
H. $$19 m/s$$
I. $$17 m/s$$
J. $$12 m/s$$
Answer: I
How many millimoles of NaOH should be added to 1L of 0.1 M $FeCl_3$ solution to just start the precipitation of $Fe(OH)_3$ ? $K_{sp}[Fe(OH)_3] = 8 \times 10^{-13}$

Options:
A. 1.2
B. 0.2
C. 0.6
D. 1.5
E. 0.3
F. 1.6
G. 1.8
H. 0.1
I. 4
J. 0.4
Answer: C
A small particle slides without friction from the top of a  hemispherical bowl of radius R as shown.If it leaves the surface of the cup at a vertical distance h below the highest point, then.

Options:
A. h=3R/4
B. h=2R/3
C. h=R
D. h=4R/5
E. h=R/3
F. h=7R/6
G. h=R/2
H. h=R/4
I. h=5R/3
J. h=5R/6
Answer: C
A car $P$ is moving with a uniform speed of $5\sqrt{3} m/s$ towards a carriage of mass $9\,kg$ at rest kept on the rails at a point $B$ as shown in figure. The height $AC$ is $120 \,m$. Cannon balls of $1 \,kg$ are fired from the car with and initial velocity $100 \,m.s$ at an angle $30^o$ with the horizontal, The first cannon ball hits the stationary carriage after time $t_0$ and sticks to it. Determine $t_0$. At $t_0$ the second cannon ball is fired. Assume that the resistive force between the rails and the carriage is constant and ignore the vertical motion of the carriage throughout. If the second ball also hits and sticks to the carriage, what will be the horizontal velocity of the carriage just after the second impact?

Options:
A. $$t = 11s, v = 100\sqrt{3}/10 \,m/s just after the second impacts.$$
B. $$t = 11s, v = 100\sqrt{3}/11 \,m/s just after the second impacts.$$
C. $$t = 12s, v = 100\sqrt{3}/12 \,m/s just after the second impacts.$$
D. $$$t = 12s$, $v = 100\sqrt{3}/11 \,m/s$ just after the second impacts.$$
E. $t = 124s$, $v = 100\sqrt{3}/11 \,m/s$ just after the second impacts.$
F. $$t = 11s, v = 100\sqrt{3}/13 \,m/s just after the second impacts.$$
G. $$t = 11s, v = 100\sqrt{3}/12 \,m/s just after the second impacts.$$
H. $$$t = 13s$, $v = 100\sqrt{3}/11 \,m/s$ just after the second impacts.$$
I. $$$t = 22s$, $v = 100\sqrt{3}/11 \,m/s$ just after the second impacts.$$
Answer: D
One end of rod of length L=1 m is fixed to a point on the circumference pf a wheel of radius $R = 1/\sqrt{3}$ m. The othrt end is sliding freely along a straight channel passing through the centre O of the wheel as shown in the figure below. The wheel is rotating with a constant velocity $\omega$ about O.
The speed of the sliding end P when $\theta = 60^o$ is

Options:
A. $$\dfrac{2\omega}{\sqrt{3}}$$
B. $$\dfrac{3\omega}{2\sqrt{3}}$$
C. $$\dfrac{\omega}{2\sqrt{3}}$$
D. $$\dfrac{\omega}{\sqrt{2}}$$
E. $$\dfrac{\omega}{\sqrt{6}}$$
F. $$\dfrac{\omega}{3}$$
G. $$\dfrac{3\omega}{\sqrt{3}}$$
H. $$\dfrac{2\omega}{3}$$
I. $$\dfrac{4\omega}{\sqrt{3}}$$
J. $$\dfrac{\omega}{\sqrt{3}}$$
Answer: H
In order to prepare a buffer of $P^H$ 8.26, the amount of $(NH_4)_2SO_4$ required to be mixed with 1L of 0.1 M $NH_3(pK_b=4.74)$ is:

Options:
A. 1 mol
B. 1.5 mol
C. 0.5 mol
D. 0.3 mol
E. 2 mol
F. 5 mol
G. 0.4 mol
H. 0.2 mol
I. 0.1 mol
J. 10 mol
Answer: C
A short bar magnet with the north pole facing north forms a neutral point at P in the horizontal plane. If the magnet is rotated by $90^o$ in the horizontal plane, the net magnetic induction at $P$ is ( Horizontal component of earth's magnetic field $= B_H$):

Options:
A. $$\sqrt{5}B_H$$
B. $$\dfrac{\sqrt{5}}{8} B_H$$
C. $$\dfrac{\sqrt{5}}{3} B_H$$
D. $$\dfrac{3\sqrt{5}}{4} B_H$$
E. $$2 B_H$$
F. $$\displaystyle \dfrac{\sqrt{5}}{2} B_H$$
G. $$\dfrac{3\sqrt{5}}{8} B_H$$
H. $$\dfrac{\sqrt{5}}{4} B_H$$
I. $$\dfrac{\sqrt{5}}{2} B_H$$
J. $$zero$$
Answer: A
A steady current I goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x and QR = 5x. If the magnitude of the magnetic field at P due to this loop is k$\left (\dfrac{\mu_0I}{48\pi x} \right )$ find the value of k.

Options:
A. 13
B. 14
C. 11
D. 9
E. 10
F. 7
G. 5
H. 8
I. 12
J. 6
Answer: F
What is the equilibrium constant of the reaction: $2Fe^{3+} + Au^{+}\rightarrow 2Fe^{2+}+Au^{3+}?$ 
Given $ E_{Au^{+}|Au}^{0} = 1.68 V,E_{Au^{3+}|Au}^{0} = 1.50 V,$
$ E_{Fe^{3+}|Fe^{2+}}^{0} = 0.75V $ and $ 2.303RT/F = 0.06.$

Options:
A. $$1 \times 10^{-32}$$
B. $$ 1 \times 10^{-22} $$
C. $$1 \times 10^{-24}$$
D. $$1 \times 10^{-23}$$
E. $$ 1 \times 10^{-11} $$
F. $$1 \times 10^{-20}$$
G. $$ 1 \times 10^{22} $$
H. $$ 1 \times 10^{-72}$$
I. $$1 \times 10^{-25}$$
J. $$1 \times 10^{-21}$$
Answer: B
Find the capacitance of a system of two identical metal balls of radius $  \cdot a  $ " if the distance between their centres is equal to $b^{*},  $ with $  b>>a $. The system is located in a uniform dielectric with relative permittivity K

Options:
A. $$C = 2 \pi \varepsilon_0 \mathrm{E} a$$
B. $$C = \pi a \varepsilon_0 \varepsilon a$$
C. $$C = 3 \pi \varepsilon_0 \varepsilon \mathrm{a}$$
D. $$\mathrm{C} = 5 \pi \varepsilon_0 \mathrm{E} a$$
E. $$C = 2 \pi \varepsilon_0 \varepsilon a$$
F. $$C = 4 \pi \varepsilon_0 \varepsilon \mathrm{a}$$
G. $$C = 2 \pi \varepsilon_0 \varepsilon \mathrm{b}$$
H. $$C = 2 \pi \varepsilon_0 \varepsilon \mathrm{a}^2$$
I. $$C = \frac{2 \pi \varepsilon_0 \varepsilon}{a}$$
J. $$C = 3 \pi \varepsilon_0 \varepsilon a$$
Answer: B
A body of area $ 0.8 \times 10^{-2} m{-2}$ and mass $ 5 \times 10^{-4}kg$ directly faces the sun on a clear day. The body has an emissivity of 0.8 and specific heat of 0.8 cal/kg K. The surroundings are at $27^{0}C$. (solarconstant = 1.4 kW/$m^{2}$).     The temperature that the body would reach if it lost all its heat by radiation is

Options:
A. $$196^{0}C$$
B. $$85^{0}C$$
C. $$85 K$$
D. 95^{0}C
E. 185^{0}C
F. $$186^{0}C$$
G. $$87^{0}C$$
H. 396 K
I. $$96^{0}C$$
J. $$296^{0}C$$
Answer: B
A metallic ring is held horizontally and a magnet is allowed to fall vertically through it with N-pole pointing upwards. The acceleration of a magnet near the ring is $a$. Then:

Options:
A. $a g$ while approaching.
B. $$a=g$$
C. $a > g$ while approaching and $a < g$ while receding.
D. None.
E. $a > g$ while approaching and $a = g$ while receding.
F. $a < g$ while approaching as well as receding.
G. $a < g$ while approaching and $a = g$ while receding.
H. $a = g$ while approaching and $a = g$ while receding.
I. $a > g$ while approaching and $a > g$ while receding.
Answer: F
A sphere is moving on a smooth surface with linear speed $v_o$ and angular velocity $\omega_0$. It finds a rough inclined surface and it starts climbing up:

Options:
A. if $v_0 < R\omega_0$, no friction force will act
B. if $v_0 > R\omega_0$, friction force will act upwards
C. if $v_0< R\omega_0$, friction force will act upwards
D. if $v_0 = R\omega_0$, friction force will act upwards
E. if $v_0 = R\omega_0$, friction force will act downwards
F. if $v_0 = R\omega_0$, no friction force will act
G. if all above case friction will act upwards
H. if $v_0 < R\omega_0$, friction force will act downwards
I. if $v_0 > R\omega_0$, friction force will act downwards
J. if $v_0=R\omega_0$, no friction force will act upwards
Answer: I
The ends of a stretched string of length $L$ are fixed at $x=0$ and $x=L$. In one experiment, the displacement of the wire is $y_{1}=2A\sin\left(\dfrac{\pi x}{L}\right)\sin\omega t$ and energy $E_1$ and in another experiment, its displacement is $y_2 = A\sin\left({\displaystyle\frac{2\pi x}{L}}\right)\sin{2\omega t}$ and energy $E_2$ then

Options:
A. $$E_2 = 6E_1$$
B. $$E_2 = 2E_1$$
C. $$E_2 = 5E_1$$
D. $$ 16E_1$$
E. $$E_2 = 8E_1$$
F. $$E_2 = 7E_1$$
G. $$E_2 = E_1$$
H. $$E_2 = 3E_1$$
Answer: G
What is the total entropy change $(\Delta S_{system}+\Delta S_{surrounding})$ for the following process $4Fe(s)+3O_{2}(g)\rightarrow 2Fe_{2}O_{3}(s)?$
Given : $\Delta S_{reaction}^{0}=-549.4 J/K$
$$\Delta H_{reaction}^{0}= 1648 kJ/mol$$

Options:
A. 5450 $JK^{-1}mol^{-1}$
B. 5530 $JK^{-1}mol^{-1}$
C. 5540 $JK^{-1}mol^{-1}$
D. 5525 $JK^{-1}mol^{-1}$
E. 4980.6 $JK^{-1}mol^{-1}$
F. 6069 $JK^{-1}mol^{-1}$
G. 5550 $JK^{-1}mol^{-1}$
H. 5520 $JK^{-1}mol^{-1}$
I. 5510 $JK^{-1}mol^{-1}$
J. 5907 $JK^{-1}mol^{-1}$
Answer: B
How many grams of $ H_{2}S $ will react with $6.32 g$ $ KMnO_{4} $ to produce $ K_{2}SO_{4} $ and $ MnO_{2} $? $(K = 39, Mn = 55)$

Options:
A. $$4.08 g$$
B. 1.04 g
C. $$0.85 g$$
D. 0.61 g
E. 1.02 g
F. 1.03 g
G. $$0.51 g$$
H. $$2.04 g$$
I. 0.76 g
J. 1.53 g
Answer: G
Two masses $m_1=2kg$ and $m_2=5 kg$ are moving on a frictionless surface with velocities 10 m/s and 3 m/s respectively. $m_2$ is ahead of $m_1$. An ideal spring of spring constant $k=1120 N/m$ is attached on the back side of $m_2$.The maximum compression of the spring will be

Options:
A. $$0.37 m$$
B. $$0.062 m$$
C. 0.42 m
D. 0.36 m
E. 0.38 m
F. 0.39 m
G. 0.41 m
H. $$0.72 m$$
I. $$0.51 m$$
Answer: A
The order of reactivity of ammonia with the following compound is :
(I) $CH_{2}=CHBr$
(II) $CH_{3}-CH_{2}-COCl$
(III) $CH_{3}-CH_{2}-CH_{2}-Cl$
(IV) $(CH_{3})_{3}C-Br$

Options:
A. $$I>IV>II>III$$
B. $$II>IV>III>I$$
C. IV>III>II>IV
D. II>IV>I>III
E. $$II>I>III>IV$$
F. $$IV>II>I>III$$
G. $$IV>III>II>I$$
H. II>III>I>IV
I. $$IV>III>I>II$$
J. $$III>IV>II>I$$
Answer: F
A block of mass $m_1$ undergoes a completely inelastic collision with an initially stationary block of mass $m_2$. Let K be the kinetic energy transferred to $m_2$ after collision and $K_1$ be the initial kinetic energy. Consider all motion to be in one dimension. The ratio $K/K_1$

Options:
A. has a maximum value of 1/4
B. attains maximum value when $m_2/m_1 = 1$
C. has a maximum value of 1/7
D. has a maximum value of 1/3
E. has a maximum value of 1/5
F. is constant for all values of $m_2/m_1 $
G. has a maximum value of 1/8
H. decreases for $m_2/m_1 > 1$
I. has a maximum value of 1/2
J. has a maximum value of 1/6
Answer: A
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C  remains constant. If during this process the relation of pressure p and volume V is given by $pV^n = constant$, then n is given by (Here, $C_p$ and $C_V$ are molar specific heat at constant pressure and constant volume, repsectively)

Options:
A. $$n = \dfrac{C - C_V}{C - C_p}$$
B. $$n = \dfrac{C+p}{C_V}$$
C. $$n = \dfrac{C - C_V}{C_p - C}$$
D. $$n = \dfrac{C_p - C}{C - C_V}$$
E. $$n = \dfrac{C_p - C_V}{C - C_V}$$
F. $$n = \dfrac{C_p - C_V}{C - C_p}$$
G. $$n = \dfrac{C_p - C}{C_V - C}$$
H. $$n = \dfrac{C - C_p}{C - C_V}$$
I. $$n = \dfrac{C_p - C_V}{C_V - C}$$
Answer: H
A charged spherical drop of mercury is in equilibrium in a plane horizontal air capacitor and the intensity of the electric field is $6\times 10^4 \;Vm^{-1}$. If the charge on the drop is $8\times 10^{-18} \;C$, the radius of the drop is :$\left [\rho_{air} = 1.29 \;kg/m^3 ; \rho_{Hg} = 13.6\times 10^3 \;kg/m^3 \right ]$

Options:
A. $$1.90 \times 10^{-8} \;m$$
B. $$2.7 \times 10^{-10} \;m$$
C. $$1.90 \times 10^{-10} \;m$$
D. $$0.95\times 10^{-8} \;m$$
E. $$0.95\times 10^{-6} \;m$$
F. $$3.80 \times 10^{-8} \;m$$
G. $$1.90 \times 10^{-6} \;m$$
H. $$2.7\times 10^{-8} \;m$$
I. $$1.35 \times 10^{-10} \;m$$
Answer: E
Two point charges ${ Q }_{ 1 }=2\mu C$ and ${ Q }_{ 2 }=1\mu C$ are placed as shown. The coordinates of the point $P$ are $\left( 2cm,1cm \right) $. The electric intensity vector at $P$ subtends an angle $\theta $ with the positive $X$ axis. The value of $\theta $ is given by:

Options:
A. $$\tan { \theta } =3$$
B. $$\tan { \theta } = 2.25$$
C. $$\tan { \theta } = 1.85$$
D. $$tan { \theta } = 2.5$$
E. \( \tan { \theta } = 1.75 \)
F. $$\tan { \theta } = 2.75$$
G. $$\tan { \theta } = 1.5$$
H. $$\tan { \theta } =1$$
I. $$\tan { \theta } =2$$
J. $$\tan { \theta } =4$$
Answer: I
During the adiabatic change of ideal gas, the relation between the pressure ad the density will be-

Options:
A. $$\dfrac{P_1}{P_2}=(\dfrac{d_2}{d_1})^\dfrac{y}{1}$$
B. \(\dfrac{P_1}{P_2}=(\dfrac{d_1}{d_2})^\dfrac{y}{1} \)
C. $$\dfrac{P_1}{P_2}=(\dfrac{d_2}{d_1})^\dfrac{1}{y}$$
D. \(\dfrac{P_1}{P_2}=(\dfrac{d_1}{d_2})^\dfrac{2}{y}\)
E. $$\dfrac{P_1}{P_2}=(\dfrac{d_1}{d_2})^\dfrac{1}{y}$$
F. $$P_1d^{-y}_1=P_2d^{-y}_2$$
G. $$\dfrac{P_1}{P_2}=(\dfrac{d_2}{d_1})^\dfrac{1}{y^2}$$
H. $$P_1d^y_1=P_2d^y_2$$
I. $$\dfrac{P_1}{P_2}=(\dfrac{d_1}{d_2})^\dfrac{1}{y^2}$$
J. $$\dfrac{P_1}{P_2}=(\dfrac{d_2}{d_1})^\dfrac{y}{2}$$
Answer: C
The time period of a simple pendulum in a stationary train is T. The time period of a mass attached to a spring is also T. The train accelerates at the rate $5m/s^2$. If the new time periods of the pendulum and spring be $T_p$ and $T_s$ respectively

Options:
A. $$T_p > T_s$$
B. $$T_p = \frac{T_s}{2}$$
C. $$T_p = T_s$$
D. $$cannot be predicated$$
E. $$T_p \neq 2T_s$$
F. $$T_p = \frac{T_s}{3}$$
G. $$T_p < T_s$$
H. $$T_p = \frac{2T_s}{3}$$
I. $$T_p = \frac{3T_s}{2}$$
J. $$T_p \neq T_s$$
Answer: C
A buffer solution contains $500$ mL of $0.2\,M\,CH_3COONa$ and $500$mL of $0.1\,M\, CH_3COOH$.$1$L of water is added to this added to this buffer . pH before and after dilution:(pKa of $CH_3COOH=4.75$)

Options:
A. 5.04, 4.04
B. 5.04, 4.24
C. $$5.04,3.74$$
D. 5.04, 4.59
E. 5.04, 5.04
F. $$5.04, 4.89$$
G. $$9.56, 9.56$$
H. 5.04, 3.54
I. 5.04, 4.49
Answer: C
The heat of ionisation of formic acid is $1.5$ $kJ/mol$. $9.2$ $g$ formic acid on reaction with $7$ $g$ ammonium hydroxide gives $10.8$ $kJ$ of heat. Calculate the heat of ionisation of ammonium hydroxide. ($1$$cal$=$4.2$$J$)

Options:
A. $-160$ $kJ/mol$
B. $-200$ $kJ/mol$
C. -280 kJ/mol
D. $-180$ $kJ/mol$
E. $-240$$kJ/mol$
F. -260 kJ/mol
G. -230 kJ/mol
H. -220 kJ/mol
I. -210 kJ/mol
Answer: E
Given : 
Enthalpy of combustion of methane           $\Delta_rH = -890.36 kJ \, mol^{-1}$
Enthalpy of combustion of C (graphite)       $\Delta_rH = -393.51 kJ \, mol^{-1}$
$H_2 (g) + \dfrac{1}{2} O_2 (g) \rightarrow H_2 O (l)$   $\Delta_r H = -285.85 kJ \, mol^{-1}$
Enthalpy of dissociation of $H_2 (g)$       $\Delta_rH = 435.93 kJ \, mol^{-1}$
Enthalpy of sublimation of C (graphite)        $\Delta_r H = 716.68 kJ \, mol^{-1}$
The bond enthalpy of C - H from the following data at 298 K is

Options:
A. $$415.85 \, kJ \, mol^{-1}$$
B. $$414.85 \, kJ \, mol^{-1}$$
C. $$166.339 \, kJ \, mol^{-1}$$
D. $$416.85 \, kJ \, mol^{-1}$$
E. $$413.85 \, kJ \, mol^{-1}$$
F. $$415.35 \, kJ \, mol^{-1}$$
G. $$-415.85 \, kJ \, mol^{-1}$$
H. $$415.34 \, kJ \, mol^{-1}$$
I. $$166.34 \, kJ \, mol^{-1}$$
J. $$1663.39 \, kJ \, mol^{-1}$$
Answer: G
Nonpolar solvents such as hexane, $C_{6}H_{14}$, generally are immiscible with water, thus, when water and hexane are placed together in the same container, they will separate into two layers as shown below. Hexane, $density = 0.69\ g/ mL$, will form the top layer.Some substances are soluble in both solvents, to differing extents. Consider the following organic acids, all of which exhibit some ability to dissolve in both water and in hexane.Ethanoic acid, $CH_{3}COOH$, propanoic acid, $CH_{3}CH_{2}COOH$, butanoic acid, $CH_{3}CH_{2}CH_{2}COOH$, and pentanoic acid, $CH_{3}(CH_{2})_{3}COOH$.A student designs an experiment to determine the relative solubilities of the four acids in hexane and in water. She wants to express her results in the form of a ratio, represented by k, defined as follows.$k = \dfrac {\text {mole concentration of acid in hexane}}{\text {molar concentration of acid in water}}$Of the four organic acids, which is expected to have the GREATEST value of k, for the reason given?

Options:
A. All will behave alike. All four compounds contain a highly polar carboxylic acid group, which is attracted to water, and all have a nonpolar "tail" that is attracted to hexane.
B. Ethanoic acid, $CH_{3}COOH$, because it is the strongest acid.
C. Propanoic acid, $CH_{3}CH_{2}COOH$, because it has the smallest nonpolar portion.
D. Propanoic acid, $CH_{3}CH_{2}COOH$, because it has the smallest polar portion.
E. Pentanoic acid, $CH_{3}(CH_{2})_{3}COOH$, because it is weakest acid of the four.
F. Propanoic acid, $CH_{3}CH_{2}COOH$, because it is the strongest acid.
G. Pentanoic acid, $CH_{3}(CH_{2})_{3}COOH$, because it has the largest nonpolar portion.
H. Ethanoic acid, $CH_{3}COOH$, because it has the smallest nonpolar portion.
I. Pentanoic acid, $CH_{3}(CH_{2})_{3}COOH$, because it is the weakest acid.
J. Propanoic acid, $CH_{3}CH_{2}COOH$, because it has the largest polar portion.
Answer: B
A capacitor of capacity $C_1=2 \mu F$ is charged to a potential of $100V$. The charging battery is then removed and is connected to another uncharged capacitor of capacity $C_2 = 4 \mu F$. One plate of $C_2$ is earthed as shown. In steady state, the charges on $C_1$ and $C_2$ will be

Options:
A. $$\frac{200}{3}\, \mu C,\, \frac{800}{3}\, \mu C$$
B. $$\frac{100}{3}\, \mu C,\, \frac{600}{3}\, \mu C$$
C. $$200\, \mu C,\, 100\, \mu C$$
D. $$\frac{200}{3}\, \mu C,\, \frac{600}{3}\, \mu C$$
E. $$200\, \mu C,\, \rm {zero}$$
F. $$200\, \mu C,\, 200\, \mu C$$
G. $$\rm {zero,\, zero}$$
H. \frac{100}{3}\, \mu C,\, \frac{500}{3}\, \mu C
I. \frac{100}{3}\, \mu C,\, \frac{400}{3}\, \mu C
J. $$\frac {200}{3}\, \mu C,\, \frac {400}{3}\, \mu C$$
Answer: F
In an analytical determination of arsenic, a solution containing arsenious acid, $ H_{3}AsO_{4},$ KI and a small amount of starch is electrolysed. The electrolysis produces free $I_{2}$ from $ I^{-}$ ions and the $ I_{2}$ immediately oxidizes the arsenious acid to hydrogen arsenate ion, $ HAsO_{4}^{2-}.$
$$ I_{2}+H_{3}AsO_{3}+H_{2}O \rightarrow 2I^{-} + HAsO_{4}^{2-}+4H^{+}$$
When the oxidation of arsenic is complete, the free iodine blue colour. If during a particular run, it takes 96.5 s for a current of 10.0 mA to give an end point (indicated by the blue colour),how many g of arsenic are present in the solution. (As = 75)

Options:
A. $$ 6.3 \times 10^{-5}g$$
B. $$1.26 \times 10^{-4}g$$
C. $$6.3 \times 10^{-6}g$$
D. $$ 3.15 \times 10^{-5}g$$
E. $$3.15 \times 10^{-6}g$$
F. $$3.15 \times 10^{-4}g$$
G. $$1.26 \times 10^{-6}g$$
H. $$ 6.3 \times 10^{-4} g$$
I. $$1.26 \times 10^{-5}g$$
Answer: A
For the reaction in equilibrium,
(I) $CO(g)+\dfrac{1}{2}O_2(g)\rightleftharpoons CO_2(g)$
(II) $2CO(g)+O_2(g)\rightleftharpoons 2CO_2(g)$Equilibrium constantFree energy change$K_1$$\Delta G_1$$K_2$$\Delta G_2$Then?

Options:
A. $$K_1=K^2_2, \Delta G_1=2\Delta G_2$$
B. $$K_2=K_1^2, \Delta G_2=\Delta G_1$$
C. $$K_2=2K_1, \Delta G_2=2\Delta G_1$$
D. $$K_2=K^2_1, \Delta G_2=2\Delta G^2_1$$
E. $$K_1=2K_2, \Delta G_1=2\Delta G_2$$
F. $$K_2=2K_1, \Delta G_2=\Delta G_1$$
G. $$K_1=K_2^2, \Delta G_1=2\Delta G_2$$
H. $$K_2=K^2_1, \Delta G_2=2\Delta G_1$$
I. $$K_2=\sqrt{K_1}, \Delta G_2=\Delta G_1$$
J. $$K_1=K_2, \Delta G_1=2\Delta G_2$$
Answer: E
A particle of m = 2kg is moving with constant speed 15 m/s along a curved path $y={ x }^{ 2 }.$ Find its angular momentum about origin O when its velocity vector is at $53^{ \circ  }$ whit x-axis $(in kg m^{ 2 }/s)$

Options:
A. 16
B. 8
C. 14
D. 18
E. 15
F. 12
G. 4
H. 13
I. 11
J. 10
Answer: F
When a polyatomic gas undergoes an adiabatic process, its temperature and volume are related by the equation $TV^n=$ constant, the value of n will be?

Options:
A. 1.44
B. 1.25
C. $$1$$
D. $$1.33$$
E. 1.40
F. 1.20
G. $$2.33$$
H. $$0.33$$
I. 1.50
J. 1.67
Answer: D
A cylinder is rotating about vertical axis, Radius of cylinder is $5cm$. If it is rotating with an angular velocity of $2$ revolution per second, then the height difference between minimum and maximum height of water is : $(g=10m/s^2)$

Options:
A. $$2.6cm$$
B. $$0.8cm$$
C. $$1.2cm$$
D. $$2cm$$
E. $$2.5cm$$
F. $$1.6cm$$
G. $$1.8cm$$
H. $$2.2cm$$
I. $$2.0cm$$
J. $$2.4cm$$
Answer: G
Equation of SHM is $x=10\sin{10\pi t}$. Find the distance between the two points where speed is $50\pi$ $cm/sec$. $x$ is in cm and $t$ is in seconds

Options:
A. $$5(\sqrt{3}+1) cm$$
B. $$10cm$$
C. $$5(\sqrt{3}+3) cm$$
D. $$17.32cm$$
E. $$5(\sqrt{3}-1) cm$$
F. $$5(\sqrt{3}-2) cm$$
G. $$5(\sqrt{3}+4) cm$$
H. $$5(\sqrt{3}-3) cm$$
I. $$20cm$$
J. $$5(\sqrt{3}+2) cm$$
Answer: A
$10\ lt$. box contains $O_{3}$ and $O_{2}$ at equilibrium at $2000\ K$. The $\triangle G^{\circ} = -534.52\ kJ$ at $8$ atm equilibrium pressure. The following equilibrium is present in the container$2O_{3}(g)\rightleftharpoons 3O_{2}(g)$. The partial pressure of $O_{3}$ will be : $(ln\ 10 = 2.3, R = 8.3\ Jmol^{-1} K^{-1})$.

Options:
A. $$6.48 \times 10^{-6}$$
B. $$7.56 \times 10^{-6}$$
C. $$9.71\times 10^{-2}$$
D. $$22.62\times 10^{-7}$$
E. $$2.43 \times 10^{-6}$$
F. $$7.21 \times 10^{-6}$$
G. $$9.71\times 10^{-6}$$
H. $$3.71 \times 10^{-6}$$
I. $$4.87 \times 10^{-6}$$
J. $$8\times 10^{-6}$$
Answer: J
A vertical glass capillary tube of radius r open at both ends contains some water ( surface tension T, a density $\rho$). If L be the length of the water column then :

Options:
A. $$\displaystyle L = \frac{5T}{2r{\rho}g}$$
B. $$\displaystyle L = \frac{5T}{4r{\rho}g}$$
C. $$\displaystyle L = \frac{T}{2r{\rho}g}$$
D. $$\displaystyle L = \frac{4T}{r{\rho}g}$$
E. $$\displaystyle L = \frac{7T}{4r{\rho}g}$$
F. $$\displaystyle L = \frac{3T}{2r{\rho}g}$$
G. $$\displaystyle L = \frac{2T}{r{\rho}g}$$
H. $$\displaystyle L = \frac{3T}{4r{\rho}g}$$
I. $$\displaystyle L = \frac{9T}{4r{\rho}g}$$
J. $$\displaystyle L = \frac{T}{4r{\rho}g}$$
Answer: D
A cubical block of plastic of edge 3 cm floats in water. The lower surface of the cube just touches the free end of a vertical spring fixed at the bottom of the pot. Find the maximum weight that can be put on the block without wetting it. Density of plastic = 800 kg / $m^3$ and spring constant of the spring = 100 N/m. Take g = 10 $m/s^2$.

Options:
A. $$0.60N$$
B. $$0.54N$$
C. $$0.57N$$
D. 0.63N
E. $$0.64N$$
F. $$5.5N$$
G. $$0.65N$$
H. $$0.59N$$
I. $$0.58N$$
J. $$6.5N$$
Answer: G
${ CaF }_{ 2 }$ dissolved in 0.5 M solution of ${ CaCl }_{ 2 }$ and water. How many times is solubility in the second solution of $Ca^{2+}$ greater than in the first? Given: $K_{ sp }\left( { CaF }_{ 2 } \right) =3.2\times { 10 }^{ -11 }$

Options:
A. 25 times
B. $$10 times$$
C. 50 times
D. 30 times
E. 200 times
F. 100 times
G. 500 times
H. 40 times
I. 20 times
J. 400 times
Answer: I
An achromatic convergent lens of focal length 20 cms is made of two lenses (in contact) of materials having dispersive powers in the ratio of 1 : 2 and having focal lengths f$_1$ and f$_2$. Which of the following is true?

Options:
A. $$f_1 = 10 cms, f_2 = 10 cms$$
B. $$f_1 = 20 cms, f_2 = - 20 cms$$
C. $$f_1 = 20 cms, f_2 = 40 cms$$
D. $$f_1 = 20 cms, f_2 = 10 cms$$
E. $$f_1 = -20 cms, f_2 = 10 cms$$
F. $$f_1 = -10 cms, f_2 = 20 cms$$
G. $$f_1 = 10 cms, f_2 = -20 cms$$
H. $$f_1 = -10 cms, f_2 =- 20 cms$$
I. $$f_1 = 10 cms, f_2 = 40 cms$$
Answer: G
A uniform bar of length $6a$ and mass $8m$ lies on a smooth table. Two point masses $m$ and $2m$ moving in a horizontal plane with speeds $2V$ and $V$ strike the bar as shown and stick to the bar after collision.lf $\omega $  is the angular Velocity about center of mass, $E$ is total energy and $v_{c}$ is velocity of the center of mass, after collision, then :

Options:
A. \displaystyle {E}= \frac{7{m}{v}^{2}}{5}
B. \displaystyle {E}= \frac{5{m}{v}^{2}}{5}
C. $$\displaystyle {E}=\frac{3{m}{v}^{2}}{5}$$
D. $$\displaystyle {E}=\frac{{m}{v}^{2}}{5}$$
E. \displaystyle {E}= \frac{8{m}{v}^{2}}{5}
F. $$\displaystyle {v}_{{c}}=\frac{3{v}}{5{a}}$$
G. \displaystyle {E}= \frac{6{m}{v}^{2}}{5}
H. $${v}_{\mathrm{c}}=0$$
I. \displaystyle {E}= \frac{2{m}{v}^{2}}{5}
J. \displaystyle {E}=\frac{4{m}{v}^{2}}{5}
Answer: C
A machine in an amusement part consists of a cage at the end of one arm hinged at $O$ The cage revolves along a vertical circle of radius $r(ABCDFGH)$ about its hinged $O$, at constant linear speed 
$\displaystyle v=\sqrt{gr}$  The cage is so attached that the man of weight $'w'$ 
standing on a weighing machine inside the cage is always vertical.
Then which of the following is correct.

Options:
A. the ratio of the weight reading at $A$ to that at $C=4$
B. the weight reading at $A$ is greater than the weight reading at $E$ by $w$.
C. the ratio of the weight reading at $E$ to that at $A=0$.
D. the weight reading at $A$ is greater than the weight reading at $E$ by $2\ w$.
E. the weight reading at $G$ is equal to $w$.
F. the ratio of the weight reading at $A$ to that at $C=1.5$
G. the ratio of the weight reading at $A$ to that at $C=1$
H. the ratio of the weight reading at $A$ to that at $C=3$
I. the ratio of the weight reading at $A$ to that at $C=2$.
J. the ratio of the weight reading at \( A \) to that at \( C = \frac{3}{2} \)
Answer: I
A board of mass $m = 1 \ kg$ lies on a table and a weight of $M = 2 \ kg$ on the board. What minimum force $F$ must be applied on the board in order to pull it out from under the load? The coefficient of friction between the load and the board is $\mu_1 = 0.25$ and that between  board and table is $\mu_2 = 0.5$:-

Options:
A. $$10 \ N$$
B. $$25 \ N$$
C. $$17.5 \ N$$
D. $$30 \ N$$
E. $$12.5 \ N$$
F. $$22.5 \ N$$
G. $$20 \ N$$
H. $$7.5 \ N$$
I. $$27.5 \ N$$
J. $$15 \ N$$
Answer: F
For an ideal monoatomic gas during any process $T=kV$, find out the molar heat capacity of the gas during the process. (Assume vibrational degree of freedom, to be active)

Options:
A. $$\frac{15}{2}R$$
B. $$\frac{9}{2}R$$
C. $$\frac{11}{2}R$$
D. $$4R$$
E. $$\frac {5 }{2 }R $$
F. $$\frac{3}{2}R$$
G. $$\frac{6}{2}R$$
H. $$3R$$
I. $$\frac{13}{2}R$$
J. $$\frac{7}{2}R$$
Answer: E
2 gm iron pyrite $(FeS_2)$ is burnt with $O_2$ to form $Fe_2O_3$ and $SO_2$ the mass of $SO_2$ produced is
(Fe=56, S=32, O=16)

Options:
A. 2.03 gm
B. 4.26 gm
C. 4 gm
D. 2.26 gm
E. 2.36 gm
F. 2.13 gm
G. 2 gm
H. 2.53 gm
I. 2.06 gm
J. 2.46 gm
Answer: G
A particle of mass $'m'$ is moving along a circle of radius $'r'$. At some instant, its speed is $'v'$ and it is gaining speed at a uniform rate $'a'$, then, at the given instant, acceleration of the particle is :

Options:
A. inclined to radius at $\theta = \sin ^ { - 1 } \dfrac { 1 } { \left[ 1 + \dfrac { v ^ { 4 } } { 2 a ^ { 2 } r ^ { 2 } } \right] ^ { 1 / 2 } }$
B. inclined to radius at $\theta = \tan ^ { - 1 } \dfrac { v ^ { 2 } } { a r }$
C. inclined to radius at $\theta = \sin ^ { - 1 } \dfrac { 1 } { \left[ 1 + \dfrac { v ^ { 4 } } { 6 a ^ { 2 } r ^ { 2 } } \right] ^ { 1 / 2 } }$
D. along the radius
E. inclined to radius at $\theta = \sin ^ { - 1 } \dfrac { 1 } { \left[ 1 + \dfrac { v ^ { 4 } } { 3 a ^ { 2 } r ^ { 2 } } \right] ^ { 1 / 2 } }$
F. inclined to radius at $\theta = \cos ^ { - 1 } \dfrac { a r } { v ^ { 2 } }$
G. inclined to radius at $\theta = \cos ^ { - 1 } \dfrac { v ^ { 2 } } { a r } $
H. inclined to radius at $\theta = \sin ^ { - 1 } \dfrac { 1 } { \left[ 1 + \dfrac { v ^ { 4 } } { 5 a ^ { 2 } r ^ { 2 } } \right] ^ { 1 / 2 } }$
I. inclined to radius at $\theta = \sin ^ { - 1 } \dfrac { 1 } { \left[ 1 + \dfrac { v ^ { 4 } } { a ^ { 2 } r ^ { 2 } } \right] ^ { 1 / 2 } }$
J. inclined to radius at $\theta = \sin ^ { - 1 } \dfrac { 1 } { \left[ 1 + \dfrac { v ^ { 4 } } { 4 a ^ { 2 } r ^ { 2 } } \right] ^ { 1 / 2 } }$
Answer: I
An organic compound $C_5H_9Br$ (A) which readily decolourises bromine water and cold alkaline $KMnO_4$ solution gives $C_5H_{11}Br$ (B) on catalytic hydrogenation. The reaction of A with alcoholic KOH first and then with $NaNH_2$ produces C with an evolution of $NH_3$. C reacts with Lindlar's catalyst to give D and on reaction with Na in liquid $NH_3$ produces E. D and E are isomers. The compound A is:

Options:
A. $$CH_3CH_2-\overset{\overset{\displaystyle Br}{|}}{C}-CH=CH_2$$
B. $$CH_3CH_2CH=\overset{\overset{\displaystyle Br}{|}}{C}H$$
C. $$CH_3CH_2-\overset{\overset{\displaystyle Br}{|}}{C}=CHCH_3$$
D. $$both (a) and (b)$$
E. $$CH_3CH_2-\overset{\overset{\displaystyle Br}{|}}{C}=CH_2$$
F. $$neither (a) nor (b)$$
G. $$CH_3CH_2-\overset{\overset{\displaystyle Br}{|}}{C}-CH_2CH_3$$
H. $$CH_3CH_2CH=\overset{\overset{\displaystyle Br}{|}}{C}-CH_3$$
I. $$CH_3CH_2CH=\overset{\overset{\displaystyle Br}{|}}{C}-CH_2CH_3$$
J. $$CH_3CH=\overset{\overset{\displaystyle Br}{|}}{C}-CH_2CH_3$$
Answer: H
In the Young's arrangement, screen starts moving towards right with constant speed $v$. Initial distance between screen and plane of slits is $x$. At $t=0$ $1st$ order maxima is lying at point $A$. After how much time first order minima lies at point $A$?

Options:
A. $$\dfrac{2x}{3v}$$
B. $$\dfrac{x}{2v}$$
C. $$\dfrac{x}{v}$$
D. $$\dfrac{x}{3v}$$
E. $$\dfrac{3x}{2v}$$
F. $$\dfrac{4x}{3v}$$
G. $$\dfrac{3x}{v}$$
H. $$\dfrac{5x}{3v}$$
I. $$\dfrac{2x}{v}$$
J. $$\dfrac{3x}{4v}$$
Answer: C