
JEE MAINS 2013 – Code P
Test Name: JEE MAINS 2013 – Code P
Question 1:
A uniform cylinder of length L and mass M having cross – sectional area A is suspended, with its length vertical, form a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at equilibrium position. The extension x_{0} of the spring when it is in equilibrium is:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
(Here k is spring constant) 
Solutions:
The forces acting on the cylinder is as shown in the figure:
The correct answer is C.
Question 2:
A metallic rod of length ‘l’ is tied to a spring of length 2l and made to rotate with angular speed ω on a horizontal table with one end of the spring fixed. If there is a vertical magnetic field ‘B’ in the region, the e.m.f. induced across the ends of the rod is:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Let dx be the small element of the rod at a distance x from the center of rotation as shown in the figure.
Thus the e.m.f. induced in the rod,
The correct answer is D.
Question 3:
This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement − I: A point particle of mass m moving with speed ν collides with stationary point particle of mass M. if the maximum energy loss possible is given as .
Statement − II : Maximum energy loss occurs when the particles get stuck together as a result of the collision.
Option 1:  Statement − I is true, Statement − II is true, Statement − II is a correct explanation of Statement − I. 
Option 2:  Statement − I is true, Statement − II is true, Statement − II is not a correct explanation of Statement − I. 
Option 3:  Statement − I is true, Statement − II is false. 
Option 4:  Statement − I is false, Statement − II is true. 
Solutions:
Loss of energy is maximum when the collision is inelastic i.e, the bodies stick to each other.
Maximum energy loss,
Thus the statement I is wrong while the statement II is correct.
The correct answer is D.
Question 4:
Let [∈_{0}] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:
Option 1:  [∈_{0}] = [M^{−1} L^{−3} T^{2} A] 
Option 2:  [∈_{0}] = [M^{−1} L^{−3} T^{4} A^{2}] 
Option 3:  [∈_{0}] = [M^{−1} L^{2} T^{4} A^{−2}] 
Option 4:  [∈_{0}] = [M^{−1} L^{2} T^{−1} A] 
Solutions:
Since electrostatic force is given as:
The correct answer is B.
Question 5:
A projectile is given an initial velocity of , where is along the ground and is along the vertical. If g = 10 m/s^{2}, the equation of its trajectory is:
Option 1:  y = x − 5x^{2} 
Option 2:  y = 2x − 5x^{2} 
Option 3:  4y = 2x − 5x^{2} 
Option 4:  4y = 2x − 25x^{2} 
Solutions:
The velocity of the projectile is given as:
The correct answer is B.
Question 6:
The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to α times its original magnitude, where α equals:
Option 1:  0.7 
Option 2:  0.81 
Option 3:  0.729 
Option 4:  0.6 
Solutions:
The amplitude of a damped oscillator is given as:
Since, the amplitude decreases to 0.9 times of the original amplitude in 5 s
The correct answer is C.
Question 7:
Two capacitors C_{1} and C_{2} are charged to 120V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then:
Option 1:  5C_{1} = 3C_{2} 
Option 2:  3C_{1} = 5C_{2} 
Option 3:  3C_{1} + 5C_{2} = 0 
Option 4:  9C_{1} = 4C_{2} 
Solutions:
When the capacitors are connected, then for the potential to be zero,
120 C_{1} = 200 C_{2}
⇒ 3C_{1} = 5C_{2}
The correct answer is B.
Question 8:
A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 10^{3} kg/m^{3} and 2.2 × 10^{11} N/m^{2} respectively?
Option 1:  188.5 Hz 
Option 2:  178.2 Hz 
Option 3:  200.5 Hz 
Option 4:  770 Hz 
Solutions:
The fundamental frequency in the wire is given as,
Putting the values of l, Y, d and Δl/l.
We get,
The correct answer is B.
Question 9:
A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm. If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is:
Option 1:  9.1 × 10^{−11} weber 
Option 2:  6 × 10^{−11} weber 
Option 3:  3.3 × 10^{−11} weber 
Option 4:  6.6 × 10^{−9} weber 
Solutions:
The coefficient of mutual induction between the loops,
∴ The flux linked with the bigger loop,
The correct option is A.
Question 10:
Diameter of a planoconvex lens is 6 cm and thickness at the centre is 3 mm. If speed of light in material of lens is 2 × 10^{8} m/s, the focal length of the lens is:
Option 1:  15 cm 
Option 2:  20 cm 
Option 3:  30 cm 
Option 4:  10 cm 
Solutions:
From the diagram, in ΔOPQ
The correct answer is C.
Question 11:
What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
The final energy of the satellite in the circular orbit at an altitude 2R,
The initial energy of the satellite at the surface of earth,
where k is the minimum energy required to launch the satellite
The correct answer is A.
Question 12:
A diode detector is used to detect an amplitude modulated wave of 60% modulation by using a condenser of capacity 250 pico farad in parallel with a load resistance 100 kilo ohm. Find the maximum modulated frequency which could be detected by it.
Option 1:  10.62 MHz 
Option 2:  10.62 kHz 
Option 3:  5.31 MHz 
Option 4:  5.31 kHz 
Solutions:
The highest frequency that can be detected with tolerable distortion,
The correct answer is B.
Question 13:
A beam of unpolarised light of intensity I_{0} is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45° relative to that of A. The intensity of the emergent light is:
Option 1:  I_{0} 
Option 2:  
Option 3:  
Option 4: 
Solutions:
The intensity of the light passing through the polarisers are as shown in the figure:
The correct answer is C.
Question 14:
The supply voltage to a room is 120 V. The resistance of the lead wires is 6Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?
Option 1:  Zero Volt 
Option 2:  2.9 Volt 
Option 3:  13.3 Volt 
Option 4:  10.04 Volt 
Solutions:
The power of the bulb is given as,
Resistance of the heater
Voltage across the bulb, before the heater is switched on
Voltage across the bulb after the water is switched on,
The correct answer is D.
Question 15:
The above p–v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is:
Option 1:  p_{0}v_{0} 
Option 2:  
Option 3:  
Option 4:  4p_{0}v_{0} 
Solutions:
Heat is extracted from the source in the path DA and BC,
The correct answer is B.
Question 16:
A hoop of radius r and mass m rotating with and angular velocity ω_{0} is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity of the centre of the hoop when it ceases to slip?
Option 1:  
Option 2:  
Option 3:  
Option 4:  rω_{0} 
Solutions:
Let v_{0} be the initial translational velocity of the centre of the loop. When the hoop ceases to slip,
The correct answer is C.
Question 17:
An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V_{0} and its pressure if P_{0}. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Let the piston be displaced by Δx.
Since the process is adiabatic,
Now,
Mg − PA = F
Thus, the restoring force, F is given as
The correct answer is C.
Question 18:
If a piece of metal is heated to temperature θ and then allowed to cool in a room which is at temperature θ_{0}, the graph between the temperature T of the metal and time t will be closest to:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
From Newton’s law of cooling, initially the decrease in temperature will be more, the temperature goes on decreasing with time and it becomes equal to the temperature of the room after some time.
The correct answer is C.
Question 19:
This question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two Statements.
Statement − I: Higher the range, greater is the resistance of ammeter.
Statement − II: To increase the range of ammeter, additional shunt needs to be used across it.
Option 1:  Statement − I is true, Statement − II is true, Statement − II is the correct explanation of statement − I 
Option 2:  Statement − I is true, Statement − II is true, Statement − II is not the correct explanation of Statement − I. 
Option 3:  Statement − I is true, Statement − II is false. 
Option 4:  Statement − I is false, Statement − II is true. 
Solutions:
For a ammeter,
In order to increase I, the shunt resistance S should be decreased, so additional S can be connected across it.
The correct answer is D.
Question 20:
In an LCR circuit as shown below both switches are open initially. Now switch S_{1} is closed, S_{2} kept open. (q is charge on the capacitor and τ = RC is Capacitive time constant). Which of the following statement is correct?
Option 1:  Work done by the battery is half of the energy dissipated in the resistor 
Option 2:  At t = τ, q = CV/2 
Option 3:  At = t = 2τ, q = CV(1 − e^{−2}) 
Option 4: 
Solutions:
Charge on the capacitor is given as:
Q = CV (1 − e^{−t/}^{τ})
and at t = 2τ
Q = CV(1 − e^{−2})
The correct answer is C.
Question 21:
Two coherent point sources S_{1} and S_{2} are separated by a small distance ‘d’ as shown. The fringes obtained on the screen will be:
Option 1:  points 
Option 2:  straight lines 
Option 3:  semicircles 
Option 4:  concentric circles 
Solutions:
In the given situation, the locus of the points having same path difference is circle, so the interference pattern produced on the screen will be concentric circles.
The correct answer is D.
Question 22:
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is:
Option 1:  3 V/m 
Option 2:  6 V/m 
Option 3:  9 V/m 
Option 4:  12 V/m 
Solutions:
The correct answer is B.
Question 23:
The anode voltage of photocell is kept fixed. The wavelength λ of the light falling of the cathode is gradually changed. The plate current I of the photocell varies as follows:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
With increase in the wavelength λ, the frequency goes on decreasing. At same wavelength, the frequency becomes equal or less than the threshold frequency of the photocell and thus the electron ceases to comes out of it. This will result in zero photocurrent.
The correct answer is D.
Question 24:
The IV characteristic of an LED is:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
In the IV characteristics of LED, for equal value of current, higher voltage is required for higher frequency.
The correct answer is A.
Question 25:
Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop of this to be possible? The surface tension is T, density of liquid is ρ and L is its latent heat of vaporization.
Option 1:  ρL/T 
Option 2:  
Option 3:  T/ρL 
Option 4:  2T/ρL 
Solutions:
Let the radius of the drop decreases by ΔR
∴ decrease in surface energy = heat required for evaporation
The correct answer is D.
Question 26:
In a hydrogen like atom election makes transition from an energy level with quantum number n to another with quantum number (n − 1). If n >> 1, the frequency of radiation emitted is proportional to:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
When there is transition from energy level x to the energy level (n − 1), energy is released,
For n> > 1
The correct answer is D.
Question 27:
The graph between angle of deviation (δ) and angle of incidence (i) for a triangular prism is represented by:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Since angle of deviation is related to angle of incidents as:
Thus the graph between angle of deviation (δ) and angle of incidence (i) is as shown:
The correct answer is C.
Question 28:
Two charges, each equal to q, are kept at x = −a and x = a on the xaxis. A particle of mass m and charge is placed at the origin. If charge q_{0} is given a small displacement (y << a) along the yaxis, the net force acting on the particle is proportional to:
Option 1:  y 
Option 2:  −y 
Option 3:  
Option 4: 
Solutions:
Net force acting on the charge
The correct answer is A.
Question 29:
Two short bar magnets of length 1 cm each have magnetic moments 1.20 A m^{2} and 1.00 A m^{2} respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the midpoint O of the line joining their centres is close to (Horizontal component of earth’s magnetic induction is 3.6 × 10^{−5} Wb/m^{2})
Option 1:  3.6 × 10^{−5} Wb/m^{2} 
Option 2:  2.56 × 10^{−4} Wb/m^{2} 
Option 3:  3.50 × 10^{−4} Wb/m^{2} 
Option 4:  5.80 × 10^{−4} Wb/m^{2} 
Solutions:
The result and horizontal magnetic induction at point O is given as
The correct answer is B.
Question 30:
A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Let dx be an element of the rod at distance x from point O.
The correct answer is D.
Question 31:
Which of the following complex species is not expected to exhibit optical isomerism?
Option 1:  [Co(en)_{3}]^{3+} 
Option 2:  [Co(en)_{2 }Cl_{2}]^{+} 
Option 3:  [Co(NH_{3})_{3 }Cl_{3}] 
Option 4:  [Co(en)(NH_{3})_{2 }Cl_{2}]^{+} 
Solutions:
Optical isomers are the mirror images that cannot be superimposed on one another. This type of isomerism is common in octahedral complexes involving didentate ligands and complex lacking any plane of symmetry. [Co(NH_{3})_{3}Cl_{3}] is an octahedral complex which can show facial and meridional isomerism under geometrical isomerism. But due to presence of plane of symmetry in this complex, it is not expected to exhibit optical isomerism.
Question 32:
Which one of the following molecules is expected to exhibit diamagnetic behaviour?
Option 1:  C_{2} 
Option 2:  N_{2} 
Option 3:  O_{2} 
Option 4:  S_{2} 
Solutions:
On the basis of molecular orbital configuration, C_{2} and N_{2} contain paired electrons. So C_{2} and N_{2} are diamagnetic while O_{2} and S_{2} contain unpaired electrons so they are paramagnetic.
Question 33:
A solution of (−) −1 − chloro − 1 − phenylethane in toluene racemises slowly in the presence of a small amount of SbCl_{5}, due to the formation of :
Option 1:  carbanion 
Option 2:  carbene 
Option 3:  carbocation 
Option 4:  free radical 
Solutions:
In the presence of SbCl_{5} and toluene_{,} carbocation of ()1Chloro1phenylethane is formed which contains sp^{2} hybridised planar molecule. Such carbonation undergoes S_{N}1 reaction and is stabilized by racemisation. Cl^{–} attacks on either side of the plane of the molecule to give a mixture of opposite configurations (leavo and dextro) in equal amount. This results in cancellation of rotation of light by the molecule.
Question 34:
Given
Based on the data given above, strongest oxidising agent will be :
Option 1:  Cl^{−} 
Option 2:  Cr^{3+} 
Option 3:  Mn^{2+} 
Option 4: 
Solutions:
Higher the value of positive standard electrode potential, stronger is the oxidizing agent. So is a stronger oxidizing agent
Question 35:
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be :
(R = 8.314 J/mol K) (In 7.5 = 2.01)
Option 1:  q = + 208 J, w = − 208 J 
Option 2:  q = − 208 J, w = − 208 J 
Option 3:  q = − 208 J, w = + 208 J 
Option 4:  q = + 208 J, w = + 208 J 
Solutions:
Since the temperature is constant, the process is isothermal expansion i.e. ΔU = 0. It absorbs heat, so q= +208 J. Using first law of thermodynamic, q = w. So w = 208 J
Question 36:
The molarity of a solution obtained by mixing 750 mL of 0.5(M)HCl with 250 mL of 2(M)HCl will be :
Option 1:  0.875 M 
Option 2:  1.00 M 
Option 3:  1.75 M 
Option 4:  0.975 M 
Solutions:
After mixing two solution having molarities M_{1} and M_{2} and volume V_{1} and V_{2} respectively, the total number of moles of two solutions will remains equal to the number of moles of resultant solution. Let molarity of final solution is M.
Number of moles of first solution = M_{1}V_{1}
Number of moles of second solution = M_{2}V_{2}
Number of moles of resultant solution after mixing = MV_{1} + MV_{2} = M (V_{1} + V_{2})
So, M (V_{1} + V_{2}) = M_{1}V_{1} + M_{2}V_{2}
Question 37:
Arrange the following compounds in order of decreasing acidity :
Option 1:  II > IV > I > III 
Option 2:  I > II > III > IV 
Option 3:  III > I > IV > II 
Option 4:  IV > III > I > II 
Solutions:
Electron withdrawing groups (EWG) increases the acidic strength and electron releasing group (ERG) decreases the acidic strength as EWG stabilises the phenoxide anion formed and ERG destabilises the phenoxide ion. Groups which show both – R (negative resonance effect) and I (negative induction effect) effect has more acidic strength. Cl : +R and I effect CH_{3}: +I effect NO_{2} : R and I effect OCH_{3} : +R and +I effect The order of the effect of the groups in decreasing acidity is NO_{2}> Cl > OCH_{3} > CH_{3}
Question 38:
For gaseous state, if most probable speed is denoted by C*, average speed by and mean square speed by C, then for a large number of molecules the ratios of these speeds are :
Option 1:  C* : : C = 1.225 : 1.128 : 1 
Option 2:  C* : : C = 1.128 : 1.225 : 1 
Option 3:  C* : : C = 1 : 1.128 : 1.225 
Option 4:  C* : : C = 1 : 1.225 : 1.128 
Solutions:
The three molecular speeds are :
Most probable speed (u_{mp}) − The speed possessed by the maximum number of molecules of a gas at a given temperature.
Average speed (u_{av}) − The arithmetic mean of the various speeds of the molecules.
Root mean square speed (u_{rms}) − It is the square root of the mean of the squares of the speeds of all the molecules present in the given sample of the gas.
These three are related to each other as follows :
Question 39:
The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be :
(R = 8.314 JK^{−1 }mol^{−1} and log 2 = 0.301)
Option 1:  53.6 kJ mol^{−1} 
Option 2:  48.6 kJ mol^{−1} 
Option 3:  58.5 kJ mol^{−1} 
Option 4:  60.5 kJ mol^{−1} 
Solutions:
Given : Let k_{1} = k ; k_{2} = 2k ; T_{1} = 300K ; T_{2} = 310K ; E_{a} = ?
Question 40:
A compound with molecular mass 180 is acylated with CH_{3}COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is :
Option 1:  2 
Option 2:  5 
Option 3:  4 
Option 4:  6 
Solutions:
RNH_{2} + CH_{3}COCl
→HClRNHCOCH_{3}
Mass of acetyl group CH_{3}CO = 12 + 3 + 12 + 16 = 43. On acetylation it replaces one Hatom from one NH_{2} group of the compound. So, one mole of acetyl group increases molecular mass of compound by 42 unit.
Since, total increase in mass after acetylation = 390 − 180 = 210 unit
So, number of amino group = 210/42 = 5
Question 41:
Which of the following arrangements does not represent the correct order of the property stated against it ?
Option 1:  V^{2}^{+} < Cr^{2}^{+} < Mn^{2}^{+} < Fe^{2}^{+} : paramagnetic behaviour 
Option 2:  Ni^{2}^{+} < Co^{2}^{+} < Fe^{2}^{+} < Mn^{2}^{+} : ionic size 
Option 3:  Co^{3}^{+} < Fe^{3}^{+} < Cr^{3}^{+} < Sc^{3}^{+} : stability in aqueous solution 
Option 4:  Sc < Ti < Cr < Mn : number of oxidation states 
Solutions:
Magnetic moment increases with the increase in the number of unpaired electrons.
Configuration  Number of unpaired electrons  
3d^{3}  3  
3d^{4}  4  
3d^{5}  5  
3d^{4}  4 
More the number of unpaired electrons more is the paramagnetism. So order of paramagnetic behavior is <=<
Ionic size of ions of the same charge generally decreases from left to right across a period. So order of ionic size is Ni^{2}^{+}< Co^{2}^{+}< Fe^{2}^{+}< Mn^{2}^{+}
The stability of ions of dblock elements depend on the hydration enthalpy.
Hydration enthalpy (kJ/mol)  
3960  
4429  
4563  
4653 
The hydration enthalpy of Sc^{3}^{+ } is least so it is least stable while he hydration enthalpy of Co^{3}^{+ } is maximum so it is most stable.
So order is <<<
Question 42:
The order of stability of the following carbocations :
Option 1:  III > II > I 
Option 2:  II > III > I 
Option 3:  I > II > III 
Option 4:  III > I > II 
Solutions:
The order of stability of carbocation will be
Benzyl carbocation and allyl carbonations are more stable than propyl carbocation due to resonance.
Question 43:
Consider the following reaction :
The values of x, y and z in the reaction are, respectively :
Option 1:  5, 2 and 16 
Option 2:  2, 5 and 8 
Option 3:  2, 5 and 16 
Option 4:  5, 2 and 8 
Solutions:
∴ Balanced Equation :
So, x = 2, y = 5 & z = 16
Question 44:
Which of the following is the wrong statement ?
Option 1:  ONCl and ONO^{−} are not isoelectronic. 
Option 2:  O_{3} molecule is bent. 
Option 3:  Ozone is violetblack in solid state. 
Option 4:  Ozone is diamagnetic gas. 
Solutions:
(All statement are correct there is no answer)
(1) ONCl = 8 + 7 + 17 = 32e^{−}
ONO^{−} = 8 + 7 + 8 + 1 = 24e^{− }
(2) Central atom O is sp^{2} hybridised with 1 lone pair, so bent shape.
(3) Ozone is violetblack in solid state.
(4) O_{3} has no unpaired electrons, so diamagnetic.
Question 45:
A gaseous hydrocarbon gives upon combustion 0.72 g. of water and 3.08 g. of CO_{2}. The empirical formula of the hydrocarbon is :
Option 1:  C_{2}H_{4} 
Option 2:  C_{3}H_{4} 
Option 3:  C_{6}H_{5} 
Option 4:  C_{7}H_{8} 
Solutions:
Empirical formula of the hydrocarbon can be determined with the help of number of hydrogen atoms and number of oxygen atoms present in the hydrocarbon.
2 g of hydrogen is present in 18 g of water
So, amount of hydrogen present in 0.72 g of water = =0.08 g
Similarly, amount of Carbon present in 3.08 g of carbon dioxide = = 0.84 g
Element  Relative number of atoms  Whole number ratio 
Carbon  == 0.07  7 
Hydrogen  == 0.08  8 
Therefore, the empirical formula of the hydrocarbon is C_{7}H_{8}.
Question 46:
In which of the following pairs of molecules/ions, both the species are not likely to exist?
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Molecular/ ionic species 
Molecular orbital electronic configuration 
Bond order 
= =0.5 

= =0 

= =0.5 

He_{2} 
= =0 

= =1 

= =1 
Out of the given species only and He_{2} have zero bond order therefore, these species would not exist.
Question 47:
Which of the following exists as covalent crystals in the solid state?
Option 1:  Iodine 
Option 2:  Silicon 
Option 3:  Sulphur 
Option 4:  Phosphorus 
Solutions:
Silicon possesses network like structure similar to that of diamond, therefore, it exists as covalent crystals in solid state.
Question 48:
Synthesis of each molecule of glucose in photosynthesis involves:
Option 1:  18 molecules of ATP 
Option 2:  10 molecules of ATP 
Option 3:  8 molecules of ATP 
Option 4:  6 molecules of ATP 
Solutions:
Synthesis of glucose during photosynthesis from carbon dioxide, ATP and NADPH can be represented using the following balanced chemical equation:
6CO_{2} + 12NADPH + 18ATP → C_{6}H_{12}O_{6} + 12NADP + 18ADP
Therefore, 18 molecules of ATP are required to produce one molecule of glucose.
Question 49:
The coagulating power of electrolytes having ions Na^{+}, Al^{3+ }and Ba^{2+ }for arsenic sulphide sol increases in the order :
Option 1:  Al^{3+ }< Ba^{2+ }< Na^{+} 
Option 2:  Na^{+} < Ba^{2+} < Al^{3+} 
Option 3:  Ba^{2+ }< Na^{+} < Al^{3+} 
Option 4:  Al^{3+} < Na^{+} < Ba^{2+} 
Solutions:
Arsenic sulphide is a negatively charged sol, therefore greater the positive charge on the ionic species greater will be its coagulating power for arsenic sulphide sol. The cations can be arranged in the following order of increasing magnitude of the positive charge: Na^{+} < Ba^{2+}<al< font=””></al<>^{3+}.
Therefore, the coagulating power of the ions will also increase in the same direction Na^{+} < Ba^{2+}<al< font=””></al<>^{3+}.
Question 50:
Which of the following represents the correct order of increasing first ionization enthalpy for Ca, Ba, S, Se and Ar ?
Option 1:  Ca < S < Ba < Se < Ar 
Option 2:  S < Se < Ca < Ba < Ar 
Option 3:  Ba < Ca < Se < S < Ar 
Option 4:  Ca < Ba < S < Se < Ar 
Solutions:
When we move down in a group, the size of the atoms increase, due to which the effect of the nuclear charge on the valence electron decreases making the removal of valence electron relatively easy.
Therefore, the correct order of increasing first ionization energy of Ca, Ba (Group II) and S, Se (Group XVI) is Ba < Ca and Se < S.
Similarly, when we move across a period the size of the atom decreases making the removal of valence electron difficult, therefore the increasing order of ionization energy is Ba < Ca < Se < S < Ar.
Question 51:
Energy of an electron is given by. Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be : (h = 6.62 × 10^{−34})Js and c = 3.0 × 10^{8} ms^{−1})
Option 1:  1.214 ×10^{−7} m 
Option 2:  2.816 ×10^{−7} m 
Option 3:  6.500 ×10^{−7} m 
Option 4:  8.500 ×10^{−7} m 
Solutions:
Energy of the electron in the level n = 1 can be calculated as:
Similarly, the energy of the electron in the level n = 2 can be calculated as:
E2 = 2.178 × 10181222 = 0.544 × 1018 J
ΔE = E_{2} − E_{1}
ΔE = (−0.544 + 2.178) × 10^{−18} J = 1.633 × 10^{−18 }J
Therefore, the wavelength of light required to excite an electron in a hydrogen atom from level n = 1 to n = 2 is 1.24 × 10^{−7} m.
Question 52:
Compound (A), C_{8}H_{9}Br, gives a white precipitate when warmed with alcoholic AgNO_{3}.Oxidation of (A) gives an acid (B), C_{8}H_{6}O_{4}. (B) easily forms anhydride on heating. Identify the compound (A).
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
First calculate the double bond equivalents for compound A:
4^{o} of unsaturation indicates presence of three double bonds and a ring. The two possible structure of compound A can be:
From the two structures only (I) will give the characteristic reactions of compound (A) as shown in the given scheme:
Therefore, compound A is option (4).
Question 53:
Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest value ?
Option 1:  Cr(Z = 24) 
Option 2:  Mn(Z = 25) 
Option 3:  Fe(Z = 26) 
Option 4:  Co(Z = 27) 
Solutions:
The standard reduction potential values of the dblock elements increases from left to right in a period however among the elements of the third transition series some elements have exceptionally high value of E^{o} due to their stability. The standard reduction potential values are given as:
Therefore, Cobalt will have the highest standard reduction potential value.
Question 54:
How many litres of water must be added to 1 litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ?
Option 1:  0.1 L 
Option 2:  0.9 L 
Option 3:  2.0 L 
Option 4:  9.0 L 
Solutions:
For first case:
pH = 1
pH = −log[H^{+}]
−log[H^{+}] = 1
[H^{+}] = 10^{−1} = 0.1 M
For the second case:
pH = 2
pH= −log[H^{+}]
−log[H^{+}] = 2
[H^{+}] = 10^{−2} = 0.01 M
Volume of the solution before dilution = 1 L
According to molarity equation
M_{1}V_{1} = M_{2}V_{2}
0.01 × 1 = 0.001 × V_{2}
V_{2} = 10 L
Final volume of the solution after dilution = 10 L
Volume of water added = 10 − 1 = 9 L
Therefore the correct answer is 9L.
Question 55:
The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na^{+} will be :
Option 1:  − 2.55 eV 
Option 2:  − 5.1 eV 
Option 3:  − 10.2 eV 
Option 4:  + 2.55 eV 
Solutions:
Electron gain enthalpy of Na^{+} will be negative of first ionization enthalpy of Na. Since,
First ionization enthalpy of Na = 5.1 eV
Electron gain enthalpy of Na^{+} = 5.1 eV
Therefore, the correct answer is 5.1 eV.
Question 56:
An organic compound A upon reacting with NH_{3} gives B. On heating, B gives C. C in presence of KOH reacts with Br_{2} to give CH_{3}CH_{2}NH_{2}. A is :
Option 1:  CH_{3}COOH 
Option 2:  CH_{3}CH_{2}CH_{2}COOH 
Option 3:  
Option 4:  CH_{3}CH_{2}COOH 
Solutions:
CH_{3} − CH_{2} − NH_{2} can be obtained by carrying out Hofmann Bromamide reaction with CH_{3} − CH_{2} − CO − NH_{2} as shown below:
C is obtained by heating B, which is the ammonium salt of carboxylic acid A:
Therefore, A is CH_{3} − CH_{2} − COOH.
Question 57:
Stability of the species and increases in the order of :
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Ionic species 
Molecular orbital electronic configuration 
Bond order 
==0.5 

Li_{2} 
==1 

==0.5 
Li_{2} will have the highest stability due to highest bond order, and will be least stable as the last electron enters an antibonding molecular orbital.
Therefore, the correct order of increasing stability is: Li_{2} > >.
Question 58:
An unknown alcohol is treated with the “Lucas reagent” to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism :
Option 1:  secondary alcohol by S_{N}1 
Option 2:  tertiary alcohol by S_{N}1 
Option 3:  secondary alcohol by S_{N}2 
Option 4:  tertiary alcohol by S_{N}2 
Solutions:
The reaction between Lucas reagent and alcohol for determination of primary, secondary or tertiary alcohol follows S_{N}1 mechanism whose rate of the reaction is directly proportional to the carbocation formed as an intermediate (more stable the intermediate higher is the rate of S_{N}1 reaction).
As we know that tertiary carbocation is highly stable therefore, a tertiary alcohol will react fastest with Lucas reagent by S_{N}1 mechanism.
Thus, the correct answer is tertiary alcohol by S_{N}1 mechanism.
Question 59:
The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was:
Option 1:  Methylisocyanate 
Option 2:  Methylamine 
Option 3:  Ammonia 
Option 4:  Phosgene 
Solutions:
The gas leaked from a storage tank of the Union Carbide plant in Bhopal was methyl isocyanate (CH_{3} − N = C = O) also known as MIC gas.
Therefore, the correct answer is Methyl isocyanate.
Question 60:
Experimentally it was found that a metal oxide has formula M_{0.98}O. Metal M, is present as M^{2+} and M^{3+} in its oxide. Fraction of the metal which exists as M^{3+} would be :
Option 1:  7.01% 
Option 2:  4.08% 
Option 3:  6.05% 
Option 4:  5.08% 
Solutions:
Metal oxide = M_{0.98}O
One mole of metal oxide contains 0.98 moles of metal and 1 mole of oxide ion.
Suppose that moles of M^{3+} in the metal oxide = x
Moles of M^{2+} in the metal oxide = (0.98 − x)
Overall the metal oxide is a neutral species, thus,
2(0.98 − x) + 3x − 2 = 0
1.96 − 2x + 3x − 2 = 0
x = 2 − 1.96 = 0.04
Moles of M^{3+} = 0.04
Moles of M^{2+} = 0.98 − 0.04 = 0.94
Percentage of M^{3+} = = 4.08%
Therefore, the percentage of metal present in its +3 oxidation state is 4.08%.
Question 61:
Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is :
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
The equations of the given planes are
2x + y + 2z = 8 ⇒ 2x + y + 2z − 8 = 0 … (1)
4x + 2y + 4z + 5 = 0
∴ Distance between the planes (1) and (2)
The Correct Answer is C.
Question 62:
At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional number of workers x is given by If the firm employs 25 more workers, then the new level of production of items is :
Option 1:  2500 
Option 2:  3000 
Option 3:  3500 
Option 4:  4500 
Solutions:
Given,
Integrating on both sides, we have
When x = 0, P = 2000, we have
When x = 25,
Thus, the new level of production of items is 3500 units.
The Correct Answer is C.
Question 63:
Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of A × B having 3 or more elements is :
Option 1:  256 
Option 2:  220 
Option 3:  219 
Option 4:  211 
Solutions:
Consider a finite set P containing m element. Let us consider those subsets of P that have r elements each. We know that, the number of ways in which r elements can be chosen out of n elements is . Thus, the number of subsets of P having r elements is.
Given, Number of elements in set A = n (A) = 2
Number of elements in set B = n (B) = 4
Number of elements in set A × B = n (A × B) = n (A) × n (B) = 2 × 4 = 8
∴ Number of subsets of A × B having 3 or more elements
The Correct Answer is C.
Question 64:
If the lines and are coplanar, then k can have:
Option 1:  any value. 
Option 2:  exactly one value. 
Option 3:  exactly two values. 
Option 4:  exactly three values. 
Solutions:
The equations of the given lines are
Given, the lines (1) and (2) are coplanar.
Thus, k can have exactly two values.
The Correct Answer is C.
Question 65:
If the vectors and are the sides of a triangle ABC, then the length of the median through A is :
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Given, are the sides of ΔABC.
In ΔABC
Now in ΔABM,
Thus, the length of median through A is.
The Correct Answer is C.
Question 66:
The real number k for which the equation, 2x^{3} + 3x + k = 0 has two distinct real roots in [0, 1]
Option 1:  lies between 1 and 2. 
Option 2:  lies between 2 and 3. 
Option 3:  lies between −1 and 0. 
Option 4:  does not exist. 
Solutions:
⇒ f (x) is strictly increasing function.
Hence, f (x) = 0 can have only one real root in [0, 1].
So, the given equation cannot have two distinct in [0, 1].
Thus, the real value of k for which the given equation has two distinct real roots in [0, 1] does not exist.
The Correct Answer is D.
Question 67:
The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, …, is :
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
The Correct Answer is C.
Question 68:
A ray of light along gets reflected upon reaching xaxis, the equation of the reflected ray is :
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
The line intersects the xaxis at. Angle made by the line with the positive direction of xaxis is 150°.
Thus, the angle made by the reflected ray of light with the positive direction of xaxis is 30°.
∴ Equation of the reflected ray of light is
Thus, the equation of the reflected ray of light is .
The Correct Answer is B.
Question 69:
The number of values of k, for which the system of equations :
(k + 1)x + 8y = 4k
kx + (k + 3)y = 3k −1
has no solution, is :
Option 1:  infinite 
Option 2:  1 
Option 3:  2 
Option 4:  3 
Solutions:
The given system of equations is
(k + 1) x + 8y = 4k
kx + (k + 3) y = 3k −1
Given, the system of equations has no solution.
From (1) and (2), we have
From (2) and (3), we have
From (1) and (2), we have
k = 3
Thus, k can take only one value.
The Correct Answer is B.
Question 70:
If the equations x^{2} + 2x + 3 = 0 and ax^{2} + bx + c = 0, a, b, c ∈ R, have a common root, then a : b : c is:
Option 1:  1 : 2 : 3 
Option 2:  3 : 2 : 1 
Option 3:  1 : 3 : 2 
Option 4:  3 : 1 : 2 
Solutions:
The given equations are
x^{2} + 2x + 3 = 0 …(1)
ax^{2} + bx + c = 0 …(2)
The roots of equation (1) are imaginary.
We know that, complex roots occur in conjugate pairs.
Given, the equations have a common root. So, both the roots of equation (1) and (2) are same.
Hence, both the given equations are identical
The Correct Answer is A.
Question 71:
The circle passing through (1, −2) and touching the axis of x at (3, 0) also passes through the point :
Option 1:  (−5, 2) 
Option 2:  (2, −5) 
Option 3:  (5, −2) 
Option 4:  (−2, 5) 
Solutions:
Let the equation of the circle be
Circle (1) passes through (1, −2)
So, the equation of the circle is
(x − 3)^{2} + y^{2} + 4y = 0
It can be seen that, (5, −2) satisfy the equation (x − 3)^{2} + y^{2} + 4y = 0.
Thus, the circle passes through the point (5, −2).
The Correct Answer is C.
Question 72:
If x, y, z are in A.P. and tan^{−1}x, tan^{−1}y, and tan^{−1}z, are also in A.P., then :
Option 1:  x = y = z 
Option 2:  2x = 3y = 6z 
Option 3:  6x = 3y = 2z 
Option 4:  6x = 4y = 3z 
Solutions:
Given, x, y, z are in A.P.
∴ 2y = x + z …(1)
Also, tan^{−1} x, tan^{−1} y, tan^{−1} z are in A.P.
Hence, the correct option is A.
Question 73:
Consider :
Statement − I : (p ∨ ~ q) ∨ (~ p ∨ q) is a fallacy.
Statement − II : (p → q) ↔ (~ q → ~ p) is a tautology.
Option 1:  Statement − I is true; Statement − II is true; Statement − II is a correct explanation for Statement − I. 
Option 2:  Statement − I is true; Statement − II is true; Statement − II is not a correct explanation for Statement − I. 
Option 3:  Statement − I is true; Statement − II is false. 
Option 4:  Statement − I is false; Statement − II is true. 
Solutions:
Consider statement I.
Hence, (p∨∼q) ∨ (∼p∨q) is a fallacy.
So, the statement I is true.
Consider the statement II.
(p → q) ↔ (∼q → ∼p)
(p → q) ↔ (p → q), which is always true.
Hence, (p → q) ↔ (∼q → ∼p) is a tautology.
So, the statement II is true. But, statement II is not a correct explanation for statement I.
Thus, statementI is true, statementII is true; statementII is not a correct explanation for statementI.
The Correct Answer is B.
Question 74:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
The Correct Answer is C.
Question 75:
Option 1:  
Option 2:  
Option 3:  1 
Option 4:  2 
Solutions:
Question 76:
Statement − I:
Then value of the integral is equal to
Statement − II:
Option 1:  Statement − I is true; Statement − II is true; Statement − II is a correct explanation for Statement − I. 
Option 2:  Statement − I is true; Statement − II is true; Statement − II is not a correct explanation for Statement − I. 
Option 3:  Statement − I is true; Statement − II is false. 
Option 4:  Statement − I is false; Statement − II is true. 
Solutions:
Statement II is true.
Adding (1) and (2), we have
So, the statement I is false.
Thus, statementI is false, StatementII is true.
The Correct Answer is D.
Question 77:
The equation of the circle passing through the foci of the ellipse and having centre at (0, 3) is :
Option 1:  x^{2} + y^{2} − 6y − 7 = 0 
Option 2:  x^{2} + y^{2} − 6y + 7 = 0 
Option 3:  x^{2} + y^{2} − 6y − 5 = 0 
Option 4:  x^{2} + y^{2} − 6y + 5 = 0 
Solutions:
Equation of the given ellipse is .
Here, a = 4 and b = 3
Let e be the eccentricity of the ellipse
Foci of the ellipse = (± ae, 0) =
∴ Radius of the circle
Centre of the circle = (0, 3)
∴ Equation of the circle is
(x − 0)^{2} + (y − 3)^{2} = (4)^{2}
⇒ x^{2} + y^{2} − 6y − 7 = 0
The Correct Answer is A.
Question 78:
A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
Let p denote the probability of correct the correct answer.
Then,
Suppose X denote the number of correct answers. Then, X is a binomial variate with parameter n = 5 and .
∴Probability of getting 4 or more correct answers
= P (X ≥ 4)
= P (X = 4) + P (X = 5)
The Correct Answer is C.
Question 79:
The xcoordinate of the incentre of the triangle that has the coordinates of mid points of its sides as (0, 1) (1, 1) and (1, 0) is:
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
The mid points of the sides of the triangle are (0, 1), (1, 1) and (1, 0).
∴ Vertices of ΔOAB are (0, 0), (2, 0) and (0, 2).
OA = 2, OB = 2 and
∴ xcoordinate of the incentre
The Correct Answer is B.
Question 80:
The term independent of x in expansion of
Option 1:  4 
Option 2:  120 
Option 3:  210 
Option 4:  310 
Solutions:
General term of the expansion, T_{r}_{+}_{1} = (−1)^{r}^{10}C_{r} (x^{1}^{/3})^{10 − }^{r }(x^{−}^{1}^{/2})^{r}
For the term independent of x, we have
Thus, the term independent of x is 210.
Hence, the correct option is C.
Question 81:
The area (in square units) bounded by the curves xaxis, and lying in the first quadrant is :
Option 1:  9 
Option 2:  36 
Option 3:  18 
Option 4: 
Solutions:
Graph of and 2y − x + 3 = 0:
To find the intersection point P, let us solve and 2y − x + 3 = 0.
∴ Coordinates of point P = (9, 3)
Now,
Hence, the correct option is A.
Question 82:
Let T_{n} be the number of all possible triangles formed by joining vertices of an nsided regular polygon. If T_{n + 1} − T_{n} = 10, then the value of n is:
Option 1:  7 
Option 2:  5 
Option 3:  10 
Option 4:  8 
Solutions:
For forming a triangle, we need to select three vertices out of n vertices.
This can be done in ^{n}C_{r } ways.
∴ ^{n+}^{1}C_{3 }− ^{n}C_{3} = 10
⇒ We know that
⇒ ^{n}C_{2} = 10 ⇒ n = 5
Therefore, the value of n is 5.
The Correct Answer is B.
Question 83:
If z is a complex number of unit modulus and argument θ, then arg equals:
Option 1:  −θ 
Option 2:  π2θ 
Option 3:  θ 
Option 4:  π − θ 
Solutions:
Given, z = 1 and arg (z) = θ
z = 1
Thus,
The Correct Answer is C.
Question 84:
ABCD is a trapezium such that AB and CD are parallel and BC⊥CD. If ∠ADB = θ, BC = p and CD = q, then AB is equal to :
Option 1:  
Option 2:  
Option 3:  
Option 4: 
Solutions:
∠BDC = ∠DBA = α (say)
In ΔABD,
The Correct Answer is A.
Question 85:
is the adjoint of a 3 × 3 matrix A and A = 4, then α is equal to:
Option 1:  4 
Option 2:  11 
Option 3:  5 
Option 4:  0 
Solutions:
Given, Adj
We know that,
⇒ 1 (12 − 12) − α (4 − 6) + 3 (4 − 6) = 16
⇒ 2α = 22
⇒ α = 11
The Correct Answer is B.
Question 86:
The intercepts on xaxis made by tangents to the curve, which are parallel to the line y = 2x, are equal to:
Option 1:  ± 1 
Option 2:  ± 2 
Option 3:  ± 3 
Option 4:  ± 4 
Solutions:
Slope of tangents to the curve is same as the slope of line y = 2x.
There will be two tangents.
Equation of 1^{st} tangent: y − 2 = 2 (x − 2)
Equation of 2^{nd} tangent: y + 2 = 2 (x + 2)
∴ Intercepts = ±1
The Correct Answer is A.
Question 87:
Given : A circle, 2x^{2} + 2y^{2} = 5 and a parabola,
Statement − I : An equation of a common tangent to these curves is
Statement − II : If the line, is their common tangent, then m satisfies m^{4} − 3m^{2} + 2 = 0.
Option 1:  Statement − I is true; Statement − II is true; Statement − II is a correct explanation for Statement − I. 
Option 2:  Statement − I is true; Statement − II is true; Statement − II is not a correct explanation for Statement − I. 
Option 3:  Statement − I is true; Statement − II is false. 
Option 4:  Statement − I is false; Statement − II is true. 
Solutions:
The equation of tangent to a parabola y^{2} = 4ax is in the form, where ‘m’ is the slope of the tangent.
Equation of parabola
Equation of tangent:
Center of the circle 2x^{2} + 2y^{2} = 5 is (0, 0) and radius
Equation of common tangents are
∴ Statements I and II are correct.
The Correct Answer is B.
Question 88:
at x = 1 is equal to :
Option 1:  
Option 2:  
Option 3:  1 
Option 4: 
Solutions:
The Correct Answer is A.
Question 89:
The expression can be written as :
Option 1:  sinA cosA + 1 
Option 2:  secA cosecA + 1 
Option 3:  tanA + cotA 
Option 4:  secA + cosecA 
Solutions:
The Correct Answer is B.
Question 90:
All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given?
Option 1:  mean 
Option 2:  median 
Option 3:  mode 
Option 4:  variance 
Solutions:
Let the initial marks of the students be denoted by m_{i}
Here, is the mean of marks of the student and N is the total number of students.
Final marks can be written as (m_{i} + 10)
Therefore variance will not change.
The Correct Answer is D.