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Test: BITSAT Past Year Paper- 2015 - Question 1

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height (h) of the satellite above the earth’s surface is (Take radius of earth as Re)

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 1

The escape velocity from earth is given by
ve =
The orbital velocity of a satellite revolving around earth is given by

where, Me = mass of earth, Re = radius of earth, h = height of satellite from surface of earth.
By the relation GMe = gR2s
So, 
Dividing equation (i) by (ii), we get

Given, 

Squaring on both side, we get

or Re + h = 2Re i.e., h = Re

Test: BITSAT Past Year Paper- 2015 - Question 2

In figure, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block A weighs 400N, block B weighs 300N, and the strut AB weigh 200N. If µ = 0.25 under B, determine the minimum coefficient of friction under A to prevent motion.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 2

Consider FBD of structure.

Applying equilibrium equations,
Av + Bv = 200 N    … (i)
AH = BH    … (ii)
From FBD of block B,

BH + FB cos 60° – NB sin 60° = 0
NB cos 60° – BV – 300 + FB sin 60° = 0
FB = 0.25 NB
BH – 0.74 NB = 0 … (iii)
– BV + 0.71 NB = 300 … (iv)
FBD of block A

FA –  AH = 0
NA – AV = 400 … (v)
FA = µA NA
∴ µANA – AH = 0 … (vi)
On solving above equations, we get NA = 650 N, FA = 260N,  FA = µANA

Test: BITSAT Past Year Paper- 2015 - Question 3

Two tuning forks with n atur al fr equen cies 340 Hz each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards the observer at the same speed. The observer hears beats of frequency 3 Hz. Find the speed of the tuning forks.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 3

Let v = speed of sound and vS = speed of tuning forks. Apparent frequency of fork moving towards the observer is 
Apparent frequency of the fork moving away from the observer is 
If f is the number of beats heard per second. th en f = n1 – n2

putting v = 340 m/s, f = 3, n = 340 Hz we get,

Test: BITSAT Past Year Paper- 2015 - Question 4

The displacement of a particle is given at time t, by: x = A sin (-2ωt ) + B sin 2ωt Then,

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 4

The displacement of the particle is given by:

This motion represents SHM with an amplitude:  and mean position B/2.

Test: BITSAT Past Year Paper- 2015 - Question 5

A ray parallel to principal axis is incident at 30° from normal on concave mirror having radius of curvature R. The point on principal axis where rays are focussed is Q such that PQ is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 5

From similar triangles,


or 
Thus
PQ  = PC – QC = 

Test: BITSAT Past Year Paper- 2015 - Question 6

A solid sph ere of r adius R has a charge Q distributed in its volume with a charge density ρ = kra, where k and a are constants and r is the distance from its centre. If the electric field at r = R/2 is 1/8 times that at r = R, the value of a is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 6

Using Gauss's law, we have

For 
Given, E2 = E1/8
or 
∴ 
or a = 2.

Test: BITSAT Past Year Paper- 2015 - Question 7

A charged particle moving in a uniform magnetic field and losses 4% of its kinetic energy. The radius of curvature of its path changes by

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 7

As we know F = qvB = mv2/r

And KE = k = 1/2 mv2


Therefore percentage changes in radius of path,

Test: BITSAT Past Year Paper- 2015 - Question 8

Calculate th e wavelength of ligh t used in an interference experiment from the following data : Fringe width = 0.03 cm. Distance between the slits and eyepiece through which the interference pattern is observed is 1m. Distance between the images of the virtual source when a convex lens of focal length 16 cm is used at a distance of 80 cm from the eyepiece is 0.8 cm.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 8

Given: fringe with β = 0.03 cm, D = 1 m = 100 cm
Distance between images of the source = 0.8 cm.
Image distance v = 80 cm
Object distance = u
Using mirror formula, 
⇒ u = 20 cm
Magnification, 
Magnification 

Fringe width 
or, 
Therefore, wavelength of light used λ = 6000 Å

Test: BITSAT Past Year Paper- 2015 - Question 9

The masses of blocks A and B are m and M respectively. Between A and B, there is a constant frictional force F and B can slide on a smooth horizontal surface. A is set in motion with velocity while B is at rest. What is the distance moved by A relative to B before they move with the same velocity?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 9

For th e blocks A and B FBD as sh own below

Equations of motion

Relative acceleration, of A w.r.t. B,


Initial relative velocity of A w.r.t. B, uAB = v0
using equation v2 = u2 + 2as

i.e., Distance moved by A relative to B

Test: BITSAT Past Year Paper- 2015 - Question 10

An elastic string of un stretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 10

In the string elastic force is conservative in nature.
∴ W = – ΔU
Work done by elastic force of string,
W = – (UF – Ui) = Ui – UF.


Therefore, the work done against elastic force 

Test: BITSAT Past Year Paper- 2015 - Question 11

A body is thrown vertically upwards from A, the top of the tower, reaches the ground in time t1. If it is thrown vertically downwards from A with the same speed, it reaches the ground in time t2. If it is allowed to fall freely from A, then the time it takes to reach the ground is given by

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 11

Let the body is projected ver tically upwards from A with a speed u0.
Using equation , s = ut + (1/2)at2
For case (1) – h = 
For case (2) – h = 
Subtracting eq (2) from (1), we get

Putting the value of u0  in eq (2), we get

For case 3, u0 = 0, t = ?

Comparing eq. (4) and (5), we get

Test: BITSAT Past Year Paper- 2015 - Question 12

0.5 mole  of an ideal gas at constant temperature 27°C kept inside a cylinder of length L and crosssection area  A closed  by a massless piston.

The cylinder is attached with a conducting rod of length L , cross-section ar ea (1/9) m2 and thermal conductivity k, whose other end  is maintained at 0°C. If piston is moved such that rate of heat flow through the conducing rod is constant then velocity of piston when it is at height L/2 from the bottom of cylinder is : [Neglect any kind of heat loss from system ]

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 12

work done per unit time = 

⇒ 
⇒ 
⇒ 

Test: BITSAT Past Year Paper- 2015 - Question 13

A conducting square loop is placed in a magnetic field B with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate α. The induced emf in the loop at an instant when its side is ‘a’ is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 13

At any time t, the side of the square a = (a0 – α t),  where a0 = side at t = 0.
At this instant, flux through the square :

⇒  E = – B.2 (a0 – α t) (0 – α) = +2αaB

Test: BITSAT Past Year Paper- 2015 - Question 14

Th e beam of light h as th r ee wavelength s 4144Å, 4972Å and 6216 Å with a total intensity of 3.6 × 10–3 Wm2 equally distributed amongst the three wavelengths. The beam falls normally on the area 1 cm2 of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in 2s.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 14

As we know, thresh old wavelength 

⇒ λ0  = 5404 Å
Hence, wavelength 4144 Å and 4972 Å will emit electron from the metal surface.
For each wavelength energy incident on the surface per unit time = intensity of each × area of the surface wavelength 
Therefore, energy incident on the surface for each wavelength in 2s
E = (1.2 × 10–7) × 2 = 2.4 × 10–7 J
Number of photons n1 due to wavelength 4144 Å

Number of photons n2 due to the wavelength 4972 Å

Therefore total number of photoelectrons liberated in 2s,
N = n1 + n2
= 0.5 x 1012 + 0.575 x 1012
= 1.075 x 1012

Test: BITSAT Past Year Paper- 2015 - Question 15

A squar e gate of size 1 m × 1m is h inged at its mid-point. A fluid of density ρ fills the space to the left of the gate. The force F required to hold the gate stationary is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 15

The net force acting on the gate element of width dy at a depth y from the surface of the fluid, is
dy = p0 + ρgy - p0) x 1dy
= ρgydy

Torque about the hinge is

Net torque experienced by the gate is


⇒ F = ρg/6
i.e., The force F required to hold the gate stationary is pg/6

Test: BITSAT Past Year Paper- 2015 - Question 16

When 0.50 Å X-r ays strike a material, the photoelectrons from the k shell are observed to move in a circle of radius 23 mm in a magnetic field of 2 × 10–2 tesla acting perpendicularly to the direction of emission of photoelectrons. What is the binding energy of k-shell electrons?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 16

As we know,
F = qvB = m
The kinetic energy of the photoelectron 

= 2.97 × 10–15 J

Energy of the incident photon = hc/λ = 12.4/0.50 = 24.8keV
Therefore, Binding energy = 24.8 – 18.6 = 6.2  keV

Test: BITSAT Past Year Paper- 2015 - Question 17

In CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. If the base resistance is 1kΩ and the current amplification of the transistor is 100, the input signal voltage is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 17

Given : Voltage across the collector V0 = 2 V; collector resistance, Rc = 2 × 103 Ω; Base resistance RB = 1 × 103 Ω; Input signal voltage, Vi = ?

V0 = ICRC = 2

Current gain 

Test: BITSAT Past Year Paper- 2015 - Question 18

At the corners of an equilateral tri angle of side a (1 metre), three point charges are placed (each of 0.1 C). If this system is supplied energy at the rate of 1 kw, then calculate the time required to move one of the mid-point of the line joining the other two.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 18

Initial potential energy of the system 

Let charge at A is moved to mid-point O, Then final potential energy of thhe system 

Work done = Uf – Ui = 18 × 107 J
Also, energy supplied per sec = 1000 J (given)
Time required to move one of the mid-point of the line joining the other two

Test: BITSAT Past Year Paper- 2015 - Question 19

A vessel of volume 20L contains a mixture of hydrogen and helium at temperature of 27°C and pressure 2 atm. The mass of mixture is 5g. Assuming the gases to be ideal, the ratio of mass of hydrogen to that of helium in the given mixture will be

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 19

Let ther e are n1 moles of hydrogen and n2 moles of helium in the given mixture. As Pv = nRT
Then the pressure of the mixture 
⇒ 
or, 
or, n1 + n2 = 1.62 ... (1)
The mass of the mixture is (in grams)
  n1 × 2 + n2 × 4 = 5
⇒ (n1 + 2n2) = 2.5 ... (2)
Solving the eqns. (1) and (2), we get n1 = 0.74 and n2 = 0.88
Hence, 

Test: BITSAT Past Year Paper- 2015 - Question 20

The resistance of a wire is R. It is bent at the middle by 180° and both the ends are twisted together to make a shorter wire. The resistance of the new wire is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 20

Resistan ce of wire (R) = ρ(l/A)
If wire is bent in the middle then 
∴ New resistance,  

Test: BITSAT Past Year Paper- 2015 - Question 21

In a YDSE, the light of wavelength λ = 5000 Å is used, which emerges in phase from two slits a distance d = 3 × 10–7m apart. A transparent sheet of thickness t = 1.5 × 10–7m refractive index μ = 1.17 is placed over one of the slits. what is the new angular position of the central maxima of the interference pattern, from the centre of the screen? Find the value of y.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 21

The path difference when transparent sheet is  introduced Δx = (μ – 1)t
If the central maxima occupies position of nth fringe, then (μ – 1)t = n λ = d sin θ

Therefore, angular position of central maxima θ = sin-1 0.085 = 4.88°≈ 4.9
For small angles, sin θ ≈ θ ≈ tan θ
⇒ 

Test: BITSAT Past Year Paper- 2015 - Question 22

The position of a projectile launched from the origin at t = 0 is given by m at t = 2s. If the projectile was launched at an angle θ from the horizontal, then θ is (take g = 10 ms–2)

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 22

From question,
Horizontal velocity (initial), ux = 40/2 = 20m/s
Vertical velocity (initial), 50 
⇒ 
or, 50 = 2uy  – 20
or, 

Test: BITSAT Past Year Paper- 2015 - Question 23

Water is flowing on a horizontal fixed surface, such that its flow velocity varies with y (vertical direction) as  If coefficient of viscosity for water is h, what will be shear stress between layers of water at y = a.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 23

Newton’s law of viscosity, 
Stress = 
At y = a, stress = 

Test: BITSAT Past Year Paper- 2015 - Question 24

A load of mass m falls from a h eight h on to the scale pan hung from the spring as shown in the figure. If the spring constant is k and mass of the scale pan is zero and the mass m does not bounce relative to the pan, then the amplitude of vibration is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 24

According to energy conservation principle,
If, x1 is maximum elongation in the spring when the particle is in its lowest extreme position. Then,


Amplitude A = X1 – X0 (elongation in spring for equilibrium position)

Test: BITSAT Past Year Paper- 2015 - Question 25

In an ore containing uranium, the ratio of U238 to Pb206 is 3. Calculate the age of the ore, assuming that all the lead present in the ore is the final stable product of U238. Take the half-life of U238 to be 4.5 × 109 yr.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 25

Let the initial mass of uranium be M0
Final mass of uranium after time t, M = 3/4(M0)
According to the law of radioactive disintegration.


⇒ t = 1.867 × 109 yr.

Test: BITSAT Past Year Paper- 2015 - Question 26

A direct current of 5A is superposed on an alternating current I = 10 sin ωt flowing through the wire. The effective value of the resulting current will be

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 26

Total carrent, 1 = (5 + 10 sin ωt)

But,  and 
So, 

Test: BITSAT Past Year Paper- 2015 - Question 27

A planoconvex lens fits exactly in to a planoconcave lens. Their plane surface are parallel to each other. If the lenses are made of different materials of refractive indices µ1 & µ2 and R is the radius of curvature of the curved surface of the lenses, then focal length of combination is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 27

If F be the equivalent focal length, then
  

Test: BITSAT Past Year Paper- 2015 - Question 28

A thin rod of length 4l and mass 4m is bent at the points as shown in figure. What is the moment of inertia of the rod about the axis passes through point O and perpendicular to the plane of paper?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 28

Total moment of inertia
= I1 + I2 + I3 + I4 = 2I1 + 2I2
= 2(l1 + l2) [I3 = I1, I1 = I4]
Now, I2 = I3 = MI2/3
Using parallel axes theorem, we have

Putting all values we get
Moment of inertia, 

Test: BITSAT Past Year Paper- 2015 - Question 29

One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is n = 12 → n = x. Find the value of x.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 29

For 2nd line of Balmer series in hydrogen spectrum 

which is satisfied by n = 12 → n = 6.

Test: BITSAT Past Year Paper- 2015 - Question 30

Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii R1 and R2, respectively. The ratio of masses of X and Y is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 30

When a charge particle is allowed to move in a uniform magnetic field, then it describes spiral or circular path
Centripetal force, mv2/R = qvB ∴ 
Hence, 

or,   m ∝ R2
[∵ V, q and B are constant]
or, 

Test: BITSAT Past Year Paper- 2015 - Question 31

A glass capillary tube of internal radius r = 0.25 mm is immersed in water. The top end of the tube projected by 2 cm above the surface of the water.  At what angle does the liquid meet the tube? Surface tension of water = 0.7 N/m.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 31

Water wets glass and so the angle of contact is zero.
For full rise, neglecting the small mass in the meniscus


As the tube is only 2 cm above the water and so, water will rise by 2 cm and meet the tube at an angle such that,



The liquid will meet the tube at an angle,  θ ≌ 70°

Test: BITSAT Past Year Paper- 2015 - Question 32

A particle of mass 2 m is projected at an angle of 45° with the horizontal with a velocity of 20√2 m/s. After 1s, explosion takes place and the particle is broken into two equal pieces. As a result of explosion, one part comes to rest. The maximum height from the ground attained by the other part is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 32

Given : Initial velocity u0 = 20√2 m/s; angle of projection θ = 45°
Therefore horizontal and vertical components of initial velocity are ux = 20√2 cos 45°= 20m/s and uy = 20√2 sin 45°= 20m/s
After 1s, horizontal component remains unchanged while the vertical component becomes vy = uy – gt
Due to explosion, one part comes to rest.
Hence, from the conservation of linear momentum, vertical component of second part will become  v'y = 20m/s.
Therefore, maximum height attained by the second part will be H = h1 + h2
Using, h = ut + (1/2)at2

a = g = 10 m/s2

H = 20 + 15 = 35 m

Test: BITSAT Past Year Paper- 2015 - Question 33

A 2 m wide truck is moving with a uniform speed v0 = 8 m/s along a straight horizontal road. A pedestrain starts to cross the road with a uniform speed v when the truck is 4 m away from him. The minimum value of v so that he can cross the road safely is 

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 33

Let the man starts crossing the road at an angle q as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance 4 + AC or 4 + 2cot θ.
...(i)
For minimum v, 

or or 2 cos θ – sin θ = 0
or tan θ = 2
From equation (i),  

Test: BITSAT Past Year Paper- 2015 - Question 34

A neutron moving with speed v makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision takes place is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 34

Let speed of neutr on before collision = V Speed of neutron after collision = V1
Speed of proton or hydrogen atom after collision = V2
Energy of excitation = DE From the law of conservation of linear momentum, mv = mv1 + mv2 ...(1)
And for law of conservation of energy,  ...(2)
From squaring eq. (i), we get
 ...(3)
From squaring eq. (ii), we get
 ...(4)
From eqn (3) & (4)


As, v1 – v2 must be real, 
⇒ 
The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is 10.2 eV. Therefore, the maximum kinetic energy needed is  2 x 10.2 = 20.4 eV

Test: BITSAT Past Year Paper- 2015 - Question 35

Vertical displacement of a Planck with a body of mass m on it is varying according to law y = sin ωt + √3 cos ωt. The minimum value of w for which the mass just breaks off the Planck and the moment it occurs first after t = 0, are given by

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 35

From, figure, 

∴ y = 2 sin 

amax = -2ω2 = g
For which mass just breaks off the plank 
This will be happen for the first time when 
∴ 

Test: BITSAT Past Year Paper- 2015 - Question 36

A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charge to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 36

From the figure.
The net charge shared between the two capacitors
Q' = Q2 – Q2 = 4CV – CV = 3CV

The two capacitors will have some potential, say V'.
The net capacitance of the parallel combination of the two capacitors C' = C1 + C2 = C + 2C+ 3C
The potential of the capacitors 
The electrostatic energy of the capacitors 

Test: BITSAT Past Year Paper- 2015 - Question 37

In the circuit shown below, the ac source has voltage V = 20 cos(ωt) volt with ω = 2000 rad/s. The amplitude of the current will be nearest to 

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 37



= 10Ω

Test: BITSAT Past Year Paper- 2015 - Question 38

A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 38

The heat produced is given by 

Thus heat (H) is doubled if both length (l) and radius (r) are doubled.

Test: BITSAT Past Year Paper- 2015 - Question 39

The frequency of a son ometer wir e is 100 Hz. When the weights producing the tensions are completely immersed in water, the frequency becomes 80 Hz and on immersing the weights in a certain liquid, the frequency becomes 60 Hz. The specific gravity of the liquid is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 39

As we know, frequency

In water, fw = 0.8fair

In liquid, g'/g = (0.6)2 = 0.36
 ...(2)
From eq. (1) and (2)

Test: BITSAT Past Year Paper- 2015 - Question 40

A long straight wire along the Z-axis carries a current I in the negative Z-direction. The magnetic vector field at a point having coordinates (x, y) in the Z = 0 plane is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 40

The wire carries a current I in the negative z-direction. We have to consider the magnetic vector field  at (x, y) in the z = 0
plane.

Magnetic field is perpendicular to OP.

Test: BITSAT Past Year Paper- 2015 - Question 41

Which of the following pollutants is main product of automobiles exhaust?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 41

NO pollutant is the main product of automobiles exhaust.

Test: BITSAT Past Year Paper- 2015 - Question 42

The disease caused the high concentr ation of hydrocarbon pollutants in atmosphere is/are

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 42

The high concentration of hydrocarbon pollutants in atmosphere causes cancer.

Test: BITSAT Past Year Paper- 2015 - Question 43

The element, with atomic number 118, will be

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 43

Electronic configuration of element with atomic number 118 will be [Rn]5f146d10 7s27p6. Since its elctronic configuration in the outer most orbit (ns2np6) resemble with that of inert or noble gases, therefore it will be noble gas element.

Test: BITSAT Past Year Paper- 2015 - Question 44

Which law of the thermodynamics helps in calculating the absolute entropies of various substances at different temperatures?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 44

The third law helps to calculate the absolute entropies of pure substances at different temperature.
The entropy of the substance at different temperature. T may be calculated by the measurement of heat capacity change 
Where ST = Entropy at T K
S0 = Entropy at 0K = Cp . logeT
= 2.303Cp.logT

Test: BITSAT Past Year Paper- 2015 - Question 45

The color of CoCl3.5NH3.H2O is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 45

CoCl3.5NH3.H2O is pink in colour

Test: BITSAT Past Year Paper- 2015 - Question 46

The metal presen t in vitamin B12 is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 46

Cobalt is present in vitamin B12.

Test: BITSAT Past Year Paper- 2015 - Question 47

Cobalt (60) isotope is used in the treatment of :

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 47

Cobalt (60) isotope is used in the treatment of cancer.

Test: BITSAT Past Year Paper- 2015 - Question 48

Polymer used in bullet proof glass is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 48

PMMA is used in bullet proof glass

Test: BITSAT Past Year Paper- 2015 - Question 49

What is the correct increasing order of Bronsted bases?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 49

ClO4- < ClO3-< ClO2-< ClOis the correct increasing order of Bronsted base. With increase in the number of oxygen atoms in the conjugate bases, the delocalisation of the π bond becomes more and more extended. This results in decrease in the electron density. Consequently basicity also decreases.

Test: BITSAT Past Year Paper- 2015 - Question 50

The boiling point of alkyl halide are higher than those of corresponding alkanes because of

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 50

Due to dipole-dipole interaction the boiling point of alkyl halide is higher as compared to corresponding alkanes.

Test: BITSAT Past Year Paper- 2015 - Question 51

Some salts containing two different metallic elements give test for only one of them in solution, such salts are

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 51

Complex compounds contains two different metallic elements but give test only for one of them. Because complex ions such as [Fe (CN)6]4– of K4 [Fe (CN)6], do not dissociate into Fe2+ and CN ions.

Test: BITSAT Past Year Paper- 2015 - Question 52

The carbylamine reaction is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 52

Primary amines (aromatic or aliphatic) on warming with chloroform and alcoholic KOH, gives carbylamine having offensive smell. This reaction is called carbylamine reaction.

Test: BITSAT Past Year Paper- 2015 - Question 53

Laughing gas is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 53

Nitrous oxide (i.e., N2O) is the laughing gas.

Test: BITSAT Past Year Paper- 2015 - Question 54

The anthracene is purified by

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 54

Anthracene is purified by sublimation. In sublimation, a solid is converted directly into gaseous state on heating without passing through liquid phase.

Test: BITSAT Past Year Paper- 2015 - Question 55

The common name of K[PtCl32.C2H4)] is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 55

Zeise’s salt is common name of K[Pt Cl32 = C2H4)]

Test: BITSAT Past Year Paper- 2015 - Question 56

The by product of Solvay-ammonia process is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 56

CaCl2 is produced as a by product in solvay ammonia process.
(i) NaCl + CO2 + NH3 + H2O → NaHCO3 + NH4Cl
(ii) CaCO3 → CO2 + CaO
(iii) 2 NH2Cl + CaO → 2 NH3 + 

Test: BITSAT Past Year Paper- 2015 - Question 57

Semiconductor materials like Si and Ge are usually purified by

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 57

Semiconductor materials like Si and Ge are usually purified by zone refining. Zone refining is based on the principle of fractional crystallisation i.e. difference in solubilities of impurities in solid and molten states of metal, so that the zones of impurities are formed and finally removed.

Test: BITSAT Past Year Paper- 2015 - Question 58

Which of the following is a strong base?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 58

Order of basic character is NH3 > PH3 > AsH3 > SbH3. Basic-character decreases down the group from N to Bi due to increase in atomic size.

Test: BITSAT Past Year Paper- 2015 - Question 59

Ordinary glass is :

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 59

Normal glass is calcium alkali silicate glass made by fusing the alkali metal carbonate, CaCO3 and SiO2.

Test: BITSAT Past Year Paper- 2015 - Question 60

The prefix 1018 is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 60

Exa = 1018

Test: BITSAT Past Year Paper- 2015 - Question 61

Which of the following is the most basic oxide?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 61

More the oxidation state of the central atom (metal) more is its acidity. Hence SeO2 (O. S. of Se = +4) is acidic. Further for a given O.S., the basic character of the oxides increases with the increasing size of the central atom. Thus Al2O3 and Sb2O3 are amphoteric and Bi2O3 is basic.

Test: BITSAT Past Year Paper- 2015 - Question 62

Wh ich on e of th e following does not follow octate rule?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 62

BF3 does not follow octate rule because central atom, boron lacks an electron pair.
Thus, it also acts as Lewis acid.

Test: BITSAT Past Year Paper- 2015 - Question 63

Whch of the following according to LeChatelier’s principle is correct?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 63

According to Le-Chatelier’s principle increase in temperature favours the endothermic reaction while decrease in temperature favour the exothermic reaction. Increase in pressure shifts the equilibrium in that side in which number of gaseous moles decreases.

Test: BITSAT Past Year Paper- 2015 - Question 64

The efficiency of fuel cell is given by the expression, η is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 64

Efficiency of fuel cell is:

Test: BITSAT Past Year Paper- 2015 - Question 65

The mass of the substance deposited when one Faraday of charge is passed through its solution is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 65

The mass of the substance deposited when one Faraday of charge is passed through its solution is equal to gram equivalent weight.

Test: BITSAT Past Year Paper- 2015 - Question 66

The unit of rate constant for reactions of second order is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 66

Unit of rate constant for second order reaction is L mol–1 sec–1.

Test: BITSAT Past Year Paper- 2015 - Question 67

In a first order reaction with time the concentration of the reactant decreases

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 67

For first order reaction [A] = [A0]e–kt
∴ The concentration of reactants will exponentially decreases with time.

Test: BITSAT Past Year Paper- 2015 - Question 68

The P—P—P angle in P4 molecule and S—S—S angle in S8 molecule is(in degree) respectively

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 68

In P4 molecule, the four sp3-hybr idised phosphorous atoms lie at the corners of a regular tetrahedron with ÐPPP = 60°.
In S8 molecule S-S-S angle is 107° rings.

Test: BITSAT Past Year Paper- 2015 - Question 69

The number of elements present in the d-block of the periodic table is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 69

40 elements are pr esent in d-block.

Test: BITSAT Past Year Paper- 2015 - Question 70

Which of the following represents hexadentate ligand?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 70

EDTA is hexadentate ligand

Test: BITSAT Past Year Paper- 2015 - Question 71

Which one of given elements shows maximum number of different oxidation states in its compounds?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 71

Am shows maximum n umber of oxidation states, + 3, + 4, + 5, + 6

Test: BITSAT Past Year Paper- 2015 - Question 72

K4[Fe(CN)6] is used in detecting.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 72

Fe3+ ion can be detected by K4[Fe(CN)6]
4Fe3+ + 3K4 [Fe(CN)6] → Fe4 [Fe(CN)6]3+ 12K+

Test: BITSAT Past Year Paper- 2015 - Question 73

A spontaneous reaction is impossible if

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 73

ΔG = ΔH – TΔS; ΔG is positive for a reaction to be non-spontaneous when ΔH is positive and ΔS is negative.

Test: BITSAT Past Year Paper- 2015 - Question 74

Which one the following removes temporary hardness of water ?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 74

This method is known as Clark's process.
In this method temporary hardness is removed by adding lime water or milk of lime.

Test: BITSAT Past Year Paper- 2015 - Question 75

Graphite is a

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 75

Graphite is covalent solid.

Test: BITSAT Past Year Paper- 2015 - Question 76

Which of the following ionic substances will be most effective in precipitating the sulphur sol?

Test: BITSAT Past Year Paper- 2015 - Question 77

Which of the following fluorides of xenon is impossible?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 77



Structures of Xenon fluorides
XeF2 : Hybridization sp3d2

Linear
Xe F4 : Hybridization sp3d2

Square planar
XeF6 : Hybribisation sp3d3  

Pentagonal pyramidal or distorted octahedral  

Test: BITSAT Past Year Paper- 2015 - Question 78

Thomas slag is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 78

Calcium silicophosp h ate (a mixture of Ca3(PO4)2 & Ca2SiO4) is called Thomas slag.

Test: BITSAT Past Year Paper- 2015 - Question 79

A sequence of how many nucleotides in messenger RNA makes a codon for an amino acid?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 79

The sequence of bases in mRNA are read in a serial order in groups of three at a time.
Each triplet of nucleotides (having a specific sequence of bases) in known as codon.
Each codon specifies one amino acid.
Further since, there are four bases. therefore, 43 = 64 triplets or codons are possible.

Test: BITSAT Past Year Paper- 2015 - Question 80

Which of the following molecule/ion has all the three types of bonds, electrovalent, covalent and co-ordinate :

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 80

Bond structure of molecules are :
HCl = H+ – Cl

hence, clearly NH+4 ion contains all three types of bonds.

Test: BITSAT Past Year Paper- 2015 - Question 81

DIRECTIONS : Choose the word which best expreses the meaning of the underlined word in the sentence.

Q. Decay is an immutable factor of human life.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 81

‘Immutable’ means ‘unchangeable’. So, option (c) is correct choice.

Test: BITSAT Past Year Paper- 2015 - Question 82

DIRECTIONS : Choose the word which best expreses the meaning of the underlined word in the sentence.

Q. It was an ignominious defect for the team.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 82

‘ignominious’ means ‘shameful’. So, option (a) is correct choice.

Test: BITSAT Past Year Paper- 2015 - Question 83

DIRECTIONS : Choose the word which best expreses the meaning of the underlined word in the sentence.

Q. The attitude of western countries towards the third world countries is rather callous to say the least.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 83

‘callous’ means ‘showing or having an insensitive and cruel disregard for others’.
So, option (c) is correct choice.

Test: BITSAT Past Year Paper- 2015 - Question 84

DIRECTIONS : Fill in the blank.

Q. Freedom and equality are the ______ rights of every human.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 84

Option (d) institutional as the word means relating to principles esp. of law, so legally also every human has rights of freedom and equality.

Test: BITSAT Past Year Paper- 2015 - Question 85

DIRECTIONS : Fill in the blank.

Q. The team was well trained and strong, but some how their ______ was low.

Test: BITSAT Past Year Paper- 2015 - Question 86

DIRECTIONS : Fill in the blank.

Q. His speech was disappointing: it ______ all the major issues.

Test: BITSAT Past Year Paper- 2015 - Question 87

DIRECTIONS : Choose the word which is closest to the opposite in meaning of the underlined word in the sentence.

Q. Hydra is biologically believed to be immortal.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 87

Immortal means living forever, never dying or decaying. So, perishable is the correct opposite to it.

Test: BITSAT Past Year Paper- 2015 - Question 88

DIRECTIONS : Choose the word which is closest to the opposite in meaning of the underlined word in the sentence.

Q. The Gupta rulers patronised all cultural activities and thus Gupta period was called the golden era in Indian History.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 88

Opposed is the correct answer of this. To patronise means favour or pat on the back.

Test: BITSAT Past Year Paper- 2015 - Question 89

DIRECTIONS : Choose the word which is closest to the opposite in meaning of the underlined word in the sentence.

Q. The General Manager is quete tactful and handles the workers union very effectively.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 89

Tactful means having or showing skill and senstivity in dealing with others or with defficult essues.
So, in cautious is the correct opposite of factful.

Test: BITSAT Past Year Paper- 2015 - Question 90

DIRECTIONS : In each ot the following questions, out of the four alternatives, choose the one which can be substituted for the given words/ sentence.

Q. A person who does not believe in any religion

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 90

Atheist is the best alternative.

Test: BITSAT Past Year Paper- 2015 - Question 91

DIRECTIONS : In each ot the following questions, out of the four alternatives, choose the one which can be substituted for the given words/ sentence.

Q. A person who believes that pleasure is the chief good

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 91

‘Epicure’ is the best alternative.

Test: BITSAT Past Year Paper- 2015 - Question 92

DIRECTIONS : In each ot the following questions, out of the four alternatives, choose the one which can be substituted for the given words/ sentence.

Q. A person who is in charge of museum.

Test: BITSAT Past Year Paper- 2015 - Question 93

DIRECTIONS : Choose the order of the sentences marked A, B, C, D and E to form a logical paragraph.

Q.
A. Tasty and healthy food can help you bring out their best.
B. One minute they are toddlers and next you see them in their next adventure.
C. Your young ones seem to be growing so fast.
D. Being their loving custodians, you always want to see them doing well.
E. Their eye sparkle with curiosity and endless questions on their tongues.

Test: BITSAT Past Year Paper- 2015 - Question 94

DIRECTIONS : Choose the order of the sentences marked A, B, C, D and E to form a logical paragraph.

Q.
A. It is hoping that overseas friends will bring in big money and lift the morale of the people.
B. But a lot needs to be done to kick start industrial revival.
C. People had big hopes from the new government.
D. So far government has only given an incremental push to existing policies and programmes.
E. Government is to go for big time reforms, which it promised.

Test: BITSAT Past Year Paper- 2015 - Question 95

DIRECTIONS : Choose the order of the sentences marked A, B, C, D and E to form a logical paragraph.

Q.
A  : Forecasting the weather has always been a defficult business.
B : During a period of drought, steams and rivers dried up, the cattle died from thirst and were ruined.
C : Many different things affect the weather and we have to study them carefully to make accurate forecast.
D  : Ancient egyptians had no need of weather in the Nille valley hardly ever changes.
E: In early times, when there were no instruments, such as their mometer or the barometer, a man looked for tell tale signs in the sky.

Test: BITSAT Past Year Paper- 2015 - Question 96

Choose the correct answer figure which will make a complete square on joining with the problem figure
Problem figure

Test: BITSAT Past Year Paper- 2015 - Question 97

In the following question, five figures are given. Out of them, find the three figures that can be joined to form square.

Test: BITSAT Past Year Paper- 2015 - Question 98

Choose the answer figure which completes the problem figure matrix.
Problem Figures

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 98

The contents of the third figure in each row (and column) are determined by the contents of the first two figures. Lines are carried forward from the first two figures to the third one, except where two lines appear in the same position, in which they are cancelled out.

Test: BITSAT Past Year Paper- 2015 - Question 99

What is the opposite of 3, if four different positions of dice are as shown below :

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 99

From figure, (i), (iii) and (iv), we have concluded that 2, 6, 1 and 5 appear adjacent so 3. clearly, 4 will appear opposite to 3.

Test: BITSAT Past Year Paper- 2015 - Question 100

In the following questions, one or m ore dots are placed in the figure marked as (A). The figure is followed by four alternatives marked as (a), (b), (c) and (d). One out of these four options contains region(s) common to the circle, square, triangle, similar to that marked by the dot in figure (A).
Problem Figure

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 100

In figure (A), the dot is placed in the region which is common to the circle and triangle.
Now, we have to find similar common region in all the four options. Only in figure (c), we find such a region which is common to the circle and triangle.

Test: BITSAT Past Year Paper- 2015 - Question 101

Complete the series by replaing ‘? mark
G4T, J9R, M20P, P43N, S90L

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 101

Test: BITSAT Past Year Paper- 2015 - Question 102

Neeraj starts walking towards South. After walking 15 m, he turns towards North. After walking 20 m, he turns towards East and walks 10 m. He then turns towards South and walks 5 m. How far is he from his original position and in which direction?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 102

According to the given information, the direction of Neeraj is as following.

Test: BITSAT Past Year Paper- 2015 - Question 103

The average age of 8 men is increased by 2 yr when one of them whose age is 20 yr is replaced by a new man. What is the age of the new man

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 103

Let the average age of 8 men = x yr
Total age of 8 men = 8 x yr
Now, new average age = (x + 2)yr
Total age = 8(x + 2) yr
Difference of ages = 8(x + 2) – 8x
= 8x + 16 – 8x = 16 yr
∴ Age of new man = 20 + 16 = 36 yr
So, the new man is 16yr older to the man by whom the new man is replaced.

Test: BITSAT Past Year Paper- 2015 - Question 104

Shikha is mother-in-law of Ekta who is sister-inlaw of Ankit. Pankaj is father of Sanjay, the only brother of Ankit. How is Shikha related to Ankit?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 104

The relation is as following:

It is clearly shown that Shikha is the mother of Ankit.

Test: BITSAT Past Year Paper- 2015 - Question 105

In a queue of children, Arun is fifth from the left and Suresh is sixth from the right. When they interchange their places among themselves, Arun becomes thirteenth from the left. Then, what will be Suresh's position from the right?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 105

Since Arun and Suresh interchange places, so Arun's new position (13th from left) is the same as Suresh's earlier position (6th from right).
So, number of children in the queue = (12 + 1 + 5) = 18.
Now, Suresh's new position is the same as Arun's earlier position fifth from left.
Therefore Suresh's position from the right = (18 – 4) = 14th.

Test: BITSAT Past Year Paper- 2015 - Question 106

 equals

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 106

Consider 

Test: BITSAT Past Year Paper- 2015 - Question 107

If ω is the complex cube r oot of unity, then the value of  is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 107

Consider 
Which can be written as


Since  is a G. P.. therfore by sum of infinite G.P, we have

∴  Given expression = -1

Test: BITSAT Past Year Paper- 2015 - Question 108

The root of the equation 2(1+ i)x2 - 4(2 - i)x - 5 - 3i = 0 which has greater modulus is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 108

Roots = 

Test: BITSAT Past Year Paper- 2015 - Question 109

The value of  upto n terms is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 109

Consider  upto n terms
 upto n terms
 upto n terms
= (1 + 1 + 1+ ....upto n terms)


Test: BITSAT Past Year Paper- 2015 - Question 110

The period of tan 3θ is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 110

tanθ is of period π so that tan 3θ is of period π/3.

Test: BITSAT Past Year Paper- 2015 - Question 111

If a function f(x) is given by   then at x = 0, f(x)

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 111

Let 
For x = 0, we have f(x) = 0
Thus, we have 
Clearly, 
So, f(x) is not continuous at x = 0.

Test: BITSAT Past Year Paper- 2015 - Question 112

If g is the inverse of function f and f'(x) = sin x, then g'(x) is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 112

Since, g is the inverse of function  f.
Therefore, g(x) = f–1(x)
⇒ f[g(x)] = x
⇒ fog (x) = x, for all x
Differentiate both side, w.r.tx

(By defn of f'(x))

Test: BITSAT Past Year Paper- 2015 - Question 113

A bag contains (2n + 1) coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is 31/42, then n is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 113

Total number of coins= 2n+1
Consider the following events:
E1 = Getting a coin having head on both sides from the bag.
E2 = Getting a fair coin from the bag
A = Toss results in a head
Given:   and P 
Then,
P(A) = P(E1) P(A /E1) + P(E2) P(A/E2)


n = 10

Test: BITSAT Past Year Paper- 2015 - Question 114

If φ(x) is a differential function, then the solution of the differential equation dy + {y φ(x) – φ(x) φ'(x)}dx = 0, is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 114

Given differential equation is 
which is a linear differential equation with

∴ Solution is 

Test: BITSAT Past Year Paper- 2015 - Question 115

The area of the region R = {(x, y):|x| ≤ |y| and x2 + y2 ≤ 1} is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 115


Required area = 4 (Area of the shaded region in first quadrant)

Test: BITSAT Past Year Paper- 2015 - Question 116

Universal set,
U = {x | x5 – 6x4 + 11x3 – 6x2 = 0}
A = {x | x2 – 5x + 6 = 0}
B = {x | x2 – 3x + 2 = 0}
What is (A ∩ B)' equal to ?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 116

U = {x : x5 – 6x4 + 11x3 – 6x2 = 0}
Solving for values of x, we get
U = {0, 1, 2, 3}
A = {x : x2 – 5x + 6 = 0}
Solving for values of x, we get
A  = {2, 3}
and B = {x : x2 – 3x + 2 = 0}
Solving for values of x, we get B = {2, 1}
A ∩ B = {2}
∴ (A ∩ B)' = U – (A ∩ B)
= {0, 1, 2, 3} – {2} = {0, 1, 3}

Test: BITSAT Past Year Paper- 2015 - Question 117

If   then  4x2 - 4 xy cosα + y2 is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 117


⇒ 4 - y2 -4x+ x2y2 = 4 cos2 α + x2 y2 - 4 xy cosα
⇒ 4x2 + y2 - 4xycosα = 4sin2α.

Test: BITSAT Past Year Paper- 2015 - Question 118

If  then the value of  2a1 + 23a3 + 25a5+ .... is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 118

Let 

By using

= a1 = a= a= ... = 0
Hence, 2a1 + 23a3 + 25a5 + ....= 0

Test: BITSAT Past Year Paper- 2015 - Question 119

Let a, b and c be three vectors satisfying a × b = (a ×c), |a| = |c| = 1, |b| = 4 and |b × c| = √15 . If b – 2c = λa, then λ equals

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 119

Let θ be the angle between b and c.


Now given, 

Test: BITSAT Past Year Paper- 2015 - Question 120

The total number of 4-digit numbers in which the digits are in descending order, is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 120

Total number of arrangements of 10 digits 0, 1, 2, ...., 9 by taking 4 at a time = 10C4 × 4!
We observe that in every arrangement of 4 selected digits there is just one arrangement in which the digits are in descending order.
∴ Required number of 4-digit numbers.

Test: BITSAT Past Year Paper- 2015 - Question 121

The line which is parallel to X-axis and crosses the curve y = √x at an angle of 45°, is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 121

Given equation of a line parallel to X-axis is y = k.
Given equation of the curve is y = √x ,
On solving equation of line with the equation of curve, we get x = k2
Thus the intersecting point is (k2, k)
It is given that the line y = k intersect the curve y = √x at an an gle of π/4. This means that the slope of the tangent to 

Test: BITSAT Past Year Paper- 2015 - Question 122

In a ΔABC, the lengths of th e two larger sides are 10 and 9 units, respectively. If the angles are in AP, then the length of the third side can be

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 122

Let A, B and C be the three angles of ΔABC and
Let a = 10 and b = 9
It is given that the angles are in AP.
∴ 2B = A + C on adding B both the sides, we get 3B = A + B + C
⇒ 3B = 180° ⇒ B = 60°
Now, we know 

Test: BITSAT Past Year Paper- 2015 - Question 123

The arithmetic mean of the data 0, 1, 2, ...... , n with frequencies 1, nC1, nC2,....., nCn is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 123

Since,  where xi are observations with frequencies fi, i = 1, 2, ........n The required mean is given by

Test: BITSAT Past Year Paper- 2015 - Question 124

 The mean square deviation of a set of n observation x1, x2, .... xn about a point c is defined as  The mean square deviations about – 2 and 2 are 18 and 10 respectively, the standard deviation of this set of observations is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 124

We have  and 
 and 



Test: BITSAT Past Year Paper- 2015 - Question 125

Let S be the focus of the parabola y2 = 8x and PQ be the common chord of the circle x2 + y2 – 2x – 4y = 0 and the given parabola. The area of DPQS is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 125

The parametr ic equation s of the parabola y2 = 8x are x = 2t2 and y = 4t.
and the given equation of circle is x2 + y2 – 2x – 4y = 0
On putting x = 2t2 and y = 4t in circle we get
4t4 + 16t2 – 4t2 – 16t = 0
⇒ 4t2 + 12t2 – 16t = 0
⇒ 4t (t3 + 3t – 4) = 0
⇒ t(t – 1) (t2 + t + 4) = 0
⇒ t = 0, t = 1

Thus the coordinates of points of intersection of the circle and the parabola are Q (0, 0) and P(2, 4). Clearly these are diametrically opposite points on the circle.
The coordinates of the focus S of the parabola are (2, 0) which lies on the circle.
 = 4 sq. units.

Test: BITSAT Past Year Paper- 2015 - Question 126

The number of real roots of the equation ex–1 + x – 2 = 0 is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 126

Let f(x) = ex–1 + x – 2
check for x = 1
Then, f (1) = e0 + 1 – 2 = 0
So, x = 1 is a real root of the equation f(x) = 0 Let x = α be the other root such that α > 1 or α < 1. Consider the interval [1, α ] or [α,1].
Clearly f(1) = f(α)  = 0
By Rolle’s theorem f'(x) = 0 has a root in (1, α) or in (α, 1).
But f'(x) = ex–1 + 1 > 0, for all x. Thus, f'(x)≠ 0 , for any x ∈ (1, α) or x ∈ (α,1) , which is a contradiction.
Hence, f(x) = 0 has no real root other than 1.

Test: BITSAT Past Year Paper- 2015 - Question 127

Minimise 
Subject to 
 is a LPP with number of constraints

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 127

Constraints will be

So, total number of constraints = m + n

Test: BITSAT Past Year Paper- 2015 - Question 128

A bag contains 3 red and 3 white balls. Two balls are drawn one by one. The probability that they are of different colours is.

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 128

Let A ≡ event that drawn ball is red B ≡ event that  drawn ball is white Then AB and BA are two disjoint cases of the given event.
∴ P (AB + BA) = P(AB) + P (BA)

Test: BITSAT Past Year Paper- 2015 - Question 129

Let M be a 3 × 3 non-singular matrix with det (M) = α. If [M–1 adj (adj (M)] = KI, then the value of K is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 129

We know that, M (adj M) = |M| I
Replacing M by adj M, we get adj M [adj (adj M) = det (adj M) I = det (M) M–1 [adj (adj M) = α2l


Hence, K = α

Test: BITSAT Past Year Paper- 2015 - Question 130

Tangents are drawn from the origin to the curve y = cos x. Their points of contact lie on

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 130

Let (x1, y1) be one of the points of contact.
Given curve is y = cos x

Now the equation of the tangent at (x1,y1) is

⇒y -y1 = - sin x1(0- x1)
Since, it is given that equation of tangent passes through origin.
∴ 0 - y1 = - sin x1 (0-x1)
⇒ y1 = – x1 sin x1 ...(i)
Also,  point (x1, y1) lies on y = cos x.
∴ y1 = cos x1
From Eqs. (i), (ii) , we get

⇒ 
Hence, the locus of (x1, y1) is x2 = y2 + y2x2 ⇒ x2y2 = x2 – y2

Test: BITSAT Past Year Paper- 2015 - Question 131

 The slope of the tangent to the curve y = ex cos x is minimum at x = α, 0 ≤ a ≤ 2π, then the value of α is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 131

Let m be the slope of the tangent to the curve
y = ex cos x.
Then, 
Diff. w.r.t ‘x’
⇒  =-2ex sinx

Clearly, 
Thus, y is minimum at x = π.
Hence the value of α = π.

Test: BITSAT Past Year Paper- 2015 - Question 132

Two lines   are coplanar. Then, α can take value (s)

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 132

The equations of given lines can be written as

Since, these lines are coplanar.
Therefore, 

Test: BITSAT Past Year Paper- 2015 - Question 133

The eccentricity of an ellipse, with its centre at the origin, is 1/2.  If one of the directrices is x = 4 , then the equation of the ellipse is:

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 133



Equation of ellispe is

Test: BITSAT Past Year Paper- 2015 - Question 134

The function  has a local minimum at

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 134

 is of the form  and
equality holds for x = 1

Test: BITSAT Past Year Paper- 2015 - Question 135

If 

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 135

 




Squaring, 
Differentiating,

Test: BITSAT Past Year Paper- 2015 - Question 136

If  and  then which one of the following is correct?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 136

As given,


and 

⇒ B=0
∴ A = 1 and B = 0 is correct

Test: BITSAT Past Year Paper- 2015 - Question 137

If a and b are non-zero roots of x2 + ax + b = 0 then the least value of  x2 + ax + b is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 137

As given a and b are the roots of the equation x2 + ax + b = 0
⇒ sum of roots, a + b = – a
⇒ b = – 2a ...(1)
and product of roots, ab = b
⇒ ab – b = 0
⇒ b (a – 1) = 0
if b = 0 then a = 0
if b ≠ 0 then a = 1 and b = – 2
so, the expression will be, f (x) = x2 + x – 2

⇒ 
So, f (x) will be minimum, if 
i.e. when x = -(1/2)
⇒ minimum value of function = -(9/4)

Test: BITSAT Past Year Paper- 2015 - Question 138

If 0<x<π/2, then

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 138

Let us assume the functions f(x) and g(x) given by
f(x) = tanx – x and g(x) = x – sinx, for 0<x<π/2
Now, f'(x) = sec2x – 1 and
g'(x) = 1 – cos x

Test: BITSAT Past Year Paper- 2015 - Question 139

The degree of the differential equation satisfying 

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 139

Put x = sin θ and y = sin φ
⇒ cos θ + cos φ = a (sin θ – sin φ)
⇒ 2 cos 

Differentiate 
so the degree is one

Test: BITSAT Past Year Paper- 2015 - Question 140

Let f(x) be a polynomial of degree three satisfying f(0) = – 1 and f(1) = 0. Also, 0 is a stationary point of f(x). If f(x) does not have an extremum at x = 0, then the value of  is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 140

Let f(x) = ax3 + bx2 + cx + d
Put x = 0 and x = 1
Then, we get f(0) = –1 and f(1) = 0
⇒ d = – 1 and a + b + c + d = 0
⇒ a + b + c = 1 ...(i)
It is given that x = 0 is a stationary point of f(x), but it is not a point of extremum.
Therefore,  f"(0) = 0 = f " (0) and f"(0) =0
Now, f(x) = ax3 + bx2 + cx + d
⇒ f'(x) = 3ax2 + 2bx + c,
f " (x) = 6ax + 2b and f"' (x) = 6a
f' = 0, f "' (0) =0 and f "' (0) = 0≠0
⇒ c = 0, b = 0 and a ≠ 0
From Eqs. (i) and (ii), we get a = 1, b = c = 0 and d = – 1
Put these values in f(x)
we get f(x) = x3 – 1
Hence, 

Test: BITSAT Past Year Paper- 2015 - Question 141

The domain of the function 

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 141


if (i)-1 ≤ x -3 ≤ 1 ⇒ 2 ≤ x ≤ 4 and
(ii) 9-x2 > 0 ⇒ -3 < x < 3
Taking common solution of  (i) and (ii), we get 2 ≤ x < 3
∴ Domain = [2, 3)

Test: BITSAT Past Year Paper- 2015 - Question 142

If the lines p1x + q1y = 1, p2x + q2y = 1 and p3x + q3y = 1 be concurrent, then the points (p1, q1), (p2, q2) and (p3, q3)

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 142

The equations of the lines are 
p1x + q1y - 1 = 0 ...(i)
p2x + q2y - 1 = 0 ...(ii)
and p3x + q3y - 1 = 0 ...(iii)
As they are concurrent,

This is also the condition for the points (p1, q1), (p2, q2)  and (p3, q3) to be collinear.

Test: BITSAT Past Year Paper- 2015 - Question 143

Area of the circle in which a chord of length √2 makes an angle π/2 at the centre, is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 143

Let AB be the chord of length √2. Let O be the centre of the circle and let OC be the perpendicular from O on AB.
Then, AC = BC = 
In ΔOBC, we have
OB= BC cosec 45°


∴ Area of the circle = π(OB)2 = π sq units

Test: BITSAT Past Year Paper- 2015 - Question 144

If  then the value of (m2 – n2) sin2 B is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 144

cos A = n cos B and sin A = m sin B
Squaring and adding, we get
1 = n2 cos2B + m2sin2B
⇒ 1 = n2 (1 – sin2B) + m2sin2B
∴ (m2 – n2) sin2B = 1 – n2

Test: BITSAT Past Year Paper- 2015 - Question 145

If complex number z1, z2  and 0 are vertices of equilateral triangle, then  is equal to

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 145

z1, z2, 0 are vertices of an equilateral triangle, so we have 
⇒ 

Test: BITSAT Past Year Paper- 2015 - Question 146

If ρ = {(x, y) |x2 + y2 = 1; x, y ∈ R}. Then, ρ is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 146

Obviously, the relation is not reflexive and transitive, but it is symmetric, because
x2 + x2 = 2x2 ≠ 1
and x2 + y2 = 1, y2 + z2 = 1
⇒ x2 + z2 = 1
But x2 + y2 = 1 ⇒ y2 + x2 = 1

Test: BITSAT Past Year Paper- 2015 - Question 147

A line makes the same angle θ with each of the X and Z-axes. If the angle β, which it makes with Y-axis, is such that sin2 β  = 3sin2 θ, then cos2 θ equals

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 147

Let l, m and n be the direction cosines.
Then, l = cos θ, m = cos β, n = cos θ
we have l2 + m2 + n2 = 1

⇒ tan2θ = 2/3

Test: BITSAT Past Year Paper- 2015 - Question 148

If in a binomial distribution n = 4, P(X = 0) = 16/81, then P(X = 4) equals

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 148

Given n = 4 and P(X = 0) = 16/81
Let p be the probability of success and q that of failure in a trial.
Then, P(X = 0) = 4C0p0q4 = 16/81

⇒ 
∴ P(X = 4) = 4C4p4q0 = p4 = (1/3)4 = 1/81

Test: BITSAT Past Year Paper- 2015 - Question 149

Let f : R → R be a function such that
f (x + y) = f (x) + f (y),
If f (x) is differentiable at x = 0, then which one of the following is incorrect?

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 149

Let f(x + y) = f(x) + f(y), 
Put x = 0 = y
⇒ f(0) = f(0) + f(0)
⇒ f(0) = 0
Now, 

Now, 

⇒ f (x) = x f'(0) + C
But f(0) = 0
∴  C = 0
Hence, f(x) = x f'(0), 
Clearly, f(x) is everywhere continuous and differentiable and f'(x) is constant.

Test: BITSAT Past Year Paper- 2015 - Question 150

If binomial coefficients of three consecutive terms of (1 + x)n are in HP, then the maximum value of n is

Detailed Solution for Test: BITSAT Past Year Paper- 2015 - Question 150

Let the coefficients of rth, (r + 1)th, and (r + 2)th terms be in HP.
Then, 

⇒ n2 – 4nr + 4r2 + n = 0
⇒ (n – 2r)2 + n = 0 
which is not possible for any value for n.

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