Two forces P and Q of magnitude 2F and 3F, respectively, are at an angle θ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle is :
4F^{2} + 9F^{2} + 12F^{2} cos θ = R^{2}
4F^{2} + 36 F^{2} + 24 F^{2} cos θ = 4R^{2}
4F^{2} + 36 F^{2} + 24 F^{2} cos θ
= 4(13F^{2} + 12F^{2}cosθ) = 52 F^{2} + 48F^{2}cosθ
The actual value of resistance R, shown in the figure is 30Ω. This is measured in an experiment as shown using the standard formula R = V/1 , where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is:
0.95 × 30 = 0.05 R_{υ}
R_{υ} = 19 × 30 = 570 Ω
An unknown metal of mass 192 g heated to a temperature of 100ºC was immersed into a brass calorimeter of mass 128 g containing 240 g of water a temperature of 8.4ºC Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5ºC (Specific heat of brass is 394 J kg^{–1} K^{–1})
192 × S × (100 – 21.5)
= 128 × 394 × (21.5 – 8.4)
+ 240 × 4200 × (21.5 – 8.4)
⇒ S = 916
A particle starts from the origin at time t = 0 and moves along the positive xaxis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s?
S = Area under graph
1/2× 2 × 2 + 2 × 2 + 3 × 1 = 9 m
The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is:
A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11V is connected across it is :
P = I^{2}R
4.4 = 4 × 10^{–6} R
R = 1.1 × 10^{6} Ω
The diameter and height of a cylinder are measured by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ?
At some location on earth the horizontal component of earth's magnetic field is 18 x 10^{6} T. At this location, magnetic neeedle of length 0.12 m and pole strength 1.8 Am is suspended from its midpoint using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is :
The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot?
∴ Range of signal = 2250 Hz to 2750 Hz
Now check all options : for 2000 KHZ
f_{mod} = 200 Hz
∴ Range = 1800 KHZ to 2200 KHZ
A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are T_{h} and T_{c} respectively, then :
The electric field of a plane polarized electromagnetic wave in free space at time t= 0 is given by an expression
The magnetic field : (cis the velocity of light)
i.e. direction of 'c'.
Condiser the nuclear fission
Ne^{20} → 2He^{4} + C^{12}
Given that the binding energy/nucleon of Ne^{20}, He^{4} and C^{12} are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement :
Two vectors have equal magnitudes. The magnitude of is 'n' times the magnitude of The angle between is :
A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is :
Consider a Young's double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength λ such that the first minima occurs directly in front of the slit (S_{1}) ?
The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.
Half mole of an ideal monoatomic gas is heated at constant pressure of 1atm from 20ºC to 90ºC. Work done by gas is close to :
(Gas constant R = 8.31 J /mol.K)
A metal plate of area 1 × 10^{–4} m^{2} is illuminated by a radiation of intensity 16 mW/m^{2}.The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be :
[1 eV = 1.6 × 10^{–19}J]
Charges q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences and electrostatic force F. If O is now moved alons the equatorial line to P' such that OP = (y/3), the force on Q will be close to (y/3 >> 2a)
Electric field of equitorial plane of dipole
Two stars of masses 3 × 10^{31} kg each, and at distance 2 × 10^{11}m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star's rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is : (Take Gravitational constant G = 6.67 ×10^{–11} Nm^{2} kg^{–2})
By energy convervation between 0 & ∞.
[M is mass of star m is mass of meteroite)
A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be : (Assume that the highest frequency a person can hear is 20,000 Hz)
For closed organ pipe, reson ate frequency is odd multiple of fundamental frequency.
∴ (2n + 1) f_{0} ≤ 20,000
(f_{o} is fundamental frequency = 1.5 KHz)
∴ n = 6
A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be :
Applying torque equation about point P.
For the circuit shown below, the current through the Zener diode is:
Assuming zener diode doesnot undergo breakdown, current in circuit = 120/15000 = 8mA
∴ Voltage drop across diode = 80 V > 50 V.
The diode undergo breakdown.
Four equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth charge Q at the origin of the coordinate system will be :
(Potential at ∞ = 0)
∴Work required to put a fifth charge Q at origin is equal to
A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency ω. If the radius of the bottle is 2.5 cm then ω close to : (density of water = 10^{3} kg / m^{3})
Extra Boyant force = δAxg
B_{0} + B × mg = ma
B = ma
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates the work done by the capacitor on the slab is :
Two kg of a monoatomic gas is at a pressure of 4 × 10^{4} N/m^{2} . The density of the gas is 8 kg /m^{3}. What is the order of energy of the gas due to its thermal motion ?
Thermal energy of N molecule
order will 10^{4}
A particle which is experiencing a force, given by undergoes a displacement of If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement ?
The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as R_{1} has the colour code ( Orange, Red, Brown). The resistors R_{2} and R_{4} are 80Ω and 40Ω, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R_{3}, would be :
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is :
For Ball
using parallel axis theorem.
An ide al ga s under goes isothe rma l compression from 5 m^{3} against a constant external pressure of 4 Nm^{–2}. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol^{–1} K^{–1}, the temperature of Al increases by :
Work done on isothermal irreversible for ideal gas
= –P_{ext} (V_{2 }– V_{1})
= –4 N/m^{2} (1m^{3} – 5m^{3})
= 16 Nm
Isothermal process for ideal gas
ΔU = 0
q = –w
= –16 Nm
= – 16 J
Heat used to increase temperature of Aℓ
q = n C_{m} ΔT
The 71^{st }electron of an element X with an atomic number of 71 enters into the orbital :
Filling of electrons usually follow many rules, but what orbit the electron will enter is given the Aufbau rule.
This rule states that electrons will be filled in the atomic orbitals which are low in energy than the atomic orbitals which have high energy. The filling of atomic orbitals follow the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s and so on....
For the element X having atomic number 71 has 71 electrons.
The electronic configuration of this element is:
From the above configuration, it is visible that the 71st electron is entering the 5d orbital.
The number of 2c entre2electron and 3centre2electron bonds in B_{2}H_{6}, respectively, are :
The amount of sugar (C_{12}H_{22}O_{11}) required to prepare 2 L of its 0.1 M aqueous solution is :
wt (C_{12}H_{22}O_{11}) = 68.4 gram
Among the following reactions of hydrogen with halogens, the one that requires a catalyst is :
Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the formation of:
What will be the major product in the following mononitration reaction?
amine is op directing
In the cell Pt(s)H_{2}(g, 1barHCl (aq)Ag(s)Pt(s) the cell potential is 0.92 when a 10^{–6} molal HCl solution is used. THe standard electrode potential of (AgCl/Ag,Cl^{–}) electrode is :
The major product of the following recation is:
The pair that contains two P–H bonds in each of the oxoacids is
The major product of the following reaction is:
SN^{2} reaction
The difference in the number of unpaired electrons of a metal ion in its highspin and lowspin octahedral complexes is two. The metal ion is :
Co^{2+} >d^{7}
hs, n = 3,ls, n = 1
A compound of formula A_{2}B_{3} has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms :
A_{2}B_{3} has HCP lattice
If A form HCP, then 3^{th}/4 of THV must occupied by B to form A_{2}B_{2}
If B form HCP, then 1^{th}/3 of THV must occupied
by A to form A_{2}B_{3}
The reaction that is NOT involved in the ozone layer depletion mechanism in the stratosphere is:
The process with negative entropy change is :
N_{2}(g) + 3H_{2}(g) ⇔ 2NH_{3}(g) ; Δn_{g} < 0
The major product of the following reaction is:
A reaction of cobalt(III ) chloride and ethylenediamine in a 1:2 mole ratio generates two isomeric products A (violet coloured) B (green coloured). A can show optical activity, B is optically inactive. What type of isomers does A and B represent ?
[Co(Cn)_{2} Cl_{2}]Cl
cis > Optically active
rans > Optically in active
The major product obtained in the following reaction is :
Which of the following tests cannot be used for identifying amino acids ?
What is the IUPAC name of the following compound ?
Which is the most suitable reagent for the following transformation?
The correct match between item 'I' and item 'II' is :
In the reaction of oxalate with permaganate in acidic medium, the number of electrons involved in producing one molecule of CO_{2} is :
10 e^{–} trans for 10 molecules of CO_{2} so per molecule of CO_{2} transfer of e^{–} is '1'
5.1g NH_{4}SH is introduced in 3.0 L evacuated flask at 327°C. 30% of the solid NH_{4}SH decomposed to NH_{3} and H_{2}S as gases. The K_{p} of the reaction at 327°C is (R = 0.082 L atm mol^{–1}K^{–1}, Molar mass of S = 32 g mol^{/01}, molar mass of N = 14g mol^{–1})
NH_{4}SH(s)⇔ NH_{3} (g) H_{2}S(g)
α = 30% = .3
so number of moles at equilibrium
Now use PV = nRT at equilibrium
P_{total} × 3 lit = (.03 + .03) × .082 × 600
P_{total} = .984 atm
At equilibrium
The electrolytes usually used in the electroplating of gold and silver, respectively, are :
Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point of 2 molal solutions of glucose in the same solvent is 2 K. The relation between K_{b} and K_{f} is:
An aromatic compound 'A' having molecular formula C_{7}H_{6}O_{2} on treating with aqueous ammonia and heating forms compound 'B'. The compound 'B' on reaction with molecular bromine and potassium hydroxide provides compound 'C' having molecular formula C_{6}H_{7}N. The structure of 'A' is :
The ground state energy of hydrogen atom is –13.6 eV. The energy of second excited state He^{+} ion in eV is :
For an elementary chemical reaction,
the expression for is :
Haemoglobin and gold sol are examples of :
Haemoglobin→ positive sol
Ag  sol → negative sol
Let If R(z) and I[z] respectively denote the real and imaginary parts of z, then :
Let a_{1},a_{2},a_{3},....a_{10} be in G.P. with a_{i}; > 0 for i = 1,2,...., 10 and S be the set of pairs (r,k), r k∈N (the set of natural numbers) for which
Then the number of elements in S, is :
Apply
C_{3} → C_{3} – C_{2}
C_{2} → C_{2} – C_{1}
We get D = 0
The positive value of λ for whic h the coefficient of x^{2} in the expression is 720, is :
The value of is :
The value of where [t ] denotes the greatest integer less than or equal to t, is :
If the probability of hitting a target by a shooter, in any shot, is 1/3, then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than 5/6, is :
n_{min} = 5
If me an and standa rd devia tion of 5 observations x_{1}, x_{2}, x_{3},x_{4},x_{5} are 10 and 3, respectively, then the variance of 6 observations x_{1},x_{2}, ....,x_{5} and –50 is equal to :
= 507.5
The length of the chord of the parabola x^{2} = 4y having equation x √2y + 4√2 = 0 is:
Solving together we get
Similarly,
Let where b > 0. Then the minimum value of is :
A = 2(2b^{2} + 2 – b^{2}) – b(2b – b) + 1 (b^{2} – b^{2} – 1)
A = 2(b^{2} + 2) – b^{2} – 1
A = b^{2} + 3
The ta ngent to the cur ve, passing through the point (1,e) also passes through the point :
T : y – e = 3e (x – 1)
y = 3ex – 3e + e
y = (3e)x – 2e
The number of values of θ∈(0,π) for which the system of linear equations
x + 3y + 7z = 0
–x + 4y + 7z = 0
(sin 3θ)x + (cos 2θ) y + 2z = 0
has a nontrivial solution, is :
(8 – 7 cos 2θ) – 3(–2 – 7 sin 3θ) + 7 (– cos 2θ – 4 sin 3θ) = 0
14 – 7 cos 2θ + 21 sin 3θ – 7 cos 2θ – 28 sin 3θ = 0
14 – 7 sin 3θ – 14 cos 2θ = 0
14 – 7 (3 sin θ – 4 sin^{3} θ) – 14 (1 – 2 sin^{2} θ) = 0
–21 sin θ+ 28 sin^{3} θ + 28 sin^{2} θ = 0
7 sin θ [–3 + 4 sin^{2} θ + 4 sin θ] = 0
sin θ,
4 sin^{2} θ + 6 sin θ – 2 sin θ – 3 = 0
2 sin θ(2 sin θ + 3) – 1 (2 sin θ + 3) = 0
Hence, 2 solutions in (0, π) Option (4)
If then f'(1/2) is :
Differentiate w.r.t. 'x'
f(x) = 2x + 0 – x^{2} f(x)
Let f : (–1,1) → R be a function defined by If K be the set of all points at which f is not differentiable, then K has exactly :
f : (–1, 1) → R
Nonderivable at 3 points in (–1, 1) Option (1)
Let where r ≠ ±1. Then S represents :
If then K is equal to :
∴ K = 2^{25}
Let N be the set of natural numbers and two functions f and g be defined as f,g : N→N such that :
and g(n) = n–(–1)^{n}. The fog is :
∴ many one but onto Option (4)
The values of λ such that sum of the squares of the roots of the quadratic equation, x^{2} + (3 – λ) x + 2 = λ has the least value is :
α + β = λ– 3
αβ = 2 – λ
α^{2} + β^{2} = (α + β)^{2} – 2αβ = (λ – 3)^{2} – 2(2 – λ)
= λ^{2} + 9 – 6λ – 4 + 2λ
= λ^{2} – 4λ + 5
= (λ – 2)^{2} + 1
∴ λ = 2
Two vertices of a triangle are (0,2) and (4,3). If its orthocentre is at the origin, then its third vertex lies in which quadrant?
⇒ 3b – 6 = –4a ⇒ 4a + 3b = 6 ……(ii)
From (i) and (ii)
∴ II^{nd} quadrant.
Two sides of a parallelogram are along the lines, x + y = 3 and x – y + 3 = 0. If its diagonals intersect at (2,4), then one of its vertex is:
C ⇒ (4, 5)
Now equation of BC is x – y = –1
and equation of CD is x + y = 9
Solving x + y = 9 and x – y = –3
Point D is (3, 6)
Let and be two given vectors where vectors arenoncollinear. The value of λ for which vectors are collinear, is :
The value of is:
(Where tanA=20 , tanB=1)
With the usua l notation, in Δ ABC, if ∠A + ∠B = 120^{0}, a = √3 + 1 and √3  1, then the ratio ∠A :∠B, is :
The plane which bisects the line segment joining the points (–3,–3,4) and (3,7,6) at right angles, passes through which one of the following points?
p : 3(x – 0) + 5 (y – 2) + 1 (z – 5) = 0
3x + 5y + z = 15
∴ Option (2)
Consider the following three statements :
P : 5 is a prime number.
Q : 7 is a factor of 192.
R : L.C.M. of 5 and 7 is 35.
Then the truth value of which one of the following statements is true ?
It is obvious
∴ Option (4)
On which of the following lines lies the point of intersection of the line, and the plane, x + y + z = 2 ?
General point on the given line is
x = 2λ + 4
y = 2λ + 5
z = λ + 3
Solving with plane,
2λ + 4 + 2λ + 5 + λ + 3 = 2
5λ + 12 = 2
5λ = –10
∴ Option (3)
Let f be a differentiable function such that
∴ Option (1)
A helicopter is flying along the curve given by y – x^{3/2} = 7, (x ≥ 0). A soldier positioned at the point (1/2,7) wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is :
Option (3)
If where C is a constant of integration, then f(x) is equal to :
Put x^{3} = t
3x^{2} dx = dt
∴ f(x) = –1 – 4x^{3}
Option (1)
(From the given options (1) is most suitable)
The curve a mongst the family of curves, represented by the differential equation, (x^{2} – y^{2})dx + 2xy dy = 0 which passes through (1,1) is :
(x^{2} – y^{2}) dx + 2xy dy = 0
Solving we get,
ln(v^{2} + 1) = –ln x + C
(y^{2} + x^{2}) = Cx
1 + 1 = C ⇒ C = 2
y^{2} + x^{2}= 2x
∴ Option (2)
If the area of an equilateral triangle inscribed in the circle, x^{2} + y^{2} + 10x + 12y + c = 0 is 27√3 sq. units then c is equal to :
C = 25




