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The current voltage relation of diode is given by I = (e^{1000V/T} – 1) mA, where the applied V is in volts and the temperature T is in degree kelvin. If a student makes an error measuring ± 0.01 V while measuring the current of 5 mA at 300 K, what will be the error in the value of current in mA?
I = (e^{1000 V/T} – 1)mA
When I = 5 mA, e^{1000 V/T} = 6 mA
Also, dI =
= 0.2 mA
From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:
Time taken to reach highest point is t_{1} = u/g
Speed on reaching ground =
Now, v = u + at
A mass m is supported by a massless string wound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release?
a = Rα
mg – T = ma
T × R = mR^{2}α
or T = ma
a = g/2
A block of mass m is placed on a surface with a vertical crosssection given by y = x^{3}/6. f the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is
tanθ = dy/dx = x^{2}/2
At limiting equilibrium,
μ = tanθ
⇒ x = ±1
Now, y = 1/6
When a rubberband is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx^{2} where a and b are constants. The work done in stretching the unstretched rubberband by L is :
A bob of mass m attached to an inextensible string of length l is suspended from a vertical support.
The bob rotates in a horizontal circle with an angular speed ω rad/s about the vertical. About the point of suspension
τ = mg × l sin θ. (Direction parallel to plane of rotation of particle)
as perpendicular to direction of L changes but magnitude remains same.
Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is
The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100°C is :
(For steel Young's modulus is 2 × 10^{11} Nm^{–2} and coefficient of thermal expansion is 1.1 × 10^{–5} K^{–1})
As length is constant,
Strain =
Now pressure = stress = Y × strain
= 2 × 10^{11} × 1.1 × 10^{–5} × 100
= 2.2 × 10^{8} Pa
There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d_{1} and d_{2} are filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle α with vertical. Ratio is
Equating pressure at A
(Rcosα + Rsinα)d_{2}g = (Rcosα – Rsinα)d_{1}g
⇒
On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius R and making a circular contact of radius r with the bottom of the vessel. If r << R, and the surface tension of water is T, value of r just before bubbles detatch is (Density of water is ρ_{ω})
When the bubble gets detached, Buoyant force = force due to surface tension
Three rods of copper, brass and steel are welded together to form a Yshaped structure. Area of crosssection of each rod = 4 cm^{2}. End of copper rod is maintained at 100°C whereas ends of brass and steel are kept at 0°C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cm respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is
Q = Q_{1} + Q_{2
}
200 – 2T = 2T + T
⇒ T = 40°C
Q = = 4.8 cal/s
One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement
For BC, ΔT = –200 K
ΔU = –500R
An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)
P + x = P_{0}
P = (76 – x)
8 × A × 76 = (76 – x) × A × (54 – x)
x = 38
Length of air column = 54 – 38 = 16 cm.
A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s, it travels 2a in same direction then
As it starts from rest, we have
x = Acosωt. At t = 0, x = A
when t = , x = A – a
when t = , x = A – 3a
Now, A – a = Acos
A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
∴ Answer is 6.
Assume that an electric field exists in space. Then the potential difference V_{A} – V_{O}, where V_{O} is the potential at the origin and V_{A} the potential at x = 2 m is
A parallel plate capacitor is made of two circular plates separated by a distance 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 10^{4} V/m, the charge density of the positive plate will be close to
= 2.2 × 8.85 ×10^{–12 }× 3 ×10^{4} ≈ 6 × 10^{–7 } C/m^{2}
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be:
15 × 40 + 5 × 100 + 5 × 80 + 1000
= V × I
600 + 500 + 400 + 1000 = 220 I
I = 12 A.
A conductor lies along the zaxis at –1.5 ≤ z < 1.5 m and carries a fixed current of 10.0 A indirection (see figure). For a fieldT, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 10^{–3} s. Assume parallel motion along the xaxis
Average Power = work/time
= 2.97 × 10^{–3} J
The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 × 10^{3} A m^{–1}. The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is
I = 3 A.
In the circuit shown here, the point 'C' is kept connected to point 'A' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point 'A' and connected to point 'B' at time t = 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to
Applying Kirchhoff's law in closed loop, –V_{R} – V_{C} = 0
⇒ V_{R}/V_{C} = –1
During the propagation of electromagnetic waves in a medium
Energy is equally divided between electric and magnetic field
A thin convex lens made from crown glass has focal length f. When it is measured in two different liquids having refractive indices 4/3 and 5/3, it has the focal lengths f_{1} and f_{2} respectively. The correct relation between the focal lengths is
By Lens maker's formula
⇒ f_{1} = 4f & f_{2} = –5f
A green light is incident from the water to the airwater interface at the critical angle(θ). Select the correct statement
sin θ_{c }= 1/μ
For greater wavelength (i.e. lesser frequency) μ is less
So, θ_{c} would be more. So, they will not suffer reflection and come out at angles less then 90°.
Two beams, A and B, of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are I_{A} and I_{B }respectively, then equals
By law of Malus, I = I_{0}cos^{2}θ
Now, I_{A′} = I_{A}cos^{2}30
I_{B′} = I_{B}cos^{2}60
As I_{A′} = I_{B′}
I_{A}/I_{B} = 1/3
The radiation corresponding to 3→2 transitions of hydrogen atoms falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3 × 10^{–4} T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to
= 0.8 V
For transition between 3 to 2,
Hydrogen (_{1}H^{1}), Deuterium (_{1}H^{2}), singly ionised Helium (_{2}He^{4})^{+} and doubly ionised lithium (_{3}Li^{6})^{++} all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiation are λ_{1}, λ_{2}, λ_{3} and λ_{4} respectively then approximately which one of the following is correct?
The forward biased diode connection is
For forward Bias, pside must be at higher potential than nside.
Match ListI (Electromagnetic wave type) with ListII (Its association/application) and select the correct option from the choices given below the lists :
(a) Infrared rays are used to treat muscular strain
(b) Radiowaves are used for broadcasting
(c) Xrays are used to detect fracture of bones
(d) Ultraviolet rays are absorbed by ozone
A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
As measured value is 3.50 cm, the least count must be 0.01 cm = 0.1 mm
For vernier scale with 1 MSD = 1 mm and 9 MSD = 10 VSD,
Least count = 1 MSD – 1 VSD = 0.1 mm
The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is
37 →1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}5s^{1}
So last electron enters 5s orbital
Hence n = 5, l = 0, m_{1} = 0, m_{s} =
If Z is a compressibility factor, van der Waals equation at low pressure can be written as
Compressibility factor
(For one mole of real gas)
van der Waal equation
CsCl crystallises in body centred cubic lattice. If ‘a’ is its edge length then which of the following expressions is correct?
For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of M/10 sulphuric acid. The unreacted acid required 20 mL of M/10 sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is
As per question
Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 S m^{–1}. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m^{2} mol^{–1} is
For 0.2 M solution
R = 50 Ω
σ = 1.4 S m^{–1} = 1.4 × 10^{–2} S cm^{–1}
Now,
For 0.5 M solution
R = 280 Ω
σ = ?
Now,
= 5 S cm^{2} mol^{–1}
= 5 × 10^{–4} S m^{2} mol^{–1}
For complete combustion of ethanol,
the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol^{–1} at 25°C. Assuming ideality the enthalpy of combustion, Δ_{c}H, for the reaction will be (R = 8.314 kJ mol^{–1})
Bomb calorimeter gives ΔU of the reaction
So, as per question
The equivalent conductance of NaCl at concentration C and at infinite dilution are λ_{C} and λ_{∞}, respectively. The correct relationship between λ_{C} and λ_{∞} is given as
(Where the constant B is positive)
According to Debye Huckle onsager equation,
Consider separate solutions of 0.500 M C_{2}H_{5}OH(aq), 0.100 M Mg_{3}(PO_{4})_{2}(aq), 0.250 M KBr(aq) and 0.125 M Na_{3}PO_{4}(aq) at 25°C. Which statement is true about these solutions, assuming all salts to be strong electrolytes?
π= iCRT
For the reaction If K_{P} = K_{C}(RT)^{x} where the symbols have usual meaning then the value of x is (assuming ideality)
x = Δn_{g} = no. of gaseous moles in product – no. of gaseous moles in reactant
=
For the nonstoichiometre reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiments, all at 298 K.
The rate law for the formation of C is
2A + B → C+ D
Rate of Reaction =
Let rate of Reaction = k[A]^{x}[B]^{y}
Now from table,
1.2 × 10^{–3} = k [0.1]^{x}[0.1]y .....(i)
1.2 × 10^{–3} = k [0.1]^{x}[0.2]^{y} ...(ii)
2.4 × 10^{–3} = k [0.2]^{x}[0.1]^{y} ...(iii)
Dividing equation (i) by (ii)
Now Dividing equation (i) by (iii)
Among the following oxoacids, the correct decreasing order of acid strength is
Resonance produced conjugate base.
(iv) ClO^{–} is not resonance stabilized.
As per resonance stability order of conjugate base is
The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is
On electrolysis only in case of Ca^{2+} salt aqueous solution H_{2} gas discharge at Cathode.
Case of Cr
At cathode : Cr^{3+} + 2e^{– }→ Cr
So, Cr is deposited.
Case of Ag
At cathode : Ag^{+} + e^{–} → Ag So, Ag is deposited.
Case of Cu
At cathode : Cu^{2+} + 2e^{–} → Cu
Case of Ca^{2+}
At cathode :
The octahedral complex of a metal ion M^{3+} with four monodentate ligands L_{1}, L_{2}, L_{3} and L_{4} absorb wavelengths in the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is
The energy of red light is less than that of violet light.
So energy order is
Red < Yellow < Green < Blue
The complex absorbs lower energy light lower will be its strength. So order of ligand strength is
L_{1} < L_{3} < L_{2} < L_{4}
Which one of the following properties is not shown by NO?
Nitric oxide is paramagnetic in the gaseous state as it has one unpaired electron in its outermost shell.
The electronic configuration of NO is
However, it dimerises at low temperature to become diamagnetic.
Its bond order is 2.5 and it combines with O_{2} to give nitrogen dioxide.
In which of the following reactions H_{2}O_{2} acts as a reducing agent?
(a) H_{2}O_{2} + 2H^{+ }+ 2e^{– }→ 2H_{2}O
(b) H_{2}O_{2} – 2e^{–} → O_{2} + 2H^{+}
(c) H_{2}O_{2} + 2e^{–} → 2OH^{–}
(d) H_{2}O_{2}^{ }+ 2OH^{–} – 2e^{–} → O_{2 }+ 2H_{2}O
The reducing agent oxidises itself.
Note : Powers of 'O' are oxidation number of 'O' in the compound.
The correct statement for the molecule, CsI_{3}, is
It contains Cs^{+} and I_{3}^{–} ions.
The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1:4. The ratio of number of their molecule is
Let the mass of O_{2} = x
Mass of N_{2} = 4x
Number of moles of O_{2} = x/32
Number of moles of N_{2} = 4x/28 = x/7
Given below are the halfcell reactions
The E° for 3 Mn^{2+} → Mn + 2Mn^{3+} will b
(1) – 2 × (2)
3Mn^{2+} → Mn + 2Mn^{3+} ;
= [2.36 – 2(–1.51)] F
= (2.36 + 3.02) F
= 5.38 F
But
⇒ 5.38F = –2FE°
⇒ E° = –2.69 V
As E° value is negative reaction is non spontaneous.
Which series of reactions correctly represents chemical reactions related to iron and its compound?
This reaction is corresponding to the combustion of Fe.
These reactions correspond to the production of Fe by reduction of Fe_{3}O_{4 }in blast furnace.
The equation which is balanced and represents the correct product(s) is
The complex
[CoCl(NH_{3})_{5}]+ decomposes under acidic medium, so
In S_{N}2 reactions, the correct order of reactivity for the following compounds
CH_{3}Cl, CH_{3}CH_{2}Cl, (CH_{3})_{2}CHCl and (CH_{3})_{3}CCl is
Rate of S_{N}2 reaction depends on steric crowding of alkyl halide. So order is
CH_{3}Cl > (CH_{3})CH_{2} – Cl > (CH_{3})_{2}CH – Cl > (CH_{3})_{3}CCl
On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is
The most suitable reagent for the conversion of R – CH_{2 }– OH → R – CHO is
PCC is mild oxidising agent, it will convert
The major organic compound formed by the reaction of 1, 1, 1trichloroethane with silver powder is
1, 1, 1trichloroethane
Sodium phenoxide when heated with CO_{2} under pressure at 125°C yields a product which on acetylation produces C.
The major product C would be
Considering the basic strength of amines in aqueous solution, which one has the smallest pK_{b} value?
Among C_{6}H_{5}NH_{2}, CH_{3}NH_{2}, (CH_{3})_{2}NH,
(CH_{3})_{3}N. C_{6}H_{5}NH_{2} is least basic due to resonance.
Out of (CH_{3})_{3}N, CH_{3}NH_{2}, (CH_{3})_{2}NH. (CH_{3})_{2}NH is most basic due to +I effect and hydrogen bonding in H_{2}O.
For which of the following molecule significant μ ≠ 0?
Which one is classified as a condensation polymer?
Dacron is polyester formed by condensation polymerisation of terephthalic acid and ethylene glycol.
Acrylonitrile, Neoprene and Teflon are addition polymers of acrylonitrile, isoprene and tetrafluoroethylene respectively.
Which one of the following bases is not present in DNA?
DNA contains ATGC bases
A – Adenine
T – Thymine
G – Guanine
C – Cytocine
So quinoline is not present.
In the reaction,the product C is
Ethylene
If X = {4^{n} – 3n – 1 : n ∈ N} and Y = {9(n – 1) : n ∈ N}, where N is the set of natural numbers, then X ∪ Y is equal to
If z is a complex number such that z ≥ 2, then the minimum value of
So,
If a ∈ R and the equation
–3(x – [x])^{2} + 2 (x – [x]) + a^{2} = 0
(where [x] denotes the greatest integer ≤ x) has no integral solution, then all possible values of a lie in the interval
–3(x – [x])^{2} + 2[x – [x]) + a^{2} = 0
3 {x}^{2} – 2{x} – a^{2} = 0
For nonintegral solution
0 < a^{2}< 1 and a ∈ (–1, 0) ∪ (0, 1)
Alternative
–3{x}^{2} + 2{x} + a^{2} = 0
Now, –3{x}^{2} + 2{x}
to have no integral roots 0 < a^{2 }< 1
∴ a∈(–1, 0) ∪ (0, 1)
Let α and β be the roots of equation px^{2} + qx + r = 0, p ≠ 0. If p, q, r are in A.P. and then the value of α – β is
∵ p, q, r are in AP
2q = p + r ...(i)
Also
................. (ii)
From (i)
2(–4r) = p + r
p = – 9r
q = – 4r
r = r
Now
If α, β ≠ 0, and f(n) = α^{n} + β^{n} and = K(1 – α)^{2} (1 – β)^{2} (α – β)^{2}, then K is equal to
= [(1 – α)(1 – β)(1 – β)]^{2}
So, k = 1
If A is an 3 × 3 nonsingular matrix such that AA′ = A′A and B =A^{−1} A′ , then BB′ equals
BB ' =( A^{−1}.A ')( A( A^{−1}) ')
= A^{–1}.A.A'.(A^{–1})1 {as AA' = A'A}
= I(A^{–1}A)'
= I.I = I^{2} = I
If the coefficients of x^{3} and x^{4} in the expansion of (1 + ax + bx^{2}) (1 – 2x)^{18 }in powers of x are both zero, then (a, b) is equal to
(1 + ax + bx^{2}) (1 – 2x)^{18}
(1 + ax + bx^{2})[^{18}C_{0} – ^{18}C_{1}(2x) + ^{18}C_{2}(2x)^{2} – ^{18}C_{3}(2x)^{3} + ^{18}C_{4}(2x)^{4} – .......]
Coeff. of x^{3} = –^{18}C_{3}.8 + a × 4.^{18}C_{2} – 2b × 18 = 0
= –51 × 16 × 8 + a × 36 × 17 – 36b = 0
= –34 × 16 + 51a – 3b = 0
= 51a – 3b = 34 × 16 = 544
= 51a – 3b = 544 ... (i)
Only option number (2) satisfies the equation number (i).
If (10)^{9} + 2(11)^{1} (10)^{8} + 3(11)^{2} (10)^{7}+ ... + 10(11)^{9} = k(10)^{9}, then k is equal to
10^{9} + 2⋅(11)(10)^{8} + 3(11)^{2}(10)^{7} +... + 10(11)^{9} = k(10)^{9}
x = 10^{9} + 2⋅(11)(10)^{8} + 3(11)^{2}(10)^{7}+ ... +10(11)^{9}
= 11⋅10^{8} + 2⋅(11)^{2}⋅(10)^{7} +... + 9(11)^{9 }+ 11^{10
}= 10^{9} + 11(10)^{8} + 11^{2}×(10)^{7} +... +11^{9} – 11^{10}
x/10 = (11^{10} − 10^{10} ) − 11^{10} = − 10^{10}
x = 10^{11 }= k⋅10^{9}
k = 100
Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is
a, ar, ar^{2} → G.P.
a, 2ar, ar^{2} → A.P.
2 × 2ar = a + ar^{2}
4r = 1 + r^{2}
r^{2} – 4r + 1 = 0
r = 2+√3
r = 2−√3 is rejected
∵ (r > 1)
G.P. is increasing.
If g is the inverse of a function f and f'(x) = then g′(x) is equal to
If f and g are differentiable functions in [0, 1] satisfying f(0) = 2 = g(1), g(0) = 0 and f(1) = 6, then for some c ∈]0, 1[
Using, mean value theorem
so, f'(c) = 2g'(c)
If x = –1 and x = 2 are extreme points of f (x) =α logx+ βx^{2} + x then
f(x) =α logx+ βx^{2} + x
f ′(x) = α/x + 2βx + 1= 0 at x = –1, 2
−α − 2β +1= 0 ⇒ α+ 2β =1 ...(i)
α/2 + 4β +1= 0
⇒ α+ 8β =−2 ...(ii)
6β =−3 ⇒ β =− 1/2
∴ α= 2
The integral dx is equal to
= 4√3 − 4− π/3
The area of the region described by A = {(x, y) : x^{2} + y^{2} ≤ 1 and y^{2} ≤ 1 – x} is
Shaded area
Let the population of rabbits surviving at a time t be governed by the differential equation . If p(0) = 100, then p(t) equals
Using given condition p(t) = 400 – 300 e^{t/2}
Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R (7, 3). The equation of the line passing through (1, –1) and parallel to PS is
S is midpoint of QR
So S
Slope of PS =
Equation of line ⇒ y – (–1) = – 2/9 ( x – 1)
9y + 9 = – 2x + 2 ⇒ 2x + 9y + 7 = 0
Let a, b, c and d be nonzero numbers. If the point of intersection of the lines 4ax + 2ay + c = 0 and 5bx + 2by + d = 0 lies in the fourth quadrant and is equidistant from the two axes then
Let (α, α) be the point of intersection
∴ 4aα – 2aα + c = 0 ⇒ α = c/2a
and 5bα – 2bα + d = 0
⇒ α=
⇒ 3bc = 2ad
⇒ 3bc – 2ad = 0
Alternative method :
The point of intersection will be
∵ Point of intersection is in fourth quadrant so x is positive and y is negative.
Also distance from axes is same
So x = – y (∵ distance from xaxis is – y as y is negative)
2ad – 2bc = – 5bc + 4ad
⇒ 3bc – 2ad = 0 ...(i)
The locus of the foot of perpendicular drawn from the centre of the ellipse x^{2} + 3y^{2} = 6 on any tangent to it is
Here ellipse is where a^{2} = 6, b^{2} = 2
Now, equation of any variable tangent is
...... (i)
where m is slope of the tangent
So, equation of perpendicular line drawn from centre to tangent is
............... (ii)
Eliminating m, we get
Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centred at (0, y), passing through origin and touching the circle C externally, then the radius of T is equal to
Radius of T = y
T touches C externally
The slope of the line touching both the parabolas y^{2} = 4x and x^{2} = –32y is
y^{2 }= 4x …(1)
x^{2} = –32y …(2)
m be slope of common tangent Equation of tangent (1)
y = mx + 1/m …(i)
Equation of tangent (2)
y = mx + 8m^{2} …(ii)
(i) and (ii) are identical
1/m = 8m^{2}
⇒ m^{3} = 1/8
m = 1/2
Alternative method :
Let tangent to y^{2} = 4x be y =mx+ 1/m
as this is also tangent to x^{2} =−32 y
Solving x^{2 }+ 32mx + 32/m = 0
Since roots are equal
∴ D = 0
The image of the line in the plane 2x – y + z + 3 = 0 is the line
3λ + 6 = 0 ⇒ λ = – 2
a = – 3, b = 5, c = 2
So the equation of the required line is
The angle between the lines whose direction consines satisfy the equations l + m + n = 0 and l^{2} = m^{2} + n^{2} is
l + m + n = 0
l^{2} = m^{2} + n^{2}
Now, (–m – n)^{2} = m^{2} + n^{2}
⇒ mn = 0
m = 0 or n = 0
∴ cos θ= 1/2
θ= π/3
If then λ is equal to
Let A and B be two events such that wherestands for the complement of the event A. Then the events A and B are
The variance of first 50 even natural numbers is
Variance =
= 3434 – 2601
⇒σ^{2} = 833
Let where x ∈R and k ≥ 1. Then f_{4}(x) – f_{6}(x) equals
A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is
t = 1 s
From figure tan 45°= 20/x
and tan30°= 20/x+y
so, y = 20(√3− 1)
i.e., speed = 20(√3 − 1) m/s.
The statement ~(p ↔ ~q) is
Clearly equivalent to p ↔ q
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