Two plane mirrors arc inclined to each other such that a ray of light incident on the first mirror (M_{1}) and parallel to the second mirror (M_{2}) is finally reflected from the second mirror (M_{2}) parallel to the first mirror (M_{1}). The angle between the two mirrors will be :
Assuming angles between two mirrors be θ as per geometry,
sum of anlges of Δ
3θ = 180°
θ = 60°
In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range –30°≤θ≤30°is:
Pam difference dsinθ = nλ
where d = seperation of slits
λ = wave length
n = no. of maximas
0.32 × 10^{–3} sin 30 = n × 500 × 10^{–9}
n = 320
Hence total no. of maximas observed in angular range –30°≤ θ ≤ 30° is
maximas = 320 + 1 + 320 = 641
At a given instant, say t = 0, two radioactive substances A and B have equal activities. The ratio R_{B}/R_{A }of their activities after time t itself decays with time t as e^{3t }[f the halflife of A is m_{2}, the halflife of B is :
Half life of A = ℓn2
at t = 0 R_{A} = R_{B
}
Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of V_{o} changes by : (assume that the Ge diode has large breakdown voltage)
Initially Ge & Si are both forward biased so current will effectivily pass through Ge diode with a drop of 0.3 V if "Ge" is revesed then current will flow through "Si" diode hence an effective drop of (0.7 – 0.3) = 0.4 V is observed.
A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to
Frequency of torsonal oscillations is given by
f_{2} = 0.8 f_{1}
A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature of 27°C. Amount of heat transferred to the gas, so that RMS velocity of molecules is doubled, is about :
[Take R = 8.3 J/ K mole]
Q = nC_{v}ΔT as gas in closed vessel
Q = 10000 J = 10 kJ
A particle is executing simple harmonic motion (SHM) of amplitude A, along the xaxis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :
According to the question, U = k
∴ Correct answer is (3)
A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to :
Frequency of the sound produced by flute,
Velocity of observer,
∴ frequency detected by observer, f' =
= 335.56 × 2 = 671.12
∴ closest answer is (4)
In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accomodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 × 10^{8}m/s,h = 6.6 × 10^{–34} Js)
= 3.75 × 10^{14} Hz
1% of f = 0.0375 × 10^{14} Hz
= 3.75 × 10^{12} Hz = 3.75 × 10^{6} MHz
number of channels =
∴ correct answer is (4)
Two point charges q_{1 }and_{ } are placed on the xaxis at x = l m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on yaxis is,
Let are the vaues of electric field dueto q_{1} & q_{2} respectively magnitude of
E_{2} = 9 × 10^{3} V/m
∴ correct answer is (3)
A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants K_{1}, K_{2}, K_{3}, K_{4} arranged as shown in the figure. The effective dielectric constant K will be :
in the same way we get,
on comparing equation (i) to equation (ii), we get
This does not match with any of the options so probably they have assumed the wrong combination
However this is one of the four options.
It must be a "Bonus" logically but of the given options probably they might go with (4)
A rod of length 50cm is pivoted at one end. It is raised such that if makes an angle of 30° from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s^{–1}) will be (g = 10ms^{–2})
Work done by gravity from initial to final position is,
According to work energy theorem
∴ correct answer is (1)
One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop (B_{L}) to that at the centre of the coil (B_{C}), i.e. R will be :
L = 2πR L = N × 2πr
R = Nr
The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4 x 10^{3} km) is E_{1} and kinetic energy required for the satellite to be in a circular orbit at this height is E_{2}. The value of h for which E_{1} and E_{2} are equal, is:
U_{surface} + E_{1} = U_{h}
KE of satelite is zero at earth surface & at height h
Gravitational attraction
The energy associated with electric field is (U_{E}) and with magnetic field is (U_{B}) for an electromagnetic wave in free space. Then :
Average energy density of magnetic field,
is maximum value of magnetic field.
Average energy density of electric field,
now,
u_{E} = u_{B}
since energy density of electric & magnetic field is same, energy associated with equal volume will be equal.
u_{E} = u_{B}
A series AC circuit containing an inductor (20 mH), a capacitor (120 μF) and a resistor (60Ω) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is :
R = 60Ω f = 50Hz, ω = 2πf = 100 π
Q = P.t = 8.64 × 60 = 5.18 × 10^{2}
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to :
on comparing the powers of M, L, T
– x + y = 0 ⇒ x = y
3x + 2y + z = 0 ⇒ 5x + z = 0 .. ..(i)
–2x – y – z = 1 ⇒ 3x + z = –1 .. .(ii)
on solving (i) & (ii)
The magnetic field associated with a light wave is given, at the origin, by B = B_{0} [sin(3.14 × 10^{7})ct + sin(6.28 × 10^{7})ct]. If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons ?
(c = 3 × 10^{8}ms^{–1}, h = 6.6 × 10^{–34} Js)
B = B_{0}sin (π × 10^{7}C)t + B_{0}sin (2π × 10^{7}C)t
since there are two EM waves with different frequency, to get maximum kinetic energy we take the photon with higher frequency
B_{1} = B_{0}sin(π × 10^{7}C)t
B_{2} = B_{0}sin(2π × 10^{7}C)t v_{2} = 10^{7}C
where C is speed of light C = 3 × 10^{8} m/s v_{2 }> v_{1}
so KE of photoelectron will be maximum for photon of higher energy.
v_{2} = 10^{7}C Hz
hv = φ + KE_{max}
energy of photon
E_{ph} = hn = 6.6 × 10^{34} × 10^{7} × 3 × 10^{9}
E_{ph} = 6.6 × 3 × 10^{–19J}
KE_{max} = E_{ph} – φ
= 12.375 – 4.7 = 7.675 eV ≈ 7.7 eV
Charge is distributed within a sphere of radius R with a volume charge density where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :
Two Carnot engines A and B are operated in series. The first one, A, receives heat at T_{1}(= 600 K) and rejects to a reservoir at temperature T_{2}. The second engine B receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at T_{3}(= 400 K). Calculate the temperature T_{2} if the work outputs of the two engines are equal :
w_{1} = w_{2}
Δu_{1} = Δu_{2}
T_{3} – T_{2} = T_{2} – T_{1}
2T_{2} = T_{1} + T_{3}
T_{2} = 500 K
A carbon resistance has a following colour code. What is the value of the resistance ?
R = 53 × 10^{4} ± 5% = 530 kΩ ± 5%
A force acts on a 2 kg object so that its position is given as a function of time as x = 3t^{2} + 5. What is the work done by this force in first 5 seconds ?
x = 3t^{2} + 5
v = 6t + 0
at t = 0 v = 0
t = 5 sec v = 30 m/s
W.D. = ΔKE
The position coordinates of a particle moving in a 3D coordinate system is given by x = a cosωt y = a sinωt and z = aωt The speed of the particle is :
v_{x} = –aωsinωt ⇒ v_{y }= aωcosωt
v_{z} = aω
In the given circuit the internal resistance of the 18 V cell is negligible. If R_{1} = 400 Ω, R_{3} = 100 Ω and R_{4} = 500 Ω and the reading of an ideal voltmeter across R_{4} is 5V, then the value R_{2} will be :
V_{4} = 5V
V_{3} = i_{1}R_{3} = 1V
V_{3} + V_{4} = 6V = V_{2}
V_{1} + V_{3} + V_{4} = 18V
V_{1} = 12 V
A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms ^{–2})
at equation
F = 100 N
In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed 'v' more than that of car B. Both the cars start from rest and travel with constant acceleration a_{1} and a_{2} respectively. Then 'v' is equal to
For A & B let time taken by A is t_{0}
from ques.
v_{A} – v_{B} = v = (a_{1} – a_{2})t_{0} – a_{2}t .. ..(i)
.. ..(ii)
putting t_{0} in equation
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be :
The top of a water tank is open to air and its water level is maintained. It is giving out 0.74 m^{3} water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to :
In flow volume = outflow volume
The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
LC = 0.5 × 10^{–2} mm
+ve error = 3 × 0.5 × 10^{–2} mm
= 1.5 × 10^{–2} mm = 0.015 mm
Reading = MSR + CSR – (+ve error)
= 5.5 mm + (48 × 0.5 × 10^{–2}) – 0.015
= 5.5 + 0.24 – 0.015 = 5.725 mm
A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is
(Given charge of electron =1.6 × 10^{–19}C)
mv = qBR .. .. (i)
Path is straight line
it qE = qvB
E = vB .. ..(ii)
From equation (i) & (ii)
m = 2.0 × 10^{–24} kg
lood reducing nature of H_{3}PO_{2} ttributed to the presence of:
H_{3}PO_{2} is good reducing agent due to presence
of two P–H bonds.
The complex that has the highest crystal field splitting energy (Δ), is :
As complex K_{3}[Co(CN)_{6}] hav e CN^{– }ligand which is strongfield ligand amongst the given ligands in other complexes.
The metal that forms nitride by reacting directly with N_{2} of air, is :
Only Li react directly with N_{2} out of alkali metals
6Li + N_{2} → 2Li_{3}N
In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic?
The major product of the following reaction is:
The transition element that has lowest enthalpy of atomisation, is:
Since Zn is not a transition element so transition element having lowest atomisation energy out of Cu, V, Fe is Cu.
Which of the following combina tion of statements is true regarding the interpretation of the atomic orbitals?
(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.
(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.
(c) According to wave mechanics, the ground state angular momentum is h equal to h/2π
(d) The plot of ψ Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.
The tests performed on compound X and their inferences are:
Compound 'X' is:
→ 2,4 – DNP test is given by aldehyde on ketone
→ Iodoform test is given by compound having
The major product formed in the following reaction is:
Aldehyde reacts at a faster rate than keton during aldol and stericall less hindered anion will be a better nucleophile so sefl aldol at will be the major product.
For the reaction, 2A + B → products, when the concentrations of A and B both wrere doubled, the rate of the reaction increased from 0.3 mol L^{–1}s^{–1} to 2.4 mol L^{–1} s^{–1}. When the concentration of A alone is doubled, the rate increased from 0.3 mol L^{–1}s^{–1} to 0.6 mol L^{–1}s^{–1}
Which one of the following statements is correct ?
r = K[A]^{x}[B]^{y}
⇒ 8 = 2^{3} = 2^{x+y}
⇒ x + y = 3 ...(1)
⇒ 2 = 2^{x}
⇒ x = 1, y = 2
Order w.r.t. A = 1
Order w.r.t. B = 2
The correct sequence of amino acids present in the tripeptide given below is :
The correct statement regarding the given Ellingham diagram is:
According to the given diagram Al can reduce ZnO.
3ZnO+2Al → 3Zn+Al_{2}O_{3}
For the following reaction, the mass of water produced from 445 g of C_{57}H_{110}O_{6} is :
2C_{57}H_{110}O_{6}(s) + 163O_{2}(g) → 114CO_{2}(g) + 110 H_{2}OP(1)
moles of C_{57}H_{110}O_{6}(s) = 445/890 = 0.5 moles
2C_{57}H_{110}O_{6}(s) + 163 O_{2}(g) → 114 CO_{2}(g) + 110 H_{2}O(l)
= 495gm
The correct match between Item I and Item II is :
The increasing basicity order of the following compounds is :
For coagulation of arscnious sulphide sol, which one of the following salt solution will be most effective ?
Sulphide is –ve charged colloid so cation with maximum charge will be most effective for coagulation.
Al^{3+} > Ba^{2+} > Na^{+ } coagulating power.
At 100°C, copper (Cu) has FCC unit cell structure with cell edge length of x Å. What is the approximate density of Cu (in g cm^{3}) at this temperature?
[Atomic Mass of Cu = 63.55u]
FCC unit cell Z = 4
The major product obtained in th e following reaction is :
Which of the following conditions in drinking water causes methemoglobinemia?
Concentration of nitrate >50 ppm in drinking water causes methemoglobinemia
Homoleptic octahedral complexes of a metal ion 'M^{3}+' with three monodentate ligands and L_{1}, L_{2}, L_{3} absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is :
Order of λ_{abs}  L_{3}>L_{1}>L_{2}
So Δ_{O} order will be L_{2} > L_{1} > L_{3}
So order of ligand strength will be L_{2}>L_{1}>L_{3}
The product formed in the reaction of cumene with O_{2} followed by treatment with dil. HCl are:
Cummene hydroperoxide reaction
The temporary hardness of water is due to:
Ca (HCO_{3})_{2} is re ponsible for temporary hardness of water
The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is :
(Specific heat of water liquid and water vapour are 4.2 kJ K^{1} kg^{1} and 2.0 kJ K^{1} kg^{1} ; heat of liquid fusion and vapourisation of water are 344 kJ kg^{1} and 2491 kJ kg^{1}, respectively).
ΔS_{total} = 9.26 kJ kg^{–1} K^{–1}
The pH of rain water, is approximately :
pH of rain water is approximate 5.6
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction Zn(s) + Cu^{2+}(aq) ⇔ Zn^{2+}(aq) + Cu(s) at 300 K is approximately.
(R = 8 JK^{–1} mol^{–1} , F = 96000 C mol^{–1})
A solution containing 62 g ethylene glycol in 250 g water is cooled to –10°C. If K_{f} for water is 1.86 K kg mol^{–1}, the amount of water (in g) separated as ice is :
ΔT_{f }= K_{f} . m
W = 0.186 kg
ΔW = (250 – 186) = 64 gm
When the first electron gain enthalpy (Δ_{eg}H) of oxygen is –141 kJ/mol, its second electron gain enthalpy is :
Second electron gain enthalpy is always positive for every element.
O^{–}_{(g)}+ e^{–} →O^{¯2}(g) ; ΔH = positive
The major product of the following reaction is :
Which of the following compounds is not aromatic?
Do not have (4n + 2) π electron It has 4n π electrons
So it is Anti aromatic.
Consider the following reversible chemical reactions :
The relation between K_{1} and K_{2} is :
Let f be a differentiable function from R to R such that f (x)  f(y)≤2xy^{3/2}, for all x, y ε R. If f(0) = 1 then is equal to
If then the value of k is :
The coefficient of t^{4} in the expansion of is
(1 – t^{6})3 (1 – t)^{–3}
(1 – t^{18} – 3t^{6} + 3t^{12}) (1 – t)^{–3}
⇒ cofficient of t^{4} in (1 – t)^{–3} is
3+4–1C_{4} = 6C_{2} = 15
For each xεR, let [x] be the greatest integer less than or equal to x. Then
is equal to
If both the roots of the quadratic equation x^{2}  mx + 4 = 0 are real and distinct and they lie in the interval [1 5] then m lies in the interval:
No option correct : Bonus
* If we consider αβ∈ (1,5) then option (1) is correct.
If
Then A is 
The area of the region A = [(x,y) : 0 ≤ y ≤ xx +1 and 1 ≤ x ≤ 1] in sq. units, is:
The graph is a follows
Let z_{0} be a root of the quadratic equation, x^{2} + x + 1 = 0. If z = 3 + 6iz_{0}^{81}  3iz_{0}^{93}, then arg z is equal to:
z_{0} = ω or ω^{2} (where ω is a nonreal cube root of unity)
z = 3 + 6i(ω)^{81} – 3i(ω)^{93}
z = 3 + 3i
Let and be three vectors such that the projection vector of on is perpendicular to is equal to:
⇒ 5b_{1} + b_{2} = – 10 ...(2)
from (1) and (2) ⇒ b_{1} = –3 and b_{2} = 5
Let A(4,–4) and B(9,6) be points on the parabola, y^{2} + 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of ΔACB is maximum. Then, the area (in sq. units) of ΔACB, is:
The logical statementis equivalent to:
An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is :
E_{1} : Event of drawing a Red ball and placing a green ball in the bag
E_{2} : Event of drawing a green ball and placing a red ball in the bag
E : Event of drawing a red ball in second draw
If 0 ≤ x < π/2, then the number of values of x for which sin xsin2x+sin3x = 0, is
sinx – sin2x + sin3x = 0
⇒ (sinx + sin3x) – sin2x = 0
⇒ 2sinx. cosx – sin2x = 0
⇒ sin2x(2 cosx – 1) = 0
⇒ sin2x = 0 or cosx = 1/2
⇒ x = 0, π/3
The equation of the plane containing the straight line x/2 = y/3 = z/4 and perpendicular to the plane containing the straight lines x/3 = y/4 = z/2 and x/4 = y/2 = z/3 is:
Vector along the normal to the plane containing the lines
vector perpendicular to the vectors and
so, required plane is 26x – 52y + 26z = 0
x – 2y + z = 0
Let the equations of two sides of a triangle be 3x – 2y + 6 = 0 and 4x + 5y – 20 = 0. If the orthocentre of this triangle is at (1,1), then the equation of its third side is :
Equation of AB is 3x – 2y + 6 = 0
equation of AC is 4x + 5y – 20 = 0
Equation of BE is 2x + 3y – 5 = 0
Equation of CF is 5x – 4y – 1 = 0
⇒ Equation of BC is 26x – 122y = 1675
If x = 3 tan t and y = 3 sec t, then the value of
If x = sin^{–1}(sin10) and y= cos^{ –1}(cos 1 0), then y–x is equal to:
x = sin^{–1}(sin 10) = 3π – 10
y = cos^{–1}(cos10) = 4π – 10
y – x = π
If the lines x = ay+b, z = cy + d and x=a'z + b', y = c'z + d' are perpendicular, then:
Line x = ay + b, z = cy + d
⇒
Line x = a'z + b', y = c'z + d'
Given both the lines are perpendicular
⇒ aa' + c' + c = 0
The number of all possible positive integral values of α for which the roots of the quadratic equation, 6x^{2}–11x+α = 0 are rational numbers is :
6x^{2} – 11x + α = 0
given roots are rational
⇒ D must be perfect square
⇒121 – 24α = λ^{2}
⇒ maximum value of α is 5
α = 1 ⇒ λ ∉ I
α = 2 ⇒ λ ∉ I
α = 3 ⇒ λ ∈ I ⇒ 3 integral values
α = 4 ⇒ λ ∈ I
α = 5 ⇒ λ ∈ I
A hyperbola has its centre at the or igin, passes through the point (4,2) and has transverse axis of length 4 along the xaxis. Then the eccentricity of the hyperbola is :
2a = 4 a = 2
Passes through (4,2)
Let A={xεR:x is not a positive integer}
Define a function f :A→R as f(x) = 2x/x1 then f is
⇒ ƒ is oneone but not onto
If and f(0) = 0, then the value of f(1) is :
If the circles x^{2} + y^{2} – 16x–20y + 164 = r^{2} and (x–4)^{2} + (y – 7)^{2} = 36 intersect at two distinct points, then:
x^{2} + y^{2} – 16x – 20y + 164 = r^{2}
A(8,10), R_{1} = r
(x – 4)^{2} + (y – 7)^{2} = 36
B(4,7), R_{2} = 6
R_{1} – R_{2} < AB < R_{1 }+ R_{2}
⇒ 1 < r < 11
Let S be the set of all triangles in the xyplane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50sq. units, then the number of elements in the set S is:
Let A(α,0) and B(0,β) be the vectors of the given triangle AOB
⇒ αβ = 100
⇒ Number of triangles
= 4 × (number of divisors of 100)
= 4 × 9 = 36
The sum of the follwing series
+...up to 15 terms, is:
= 7820
Let a, b and c be the 7^{th}, 11^{th }and 13^{th} terms respectively of a nonconstant A.P. If these are also the three consecutive terms of a G.P., then a/c is equal to:
a = A + 6d
b = A + 10d
c = A + 12d
a,b,c are in G.P.
⇒ (A + 10d)^{2} = (A + 6d) (a + 12d)
If the system of linear equations
x–4y+7z = g
3y – 5z = h
–2x + 5y – 9z = k
is consistent, then
P_{1} ≡ x – 4y + 7z – g = 0
P_{2} ≡ 3x – 5y – h = 0
P_{3} ≡ –2x + 5y – 9z – k = 0
Here Δ = 0
2P_{1} + P_{2} + P_{3} = 0 when 2g + h + k = 0
Let f:[0,1]→R be such that f(xy) = f(x).f(y) for all x,y,ε[0,1], and f(0)≠0. If y = y(x) satisfies the differential equation, dy/dx = f(x) with y(0) = 1, then is equal to
ƒ(xy) = ƒ(x). ƒ(y)
ƒ(0) = 1 as ƒ(0) ≠ 0
⇒ ƒ(x) = 1
⇒ y = x+c
At, x = 0, y = 1 ⇒ c = 1
y = x + 1
A data consists of n observations:
x_{1}, x_{2}, ....., x_{n}. If and , then the standard deviation of this data is :
⇒ variance = 6 – 1 = 5
⇒ Standard diviation = √5
The number of natural numbers less than 7,000 which can be formed by using the digits 0,1,3,7,9 (repitition of digits allowed) is equal to :
Number of numbers = 5^{3} – 1
2 ways for a_{4}
Number of numbers = 2 × 5^{3}
Required number = 5^{3} + 2 × 5^{3} – 1
= 374




