A planet of core density 3ρ and outer curst of density ρ has small tunnel in core. A small Particle of mass m is released from end A then time required to reach end B :
At some distance from centre inside core
Now time for A to B =
A small circular wire loop of radius a is located at the centre of a much larger circular wire loop of radius b as shown above (b> >a). Both loops are coaxial and coplanar. The larger loop carries a time (t) varying current I = I0 cos ωt where I0 and ω are constants. The large loop induces in the small loop an emf that is approximately equal to which of the following.
Magnetic flux passing through inner loop due to current in outer loop-
So, emf developed,
A spring mass system is placed on a frictionless horizontal surface as shown in the figure. The spring is expanded by 1/10m and the blocks are given velocities as shown, then maximum extension of spring is :
As there is no loss of energy
Initial mechanical energy = Final mechanical energy
For initial Mechanical energy
Where and vrelative =5 m/s
= 25 + 25 = 50
For final Mechanical energy
When there is maximum extension relative velocity between block = 0
An electromagnetic wave of frequency f = 7.3MHz passes from a vacuum into a dielectric medium with permittivityε = 9. Then,
As permittivity is greater than the permittivity of free space so the refractive index of medium is
Wavelength of the electromagnetic wave in medium will be
Velocity of wave will be.
A particle is projected from point A towards a building of height h as shown at an angle of 60 ° with horizontal. It strikes the roof of building at B at an angle of 30 ° with the horizontal. The speed of projection is
Let u be the speed of projection and v be the speed at B
Applying law of conservation of energy,
Using equation (i) in equation (ii),
Three identical bulbs each of resistance 2Ω are connected as shown. The maximum power that can be consumed by individual bulb is 32W, then the maximum power consumed by the combination is :
Resistance of B1, B2 and B3 are same which is = 2Ω
As P ∝i2 (when ‘R’ constant)
So, maximum power consumed by bulb B3
Which is = 32 W
⇒i2R = 32
⇒i2 . (2) = 32
⇒i = 4 Ampere
So, current passing through B1 + B2 = i/2 = 2
Total power consumed in circuit
P = (i/2)2 (2) + (i/2)2 (2) + i2(2)
= (2)2 (2) + (2)2 (2) + (4)2 (2)
A carrier wave has power of 1675 kW. If the side band power of a modulated wave subjected to 60%. Then find the amplitude modulation level
As you know that, sideband power is, Where PC is power of carrier wave, PS s sideband power and ‘m’ is modulation rate,
Two identical conducting spheres each having radius r are placed at large distance. lnitially charge on one sphere is q, while charge on another sphere is zero when they are connected by conducting wire as shown in figure then find total heat produced when switch S is closed :
When sphere connected by wire, charge will flow from one sphere to another until potential of both sphere become same.
As both sphere is identical in dimension and material, charge will divide equally
Now, Initial potential energy of system
(Self energy of spherical conductor )
Final potential energy of system
Heat generation in conduction = Ui - uf
Four wire A, B, C and D each of length l = 10 cm and each of area of cross section is 0.1 m2 are connected in the given circuit. Then, the position of null point is
Given that resistivity
When S1 and S2 both open and jockey has no deflection then let current in upper circuit is ” i”.
So, RA = 1Ω
RB = 3Ω
RC = 6Ω
RD = 1Ω
When jockey is touched, and there is no deflection then potential difference between ‘A’ and the point should be same in both path-
So, in lower circuit as S2 is open
No current will flow in this circuit
⇒Potential difference between A and P
Will be = 2V
⇒i1 . 3= i2 . 6 and i1 + i2 = 1
Potential difference between A and P
For this point ‘P’ should be half way in wire C.
Similarly when jockey touched on wire ‘B’, Null point will be obtained at middle point of wire B.
The radius of curvature of spherical surface is 10 cm. The spherical surface separates two media of refractive indices μ2 = 1.3 and μ3 = 1.5 as shown in Figure. The medium of refractive index 1.3 extends upto 78 cm from the spherical surface. A luminous point object O is at the distance of 144 cm from the spherical surface in the medium of refractive index μ1 = 1.1. The image formed by the spherical surface is at
Distance of object from the spherical surface
For the refraction at the spherical Surface
A transformer has an efficiency of 80%. It is connected to a power input of 4 kW and 100 V. if the secondary voltage is 240 V, then the secondary current is
Magnetic moments of two identical magnets are M and 2M respectively. Both are combined in such a way that their similar poles are same side. The time period in this case is ‘T1’ .If polarity of one of the magnets is reversed its period becomes ‘T2’ then find out ratio of their time periods T1/T2.
Msystem = M + 2M = 3M
Isystem = I + I = 2I
Msystem = 2M – M = M
Isystem = 2I
The maximum current in a galvanometer can be 10 mA. It’s resistance is 10Ω. To convert it into an ammeter of 1 Amp. A resistor should be connected in
A soap bubble (surface tension = T) is charged to a maximum surface density of charge = σ, when it is just going to burst. Its radius R is given by:
Pressure due to surface tension
Pressure due to electrostatic force
Just before the bubble burst
Pressure due to surface tension = Pressure due to electrostatic force
One mole of an ideal gas heated by law p = αV where P is pressure of gas, V is volume, α is a constant.
The heat capacity of gas in the process is-
A sonometer wire resonates with a given tuning fork forming stationary waves with 5 antinodes, between 2 bridges, when a mass of 9 kg is suspended from the wire. When the mass is replaced by another mass m, the wire with the same tuning fork forms three antinodes, for the same position of the bridges. The value of m is
where μ is the mass per unit length of wire and T = weight of mass(Mg) and p is number of antinodes.
Since wire and the tuning fork are the same in both cases, so
In a damped oscillator the amplitude of vibrations of mass m = 150 grams falls by 1/e times of its initial value in time t0 due to viscous forces. The time t0 and the percentage loss in mechanical energy during the above time interval t0 respectively are (Let damping constant be 50 grams/s)
A thermodynamic process of one mole ideal monoatomic gas is shown in figure. The efficiency of cyclic process ABCA will be :
An infinite current carrying wire, carrying current I is bent in V shape, lying in x-y plane as shown in figure. Intensity of magnetic field at point P will be (take OP = 2a)
B at P due to 1 wire
due to both wire = 2( is in same direction)
A silicon diode is connected in series with resistance and battery. What should be the value of battery if the reading of ammeter 3.62 A.
Rearranging the circuit,
As we know the potential drop across silicon diode in forward biased is 0.7V. Current flowing through the circuit is
The relation between Internal energy U, pressure P and volume V of a gas in an adiabatic process is U = a + bPV where a = b = 3. The greatest integer of the ratio of specific heats [γ] is
Number of teeth in front and rear sprocket (tooth wheel) of a bicycle is 40 and 20 respectively. If angular speed of front sprocket is 4 radian per second then find the angular speed of the rear sprocket Assume that spacing between the teeth is same for both sprocket.
Consider the lens shown in fig with radii of curvature of the lens equal to 10 cm and 20 cm. Refractive index of the material of the lens is 1.5 and x axis is the principal axis of the lens. Find the magnification produced by lens for the object placed at a distance of 20 cm from the lens.
Interference pattern is formed on a screen by Young's double slits S1 and S2 Illuminated by monochromatic light of wavelength λλ. Points P and P’ are on the screen closest to and on either side of central maxima, where intensity is one fourth that at the central maxima. If . Find x where x is the coefficient of d.
When the radiation emitted by Li++ in transition from n = 4 to n = 3 is made to fall on a metal surface, the emitted photo electrons moves in a circular track of radius cm when moved in a transverse magnetic field Find work function of the metal (in eV). (Given that Rydberg's constant
Consider the following equilibrium
N2O4 (g) ⇔ 2NO2(g)
Then select the correct graph: (graph represent relation between concertation of NO2 and N2O4 at equilibrium)
[NO2]2 = keq [N2O4]
It resembles graph of y2 = 4 ax
So option (b) is correct
If two different non-axial d-orbitals having XZ nodal plane form π bond by overlapping each other, then internuclear axis will be;
Two different non-axial d-orbitals will lie in planes perpendicular to each other, such d-orbital will not formed
Hence, (d) option is correct.
The given osazone can be formed by
Formation of osazone involves only C1 and C2 carbon atoms. The rest of the carbon atoms remains uneffected.
Glucose and fructose differ only in the configurations of first and second carbon atoms remaining positions are similar. So they form same osazone.
The cyclotrimetaphosphoric acid is
Hence, option (C ) is correct.
In a face centered cubic lattice of edge length 'a' number of next to next nearest neighbor to corner atom and distance of corner atom to next to next nearest atom is respectively-
Next nearest neighbor = a
Next to next nearest neighbor
No of such neighbors = 24
Hence, correct answer is (c)
CrO42(-) (yellow) changes to Cr2O72(-) (orange) in pH= x and Cr2O72(-) (orange) changes to CrO42(-) (yellow) in ph = y. Then x and y can be:
Hence, option (a) is correct.
Identify position most favorable for aromatic substitution (EAS)
is activating group so it would activate ortho and para position for E.A.S but only para position is
empty so EAS will take place there.
Hence, option (b) is correct
Polymer of prop – 2 – ene nitrile is-
Prop- 2 – ene nitrile
Hence (b) is correct answer
Molar mass of a substance, 1g of which when dissolved in 100g of water gave a solution whose boiling point is 100.1oC at a pressure of 1 atm (kb of water = 0.52 k kg mol–1) is-
The given values
WB = 1g kb = 0.5 2 k kg mol–1
WA = 100
Tb = 100. 1–100 = 0. 1
Hence, correct answer is (b)
Arrange the following in increasing order of their Pka values
Because acidic character depends on stability of conjugate ion.
3 resonating structure
2 resonating structure
Acidic character of x > y > z
So ka → x > y > z
Hence pka x < y < z
Reduction potential of following electrode Pt, H2 (4atm) | H2SO4 (0.01M) is-
H2 (4 atm) →2H(+) (0.01M) + 2e(-)
Hence option (b) is correct
If molecular weight of As2S3 is M. Then in the following reaction
As2S3 + 7NaClO3 + 12NaOH → 2Na3AsO4 + 7NaClO + 3Na2SO4 + 6H2O.
the equivalent weight of As2S3 is
Both As and S go under oxidation
As2S3→ AsO43(–) + 4e(–) (oxidation of As)
As2S3→ 3SO42(–) + 24e(–) (oxidation of S)
Total electrons = 24 × 4 = 28 = n-factor
Which will give only one product (excluding stereoisomer) when undergoing E2 reaction ?
Which are identical product
Hence option (d) is correct
Assume that a particular amino acid has isoelectric point as 6.0. In a solution at PH 1.0 which of the following will predominate
In acidic medium z witter ion
Hence option (a) is correct
C – 2 and C – 3 configuration of
According to CIP rule
Clockwise is R configuration and Anti-clockwise is S configuration
Hence option (c) is correct
End product in the reaction is,
Acid base reaction
Hence (a) option is correct.
Which of the following does not occur in Bessemer’s converter
n Bessemer’s converter
2CuFeS2 + O2→Cu2S + 2 FeS + SO2 does not take place
Hence, option (c) is correct
For reaction CO(g) + NO2(g) CO2(g) +NO(g) Energy profile diagram is given below.
What is the 'activation energy' of the reaction?
Activation energy is energy from reactant to highest energy barrier
So activation energy is ‘x’ hence A option is correct
Which of the following is buffer-
Buffer is conjugate acid and salt of conjugate base pair.
Hence option (C) is correct answer.
Nitrogen dioxide cannot be obtained from
Hence, correct answer is (C)
Find the value of ‘x’ in the tremolite asbestos Ca2Mgx (Si4O11)2 (OH)2
The silicate is a amphibole with general formula
Ka for butyric acid is 2 x 10-5. What will be pH of 0.2 M aqueous solution of sodium butyrate?
1.575 gm of oxalic acid (COOH)2. xH2O are dissolved in water and the volume made upto 250 ml. on titration 16.68 ml of this solution requires 25mL of N/15 NaOH solution for complete neutralization. Calculate x.
Ethereal solution of how many of the following pairs of compounds can be separated by aqueous NaHCO3 solution but not by aqueous NaOH solution?
I and III
The sum of coordination number and oxidation state of central metal ion in complex formed, when excess of KCN is added to aqueous solution of copper sulphate is
There are two balls in an urn whose colours are not known (each ball can be either white or black). A white ball is put into the urn. A ball is drawn from the urn. The probability that it is white is
Let Ei (0 < i < 2) denote the event that urn contains 'i' white and '(2 – i)' black balls.
Let A denote the event that a white ball is drawn from the urn.
We have P(Ei) = 1/3 for i = 0, 1, 2
P (A|E1) = 1/3, P(A|E2) = 2/3, P(A|E3) = 1.
By the total probability rule,
P(A)= P(E1)P(A|E1) + P(E2)P(A|E2) + P(E3)P(A|E3)
The plane passing through the point (−2, −2, 2) and containing the line joining the points
(1, 1, 1) and (1, −1, 2) makes intercepts on the coordinates axes the sum of whose lengths is
Equation of any plane passing through (−2, −2, 2) is
a(x + 2) + b(y + 2) + c(z – 2) = 0
Since it contains the line joining the points (1, 1, 1) and (1, −1, 2), it contains these points as well
3a + 3b – c = 0
and 3a + b + 0 = 0
Solving we get
and thus the equation of the plane is
(x + 2) - 3(y + 2) = 0
Intercepts on axes = 4, 4/3
The required sum= 4+(4/3)=16/3
If a, b, c are in GP and are in AP, then a, b, c are the lengths of the sides of a triangle which is
a, b, c are in GP b2 = ac
are in AP.
2(log 2b – log 3c) = log a – log 2b + log 3c – log a
log 2b = log 3c 2b = 3c
∴ b2 = ac & 2b = 3c
b = 2a/3 and c = 4a/9
Since (a+b) = 5a/3 > c, (b+c) = 10a/9 > a and
(c + a) = 13a/9 > b, therefore, a, b, c are the sides
of a triangle. Also, as ‘a’ is the greatest side, let us
find angle A of ΔABC
Hence ΔABC is an obtuse angled triangle.
Sum of the series
1 + 3 + 6 + 10 + 15 + ……………………….n terms is
The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 is
Put 2x + y = X ⇒Therefore, the given equation is reduced to
= x + constant
2(2x + y) + log (2x + y – 1) = 3x + constant
x + 2y + log (2x + y – 1) = C
The area bounded by curves y = f(x), the x-axis and the ordinates x = 1 and x = b is (b - 1) sin (3b + 4). Then f(x) is-
1. Area bounded by curve y = f(x), x =1 and x = b is
Now differentiating both sides with respect to b we get
fB. = sin (3b + 4) + 3(b – 1) cos (3b + 4)
If sin 5x + sin 3 x + sin x = 0, then the value of x other than zero, lying between 0 < x < π/2 is
1. sin 5x + sin x + sin 3x = 0
2 sin 3x . cos2x + sin3x = 0
sin 3x (2 cos 2x + 1) = 0, 0 ≤ x ≤π/2
From both x = π / 3 (other than 0).
For n > 2 the product where is equal to
Let Hence the given series S is
The value of K, for which the equation (K–2)x2 + 8x + K + 4 = 0 has both the roots real distinct and negative is:
(K–2)x2 + 8x + K + 4 = 0
If the equation f(x) = 0, has real distinct and negative roots then:
D > 0 & f(0) > 0 &
(i) D > 0
64 – 4 (K2 + 2K – 8) > 0
– K2 – 2K + 24 < 0 K2 + 2K – 24 < 0
(K + 6)(K – 4) < 0
–6 < K < 4
⇒ K < –4 or K > 2
K – 2 > 0
K > 2
∴D > 0 & f(0) > 0 &
–6 < K < 4 & (K < –4 or K > 2) & K > 2
–6 < K < 4 & K < –4 & K > 2) or (K > 2 & K > 2
– 6 < K < 4 &
2 < K < 4
Among the given options
3 lies in the given range,
Hence K = 3
The equation of the common tangent touching the circle (x-3)2 + y2 = 9 and the parabola y2 = 4x above the x-axis, is
Equation of any tangent to parabola
If the tangent to parabola is also tangent to the circle (x-3)2 + y2 = 9 then:
9m4 + 6m2 + 1 = 9m4 + 9m2
3m2 – 1 = 0
From the figure it is evident that for the tangent above x-axis Hence equation of required common tangent is:
Let f(x) be a continuous function such that f(a – x) + f(x) = 0 for all x[0,a]. Then, the value of the integral is equal to
Adding (i) and (ii):
The circles which can be drawn to pass through (1,0) & (3,0) and touching the y-axis, intersect at an angle θ. The value of cos θ is equal to
Equation of line joining A (1,0) and B (3,0) is y = 0. Equation of family of circles passing through A and B is:
(x – 1)(x – 3) + (y – 0) (y – 0) + λy = 0
x2 + y2 – 4x + λy + 3 = 0
If above circle touches y-axis then x = 0 is a tangent. Substituting x = 0:
y2 + λy + 3 = 0
Discriminant of above quadratic must be zero.
λ2 – 12 = 0
Hence the circles are:
Thus, the coordinates of C1 and C2 arerespectively. Also the radii of each of the circles is r1 = r2 = 2
We know that the angle of intersection of two circles of radii r1 and r2 is given by
where d is the distance between their centers.
Since between two lines whenever there is angle θ there is always the angle π–θ, therefore,
a, b, c are positive numbers and abc2 has the greatest value 1/ 64. Then
Also for the greatest value of abc2 the numbers have to be equal, i.e a = b = c/2
Also given that maximum value = 1/64
so, a + b + c = 1
i.e. a = b = 1/4, c = 1/2
If A and B are two square matrices such that B = –A–1 BA, then (A+B)2 is equal to
B = –A–1 BA
AB = A (–A–1BA)
AB = –BA
AB + BA = 0
Now, (A + B)2 = A2 + B2 + AB + BA
(A + B)2 = A2 + B2, (∵AB + BA = 0)
The set of all values of the parameter a for which the points of minimum of the function y = 1 + a2 x – x3
Satisfy the inequality
x2 + 5x + 6 < 0 –3 < x < –2
y = 1 + a2 x – x3
(i) Let a > 0:
Hence Hence y has minima at for a > 0.
(ii) Let a < 0:
Hence Hence y has minima at for a < 0.
From (i) & (ii)
The value of is
Substitute (x – 1) = t. Hence
Hence given limit is:
f (x) is an even function.
NOTE: Integration of every even function is odd function and integration of every odd function is even function, provided the integrands are integrable and the limit is 0 to x. Similarly differentiation of every odd function is even function and that of every even function is odd function, provided the they are differentiable. Here which is an odd function, therefore is even function.
If for a variable line the condition a–2 + b–2 = c–2 (c is a constant), is satisfied, then the locus of foot of the perpendicular drawn from origin to this is:
a–2 + b–2 = c–2
bx + ay – ab = 0
The foot of perpendicular to the above line from O(0,0) is given by:
Using a–2 + b–2 = c–2
x2 + y2 = c2
The eccentricity of the hyperbola whose latus rectum is half of its transverse axis, is
Length of latus rectum of hyperbola = Length of the transverse axis
A tangent having slope ofto the ellipse intersects the major and minor axes in points A and B respectively. If C is the center of the ellipse then the area of the triangle ABC is:
Equation of the tangent to whose slope
From symmetry of ellipse it is obvious that area of the triangle ABC will be same with respect either tangent.
Let us consider
Hence area Δ of the triangle ABC is:
The number of solutions of the equation is/are
Out of these two values of x, on;y x=1 satisfies the given equation
If the circle (x - a)2 + y2 = 25 intersects the circle x2 + (y - b)2 = 16 in such a way that common chord is of maximum length, then value of a2 + b2 is
Radial axis: - 2ax + a2 + 2by - b2 = 9
which passes through (0, b)
i.e. 0 + a2 + 2b2 – b2 = 9 ⇒ a2 + b2 = 9.
Area bounded by curves y = [cos A + cos B + cos C], (where [.]) denotes the greatest integer function and A, B, C are angles of a triangle) and curve |x-1|+|y|=2 is
If a tangent of slope 2 of the ellipse is normal to the circle x2 + y2 + 4x + 1 = 0, then the maximum value of ob is
Let then n is equal to ...