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A block of mass m is suspended from a pulley in form of a circular disc of mass m & radius R. The system is released from rest, find the angular velocity of disc when block has dropped by height h. (there is no slipping between string & pulley)
v = ωR (no slipping)
Three point masses 1kg, 1.5 kg, 2.5 kg are placed at the vertices of a triangle with sides 3cm, 4cm and 5cm as shown in the figure. The location of centre of mass with respect to 1kg mass is :
Take 1kg mass at origin
In a single slit diffraction set up, second minima is observed at an angle of 60°. The expected position of first minima is
For 2^{nd }minima
d sinθ = 2λ
sinθ = (given)
......(1)
So for 1^{st} minima is
d sinθ = λ
sinθ = (from equation (i))
θ = 25.65° (from sin table)
θ ≈ 25°
There are two infinite plane sheets each having uniform surface charge density +σ C/m^{2}. They are inclined to each other at an angle 30° as shown in the figure. Electric field at any arbitrary point P is:
A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of dielectric constant given by k = k_{0}(1 + αx). Calculate capacitance of system: (given αd << 1)
Capacitance of element
Capacitance of element
Given αd << 1
A long solenoid of radius R carries a time dependent current I = I_{0} t(1 – t). A ring of radius 2R is placed coaxially near its centre. During the time interval 0 ≤ t ≤ 1, the induced current I_{R} and the induced emf V_{R} in the ring vary as:
and
If 10% of intensity is passed from analyser, then, the angle by which analyser should be rotated such that transmitted intensity becomes zero. (Assume no absorption by analyser and polarizer).
So θ > 45° and 90 – θ < 45º so only one option is correct i.e. 18.4º
angle rotated should be = 90° – 71.6° = 18.4°
Three moles of ideal gas A with is mixed with two moles of another ideal gas B with . The of mixture is (Assuming temperature is constant)
on rearranging we get
γ_{mix }= 1.42
Given magnetic field equation is B = 3 × 10^{–8} sin(ωt + kx + φ)then appropriate equation for electric field (E) will be :
(speed of light in vacuum)
E_{0} = B_{0}C = 3 × 10^{–8} × 3 × 10^{8}
= 9 N/C
So E = 9 sin (ωt + kx + φ)
There is a LCR circuit, If it is compared with a damped oscillation of mass m oscillating with force constant k and damping coefficient 'b'. Compare the terms of damped oscillation with the devices in LCR circuit.
In damped oscillation
ma + bv + kx = 0
In the circuit
Comparing equation (i) and (ii)
m = L, b = R, k = 1/c
A lift can hold 2000kg, friction is 4000N and power provided is 60HP. (1 HP = 746W) Find the maximum speed with which lift can move up.
A H–atom in ground state has time period T = 1.6 × 10^{–16} sec. find the frequency of electron in first excited state
Magnification of compound microscope is 375. Length of tube is 150mm. Given that focal length of objective lens is 5mm, then value of focal length of eyepiece is:
CaseI
If final image is at least distance of clear vision
CaseII
If final image is at infinity
f_{e} = 22 mm
1 litre of a gas at STP is expanded adiabatically to 3 litre. Find work done by the gas. Given γ = 1.40 and 3^{1.4}= 4.65
now work done
Closest ans is 90.5 J
A string of length 60 cm, mass 6gm and area of cross section 1mm^{2} and velocity of wave 90m/s. Given young's modulus is Y = 1.6 × 10^{11} N/m^{2}. Find extension in string.
after substituting value of μ,v,l,A and Y we get
Δl = 0.3 mm
Which of the following gate is reversible
A logic gate is reversible if we can recover input data from the output eg. NOT gate
A thin uniform rod is of mass M and length L. Find the radius of gyration for rotation about an axis passing through a point at a distance of L/4 from centre and perpendicular to rod.
A satellite of mass 'M' is projected radially from surface of earth with speed 'u'. When it reaches a height equal to radius of earth, it ejects a rocket of mass M/10 and itself starts orbiting the earth in circular path
of radius 2R, find the kinetic energy of rocket.
Kinetic energy
The current 'i' in the given circuit is
A current carrying circular loop is placed in an infinite plane if φ_{1} is the magnetic flux through the inner region and φ_{0} is magnitude of magnetic flux through the outer region, then
As magnetic field lines always form a closed loop, hence every magnetic field line creating magnetic flux in the inner region must be passing through the outer region. Since flux in two regions are in opposite direction,
∴ φ_{i} =  φ_{0}
Consider a loop ABCDEFA with coordinates A (0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0) E(0, 5, 5) and F(0, 0, 5). Find magnetic flux through loop due to magnetic field
φ = (3 × 25) + (4 × 25) = 175 weber
A Carnot's engine operates between two reservoirs of temperature 900K and 300K. The engine performs 1200 J of work per cycle. The heat energy delivered by the engine to the low temperature reservoir in a cycle is:
A nonisotropic solid metal cube has coefficients of linear expansion as 5 × 10^{–5}/°C along the xaxis and 5 × 10^{–6}/°C along yaxis and zaxis. If coefficient of volume expansion of the solid is C × 10^{–6}/°C then the value of C is
A particle is released at point A. Find its kinetic energy at point P. (Given m = 1 kg and all surfaces are frictionless)
As only consevative internal force acts upon the mass and earth system, thus we can say mechanical energy is conserved. Thus we get that net loss in PE = net gain in KE
Loss in PE = mg Δh = 1 x 10 x 1 = 10J
Thus gain in KE = 10J
On a photosensitive metal of area 1 cm^{2} and work function 2eV, light of intensity 6.4 × 10^{–5} W/cm^{2} and wavelength 310 nm is incident normally. If 1 out of every 10^{3} photons are successful, then number of photoelectrons emitted in one second is 10^{x}. Find x
Energy of photon. E == 4eV > 2eV (so photoelectric effect will take place)
= 4 × 1.6 × 10^{–19} = 6.4 × 10^{–19} Joule
No. of photons falling per second
No. of photoelectron emitted per second
2 × 0.34 = + 1 x 0.522
E^{º}_{1} = 0.68 – 0.522
E^{º}_{1} = 0.158
Correct order of electron gain enthalpy (kJ/mole) of F, Cl, Br, I
Arrange the following in order of their pK_{b} value
(B) CH_{3}–NH–CH_{3 }(C) CH_{3}–CH=NH
Option "A" represent Guanadine, the conjugate acid of which is resonance stabilised. The option 'B' is aliphatic amine, here the 'N' is sp^{3} whereas in option 'C' the 'N' is sp^{2}, hence B is more basic than C.
1Methylethylene oxide Product 'X' will be –
^{excess HBr}>
Correct order of Intermolecular forces
Hex3ynal (X), formed product X will be:
, formed product 'X' is used as:
Methyl orange is used as an indicator in acid base titration.
In which of the following Saytzeff product will not be formed as major product ?
(A) (B)
(C) (D)
Match the column
Atomic radius of Ag is similar to
Atomic radius of Ag is closest to Au.
Correct IUPAC name of [Pt(NH_{3})_{2}Cl(CH_{3}NH_{2} )]Cl is:
The IUPAC name of [Pt(NH3)2Cl(CH3NH2 )]Cl is diamminechlorido(methylamine)platinum(II)chloride.
The ligands present are ammine, chloro and methyl amine. The ligands are named according to the alphabetical order.
The prefix di indicates two.
The oxidation state of platinum is +2. The oxidation state is written in roman numerals inside parenthesis.
Vapour pressure of pure CS_{2} and CH_{3}COCH_{3} are 512 mm of Hg and 312 mm of Hg respectively. Total vapour pressure of mixture is 600 mm of Hg then find incorrect statement:
Above mixture of liquids show positive deviation from Raoult's Law
Purest form of commercial iron is:
Purest form is wrought iron.
Mixture of above three organic compound was subjected to aq NaHCO_{3} and followed by dil NaOH. compounds which will be soluble in given solvent will be :
Which theory can explain bonding of Ni(CO)_{4}:
n = 5, m_{s} = + 1/2 How many orbitals are possible:
In zeolites & synthetic resin method which will be more efficient in removing permanent hardness of water :
The difference between the zeolite and ion exchange process is that the zeolite process uses the mineral zeolite as the exchanging resin for cations in the hard water whereas the ion exchange process include several different resins for the ion exchange.
∴ Synthetic resins method is more efficient as it can exchange both cations as well as anions.
Oxidation state of potassium in K_{2}O, K_{2}O_{2} & KO_{2} are respectively –
Decreasing order of dipole moment in CHCl_{3}, CCl_{4} & CH_{4} is –
Amongst the following which is not a postulate of Dalton's atomic theory
1) Elements are composed of extremely small particles called atoms that are indivisible and indestructible
2)All atoms of a given element are identical; they have the same size, mass, and chemical properties
3) Atoms of 1 element are different from the atoms of all other elements
4)Compounds are composed of atoms of more than 1 element. The relative number of atoms of each element in a given compound is always the same.
5)Chemical reactions only involve the rearrangement of atoms. Atoms are not created or destroyed during chemical reactions.
Half life of _{90}Sr is 6.93 years. In a child body 1 μg of _{90}Sr dopped in place of calcium, how many years will it take to reduce its concentration by 90% (Assume no involvement of Sr in metabolism).
Each of solution A and B of 100 L containing 4 g NaOH and 9.8 g H_{2}SO_{4}. Find pH of solution which is obtain by mixing 40 L solution of A and 10 L solution of B.
A(l) → 2B(g)
ΔU = 2.1 kcal, ΔS = 20 cal/k, T = 300 K.
Find ΔG (in kcal)
ΔU = 2.1 kcal, ΔS = 20 cal/k, T = 300 K.
ΔH = ΔU + ΔngRT
= 2.1 × 10^{3} + 2(2) (300)
= 2100 + 1200
= 3300 cal
ΔG = ΔH  TΔS
= 3300 – (300) (20)
= 3300 – 6000
= –2700 cals = –2.7 kcal
Cl_{2} on reaction with hot & conc. NaOH gives two chlorine having products X and Y. On treatment with AgNO_{3}, X gives precipitate. Determine average bond order of Cl and O bond in 'Y' ?
Number of chiral centers in chloramphenicol is :
If f(x) is continuous and differentiable in x ∈ [ 7, 0] and f'(x) ≤ 2∈ [7, 0], also f(7) = 3 then range of f(1) + f(0)
Lets use LMVT for x ∈ [7, 1]
Also use LMVT for x ∈ [7, 0]
∴ f(0) + f(1) ≤ 20
If y = mx + 4 is common tangent to parabolas y^{2} = 4x and x^{2} = 2by. Then value of b is
y = mx + 4 ……(i)
y^{2} = 4x tangent
y = mx + a/m
⇒ y = mx + 1/m ………(ii)
from (i) and (ii)
4 = 1/m ⇒ m =1/4
So line y = + 4 is also tangent to parabola x^{2} = 2by, so solve
⇒ 2x^{2} – bx – 16b = 0
⇒ D = 0
⇒ b^{2}  4 × 2 × (16b) = 0
⇒ b^{2} + 32 × 4b = 0
b = –128, b = 0 (not possible)
If α and β are the roots of equation (k+1) tan^{2}x  √2λ tanx = 1  k and tan^{2} (α+β) = 50. Find value of λ
λ = 10
Find image of point (2, 1, 6) in the plane containing points (2, 1, 0), (6, 3, 3) and (5, 2, 2)
Plane is x + y – 2z = 3
⇒ (x, y, z) = (6, 5, 2)
Let (x)^{k} + (y)^{k} = (a)^{k }where a, k > 0 and then find k
k – 1 =
k == 2/3
If g(x) = x^{2} + x – 1 and g(f(x)) = 4x^{2} – 10x + 5, then find f.
.
If z = x + iy and real part = 1 then locus of z is
Let y = f(x) is a solution of differential equation and f(0) = 0 then f(1) is equal to :
e^{y} = t
Put x = 0, y = 0 then c = 1
e^{y–x} = x + 1
y = x + ln(x + 1)
at x = 1 , y = 1 + ln(2)
If α is a roots of equation x^{2} + x + 1 = 0 and A = then A^{31} equal to :
=> A^{4} = I
=> A^{30} = A^{28 }× A^{3} = A^{3}
The six digit numbers that can be formed using digits 1, 3, 5, 7, 9 such that each digit is used at least once.
1, 3, 5, 7, 9
For digit to repeat we have ^{5}C_{1} choice And six digits can be arrange in ways.
Hence total such numbers =
The area that is enclosed in the circle x^{2} + y^{2} = 2 which is not common area enclosed by y = x & y^{2} = x is
Total area – enclosed area
If sum of all the coefficient of even powers in (1– x + x^{2} – x^{3} ……..x^{2n}) (1 + x + x^{2 }+x^{3} ……….+ x^{2n}) is 61 then n is equal to
Let (1– x + x^{2} …..) (1 + x + x^{2} ……) = a_{0} + a_{1} x + a_{2}x^{2} +……
put x = 1
1(2n+1) = a0 + a1 + a2 +...,.a2n ...(i)
put x = –1
(2n+1) x 1 = a0  a1 + a2 +....a2n.....(ii)
Form (i) + (ii)
4n + 2= 2(a_{0} + a_{2} +….)
= 2 x 16
⇒ 2n+1 = 61
⇒ n = 30
If variance of first n natural numbers is 10 and variance of first m even natural numbers is 16 then the value of m + n is
⇒ n^{2}  1 = 120
⇒ n = 11
Var (2, 4, 6,.....,2m) = 16
⇒ var (1, 2,....,m) = 4
⇒ m^{2 } 1 = 48
⇒ m = 7
⇒ m + n = 18
Put 3^{x/2} = t
If A(1, 1), B(6, 5), C are vertices of ΔABC. A point P is such that area of ΔPAB, ΔPAC, ΔPBC are equal, also , then length of PQ is
P will be centroid of ΔABC
(p → q) ∧ (q → ~p) is equivalent to
Clearly (p → q) ∧ (q → ~p) is equivalent to ~p
Find greatest value of k for which 49^{k} + 1 is factor of 1 + 49 + 49^{2} …..(49)^{125}
If f(x) = 2 – x – 3 is non differentiable in x ∈ S.Then value of is
∵ f(x) is non differentiable at x = 1,3,5
∑ f(f(x)) = f(f(1) + f(f(3)) + f (f(5))
= 1 + 1 + 1
= 3
If system of equations
2x + 2ay + az = 0
2x + 3by + bz = 0
2x + 4cy + cz = 0
have nontrivial solution
then
For nontrivial solution
(3bc – 4bc) – (2ac – 4ac) + (2ab – 3ab) = 0
–bc + 2ac – ab = 0
ab + bc = 2ac
a, b, c in H.P.
in A.P
If sum of 5 consecutive terms of 'an A.P is 25 & product of these terms is 2520. If one of the terms is – 1/2 then the value of greatest term is
Let terms be a – 2d, a –d, a, a + d , a + 2d .
sum = 25
⇒ 5a = 25
⇒ a = 5
Product = 2520
⇒ (5–2d) (5 – d) 5(5+d) (5+2d) =2520
⇒ (25  4d^{2}) (25  d^{2}) = 504
⇒ 625 100d^{2 } 25d^{2} + 4d^{4} = 504
⇒ 4d^{4}  125d^{2} + 625 504 = 0
⇒ 4d^{4}  125d^{2} + 121 = 0
⇒ 4d^{4}  121d^{2}  4d^{2} +121 = 0
⇒ (d^{2}1) (4d^{2}  121) = 0
d = ±1, d = ± 11/2
d = ±1, does not give 1/2 as a term
∴ d = 11/2
∴ Largest term = 5 + 2d = 5 + 11 = 16
Let
lies in plane of&
of a bisectors angle between& then
angle bisector can be or
Compare with
Not in option so now consider
Compare with
Given f(a + b + 1 – x) = f(x) R then the value of is equal to
x → a + b  x
[∵ put x → x + 1 in given equation]
(1) + (2)
2I =
If distance between the foci of an ellipse is 6 and distance between its directrices is 12, then length of its latus rectum is
2ae = 6 and
⇒ ae = 3 and a/e = 6
⇒ a^{2} = 18
⇒ b^{2} = a^{2}  a^{2}e^{2} = 18  9 = 9
An unbiased coin is thrown 5 times. Let X be a random variable and k be the value assigned to X for k = 3, 4, 5 times Head occurs consecutively and otherwise the value of X is assigned –1. What is value of expectation.
k = no. of times head occur consecutively
Now expectation
If when then find dy/dα at α = 5π/6
= –1 – cota
⇒ will be = 4
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