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A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position of equilibrium. The kinetic energytime graph of the particle will look like
Time taken to reach the extreme position from equilibrium position is T/4.
Velocity is maximum at equilibrium position and zero at extreme position.
(m is the mass of particle and v is the velocity of particle
Hence graph of K.E. v/s time is square cos function
The temperature of an open room of volume 30 m^{3} increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 10^{5} Pa. If n_{i} and n_{f} are the number of molecules in the room before and after heating, then n_{f} – n_{i} will be :
Using ideal gas equation
PV = NRT
(N is number of moles)
P_{0}V_{0} = N_{i}R x 290 ...... (1)
[T_{i} = 273 + 17 = 290 K]
After heating
P_{0}V_{0} = N_{f}R x 300 ...... (2)
[T_{f} = 273 + 27 = 300 K]
from equation (1) and (2)
difference in number of moles
Hence n_{f}  n_{i} is
putting P = 10^{5} P_{A} and V_{0} = 30 m^{3}
Number of molecules n_{f}  n_{i} =  2.5 x 10^{25}
Which of the following statements is false ?
On balancing condition
On balancing condition
.... (2)
As we see both equation (1) & (2) are same. So 4^{th} statement is false.
The following observations were taken for determining surface tensiton T of water by capillary method : Diameter of capilary, D = 1.25 × 10^{–2} m rise of water, h = 1.45 × 10^{–2} m
Using g = 9.80 m/s^{2} and the simplified relation N/m, the possible error in surface tension is closest to :
In amplitude modulation , sinusoidal carrier frequency used is denoted by ω_{c} and the signal frequency is denoted by ω_{m}. The bandwidth (Δω_{m}) of the signal is such that Δω_{m} << ω_{c}. Which of the following frequencies is not contained in the modulated wave ?
Three frequencies are contained
A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image formed is :
As parallel beam incident on diverging lens if forms virtual image at v_{1} = –25 cm from the diverging lense which works as a object for the converging lense (f = 20 cm)
So for converging lens u = –40 cm, f = 20 cm
V = 40 cm from converging lenses.
The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum ?
or .....(1)]
Also m = πR^{2}ℓρ
⇒ Put inequation (1)
For maxima & minima
⇒
or
⇒
or
An electron beam is accelerated by a potential difference V to hit a metallic target to produce Xrays. It produces continuous as well as characteristic X  rays. If λ_{min} is the smallest possible wavelength of Xray in the spectrum, the variation of log λ_{min} with log V is correctly represented in :
It is a straight line with –ve slope.
A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :
also let initially there are total N0 number of nuclei
An electric dipole has a fixed dipole moment which makes angle q with respect to xaxis. When subjected to an electric field it experiences a torque When subjected to another electric field it experiences torque . The angle q is :
In a common emitter amplifier circuit using an npn transistor, the phase difference between the input and the output voltages will be :
In common emitter amplifier circuit input and out put voltage are out of phase. When input voltage is increased then i_{b} is increased, i_{c} also increases so voltage drop across R_{c} is increased. However increase in voltage across R_{C} is in opposite sense.
C_{p} and C_{v} are specific heats at constant pressure and constant volume respectively. It is observed that C_{p} – C_{v} = a for hydrogen gas C_{p} – C_{v} = b for nitrogen gas The correct relation between a and b is :
C_{P} – C_{V} = R
where C_{P} and C_{V} are molar specific heat capacities As per the question
a = R/2
b = R/28
a = 14 b
A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by : (Given : room temperature = 30° C, specific heat of copper = 0.1 cal/gm°C
Heat given = Heat taken
(100) (0.1)(T  75) = (100)(0.1)(45) + (170)(1)(45)
10(T  75) = 450 + 7650 = 3100
T75 = 810
T = 885°C
A body of mass m = 10^{–2} kg is moving in a medium and experiences a frictional force F = –kv^{2}. Its intial speed is v_{0 }= 10 ms^{–1}. If, after 10 s, its energy is the value of k will be:
When a current of 5 mA is passed through a galvanometer having a coil of resistance 15 Ω, it shows full scale deflection. The value of the resistance to be put in series with the galvanometer to convert it into to voltmeter of range 0 10 V is:
10 = (5 × 10^{–3}) (15 + R)
r = 1985 Ω
A slender uniform rod of mass M and length ℓ is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is :
Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = λ_{1}/λ_{2}, is given by :
A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of :
In a coil of resistance 100 Ω, a current is induced by changing the magnetic flux through it as shown in the figure. The magnitude of change in flux through the coil is :
In a Young's double slit experiment, slits are separated by 0.5 mm, and the screen is placed 150 cm away. A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is :
A magnetic needle of magnetic moment 6.7 × 1 0 ^{–2} Am^{2} and moment of inertia 7.5 × 10^{–6} kg m^{2} is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is :
By substituting value in the formula T = .665 sec for 10 oscillation, time taken will be Time = 10 T = 6.65 sec
The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius):
inside the Earth (straight line)
out side the Earth
where M is Mass of E arth
In the above circuit the current in each resistance is :
Taking voltage of point A as = 0
Then voltage at other points can be written as shown in figure
Hence voltage across all resistance is zero.
Hence current = 0
A particle A of mass m and initial velocity v collides with a particle B of mass (m/2) which is at rest. The collision is head on, and elastic. The ratio of the de–Broglie wavelengths λ_{A} to λ_{B} after the collision is :
By conservation of linear momentum
2v = 2v_{1} + v_{2} .......(1)
by law of collision
u = v_{2}  v_{1} ......(2)
By equation (1) and (2)
An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and a is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by
A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be :
An observer is moving with half the speed of light towards a stationary microwave source emitting waves at frequency 10 GHz. What is the frequency of the microwave measured by the observer? (speed of light = 3 × 108 ms^{–1})
Doppler effect in light (speed of observer is not very small compare to speed of light)
In the given circuit diagram when the current reaches steady state in the circuit, the charge on the capacitor of capacitance C will be :
It steady state, current through AB = 0
A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is :
To hold 1 KV potential difference minimum four capacitors are required in series
So for Ceq to be 2μF, 8 parallel combinations are required.
⇒ Minimum no. of capacitors = 8 × 4 = 32
A body is t hrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?
Velocity at any time t is given by
⇒ straight line with negative slope
Let k be an integer such that triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point :
If , for apositive integer n , the quadratic equation, x(x + 1) + (x + 1) (x + 2) + ..... + (x + n) = 10n has two consecutive integral solutions, then n is equal to :
From above diagram of f (x) , f (x) is surjective but not injective.
The following statement (p → q) → [(~p → q) → q] is :
(p → q) [(~p → q) → q]
It is tautology
If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is :
and at a = 1 D_{1} = D_{2} = D_{3} = 0
but at a = 1 and b = 1
The area (in sq. units) of the region {(x,y) : x > 0, x + y < 3, x^{2} < 4y and y < 1 + is :
= 5/2
For any three positive real numbers a, b and c, 9(25a^{2} + b^{2}) + 25(c^{2} – 3ac) = 15b(3a + c).Then:
(15a)^{2} + (3b)^{2} + (5c)^{2} – (15a) (5c) – (15a) (3b) – (3b) (5c) = 0
(1/2) [(15a – 3b)^{2} + (3b – 5c)^{2} + (5c – 15a)^{2}] = 0
it is possible when 15a = 3b = 5c
a + b = 2c
⇒ b, c, a in A.P
A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in this party, is :
The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the yaxis passes through the point :
y` = 1 at point (0, 1)
Slope of normal is 1
Hence equation of normal is x + y = 1
A hyperbola passes through the point and has foci at (± 2, 0). Then the tangent to this hyperbola at P also passes through the point :
Let a. b, c ∈ R. If f(x) = ax^{2} + bx + c is such tliat a + b + c = 3 and f(x + y) = f(x) + f(y) + xy, V x, y e R. is equal to :
f(x) = ax^{2} + bx + c
f(1) = a + b + c = 3
Now f(x + y) = f(x) + f(y) + xy
put y = 1
f(x + 1) = f(x) + f(1) + x
f(x + 1) = f(x) + x + 3
Let a vertical tower AB have its end A on the level ground. Let C be the midpoint of AB and P be a point on the ground such that AP = 2AB. If ∠BPC = β, then tanβ is equal to :
Twenty meters of wire is available for fencing off a flowerbed in the form of a circular sector.Then the maximum area (in sq. m) of the flower bed, is :
Total length = r + r + rθ = 20
..........(1)
.......(2)
. I_{4} + I_{6} = a tan^{5}x + bx^{5} + C, where C is a constant of integration, then the ordered pair (a, b) is equal to :
Let ω be a complex number such that 2ω + 1 = z where z = √3 . If
then k is equal to :
Here w is complex cube root of unity R_{1} → R_{1} + R_{2} + R_{3}
The value of (^{21} C_{1} – ^{10}C_{1}) + (^{21}C_{2} – ^{10}C_{2}) + (^{21}C_{3} – ^{10}C_{3}) + (^{21}C_{4} – ^{10}C_{4}) + .... + (^{21}C_{10} – ^{10}C_{10}) is :
(^{21}C_{1 }+ ^{21}C_{2} .........+ ^{21}C_{10}) – (^{10}C1 + ^{10}C_{2} ..... .... ^{10}C_{10})
If 5(tan^{2}x – cos^{2}x) = 2cos 2x + 9, then the value of cos4x is :
If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to line, Q, then PQ is equal to :
The distantce of the point (1, 3, –7) from the plane passing through the point (1, –1, –1), having normal perpendicular to both the lines
So plane is 5(x – 1) + 7(y + 1) + 3(z + 1) = 0
⇒ 5x + 7y + 3z + 5 = 0
The radius of a circle, having minimum area, which touches the curve y = 4 – x^{2} and the lines, y = x is :
which is not in options therefore it must be bonus. But according to history of JEEMains it seems they had following line of thinking.
Given curves are y = 4 – x^{2} and y = x
There are two circles satisfying the given conditions. The circle shown is of least area.
Let radius of circle is 'r'
coordinates of centre = (0, 4 – r)
circle touches the line y = x in first quadrant
A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one–by–one, with replacement, then the variance of the number of green balls drawn is :
We can apply binomial probability distribution Variance = npq
The eccentricity of an ellipse whose centre is at the origin is 1/2 . If one of its directices is x = –4, then the equation of the normal to it at
Eccentricity of ellipse = 1/2
If two different numbers are taken from the set {0, 1, 2, 3, ......., 10), then the probability that their sum as well as absolute difference are both multiple of 4, is :
Let A ≡ {0, 1,2 ,3 ,4 ... .. ., 10}
n(s) = ^{11}C_{2} (where 'S' denotes sample space)
Let E be the given event
E ≡ {(0, 4), (0, 8), (2, 6), (2, 10), (4, 8), (6, 10)}
n(E) = 6
∴ P(E) = 6/35
For three events A, B and C, P(Exactly one of A or B occurs)
= P(Exactly one of B or C occurs)
= P(Exactly one of C or A occurs)
= 1/4 and P(All the three events occur simultaneously) = 1/16
Then the probability that at least one of the events occurs, is :
Which of the following compounds will significant amont of meta product during mononitration reaction?
(i) Nitration is carried out in presence of concentrated HNO_{3} + concentrated H_{2}SO_{4}.
(ii) Aniline acts as base. In presence of H_{2}SO_{4} its protonation takes place and anilinium ion is formed
(iii) Anilinium ion is strongly deactivating group and meta directing in nature so it give meta nitration product in significant amount.
From 1^{st} law :
ΔU = q + w
For adiabatic process :
q = 0
ΔU = w
Work involve in adiabatic process is at the expense of change in internal energy of the system.
The increasing order of t he reactivity of the following halides for the S_{N}1 reaction is
For any S_{N}1 reaction reactivity is decided by ease of dissociation of alkyl halide
Higher the stability of (carbocation) higher would be reactivity of S_{N}1 reaction. Since stability of cation follows order.
The radius of the second Bohr orbit for hydrogen atom is : (Plank's const. h = 6.6262 × 10^{–34} Js ; mass of electron = 9.1091 × 10^{–31} kg ; charge of electron e = 1.60210 × 10^{–19} C ; permittivity of vaccum ∈_{0} = 8.854158 x 10^{12} kg^{1}m^{3}A^{2})
Radius of n^{th} Bohr orbit in Hatom = 0.53 n^{2}Å Radius of II Bohr orbit = 0.53 ×(2)^{2} = 2.12 Å
pK_{a} of a weak acid (HA) and pK_{b} of a weak base (BOH) are 3.2 and 3.4, respectively.The _{p}H of their salt (AB) solution is :
Given
pK_{a} (HA) = 3.2
pK_{b} (BOH) = 3.4
As given salt is of weak acid and weak bas
= 6.9
The formation of which of the following polymers involves hydrolysis reaction?
Formation of Nylon6 involves hydrolysis of its monomer (caprolactum) in initial state.
The most abundant elements by mass in the body of a healthy human adult are : Oxygen (61.4%); Carbon (22.9%), Hydrogen (10.0%); and Nitrogen (2.6%).The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is :
Mas s in the body of a healthy human adult has :
Oxygen = 61.4%, Carbon = 22.9%,
Hydrogen = 10.0% and Nitrogen = 2.6%
Total weight of person = 75 kg
Mas dur to ^{1}H is
^{1}H atoms are replaced by ^{2}H atoms.
So mass gain by person = 7.5 kg
Which of the following, upon treatment with tertBuONa followed by addition of bromine water, fails to decolourize the colour of bromine ?
In the following reactions, ZnO is respectively acting as a/an :
(a) ZnO + Na_{2}O → Na_{2}ZnO_{2}
(b) ZnO + CO_{2} → ZnCO_{3}
ZnO is amphoteric oxide,in given reaction,
(a) ZnO + Na_{2}O → Na_{2}ZnO_{2}
In this reaction ZnO act as Acid.
(b) ZnO + CO_{2} → ZnCO_{3}
In this reaction ZnO act as Base.
Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect, is :
Mg can form basic carbonate like
While Li can form only carbonate (Li_{2}CO_{3}) not basic carbonate.
3Methylpent2ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is :
A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be :
In FCC unit cell atoms are in contant along face diagonal
Two reactions R_{1} and R_{2} have identical preexponential factors. Activation energy of R_{1} exceeds that of R_{2 }by 10 kJ mol^{−1}. If k_{1} and k_{2} are rate constants for reactions R_{1} and R_{2 }respectively at 300 K, then ln(k_{2}/k_{1}) is equal to :
(R = 8.314 J mol^{−1}K^{−1})
From arrhenius equation.
......(1)
.......(2)
so, equation
(As preexponential factors of both reactions is same)
The correct sequence of reagents for the following conversion will be :
The Tyndall effect is observed only when following conditions are satisfied :
(a) The diameter of the dispersed particles is much smaller than the wavelength of the light used.
(b) The diameter of the dispersed particle is not much smaller than the wavelength of the light used.
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude.
(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.
Colloidal solutions show Tyndall effect due to scattering of light by colloidal particles in all directions in space. lt is observed only under the following conditions.
(i) The diameter of the colloids should not be much smaller than the wavelength of light used.
(ii) The refractive indices of the dispersed phase and dispersion medium should differ greatly in magnitude.
Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution ?
Ester in presence of Aqueous KOH solution give SN^{AE} reaction so following reaction takes place
In above compound in presence of Aq.
KOH (SN^{AE}) reaction takes place & Hydroxy carbonyl compound is formed which give Tollen's test So this compound behave as reducing sugar in an aqueous KOH solution.
Given C(graphite)+O_{2}(g) → CO_{2}(g) ;
∆_{r}H^{°}=−393.5 kJ mol^{−1}
H_{2}(g)+ 1/2 O_{2}(g) → H_{2}O(l) ;
∆_{r}H^{°} = −285.8 kJ mol^{−1}
CO_{2}(g)+2H_{2}O(l) → CH_{4}(g)+2O_{2}(g) ;
∆_{r}H^{°}=+890.3 kJ mol^{−1}
Based on the above thermochemical equations, the value of ∆_{r}H^{°} at 298 K for the reaction C(graphite)+2H_{2}(g) → CH_{4}(g) will be :
Which of the following reactions is an example of a redox reaction ?
In the reaction
Xenon undergoes oxidation while oxygen undergoes reduction.
The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are :
The major product obtained in the following reaction is :
Elimination reaction is highly favoured if
(a) Bulkier base is used
(b) Higher temperature is used Hence in given reaction biomolecular ellimination reaction provides major product.
Sodium salt of an organic acid ‘X’ produces effervescence with conc. H_{2}SO_{4}. ‘X’ reacts with the acidified aqueous CaCl_{2} solution to give a white precipitate which decolourises acidic solution of KMnO_{4}. ‘X’ is :
Which of the following species is not paramagnetic?
NO ⇒ One unpaired electron is present in π* molecular orbital.
CO ⇒ No unpaired electron is present O_{2} ⇒ Two unpaired electrons are present in π* molecular orbitals.
B_{2} ⇒ Two unpaired electrons are present in π bonding molecular orbitals.
The freezing point of benzene decreases by 0.458C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be :
(K_{f} for benzene=5.12 K kg mol^{−1})
Here α is degree of association
α = 0.945
% decree of association = 94.5%
Which of the following molecules is least resonance stabilized ?
is nonaromatic and hence least reasonance stabilized
On treatment of 100 mL of 0.1 M solution of CoCl_{3}.6H_{2}O with excess AgNO_{3}; 1.2×10^{22} ions are precipitated. The complex is :
Moles of ions precipitated with excess of
= 0.02 moles
Number of Cl^{} present in ionization sphere =
It means 2Cl^{} ions present inionization sphere
∴ complex is [Co(H_{2}O)_{5}Cl]Cl_{2}.H _{2}O
The major product obtained in the following reaction is:
DIB AL – H is electrophilic reducing agent reduces cynide, esters, lactone, amide, carboxylic acid into corresponding Aldehyde (partial reduction)
A water sample has ppm level concentration of following anions F = 10 ; = 50 . The anion/anions that make/makes the water sample unsuitable for drinking is/ are :
NO_{3}^{}: The maximum limit of nitrate in drinking water is 50 ppm. Excess nitrate in drinking water can cause disease. Such as methemoglobinemia.
SO_{4}^{2} : above 500 ppm of SO_{4}^{2} ion in drinking water causes laxative effect otherwise at moderate levels it is harmless
F^{} : Above 2ppm concentration of F^{} in drinking water cause brown mottling of teeth.
∴ The concentration given in question of 3O_{4}^{2} & NO_{3}^{} in water is suitable for drinking but the concentration of F^{} (i.e 10 ppm) make water unsuitable for drinking purpose :
1 gram of a carbonate (M_{2}CO_{3}) on treatment with excess HCl produces 0.01186 mole of CO_{2}. The molar mass of M_{2}CO_{3} in g mol^{−1} is :
⇒ from the balanced chemical eqn.
Given
Among the following, the strongest reducing agent is :
Since Cr^{+3} is having least reducing potential, so Cr is the best Reducing agent.
The group having isoelectronic species is :
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