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Correct answer is (3)
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At equilibrium
Kx0 + FB = Mg
KX0 + σ(L/2)Ag = Mg
X0 = (Mg - σ(L/2)Ag)/K = (Mg/K)( 1 – σLA/2M)
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Correct answer is (4)
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As given, angular speed of rotation is ω, and length of spring = 2Lwww.1
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Correct answer is (4)
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Correct answer is (2)
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Correct answer is (2)
Given. , ux = 1 m/s
uy = 2 m/s
|u| = sqrt(1+4) = sqrt(5)
tanθ = uy/ux = 2/1= 2
So equation of its trajectory is
Y = xtanθ – (gx2/2u2) (1+tan2θ) = 2x – (10x2/10)(1+4) = 2x – 5x2
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Correct answer is (3)
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Amplitude of damped oscillator is given by:
A = A0 e-(bt/2m)
Given that, after 5 seconds, A = 0.9 A0
0.9A0 = A0 e-(5b/2m)
e-(5b/2m) = 0.9
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-(5b/2m) = -0.10536
b/2m = 0.02107 …………….(1)
After another 10 seconds it means after 15 s, let us consider amplitude is X
X = A0 e-(15b/2m)
Put value from Eqn(1)
X = 0.729 A0
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Correct answer is (2)
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As given that when they are connected, potential on each one will be zero
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120C1 – 200C2 = 0
120C1 = 200C2
3C1 = 5C2
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Correct answer is (2)
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Correct answer is (1)
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Correct answer is (3)
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Correct answer is (1)
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Initial energy, when satellite was on surface, Ei = -GMm/R
Final energy, when satellite is in orbit, Ef = -GMm/2(3R) = -GMm/6R
Energy required = Ef – Ei = -GMm/6R + GMm/R = 5GMm/6R
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Correct answer is (1)
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We know that, for RC circuit, time constant Ϯ = RC = 100 * 103 * 250 * 10-12 = 2.5 * 10-5 s
Given that amplitude modulation factor, ma = 60% = 0.6
The maximum detectable frequency can be obtained by applying condition that t rate of decay ofcapacitor voltage must be equal or less then the rate of decay modulated singnal voltage for properdetection of modulated signal.
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f = 1/(2Пma Ϯ)= 1/(2*3.14 * 0.6 * 2.5 * 10-5) = 10615 Hz = 10.62 KHz
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Correct answer is (3)
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Correct answer is (4)
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Consider initial situation when 60W bulb is ON
Power of bulb = 60W
V2/Rb = 60
Rb = 240 ohm
So total resistance, Rtotal = 240 + 6 = 246 Ohm
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Voltage across bulb = 120 *(Rb/246) = 120 *(240/246) = 117.07
Now consider when 240 W heater is swithched on in parallel to bulb
Voltage across bulb = 120 *(Rb/246) = 120 *(240/246) = 117.07
Power of heater = 240 W
Rh = V2/ Pheater = 1202/240 = 60 Ohm
So we get,
In this condition, voltage across bulb, = 120 * (48/54) = 106.67
So voltage decreased by = 117.07 – 106.67 = 10.40 ~ 10 V
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Correct answer is (2)
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At constant voulume, Q1 = nCv∆T = (3/2)(pfvf – p1v1) = (3/2)p0(2v0-v0) = (3/2)p0v0
At consatnt pressure, Q2 = nCp∆T = (5/2)(pfvf – p1v1) = (5/2)*2p0v0 = 5 p0v0
Total heat extracted = Q1 + Q2 = (3/2 + 5) = (13/2) p0v0
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Correct answer is (3).
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The angular momentum of the loop will be conserved about its initial point of contact. Let v be thespeed of centre of mass when slipping ceases,
LI = Lf
mr2ω0 = mvr + mr2(v/r)
v = ω0r/2
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Correct answer is (3)
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Correct answer is (3)
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By Newtons cooling law
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Simple theoretical question, Correct answer is (4)
Correct answer is (3)
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Initially both switches are open, so initial charge across cacitor, q0 = CV
When switch S1 is closed
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Current will flow throogh RC
Time constant Ϯ = RC
Charge in capacitor, q = q0(1-et/ Ϯ) = CV(1-e-t/ Ϯ)
At t = 2 Ϯ
Q = CV(1-e-2)
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Correct answer is (4)
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The locus of points having same path difference is circle in the given situation so the interferencepattern produced on the screen will consist of concentric circles .
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Correct answer is (2)
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Given that, peak values of magnetic field, Bm = 20 nT
Em = ?
In an EM wave, Em/Bm = c
Em = c*Bm = 3*108*20*10-9 = 6 V/m
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Correct answer is (4).
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When ƛ is increased, there will beone value of ƛ above which photoelectrons will be stop to come outso photocurrent will become zero.
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Correct answer is option (1), as for same value of current higher value of voltage is required for higherfrequency.
Correct answer is (4).
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For a small change of radius ∆R, the energy change due to surface tension,
∆E = 4П(R+∆R) 2T - 4 ПR2T = 8 П R. ∆RT
Here this energy change will be equal to energy required for vaopurization
8 П R. ∆RT = ρ[4 ПR2(∆R)L]
R = 2T/ρL
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Correct answer is (4)
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1/n3
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Correct answer is (3)
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We know that angle of deviation will be minimum only for one incidence angle (i) when the path of lightray through the prism is symmetrical.
And same values of angle of deviation are possible for two values of incidence angles.
From above two points Graph 3 is correct.
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Correct answer is (1)
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Correct answer is (2)
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Bnet = B1 + B2 + BH
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Correct answer is (4)
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Correct answer is (3)
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1 and 2 are correct options
Correct answer is (3)
In presence of SbCl5 it gives a carbocation which is planar hence racemisation takes place
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Correct answer is (3)
Standard reduvtion potential is maximum for Mn2+, so it will be the strongest oxidising agent
Correct answer is (1)
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As the process is isothermal
q = -w Δ u =0
q = + 208 J, w = -208 J
Correct answer is (1)
Correct order is (3)
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Correct answer is (3)
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Correct answer is (1)
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www.100
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Correct answer is (2)
Correct answer is (1)
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Correct order is (4)www.1
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Correct answer is (3)
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None of them is wrong
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Correct answer is (4)
Number of moles of CO2 = 3.08/(12 + 32) = 3.08/44 = 0.07 moles
Number of moles of H2O = 0.72/(2+16) = 0.04 moles
1 mole of CO2 has 1 mole of C, so number of moles of C in 0.07 moles of CO2, nC = 0.07
1 mole of H2O has 2 moles of Hydrogen, so number of moles of H in 0.04 moles of H2O, nH = 0.04*2 =0.08
nC/nH = 0.07/0.08 = 7/8
So Empirical formula is C7H8.
Correct answer is (3)
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Correct answer is (2)
Silicon exists as covalent crystal in solid state.
Correct answer is (1)
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Correct answer is (2)
According to Hardy Schulze rule, greater the charge on cat ion, greater is its coagulating power fornegatively charged solution. So, order of coagulating power : Na+ < Ba2+ < Al3+
.
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Correct answer is (3)
Ionization enthalpy is the energy required to remove 1st electron from the outer shell of the atom.
General trend of I.E
Across period – I.E increases as we move from left to right in a period, because distance of the outerelectron remain same from nucleus but effective nuclear charge increases, so nucleus attractive forcefor outer electron increases and therefore I.E also increases.
In group – when we move down in a group, I.E decreases, distance of the outer electron increases fromnucleus when we move down in a group, so less energy is required to remove the electron.
In all given options Ar is a noble gas, so its I.E will be maximum.
Ca and Se are in same period, so I.E of Se > I.E of Ca
Also S and Se are from same group, so I.E of S > I.E of Se
So correct order is
Ba < Ca < Se < S < Ar
Correct answer is (1)
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Correct answer is (4) www.100
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Correct answer is (4)
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Correct answer is (4)
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Correct answer is (2)
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Correct answer is (4)
A is propanic acid
Propanic acid and ammonia reacts with each other, propanic acid is a weak acid, it gives H+ ion toammonia and gives Ammonium; propionate, complete mechanism is shown below
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Correct answer is (2)
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Correct answer is (2)
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Correct answer is (1)
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In Bhopal gas tragedy over 500,000 people were exposed to methyl isocyanate gas and other chemicals.
Correct answer is (2)
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Correct answer is (3)
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Correct answer is (3)
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Correct answer is (3)
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Correct answer is (3)
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Correct answer is (3)
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Correct Answer is (4)
Correct answer is (3)
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Correct answer is (2)
Correct answer is (2)
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Correct answer is (1)
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Correct answer is (3)www.1
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Correct answer is (1)
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Correct answer is (2) www.100
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Correct answer is (3)
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www.100
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Correct answer is (4)
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Correct answer is (4) www.100
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www.100
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Correct answer is (1)
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Correct answer is (3)
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Correct answer is (2)
Correct answer is (3)
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Correct answer is (1) www.100
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Correct answer is (2)
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Correct answer is (3)
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Correct answer is (1) www.100
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www.100
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Correct answer is (2)
Correct answer Is (1)
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www.100
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Correct answer is (2) www.100
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www.100
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Correct answer is (1)
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Correct answer is (2)
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Correct answer is (4)
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