Solution JEE Main 2013 p set

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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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As given, angular speed of rotation is ω, and length of spring = 2L


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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

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Y = xtanθ – (gx2/2u2) (1+tan2θ) = 2x – (10x2/10)(1+4) = 2x – 5x2

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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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As given that when they are connected, potential on each one will be zero


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120C1 – 200C2 = 0

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120C1 = 200C2

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3C1 = 5C2

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Initial energy, when satellite was on surface, Ei = -GMm/R
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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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We know that, for RC circuit, time constant Ϯ = RC = 100 * 10 3 * 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 of
capacitor voltage must be equal or less then the rate of decay modulated singnal voltage for proper
detection 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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Consider initial situation when 60W bulb is ON
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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

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In this condition, voltage across bulb, = 120 * (48/54) = 106.67
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So voltage decreased by = 117.07 – 106.67 = 10.40 ~ 10 V
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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 the
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speed of centre of mass when slipping ceases,

LI = Lf
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mr2ω0 = mvr + mr2(v/r)
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v = ω0r/2

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By Newtons cooling law

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Simple theoretical question, Correct answer is (4)

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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

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Charge in capacitor, q = q0(1-et/ Ϯ) = CV(1-e-t/ Ϯ)

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At t = 2 Ϯ

Q = CV(1-e-2)
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The locus of points having same path difference is circle in the given situation so the interference
pattern produced on the screen will consist of concentric circles .

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Given that, peak values of magnetic field, Bm = 20 nT
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Em = ?
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In an EM wave, Em/Bm = c

Em = c*Bm = 3*108*20*10-9 = 6 V/m

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When ƛ is increased, there will beone value of ƛ above which photoelectrons will be stop to come out
so 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 higher
frequency.

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For a small change of radius ∆R, the energy change due to surface tension,
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∆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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1/n3

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We know that angle of deviation will be minimum only for one incidence angle (i) when the path of light
ray 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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Bnet = B1 + B2 + BH


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1 and 2 are correct options

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In presence of SbCl5 it gives a carbocation which is planar hence racemisation takes place


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Standard reduvtion potential is maximum for Mn2+, so it will be the strongest oxidising agent
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As the process is isothermal

q = -w Δ u =0

q = + 208 J, w = -208 J

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Correct answer is (1) ar
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Correct order is (3)


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Correct order is (4)


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None of them is wrong


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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 ar
1 mole of CO2 has 1 mole of C, so number of moles of C in 0.07 moles of CO2, nC = 0.07
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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 =
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0.08

nC/nH = 0.07/0.08 = 7/8
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So Empirical formula is C7H8.
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Silicon exists as covalent crystal in solid state.



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According to Hardy Schulze rule, greater the charge on cat ion, greater is its coagulating power for
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negatively charged solution. So, order of coagulating power : Na+ < Ba2+ < Al3+

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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 outer
electron remain same from nucleus but effective nuclear charge increases, so nucleus attractive force
for 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 from
nucleus when we move down in a group, so less energy is required to remove the electron.

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In all given options Ar is a noble gas, so its I.E will be maximum.

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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
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So correct order is
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Ba < Ca < Se < S < Ar
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A is propanic acid

Propanic acid and ammonia reacts with each other, propanic acid is a weak acid, it gives H+ ion to
ammonia and gives Ammonium; propionate, complete mechanism is shown below


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In Bhopal gas tragedy over 500,000 people were exposed to methyl isocyanate gas and other chemicals.

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Correct answer is (2) ar
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Correct Answer is (4)
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Correct answer Is (1)


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Solution JEE Main 2013 p set

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Solution JEE Main 2013 p set