NEET-UG 2009

Velocity = [LT^-1] so a = 0, b = 1, c = -1

Acceleration = [LT^-2] so a = 0, b = 1, c = -2

Force = [MLT^-2] so a = 1, b = 1, c = -2

Pressure = $\frac{force}{Area}=\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}=\left[M{L}^{-1}{T}^{-2}\right]$,
so a = 1, b = - 1, c = - 2

Relative velocity of the scooter with respect to the bus = (v_s - 10)

$\therefore$ $\frac{1000}{v_s-10}$ = 100

$\Rightarrow$ v_s = 20 m/s

Given u = 0.

S₁ = $\frac{1}{2}\times a\times t_1^2$

and S₂ = $\frac{1}{2}\times a\times t_2^2$

$\frac{S_1}{S_2}={\left(\frac{t_1}{t_2}\right)}^2$ = ${\left(\frac{10}{20}\right)}^2$ = $\frac{1}{4}$

S₂ = 4S₁

$\overrightarrow{F}=6\hat{i}-8\hat{j}+10\hat{k}$

$\left|\overrightarrow{F}\right|=\sqrt{36+64+100}$

$=\sqrt{200}N=10\sqrt{2}N$

Acceleration, a = 1 m s^–2

$\therefore$ $Mass,M=\frac{10\sqrt{2}}{1}=10\sqrt{2}kg$

F – Mg = Ma

8000 = 2000a

$\therefore$ Acceleration is 4 ms^–2 upwards.

When the mass attached to a spring fixed at the other end is allowed to fall suddenly, it extends the spring by x. Potential energy lost by the mass is gained by the spring.

$Mgx=\frac{1}{2}k{x}^2\Rightarrow x=\frac{2Mg}{k}$

Initial velocity u = 20 m/s; m = 1 kg

Kinetic energy = maximum potential energy

Initial kinetic energy = $\frac{1}{2}\times 1\times 20^2=200J$

Mgh (max) = 200 J

$\therefore$ h = 20 m

The height travelled by the body, h' = 18 m

$\therefore$ Loss of energy due to air friction = mgh – mgh'

$\Rightarrow$ Energy lost = 200 J – 1 × 10 × 18 J = 20 J.

Let m be the mass per unit length, so the rate of mass leaving the hose per sec = mv

Rate of kinetic energy K.E = $\frac{1}{2}\left(mv\right)v^2=\frac{1}{2}m{v}^3$

The position vector of the centre of mass of two particle system is given by

$\overrightarrow{R}=\frac{m_1\overrightarrow{R_1}+m_2\overrightarrow{R_2}}{\left(m_1+m_2\right)}$

$=\frac{1}{4}\left[-8\hat{i}-4\hat{j}+4\hat{k}\right]=-2\hat{i}-\hat{j}+\hat{k}$

When an explosion breaks a rock, by the law of conservation of momentum, initial momentum is zero and for the three pieces,


Total momentum of the two pieces 1 kg and 2 kg

$=\sqrt{{12}^2+{16}^2}=20kgm{s}^{-1}$

The third piece has the same momentum and in the direction opposite to the resultant of these two momentum.

$\therefore$ Momentum of the third piece = 20 kg ms^–1
Velocity = 4 ms^–1

$\therefore$ Mass of the 3^rd piece = $\frac{mv}{v}=\frac{20}{4}$ = 5 kg

As the masses are added to the ring gently, there is no torque and angular momentum is conserved.

$I\omega =I^{\prime }{\omega }^{\prime }$

$\Rightarrow M{R}^2\omega =\left(M{R}^2+2m{R}^2\right){\omega }^{\prime }$

$\Rightarrow {\omega }^{\prime }=\frac{M{R}^2\omega }{\left(M+2m\right)R^2}$

$\Rightarrow {\omega }^{\prime }=\frac{M\omega }{M+2m}$

Torque is always perpendicular to $\overrightarrow{F}$ as well as $\overrightarrow{r}$.

$\therefore$ $\overrightarrow{r}.\overrightarrow{\tau }=0$ as well as $\overrightarrow{F}.\overrightarrow{\tau }=0$

As per theorem of parallel axes :

I = I_cm + Ma²

where I_cm = moment of inertia about an axis through centre of mass of rod
I = moment of inertia about parallel axis at distance ‘a’

Now moment of inertia of thin rod of mass M and length L about an axis through the midpoint of the rod and perpendicular to its length is ML²/12.
The moment of inertia of a rod about parallel axis at distance L/2 will be :

ML²/12 + M(L/2)² = ML²/3

As there are 4 rods in square frame, so moment of inertia of entire frame is 4 times the moment of inertia of 1 rod, so it will be (4/3) ML²

According to Kepler’s law, the areal velocity of a planet around the sun always remains constant.

SCD : A₁– t₁ (areal velocity constant)
SAB : A₂ – t₂

$\frac{A_1}{t_1}=\frac{A_2}{t_2}$

$t_1=t_2.\frac{A_1}{A_2}$   (given A₁ = 2A₂)

$=t_2.\frac{2A_2}{A_2}$

$\therefore$ $t_1=2t_2$

Rate of heat radiated at (227 + 273) K = 7 cals/(cm²s)

Let rate of heat radiated at (727 + 273) K = x cals/(cm²s)

By Stefan’s law, 7 $\propto$ (500)⁴ and x $\propto$ (1000)⁴

$\therefore$ $\frac{x}{7}=2^4\Rightarrow x=7\times 2^4=112$ cals/(cm²s)

Similar to I = V/R

$\frac{dQ}{dt}=\frac{kA}{L}\left(T_1-T_2\right)$

k = conductivity of the rod.

The only statement which is not true is statement (a). A process in which the pressure remains constant is called isobaric process and not isochoric as in isochoric process the volume remains constant.

When a quantity Q of heat is supplied to a system it is used to do an amount of work W by the system and to increase the internal energy of the system by ∆U :

Q = ∆U + W

Here, Q given in kilo calories is converted into joule.
Therefore Q = 2×1000×4.2 J = 8400 J

The increase the internal energy

∆U = Q – W = 8400 – 500 = 7900 J

Simple harmonic motion is defined as follows

Acceleration $\frac{d^{2}y}{d{t}^2}=-{\omega }^{2}x$

The negative sign is very important in simple harmonic motion. Acceleration is independent of any initial displacement of equilibrium position.

Then acceleration = $-{\omega }^{2}x$.

Speed $v=\omega \sqrt{a^2-x^2},x=\frac{a}{2}$

$\therefore$ $v=\omega \sqrt{a^2-\frac{a^2}{4}}=\omega \sqrt{\frac{3a^2}{4}}$

$=\frac{2\pi }{T}\frac{a\sqrt{3}}{2}=\frac{\pi a\sqrt{3}}{T}$

Car is the source and the hill is observer. Frequency heard at the hill, ${\upsilon }_1$

$\therefore {\upsilon }_1=\frac{\upsilon \times v}{\left(v-V\right)}=\frac{600\times 330}{330-30}$

Now for reflection, the hill is the source and the driver the observer.

$\therefore$ ${\upsilon }_2={\upsilon }_1\times \frac{\left(330+30\right)}{330}$

$\Rightarrow {\upsilon }_2=\frac{600\times 330}{300}\times \frac{360}{330}\Rightarrow {\upsilon }_2=720$ Hz.

$l_1$ = 0.516 m, $l_2$ = 0.491 m, T = 20 N.

Mass per unit length, $\mu$ = 0.001 kg/m.

Frequency, $\upsilon =\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

${\upsilon }_1=\frac{1}{2\times 0.516}\sqrt{\frac{20}{0.001}}$

${\upsilon }_2=\frac{1}{2\times 0.491}\sqrt{\frac{20}{0.001}}$

$\therefore$ Number of beats = ${\upsilon }_1-{\upsilon }_2=7$.

As $A=2cm,\frac{\omega }{k}=128m{s}^{-1}$,

$5\lambda =4$, $\lambda =\frac{4}{5}m$

y = A sin (kx – $\omega$t)

$k=\frac{2\pi }{\lambda }=\frac{2\pi \times 5}{4}=\frac{31.4}{4}=7.85$

$\therefore$ $\omega$ = 128 × 7.85 = 1005

so y = 0.02 m sin (7.85x – 1005 t)

$V_A=\frac{1}{4\pi {\epsilon }_0}\left\{\frac{qA}{a}+\frac{qB}{b}+\frac{qC}{c}\right\}$

$=\frac{4\pi }{4\pi {\epsilon }_0}\left\{\frac{a^2\sigma }{a}-\frac{b^2\sigma }{b}+\frac{c^2\sigma }{c}\right\}$

$V_A=\frac{1}{{\epsilon }_0}\left\{\frac{a^2\sigma }{a}-\frac{b^2\sigma }{b}+\frac{c^2\sigma }{c}\right\}$

$V_B=\frac{1}{{\epsilon }_0}\left\{\frac{a^2\sigma }{a}-\frac{b^2\sigma }{b}+\frac{c^2\sigma }{c}\right\}$

$V_C=\frac{1}{{\epsilon }_0}\left\{\frac{a^2\sigma }{a}-\frac{b^2\sigma }{b}+\frac{c^2\sigma }{c}\right\}$

Given c = a + b.
If a = a, b = 2a and c = 3a for example, as c > b > a,

$V_A=\frac{1}{{\epsilon }_0}\left\{\frac{a^2\sigma }{a}-\frac{4b^2\sigma }{2a}+\frac{c^2\sigma }{c}\right\}$

$V_B=\frac{1}{{\epsilon }_0}\left\{\frac{a^2\sigma }{2a}-\frac{4a^2\sigma }{2a}+\frac{c^2\sigma }{c}\right\}$

$V_C=\frac{1}{{\epsilon }_0}\left\{\frac{a^2\sigma }{3a}-\frac{4a^2\sigma }{3a}+\frac{c^2\sigma }{c}\right\}$

It can seen by taking out common factors that
V_A = V_C > V_B i.e., V_A = V_C $\ne$ V_B

The electric potential at a point,
V = –x²y – xz³ + 4.

The field
$\overrightarrow{E}=-\overrightarrow{\nabla }V=-\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right)$

$\therefore \overrightarrow{E}=\hat{i}\left(2xy+z^3\right)+\hat{j}{x}^2+\hat{k}\left(3x{z}^2\right)$

The resistance of length 2$\pi$R is 12$\Omega$. Hence the resistance of length $\pi$R is 6$\Omega$. Thus two resistances of 6$\Omega$ can be represented as shown in fig.

$\therefore$ Equivalent resistance R = $\frac{6\times 6}{12}=3$

Given, Energy = 2 eV = eE$\lambda$

$\therefore$ E = $\frac{2eV}{e\lambda }=\frac{2}{4\times {10}^{-8}}=5\times 10^7$ V/m

Applying Kirchhoff’s equation to the loop ABFE,

– (i₁ + i₂)R – i₁r₁ + ${\epsilon }_1$ = 0

$\Rightarrow$ ${\epsilon }_1$ – (i₁ + i₂)R – i₁r₁ = 0

The terminal potential difference of a cell is given by V + Ir = $\epsilon$

V = V_A – V_B

$\Rightarrow$ V = $\epsilon$ – Ir

$\Rightarrow \frac{dV}{dI}=-r$, Also for, i = 0 then V = $\epsilon$

$\therefore$ slope = – r, intercept = $\epsilon$

In series combination of capacitors
V_eff = V + V + V = 3V

$\frac{1}{C_{eff}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$

$\Rightarrow C_{eff}=\frac{C}{3}$

Thus, the capacitance and breakdown voltage of the combination will be $\frac{C}{3}$ and 3V.

For the circular motion in a cyclotron,

$qvB=\frac{m{v}^2}{r}\Rightarrow qB=m\omega =\frac{m\times 2\pi }{T}$

$\therefore T=\frac{2\pi m}{qB}$ is independent of v and r.

The magnetic force acting on the charged paraticle is given by

$\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

$=\left(-2\times {10}^{-6}\right)\left[\left\{\left(2\hat{i}+3\hat{j}\right)\times {10}^6\right\}\times \left(2\hat{j}\right)\right]$

$=-4\left(2\hat{k}\right)$

$=-8\hat{k}$

$\therefore$ Force is of 8N along – z-axis.

G = 60 $\Omega$ , I_g = 1.0A, I = 5A.

Let S be the shunt resistance connected in parallel to galvanometer

I_g G = (I – I_g) S,

$S=\frac{I_{g}G}{I-I_g}=\frac{1}{5-1}\times 60=\frac{1.0\times 60}{4}=15\Omega$

Thus by putting 15 $\Omega$ in parallel, the galvanometer can be converted into an ammeter.

Since, diamagnetic substance is weakly magnetised in a direction opposite to the applied magnetic field. So, they are repelled in an external magnetic field.

Now work done = MB(cosθ₁ – cosθ₂)

= MB (1 – 1/2) = $\frac{2\times {10}^4\times 6\times {10}^{-4}}{2}$ = 6 J

Once a rectangular loop or a square loop is being drawn out of the field, the rate of cutting the lines of field will be a constant for a square and rectangle, but not for circular or elliptical areas.

Constant rate at which radius of the loop shrinks

$\frac{dr}{dt}=2\times 10^{-3}$ m s^-1

Magnetic flux linked with the loop is

$\varphi$ = BAcos$\varphi$ = B($\pi$r²)cos0^o = B$\pi$r²

The magnitude of the induced emf is

|e| = $\frac{d\varphi }{dt}=\frac{d}{dt}\left(B\pi r^2\right)=B\pi 2r\frac{dr}{dt}$

= 0.04 $\times$ $\pi$ $\times$ 2 $\times$ 2 $\times$ 10^-2 $\times$ 2 $\times$ 10^-3

= 3.2$\pi$ $\times$ 10^-6 V

= 3.2$\pi \mu V$

Power dissipated, P = E_rmsI_rms cos$\varphi$

cos$\varphi$ = $\frac{R}{Z}$

But I_rms = $\frac{E_{rms}}{Z}$

$\therefore$ P = $E_{rms}^2.\frac{R}{Z^2}$

Also we know, Z = $\sqrt{R^2+{\left(X_L-X_C\right)}^2}$

So, P = $E_{rms}^2.\frac{R}{R^2+{\left(X_L-X_C\right)}^2}$

= $\frac{{\epsilon }^{2}R}{\left[R^2+{\left(L\omega -\frac{1}{C\omega }\right)}^2\right]}$

Since coefficient of x is negative, but is moving along +ve x-axis,

so This is the form E_y = E₀($\omega$t – kx)

$\omega$ = $2\pi \times {10}^6$

f = $\frac{\omega }{2\pi }$ = 10⁶ Hz

k = $\pi \times 10^{-2}$

Also, $\lambda =\frac{2\pi }{k}$ = $\frac{2\pi }{\pi \times {10}^{-2}}$ = 200 m

{}_Z^AX\buildrel {2{\beta ^ - }} \over \longrightarrow {}_{Z + 2}^A{Y_1}\buildrel \alpha \over \longrightarrow {}_Z^{A + 4}{Y_2}

The resultant daughter is an isotope of the original parent nucleus.

n(n – 1)/2 = 6

$\Rightarrow$ n² – n –12 = 0

$\Rightarrow$ (n – 4) (n + 3) = 0 or n = 4

So, n = 4 to n = 3 states.

Now energy of the projectile :

= $\frac{1}{4\pi {\epsilon }_0}\frac{z_1{z}_2}{r}$

Therefore energy $\propto$ z₁z₂

First X decays by $\beta$^–. Y emits $\alpha$ resulting in the excited level of B which in turn emits a $\gamma$ ray.

$\therefore$ $\beta$, $\alpha$, $\gamma$ is the answer.

Number of photons = $\frac{E}{\frac{hc}{\lambda }}$

= $\frac{9\times {10}^{-3}\times 6.67\times {10}^{-7}}{6.6\times {10}^{-34}\times 3\times {10}^8}$

= 3 $\times$ 10¹⁶

Retarding potential depends on the frequency of incident radiation but is independent of intensity.

The number of photoelectrons emitted is proportional to the intensity of incident light.

Saturation current $\propto$ intensity.

Distance between nearest atoms in body centred cubic lattice (bcc),

d = $\frac{\sqrt{3}}{2}a$

Given d = 3.7 $\stackrel{\circ }{A}$

$\therefore$ $a=\frac{3.7\times 2}{\sqrt{3}}$ = 4.3 $\stackrel{\circ }{A}$

$$\lambda$$_max = $\frac{hc}{E}$

= $\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{2.5\times 1.6\times {10}^{-19}}$

= 5000 $\stackrel{o}{A}$

Now the wavelength detected by photodiode be less than $\lambda$_max, hence it can detect a signal of wave length 4000 $\stackrel{o}{A}$.

Current gain, $\beta$ = $\frac{\Delta I_C}{\Delta I_B}$

= $\frac{\left(10-5\right)\times {10}^{-3}}{\left(200-100\right)\times {10}^{-6}}$ = 50

OR gate, NOT gate and NAND gates are (iv), (ii) and (i) respectively.

H2	+	1/2O2	$$\to$$	H2O
2 g		16 g		18 g
1 mol		0.5 mol		1 mol

10 g of H₂ = 5 mol and 64 g og O₂ = 2 mol

$\therefore$ In this reaction, oxygen is the limiting reagent so amount of H₂O produced depends on that of O₂

Since 0.5 mol of O₂ gives 1 mol H₂O

$\therefore$ 2 mol of O₂ will give 4 mol H₂O

In an atom for any value of n, the values of l = 0 to (n-1)

For a given value of l, the values of m_l = -$l$ to 0 to +$l$

and the value of s = +1/2 or -1/2

In option(b), l = 2 and m_l = -3

This is not possible. as for l = 2 min value of m_l = -2

Maximum number of electrons in a subshell = 2(2l+1) = 4l + 2

Let oxidation number of P in PO₄ ^3– be x.

$\therefore$ x + 4(–2) = –3

$\Rightarrow$ x = +5

Let oxidation number of S in SO₄ ^2– be y.

$\therefore$ y + 4(–2) = –2

$\Rightarrow$ y = +6

Let oxidation number of Cr in Cr₂O₇ ^2– be z.

$\Rightarrow$ 2z + 7(–2) = –2

$\Rightarrow$ z = +6

Energy absorbed by each molecule = Bond energy per molecule + Kinetic energy per molecule

$\Rightarrow$ 4.4 × 10^–19 J = 4.0 × 10^–19 J + Kinetic energy per molecule

$\Rightarrow$ 0.4 × 10^–19 = Kinetic energy per molecule

Kinetic energy per atom =

Kinetic energy per molecule

2

=

0.4 × 10^–19

2

= 0.2 × 10^–19 J

= 2 × 10^–20 J

CH₃COOH $\rightleftharpoons$ CH₃COO^– + H⁺,   K₁ = 1.5 $\times$ 10^$-$5

$\Rightarrow$ K₁ = $\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$ = 1.5 $\times$ 10^$-$5

HCN $\rightleftharpoons$ CN^– + H⁺,     K₂ = 4.5 × 10^–10

K₂ = $\frac{\left[C{N}^{-}\right]\left[H^{+}\right]}{\left[HCN\right]}$ = 4.5 × 10^–10

CN^$$-$$ + CH₃COOH $\rightleftharpoons$ HCN + CH₃COO^$$-$$

K = $\frac{\left[HCN\right]\left[C{H}_{3}CO{O}^{-}\right]}{\left[C{N}^{-}\right]\left[C{H}_{3}COOH\right]}$

$\Rightarrow$ K = $\frac{K_1}{K_2}$ = $\frac{1.5\times {10}^{-5}}{4.5\times {10}^{-10}}$

= 3.33 × 10⁴

$\simeq$ 3.0 × 10⁴

Ammonium chloride is a salt of weak base and strong acid. In this case hydrolysis constant K_h can be calculated as

K_h = $\frac{K_w}{K_b}$ = $\frac{{10}^{-14}}{1.77\times {10}^{-5}}$ = 5.65 $\times$ 10^$-$10

Number of equivalents of H⁺ = 20.0 × 0.050 milliequivalents

= 1.0 milliequivalents

Number of equivalents of OH^– = 2 × 30.0 × 0.10

= 6.0 milliequivalents

$\therefore$ Equivalents of OH^- left after neutralization

= 6 – 1 = 5 milliequivalents

Total volume after neutralization

= 20.0 + 30.0 mL

= 50 mL

[OH^-] = $\frac{5}{50}$ = 0.1 M

CH₃)₃ B – is an electron deficient compound due to inclete octate of B, thus behave as a lewis acid.

The number of moles of ions produced by 1 mol of ionic compound = i

Given, m = 0.0020 m

$\Delta$T_f = 0^oC – 0.00732^oC = – 0.00732^oC

K_f = – 1.86 ^oC

$\Delta$T_f = ik_fm

$\Rightarrow$ i = $\frac{0.00732}{1.86\times 0.0020}$ = 2

$\Delta$H_reaction = Σ(Bond enthalpy)_reactants

– Σ(Bond enthalpy)_products

= [B.E_(C-C) + B.E_(H-H) + 4$\times$B.E_(C-H)]

- [B.E_(C-C) + 6$\times$B.E_(C-H)]

= [606.10 + 4(410.50) + 431.37]

– [336.49 + 6(410.50)]

= 2679.47 – 2799.49

= – 120.02 kJ mol^–1

We know, $\Delta$G = $\Delta$H – T$\Delta$S

For reaction to be spontaneous, $\Delta$G < 0

$\Rightarrow$ $\Delta$H – T$\Delta$S < 0

$\Rightarrow$ 170 $\times$ 10³ - T(170) < 0

$\Rightarrow$ T > 1000 K

Among the given temperatures, only 1110 K is greater than 1000 K thus, at this temperate the reaction will be spontaneous.

E = Z × 96500

$\Rightarrow$ $\frac{27}{3}$ = Z $\times$ 96500

$\Rightarrow$ Z = $\frac{9}{96500}$

Now applying the formula, W = Z × I × t

W = $\frac{9}{96500}$ $\times$ 4 $\times$ 10⁴ $\times$ 6 $\times$ 60 $\times$ 60

= 8.1 × 10⁴ g

The degree of dissociation ($\alpha$)

$\alpha$ = $\frac{8}{400}$ = 2 $\times$ 10^-2

From Ostwald’s dilution law for weak monobasic acid, we have

k_c = C$\alpha$²

= $\frac{1}{32}\times {\left(2\times {10}^{-2}\right)}^2$

= 1.25 $\times$ 10^-5

For the reaction,

Cu²⁺ + 2e^$$-$$ $\to$ Cu,  E^o = 0.337 V

$\Delta$G^o = - nFE^o

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu²⁺ + e^$$-$$ $\to$ Cu⁺,  E^o = 0.153 V

$\Delta$G^o = - nFE^o

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu²⁺ + e^$$-$$ $\to$ Cu⁺

$\Delta$G^o = –0.521 F = –nFE°

$\Rightarrow$ E° = 0.52 V

N₂ + 3H₂ $\to$ 2NH₃

$-\frac{1}{3}\frac{d\left[H_2\right]}{dt}=\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$

$\Rightarrow$ $-\frac{d\left[H_2\right]}{dt}=\frac{3}{2}\frac{d\left[N{H}_3\right]}{dt}$

= $\frac{3}{2}\times 2\times 10^{-4}$

= $3\times 10^{-4}$ mol L^$-$1 s^$-$1

Rate = $\frac{1}{3}\frac{d\left[B{r}_2\right]}{dt}=-\frac{1}{5}\frac{d\left[B{r}^{-}\right]}{dt}$

$\therefore$ $\frac{d\left[B{r}_2\right]}{dt}=-\frac{3}{5}\frac{d\left[B{r}^{-}\right]}{dt}$

R = k[A]^m[B]ⁿ ... (i)

2R = k[2A]^m[B]ⁿ ... (ii)

8R = k[2A]^m[2B]ⁿ ... (iii)

from (i), (ii) and (iii), m = 1, n = 2

So, rate = k[A][B]²