NEET-UG 2015

Let surface tension

S =kE^$$a$$V^bT^c
where k is a dimensionless constant

Writing the dimensions on both sides,

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

$\left[M{L}^0{T}^{-2}\right]$ = $\left[M^a{L}^{2a+b}{T}^{-2a-b+c}\right]$

Comparing both sides of the equation we get,

$a$ = 1                   ....(1)
2$a$ + b = 0            ....(2)
-2$a$ - b + c = - 2            ....(3)

Solving equation (1), (2) and (3), we get

$a$ = 1, b = - 2, c = - 2

$\therefore$ Dimension of surface tension = $\left[E{V}^{-2}{T}^{-2}\right]$

Given $v\left(x\right)=\beta x^{-2n}$

$\therefore$ $\frac{dv}{dx}=-2n\beta x^{-2n-1}$

So acceleration of the particle is
$a$ = $v\frac{dv}{dx}$ = ($\beta x^{-2n}$) $\times$ ($-2nb{x}^{-2n-1}$)

              = $-2n{\beta }^2{x}^{-4n-1}$

${\overrightarrow{V}}_A=10\left(-\hat{i}\right)$

${\overrightarrow{V}}_B=10\left(\hat{j}\right)$

${\overrightarrow{V}}_{BA}=10\hat{\mathrm{j}}+10\hat{i}=10\sqrt{2}km/h$

Distance OB = 100 cos 45° = 50$\sqrt{2}$ km


Time taken to reach the shortest distance between
A and B = $\frac{OB}{\overrightarrow{V_{BA}}}=\frac{50\sqrt{2}}{10\sqrt{2}}=5h$

Here, M_A = 4 kg, M_B = 2 kg, M_C = 1 kg, F = 14 N

Net mass, M = M_A + M_B + M_C = 4 + 2 + 1 = 7 kg

Let a be the acceleration of the system.
Using Newton’s second law of motion,

F = Ma
14 = 7a    $\therefore$ a = 2 ms^–2

Let F' be the force applied on block A by block B i.e. the contact force between A and B. Free body diagram for block A

Again using Newton’s second law of motion,
F – F' = 4a
14 – F' = 4 × 2 $\Rightarrow$ 14 – 8 = F' $\therefore$ F' = 6 N

For the motion of both the blocks
m₁a = T – ${\mu }_k$m₁g
m₂g – T = m₂a

$a=\frac{m_{2}g-{\mu }_k{m}_{1}g}{m_1+m_2}$

$m_{2}g-T=\left(m_2\right)\left(\frac{m_{2}g-{\mu }_k{m}_{1}g}{m_1+m_2}\right)$

solving we get tension in the string

$T=\frac{m_1{m}_2\left(1+{\mu }_k\right)g}{m_1+m_2}$

Here, m = 10 kg, v_i = 10 ms^–1
Initial kinetic energy of the block is
$K_i=\frac{1}{2}m{v}_i^2=\frac{1}{2}\times \left(10kg\right)\times {\left(10m{s}^{-1}\right)}^2=500J$

Work done by retarding force

$W=\underset{x_1}{\overset{x_2}{\int }}{F_{r}dx=\underset{20}{\overset{30}{\int }}-0.1xdx=-0.1\left[\frac{x^2}{2}\right]}_{20}^{30}$

$=-0.1\left[\frac{900-400}{2}\right]=-25J$

According to work-energy theorem,

W = K_f – K_i

K_f = W + K_i = – 25 J + 500 J = 475 J

Here, K_P > K_Q

Case (a) : Elongation (x) in each spring is same.

$W_P=\frac{1}{2}{K}_P{x}^2,W_Q=\frac{1}{2}{K}_Q{x}^2$
$\therefore W_P>W_Q$

Case (b) : Force of elongation is same.

So, $x_1=\frac{F}{K_P}$ and $x_2=\frac{F}{K_Q}$

$W_P=\frac{1}{2}{K}_P{x}_1^2=\frac{1}{2}\frac{F^2}{K_P}$

$W_Q=\frac{1}{2}{K}_Q{x}_2^2=\frac{1}{2}\frac{F^2}{K_Q}$
$\therefore W_P<W_Q$

As we know power P = $\frac{dw}{dt}$

$\Rightarrow w=Pt=\frac{1}{2}m{V}^2$

So, $v=\sqrt{\frac{2Pt}{m}}$

Hence, acceleration $a=\frac{dV}{dt}=\sqrt{\frac{2P}{m}}.\frac{1}{2\sqrt{t}}$

Therefore, force on the particle at time ‘t’

$=ma=\sqrt{\frac{2K{m}^2}{m}}.\frac{1}{2\sqrt{t}}=\sqrt{\frac{Km}{2t}}=\sqrt{\frac{mK}{2}}{t}^{-1/2}$

By law of conservation of energy,

K.E_f = K.E_i – excitation energy (e)

$\therefore$ $\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2-\epsilon$

Let the distance moved by spherical body of mass M is x₁ and by spherical body of mass 5m is x₂.

As their C.M. will remain stationary

$\therefore$ (M) (x₁) = (5M) (x₂) $\Rightarrow$ x₁ = 5x₂

x₁ + x₂ = 9R

So, x₁ = 7.5 R

Given situation is shown in figure.

N₁ = Normal reaction on A

N₂ = Normal reaction on B

W = Weight of the rod

In vertical equilibrium,
N₁ + N₂ = W …(i)

Torque balance about centre of mass of the rod,

N₁x = N₂(d – x)

Putting value of N₂ from equation (i)

N₁x = (W – N₁)(d – x)

$\Rightarrow$ N₁x = Wd – Wx – N₁d + N₁x

$\Rightarrow$ N₁d = W(d – x)

$\therefore$ $N_1=\frac{W\left(d-x\right)}{d}$

Net moment of inertia of the system,

I = I₁ + I₂ + I₃

The moment of inertia of a shell about its diameter,

I₁ = $\frac{2}{3}$mr²

The moment of inertia of a shell about its tangent is given by

$I_2=I_3=I_1+m{r}^2=\frac{2}{3}m{r}^2+m{r}^2=\frac{5}{3}m{r}^2$

$\therefore$ $I=2\times \frac{5}{3}m{r}^2+\frac{2}{3}m{r}^2=\frac{12m{r}^2}{3}=4m{r}^2$

According to law of conservation of angular momentum

mvr = mv'r'

$v_0{R}_0=v\left(\frac{R_0}{2}\right);v=2v_0$   …(i)

$\therefore$ $\frac{K_0}{K}=\frac{\frac{1}{2}m{v}_0^2}{\frac{1}{2}m{v}^2}={\left(\frac{v_0}{v}\right)}^2$

$\Rightarrow$ $\frac{K}{K_0}={\left(\frac{v}{v_0}\right)}^2={\left(2\right)}^2$    (Using (i))

K = 4K₀ = 2$m{v}_0^2$

Gravitational force of attraction between planet and sun gives centripetal force,

GMm/r² = mv²/r

Now velocity v = $\sqrt{GM/r}$

Time period of planet T = 2$\pi$r/v
$\Rightarrow$ T² = 4${\pi }^2$r³/GM

From Kepler’s third law, T² = Kr³

Using equations, we see 4$\pi$²r³/GM = Kr³,

so relation between G and K is GMK = 4${\pi }^2$

As the temperature difference $\Delta T$ = 10°C as well as the thermal resistance is same for both the cases, so thermal current or rate of heat flow will also be same for both the cases.

According to Wein’s displacement law

${\lambda }_{m}T$ = constant    …(i)

For star P, intensity of violet colour is maximum.
For star Q, intensity of red colour is maximum.
For star R, intensity of green colour is maximum.

Also, ${\lambda }_r>{\lambda }_g>{\lambda }_v$

Using equation (i), T_r < T_g < T_v

T_Q < T_R < T_P

Using Bernoulli's theorem and assuming density constant,

P₁+1/2ρv₁² = P₂+1/2ρv₂²

where:
P₂ = pressure outside house
P₁ = pressure inside house
v₁ = speed of air inside house
v₂ = speed of air outside house

Pressure difference,

P₁ – P₂ = 1/2ρ [v₂² – v₁²]

Now, P₁ – P₂ = 1/2 × 1.2 [40² – 0²]

P₁ – P₂ = 960 N/m²

Since Pressure P = Force/Area, so force acting on roof will be :

F = P × A

F = 960 × 250

F = 960 × 1000/4

F = 24×10⁴ or 2.4×10⁵ N

which is acting upward.

Compressibility of water, K = 45.4 × 10^–11 Pa^–1

density of water P = 10³ kg/m³

depth of ocean, h = 2700 m

We have to find $\frac{\Delta V}{V}=?$

As we know, compressibility,

$K=\frac{1}{B}=\frac{\left(\Delta V/V\right)}{P}\left(P=\rho gh\right)$

So, $\left(\frac{\Delta V}{V}\right)=K\rho gh$

= 45.4 × 10^–11 × 10³ × 10 × 2700 = 1.2258 × 10^–2

Considering the cyclic process ABCA
Q_cyclic = W = area of $\Delta$ABC

$Q_{AB}+Q_{BC}+Q_{CA}=\frac{1}{2}\times BC\times AB$

$400+100+Q_{CA}=\frac{1}{2}\times \left(2\times {10}^{-3}\right)\times 4\times 10^4$

$400+100-Q_{AC}=40$

Hence, Q_AC = 460 J

Given, efficiency of engine, $\eta =\frac{1}{10}$

work done on system W = 10J

Coefficient of performance of refrigerator

$\beta =\frac{Q_2}{W}=\frac{1-\eta }{\eta }=\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}}=9$

Energy absorbed from reservoir

$Q_2=\beta w$

$Q_2=9\times 10=90J$

For n degrees of freedom, $C_v=\frac{n}{2}R$

Also, $C_p-C_v=R$

$C_p=C_v+R=\frac{n}{2}R+R=\left(\frac{n}{2}+1\right)R$

$\gamma =\frac{C_p}{C_v}=\frac{\left(\frac{n}{2}+1\right)R}{\frac{n}{2}R}=\frac{n+2}{n}$

$\therefore$ $\gamma =1+\frac{2}{n}$

Change in internal energy from A $\to$ B

$\Delta U=\frac{f}{2}nR\Delta T=\frac{f}{2}nR\left(T_f-T_i\right)$

$=\frac{5}{2}\left\{P_f{V}_f-P_i{V}_i\right\}$

(As gas is diatomic $\therefore$ f = 5)

$=\frac{5}{2}\left\{2\times {10}^3\times 6-5\times {10}^3\times 4\right\}$

$=\frac{5}{2}\left\{12-20\right\}\times 10^{3}J=5\times \left(-4\right)\times 10^{3}J$

$\therefore \Delta U=-20KJ$

Here, $y_1=a\sin \omega t$

$y_2=b\cos \omega =b\sin \left(\omega t+\frac{\pi }{2}\right)$


Hence, resultant motion is SHM with amplitude $\sqrt{a^2+b^2}$.

As we know, for particle undergoing SHM,

$V=\omega \sqrt{A^2-X^2}$

$V_1^2={\omega }^2\left(A^2-x_1^2\right)$

$V_2^2={\omega }^2\left(A^2-x_2^2\right)$

Substracting we get,

$\frac{V_1^2}{{\omega }^2}+x_1^2=\frac{V_2^2}{{\omega }^2}+x_2^2$

$\Rightarrow \frac{V_1^2-V_2^2}{{\omega }^2}=x_2^2-x_1^2$

$\Rightarrow w=\sqrt{\frac{V_1^2-V_2^2}{x_2^2-x_1^2}}$

$\Rightarrow T=2\pi \sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}}$

According to question, electric field varies as

E = Ar

Here r is the radial distance.

At r = a, E = Aa     …(i)

Net flux emitted from a spherical surface of radius a is ${\varphi }_{net}=\frac{q_{en}}{{\epsilon }_0}$

$\Rightarrow \left(Aa\right)\times \left(4\pi a^2\right)=\frac{q}{{\epsilon }_0}$   [Using equation (i)]

$\therefore q=4\pi {\epsilon }_{0}A{a}^3$

The area of cross section of conductor is non uniform so current density will be different but the flow of electrons will be uniform so current will be constant.

Total potential difference across potentiometer wire

= 10^–3 × 400 volt = 0.4 volt

potential gradient = $\frac{1mv}{cm}$

$=10^{-3}v/cm=10^{-1}\frac{v}{m}$

Let resistance of R$\Omega$ connected in series.



$\therefore$ $\frac{2}{R+8}=\frac{{10}^{-1}\times 4}{8}=\frac{1}{20}$

$\Rightarrow$ R + 8 = 40 or, R = 32 $\Omega$

Effective resistance of
B & C = (1.5R)(3R)/(1.5R + 3R) = R

In series sequence voltage across A = voltage across B & C

Now B & C are parallel, so V_B = V_C,

$\therefore$ V_A = V_B = V_C

q = CV $\Rightarrow$ V = q/C

Due to dielectric insertion, new capacitance C₂ = CK

Initial energy stored in capacitor, $U_1=\frac{q^2}{2C}$

Final energy stored in capacitor, $U_2=\frac{q^2}{2KC}$

Change in energy stored, $\Delta$U = U₂ – U₁

$\Delta U=\frac{q^2}{2C}\left(\frac{1}{K}-1\right)=\frac{1}{2}C{V}^2\left(\frac{1}{K}-1\right)$

New potential difference between plates

$V^{\prime }=\frac{q}{CK}=\frac{V}{K}$

Magnetic field due to segment ‘1’

$\overrightarrow{B_1}=\frac{{\mu }_{0}I}{4\pi R}\left[\sin 90^{\circ }+\sin 0^{\circ }\right]\left(-\hat{k}\right)$

$=\frac{-{\mu }_{0}I}{4\pi R}\left(\hat{k}\right)=\overrightarrow{B_3}$

Magnetic field due to segment 2

$\overrightarrow{B_2}=\frac{{\mu }_{0}I}{4R}\left(-\hat{i}\right)=\frac{-{\mu }_{0}I}{4\pi R}\left(-\pi \hat{i}\right)$


$\therefore \overrightarrow{B}$ at centre

$\overrightarrow{B_c}=\overrightarrow{B_1}+\overrightarrow{B_2}+\overrightarrow{B_3}$

$=\frac{-{\mu }_{0}I}{4\pi R}\left(\pi \hat{i}+2\hat{k}\right)$

The emf in AD:

e₁ = (a × μ₀iv)/2π(x −a/2)

The emf in EF :

e₂ = (a × μ₀i × v)/2π(x + a/2)

Net emf = e₁ – e₂

$\Rightarrow e=\frac{{\mu }_0}{2\pi }a\times i\times v\left[\frac{1}{x-\frac{a}{2}}-\frac{1}{x+\frac{a}{2}}\right]$

$e\propto \frac{1}{\left(2x-a\right)\left(2x+a\right)}$

Current in the orbit, $I=\frac{e}{T}$

$I=\frac{e}{\left(2\pi /\omega \right)}=\frac{\omega e}{2\pi }$

$=\frac{\left(2\pi n\right)e}{2\pi }=ne$

Magnetic field at centre of current carrying circular coil is given by

$B=\frac{{\mu }_{0}I}{2r}=\frac{{\mu }_{0}ne}{2r}$

For pure resistor circuit, power

P = $\frac{V_{rms}^2}{R}$

$\Rightarrow$ $V_{rms}^2$ = PR

For L-R series circuit, power

P' = V_rms I_rms cos$\varphi$

= $V_{rms}\times \frac{V_{rms}}{Z}\times \frac{R}{Z}$

= $\frac{V_{rms}^{2}R}{Z^2}$ = $\frac{PR\times R}{Z^2}$

$\therefore$ P' = $P{\left(\frac{R}{Z}\right)}^2$

Momentum transferred to the surface

= change in momentum

= P_f - P_i

= $+\frac{E}{C}-\left(-\frac{E}{C}\right)$

= $\frac{2E}{C}$

As $\mu$ = $\frac{\sin \left(\frac{A+\delta }{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\Rightarrow$ $\cot \frac{A}{2}$ = $\frac{\sin \left(\frac{A+\delta }{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\Rightarrow$ $\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+\delta }{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\Rightarrow$ $sin\left(\frac{\pi }{2}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta }{2}\right)$

$\Rightarrow$ $\frac{\pi }{2}-\frac{A}{2}$ = $\frac{A}{2}+\frac{\delta }{2}$

$\Rightarrow$ $\delta$ = 180^o - 2A

We know, $\frac{1}{f}=(\mu -1)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

For the lens on the left,

$$\frac{1}{f_1}=(1.5-1)\left(\frac{1}{\infty }-\frac{1}{-20}\right)=\frac{1}{40}$$

For the lensh on the right,

$$\frac{1}{f_2}=(1.5-1)\left(\frac{1}{+20}-\frac{1}{\infty }\right)=\frac{1}{40}$$

For the oil in the intervening space,

$$\frac{1}{f_3}=(1.7-1)\left(\frac{1}{-20}-\frac{1}{20}\right)=-\frac{7}{100}$$

Hence, if the focal length of the combination is F, then

$$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{1}{40}+\frac{1}{40}-\frac{7}{100}=-\frac{1}{50}$$

i.e., F = $-$50 cm

For central maxima, sin $\theta$ = $\frac{\lambda }{a}$

Also, $\theta$ is very-very small so

sin $\theta$ $\approx$ tan$\theta$ = $\frac{y}{D}$

$\therefore$ $\frac{y}{D}$ = $\frac{\lambda }{a}$

$\Rightarrow$ y = $\frac{\lambda D}{a}$

Width of central maxima = 2y = $\frac{2D\lambda }{a}$

Fringe width, $\beta$ = $\frac{D\lambda }{d}$

As per question, width of central maxima of single slit pattern = width of 10 maxima of double slit pattern

$\frac{D2\lambda }{a}=10\left(\frac{\lambda D}{d}\right)$

$\Rightarrow$ $a$ = $\frac{2d}{10}$ = $\frac{2\times {10}^{-3}}{10}$ = 0.2 $\times$ 10^-3 m = 0.2 mm

Energy of electron in He⁺ 3^rd orbit

E₃ = -13.6 $\times$ $\frac{4}{9}eV$

= -13.6 $\times$ $\frac{4}{9}$ $\times$ 1.6 $\times$ 10^-19 J

= - 9.7 $\times$ 10^-19 J

According to Bohr’s model,

Kinetic energy of electron in the 3^rd orbit = – E₃

$\therefore$ 9.7 $\times$ 10^-19 = $\frac{1}{2}{m}_e{v}^2$

v = $\sqrt{\frac{2\times 9.7\times {10}^{19}}{9.1\times {10}^{-31}}}$ = 1.46 $\times$ 10⁶ m/s

As we know, R = R₀ (A)^1/3

where A = mass number

R_Al = R₀ (27)^1/3 = 3R₀

R_Te = R₀ (125)^1/3 = 5R₀ = $\frac{5}{3}{R}_{Al}$

de-Broglie wavelength, $\lambda$ = $\frac{h}{p}$

$\therefore$ $\lambda$ $\propto$ $\frac{1}{p}$

$\Rightarrow$ $\lambda$p = constant

This represents a rectangular hyperbola.

Stopping potential, $\varphi$ = E - K_max

3eV₀ + $\varphi$ = $\frac{hc}{\lambda }$ ......(1)

eV₀ + $\varphi$ = $\frac{hc}{2\lambda }$ ......(2)

Solving equation (1) and (2), we get

$\varphi$ = $\frac{hc}{4\lambda }$

So, threshold wavelength,

$\lambda$_th = $\frac{hc}{\varphi }$ = $\frac{hc}{\frac{hc}{4\lambda }}$ = 4$\lambda$

Here P-N junction diode rectifies half of the ac wave i.e., acts as half wave rectifier.
During + ve half cycle Diode is forward biased output across R_L will be

During –ve half cycle Diode is reverse biased output will not obtained.

The Boolean expression of this arrangement is

Y = $\overline{\overline{A}+\overline{B}}$ = $\overline{\overline{A}}+\overline{\overline{B}}$ = A.B

Thus, the combination represents AND gate.

Ratio of weight of gases = w_H2 : w_O2 = 1 : 4

$\therefore$ Number of moles of H₂ = $\frac{1}{2}$

$\therefore$ Number of moles of O₂ = $\frac{4}{32}$

Ratio of moles of gases

= n_H2 : n_O2

= $\frac{1}{2}$ : $\frac{4}{32}$

= $\frac{1}{2}$ $\times$ $\frac{32}{4}$ = 4 : 1

We know Anglular momentum = $\sqrt{l(l+1)}h$

For d obtial l = 2

$\therefore$ Angular momentum = $\sqrt{6}h$

Number of d-electrons in Fe²⁺ = 6

Number of p-electrons in Cl = 11

As, 1.6 × 10¹² is very high value of K. Thus, the reaction proceeds almost to completion and mixture must contain mostly products.

$\Delta$G = $\Delta$G^o + 2.303 RT log Q

At equilibrium, when $\Delta$G = 0 and Q = K

then 0 = $\Delta$G^o + 2.303 RT log K

$\Delta$G^o = – 2.303 RT log K

From the Ksp values of the given salts calculate the solubility values. Salt having highest solubility will precipitate at last.

AgCrO4	⇌	2Ag+	+	CrO42-
s		2s		s

K_sp = (2s)²(s) = 1.1 × 10^–12

$\Rightarrow$ s = 0.65 × 10^–4

AgCl	⇌	Ag+	+	Cl-
s		s		s

K_sp = s × s

$\Rightarrow$ 1.8 × 10^–10 = s ²

$\Rightarrow$ s = 1.34 × 10^–5

AgBr	⇌	Ag+	+	Br-
s		s		s

K_sp = s × s

$\Rightarrow$ 5 × 10^–13 = s²

$\Rightarrow$ s = 0.71 × 10^–6

AgI	⇌	Ag+	+	I-
s		s		s

K_sp = s × s

$\Rightarrow$ 8.3 × 10^–17 = s²

$\Rightarrow$ s = 0.9 × 10^–8

$\therefore$ Solubility of Ag₂CrO₄ is maximum so, it will precipitate at last.

Ai₂(SO₄)₃ ⇌ 2Al⁺³ + 3SO₄^2–

van't Hoff factor, i = 5

K₂SO₄ ⇌ 2K⁺ + SO₄^2–

van't Hoff factor, i = 3

K₃[Fe(CN)₆] ⇌ 3K⁺ + [Fe(CN)₆]^2-

van't Hoff factor, i = 4

Al(NO₃)₃ ⇌ Al³⁺ + 3NO₃^–

van't Hoff factor, i = 4

K₄[Fe(CN)₆] ⇌ 4K⁺ + [Fe(CN)₆]^4–

van't Hoff factor, i = 5

$\Delta$T_b = iK_bm

Given, ($\Delta$T_b)_x > ($\Delta$T_b)_y

$\therefore$ i_xK_bm > i_yK_bm

$\Rightarrow$ i_x > i_y

(K_b is same for same solvent)

So, x is undergoing dissociation in water.