NEET (UG) 2018 — Code ZZ

Question 1

A tuning fork is used to produce resonance in a glass tube. The length of the air column in this tube can be adjusted by a variable piston. At room temperature of 27$^{\circ}$C two successive resonances are produced at 20 cm and 73 cm of column length. If the frequency of the tuning fork is 320 Hz,

Accepted Answer

(B) 339 m/s

Two successive resonance lengths differ by half a wavelength, so

$$v = 2
u(L_2 - L_1)$$

$$= 2 	imes 320 	imes (73 - 20) 	imes 10^{-2}$$

$$= 339.2\ 	ext{ms}^{-1} \approx 339\ 	ext{m/s}$$

Question 2

An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall

Accepted Answer

(A) Smaller

The electric force gives an acceleration $a = \dfrac{eE}{m}$, so

$$h = \frac{1}{2}\frac{eE}{m}t^{2} \Rightarrow t = \sqrt{\frac{2hm}{eE}}$$

So $t \propto \sqrt{m}$, as 'e' is the same for the electron and the proton.

The electron has a smaller mass, so it will take a smaller time.

Question 3

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s$^{2}$ at a distance of 5 m from the mean position. The time period of oscillation is

Accepted Answer

(B) $\pi$ s

$$|a| = \omega^{2}y$$

$$20 = \omega^{2}(5) \Rightarrow \omega = 2\ 	ext{rad/s}$$

$$T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi\ 	ext{s}$$

Question 4

The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is

Accepted Answer

(A) Independent of the distance between the plates

For an isolated capacitor the charge Q is constant, and the force on one plate due to the field of the other is

$$F_{plate} = \frac{Q^{2}}{2A\epsilon_0}$$

This expression contains no separation term, so F is independent of the distance between the plates.

Question 5

Current sensitivity of a moving coil galvanometer is 5 div/mA and its voltage sensitivity (angular deflection per unit voltage applied) is 20 div/V. The resistance of the galvanometer is

Accepted Answer

(D) 250 $\Omega$

Current sensitivity $I_S = \dfrac{NBA}{C}$ and voltage sensitivity $V_S = \dfrac{NBA}{CR_G}$.

So the resistance of the galvanometer is

$$R_G = \frac{I_S}{V_S} = \frac{5 	imes 1}{20 	imes 10^{-3}} = \frac{5000}{20} = 250\ \Omega$$

Question 6

A thin diamagnetic rod is placed vertically between the poles of an electromagnet. When the current in the electromagnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes

Accepted Answer

(A) The current source

Energy of the current source will be converted into the potential energy of the rod.

The magnetic field itself does no work; the source maintaining the current supplies the energy.

Question 7

An inductor 20 mH, a capacitor 100 $\mu$F and a resistor 50 $\Omega$ are connected in series across a source of emf, V = 10 sin 314 t. The power loss in the circuit is

Accepted Answer

(A) 0.79 W

$$Z = \sqrt{R^{2} + \left(\omega L - \frac{1}{\omega C}ight)^{2}} = 56\ \Omega$$

$$P_{av} = \left(\frac{V_{RMS}}{Z}ight)^{2}R = \left(\frac{10}{\sqrt{2}	imes 56}ight)^{2}	imes 50 = 0.79\ 	ext{W}$$

Question 8

A metallic rod of mass per unit length 0.5 kg m$^{-1}$ is lying horizontally on a smooth inclined plane which makes an angle of 30$^{\circ}$ with the horizontal. The rod is not allowed to slide down by flowing a current through it when a magnetic field of induction 0.25 T is acting on it in the vert

Accepted Answer

(C) 11.32 A

For equilibrium along the incline,

$$mg\sin 30^{\circ} = IlB\cos 30^{\circ}$$

$$I = \frac{m}{l}\frac{g}{B}	an 30^{\circ} = \frac{0.5 	imes 9.8}{0.25 	imes \sqrt{3}} = 11.32\ 	ext{A}$$

Question 9

A carbon resistor of (47 $\pm$ 4.7) k$\Omega$ is to be marked with rings of different colours for its identification. The colour code sequence will be

Accepted Answer

(B) Yellow – Violet – Orange – Silver

$(47 \pm 4.7)$ k$\Omega$ $= 47 	imes 10^{3} \pm 10\%$

4 is Yellow, 7 is Violet, the multiplier $10^{3}$ is Orange, and $\pm 10\%$ tolerance is Silver.

So the sequence is Yellow – Violet – Orange – Silver.

Question 10

A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I. The value of 'n' is

Accepted Answer

(A) 10

In series,

$$I = \frac{E}{nR + R}$$

In parallel,

$$10I = \frac{E}{\frac{R}{n} + R}$$

Dividing the second by the first,

$$10 = \frac{(n + 1)R}{\left(\frac{1}{n} + 1\right)R}$$

Solving this equation gives $n = 10$.

Question 11

A battery consists of a variable number 'n' of identical cells (having internal resistance 'r' each) which are connected in series. The terminals of the battery are short-circuited and the current I is measured. Which of the graphs shows the correct relationship between I and n?

Accepted Answer

(A) Graph (1)

When n identical cells are in series and the terminals are short-circuited,

$$I = \frac{n\varepsilon}{nr} = \frac{\varepsilon}{r}$$

So I is independent of n and I is constant. The correct graph is a horizontal straight line.

Question 12

In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength $\lambda$ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20$^{\circ}$. To increase the fringe angular width to 0.21$

Accepted Answer

(B) 1.9 mm

Angular width $= \dfrac{\lambda}{d}$

$$0.20^{\circ} = \frac{\lambda}{2\ 	ext{mm}}\qquad \ldots(i)$$

$$0.21^{\circ} = \frac{\lambda}{d}\qquad \ldots(ii)$$

Dividing we get

$$\frac{0.20}{0.21} = \frac{d}{2\ 	ext{mm}}$$

$$d = 1.9\ 	ext{mm}$$

Question 13

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

Accepted Answer

(D) Large focal length and large diameter

For a telescope, angular magnification $= \dfrac{f_0}{f_E}$, so the focal length of the objective lens should be large.

Angular resolution $= \dfrac{D}{1.22\lambda}$ should be large, so the diameter should be large.

So the objective should have a large focal length $f_0$ and a large diameter D.

Question 14

Unpolarised light is incident from air on a plane surface of a material of refractive index '$\mu$'. At a particular angle of incidence 'i', it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

Accepted Answer

(B) Reflected light is polarised with its electric vector perpendicular to the plane of incidence

When the reflected and refracted rays are perpendicular, the incidence is at the Brewster angle and the reflected light is completely plane polarised with its electric field vector perpendicular to the plane of incidence.

Also $	an i = \mu$ at the Brewster angle, so options (3) and (4) are wrong.

Question 15

An em wave is propagating in a medium with a velocity $\vec{V} = V\hat{i}$. The instantaneous oscillating electric field of this em wave is along +y axis. Then the direction of oscillating magnetic field of the em wave will be along

Accepted Answer

(B) +z direction

For an electromagnetic wave, $\vec{E} 	imes \vec{B}$ points along the direction of propagation.

$$(E\hat{j}) 	imes \vec{B} = V\hat{i}$$

Since $\hat{j} 	imes \hat{k} = \hat{i}$, we get $\vec{B} = B\hat{k}$.

So the magnetic field oscillates along the +z direction.

Question 16

The refractive index of the material of a prism is $\sqrt{2}$ and the angle of the prism is 30$^{\circ}$. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after ref

Accepted Answer

(B) 45$^{\circ}$

For retracing its path, the light ray should be normally incident on the silvered face.

Then the angle of refraction at the first face is 30$^{\circ}$, equal to the prism angle.

Applying Snell's law at the first face M,

$$\frac{\sin i}{\sin 30^{\circ}} = \frac{\sqrt{2}}{1}$$

$$\sin i = \sqrt{2}	imes\frac{1}{2} = \frac{1}{\sqrt{2}}$$

$$i = 45^{\circ}$$

Question 17

An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be

Accepted Answer

(B) 36 cm away from the mirror

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

For $u_1 = -40$ cm and $f = -15$ cm,

$$\frac{1}{v_1} = \frac{1}{-15} + \frac{1}{40} \Rightarrow v_1 = -24\ 	ext{cm}$$

When the object is displaced by 20 cm towards the mirror, $u_2 = -20$ cm,

$$\frac{1}{v_2} = \frac{1}{-15} + \frac{1}{20} \Rightarrow v_2 = -60\ 	ext{cm}$$

So the image shifts away from the mirror by $60 - 24 = 36$ cm.

Question 18

The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA. This inductor is of inductance

Accepted Answer

(C) 13.89 H

Energy stored in an inductor is

$$U = \frac{1}{2}LI^{2}$$

$$25 	imes 10^{-3} = \frac{1}{2}	imes L 	imes (60 	imes 10^{-3})^{2}$$

$$L = \frac{25 	imes 2 	imes 10^{6} 	imes 10^{-3}}{3600} = \frac{500}{36} = 13.89\ 	ext{H}$$

Question 19

For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is

Accepted Answer

(A) 20

Number of nuclei remaining $= 600 - 450 = 150$

$$\frac{N}{N_0} = \left(\frac{1}{2}ight)^{n}$$

$$\frac{150}{600} = \left(\frac{1}{2}ight)^{t/t_{1/2}}$$

$$\left(\frac{1}{2}ight)^{2} = \left(\frac{1}{2}ight)^{t/t_{1/2}}$$

$$t = 2t_{1/2} = 2 	imes 10 = 20\ 	ext{minutes}$$

Question 20

The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is

Accepted Answer

(B) 1 : –1

In a Bohr orbit, KE $= -$(total energy).

So kinetic energy : total energy $= 1 : -1$.

Question 21

An electron of mass m with an initial velocity $\vec{V} = V_0\hat{i}$ (V$_0$ > 0) enters an electric field $\vec{E} = -E_0\hat{i}$ (E$_0$ = constant > 0) at t = 0. If $\lambda_0$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is

Accepted Answer

(A) $\dfrac{\lambda_0}{\left(1 + \dfrac{eE_0}{mV_0}t\right)}$

Initial de-Broglie wavelength $\lambda_0 = \dfrac{h}{mV_0}$.

The electron carries a negative charge, so in the field $-E_0\hat{i}$ the force on it is along $+\hat{i}$ and the acceleration is

$$a = \frac{eE_0}{m}$$

Velocity after time t is

$$V = V_0 + \frac{eE_0}{m}t$$

$$\lambda = \frac{h}{mV} = \frac{h}{mV_0\left[1 + \dfrac{eE_0}{mV_0}t\right]} = \frac{\lambda_0}{\left[1 + \dfrac{eE_0}{mV_0}t\right]}$$

Question 22

When the light of frequency 2$
u_0$ (where $
u_0$ is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v$_1$. When the frequency of the incident radiation is increased to 5$
u_0$, the maximum velocity of electrons emitted from the same plate is v$_2$

Accepted Answer

(A) 1 : 2

$$E = W_0 + \frac{1}{2}mv^{2}$$

$$h(2
u_0) = h
u_0 + \frac{1}{2}mv_1^{2} \Rightarrow h
u_0 = \frac{1}{2}mv_1^{2}\qquad \ldots(i)$$

$$h(5
u_0) = h
u_0 + \frac{1}{2}mv_2^{2} \Rightarrow 4h
u_0 = \frac{1}{2}mv_2^{2}\qquad \ldots(ii)$$

Dividing (i) by (ii),

$$\frac{1}{4} = \frac{v_1^{2}}{v_2^{2}} \Rightarrow \frac{v_1}{v_2} = \frac{1}{2}$$

Question 23

In the combination of the following gates the output Y can be written in terms of inputs A and B as

Accepted Answer

(B) $A \cdot \overline{B} + \overline{A} \cdot B$

The upper AND gate receives A and $\overline{B}$, so its output is $A\cdot\overline{B}$.

The lower AND gate receives $\overline{A}$ and B, so its output is $\overline{A}\cdot B$.

The two outputs feed an OR gate, so

$$Y = (A\cdot\overline{B} + \overline{A}\cdot B)$$

Question 24

In the circuit shown in the figure, the input voltage V$_i$ is 20 V, V$_{BE}$ = 0 and V$_{CE}$ = 0. The values of I$_B$, I$_C$ and $\beta$ are given by

Accepted Answer

(C) I$_B$ = 40 $\mu$A, I$_C$ = 5 mA, $\beta$ = 125

With $V_{CE} = 0$, the whole 20 V appears across R$_C$:

$$I_C = \frac{20 - 0}{4 	imes 10^{3}} = 5 	imes 10^{-3} = 5\ 	ext{mA}$$

With $V_{BE} = 0$,

$$V_i = V_{BE} + I_BR_B \Rightarrow 20 = I_B 	imes 500 	imes 10^{3}$$

$$I_B = \frac{20}{500 	imes 10^{3}} = 40\ \mu	ext{A}$$

$$\beta = \frac{I_C}{I_B} = \frac{5 	imes 10^{-3}}{40 	imes 10^{-6}} = 125$$

Question 25

In a p-n junction diode, change in temperature due to heating

Accepted Answer

(C) Affects the overall V - I characteristics of p-n junction

Due to heating, the number of electron-hole pairs will increase, so the overall resistance of the diode will change.

Because of this, both forward biasing and reverse biasing characteristics are changed, i.e. the overall V-I characteristic is affected.

Question 26

A solid sphere is rotating freely about its symmetry axis in free space. The radius of the sphere is increased keeping its mass same. Which of the following physical quantities would remain constant for the sphere?

Accepted Answer

(C) Angular momentum

The sphere is rotating freely in free space, so no external torque acts on it.

Hence the angular momentum of the sphere remains constant.

Since the moment of inertia increases with radius, the angular velocity and the rotational kinetic energy both decrease.

Question 27

The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K$_A$, K$_B$ and K$_C$, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then

Accepted Answer

(B) K$_A$ > K$_B$ > K$_C$

Point A is the perihelion and C is the aphelion, and B lies on the perpendicular to AC through the Sun.

By conservation of angular momentum the planet moves fastest at the perihelion and slowest at the aphelion, so

$$V_A > V_B > V_C$$

Hence

$$K_A > K_B > K_C$$

Question 28

If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

Accepted Answer

(C) 'g' on the Earth will not change

If the universal gravitational constant becomes ten times larger, then $G' = 10G$.

Since $g = \dfrac{GM_e}{R_e^{2}}$ depends on G and on the mass of the Earth (not of the Sun), the acceleration due to gravity increases.

So the statement that 'g' on the Earth will not change is the wrong option.

Question 29

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy (K$_t$) as well as rotational kinetic energy (K$_r$) simultaneously. The ratio K$_t$ : (K$_t$ + K$_r$) for the sphere is

Accepted Answer

(B) 5 : 7

$$K_t = \frac{1}{2}mv^{2}$$

$$K_t + K_r = \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} = \frac{1}{2}mv^{2} + \frac{1}{2}\left(\frac{2}{5}mr^{2}\right)\left(\frac{v}{r}\right)^{2} = \frac{7}{10}mv^{2}$$

$$\frac{K_t}{K_t + K_r} = \frac{\frac{1}{2}mv^{2}}{\frac{7}{10}mv^{2}} = \frac{5}{7}$$

Question 30

A small sphere of radius 'r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

Accepted Answer

(D) r$^{5}$

Rate of production of heat is the power dissipated against the viscous force:

$$	ext{Power} = 6\pi\eta r V_T \cdot V_T = 6\pi\eta r V_T^{2}$$

The terminal velocity satisfies $V_T \propto r^{2}$, so

$$	ext{Power} \propto r 	imes (r^{2})^{2} = r^{5}$$

Question 31

A sample of 0.1 g of water at 100$^{\circ}$C and normal pressure ($1.013 	imes 10^{5}$ Nm$^{-2}$) requires 54 cal of heat energy to convert to steam at 100$^{\circ}$C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

Accepted Answer

(B) 208.7 J

$$\Delta Q = \Delta U + \Delta W$$

$$54 	imes 4.18 = \Delta U + 1.013 	imes 10^{5}(167.1 	imes 10^{-6} - 0)$$

$$\Delta U = 208.7\ 	ext{J}$$

Question 32

Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by $\Delta l$ on applying a force F, how much force is needed to stretch the second wire by the sa

Accepted Answer

(A) 9 F

Equal volumes mean wire 1 has area A and length $3l$, while wire 2 has area $3A$ and length $l$.

For wire 1,

$$\Delta l = \left(\frac{F}{AY}\right)3l\qquad \ldots(i)$$

For wire 2,

$$\frac{F'}{3A} = Y\frac{\Delta l}{l} \Rightarrow \Delta l = \left(\frac{F'}{3AY}\right)l\qquad \ldots(ii)$$

From (i) and (ii),

$$\left(\frac{F}{AY}\right)3l = \left(\frac{F'}{3AY}\right)l$$

$$F' = 9F$$

Question 33

The power radiated by a black body is P and it radiates maximum energy at wavelength, $\lambda_0$. If the temperature of the black body is now changed so that it radiates maximum energy at wavelength $\dfrac{3}{4}\lambda_0$, the power radiated by it becomes nP. The value of n is

Accepted Answer

(D) $\dfrac{256}{81}$

By Wien's law, $\lambda_{max}T = $ constant, so

$$\lambda_0 T = \frac{3\lambda_0}{4}T' \Rightarrow T' = \frac{4}{3}T$$

By Stefan's law $P \propto T^{4}$, so

$$\frac{P_2}{P_1} = \left(\frac{T'}{T}\right)^{4} = \left(\frac{4}{3}\right)^{4} = \frac{256}{81}$$

Question 34

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere?

(Given : Mass of oxygen molecule (m) = $2.76 	imes 10^{-26}$ kg, Boltzmann's constant k$_B$ = $1.38 	imes 10^{-23}$ JK$^{-1}$)

Accepted Answer

(B) $8.360 	imes 10^{4}$ K

$V_{escape} = 11200$ m/s.

Say at temperature T the rms speed attains $V_{escape}$:

$$\sqrt{\frac{3k_BT}{m_{O_2}}} = 11200\ 	ext{m/s}$$

On solving,

$$T = 8.360 	imes 10^{4}\ 	ext{K}$$

Question 35

The volume (V) of a monatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is

Accepted Answer

(A) $\dfrac{2}{5}$

The graph of V against T is a straight line through the origin, so $\dfrac{V}{T}$ is constant and the process is isobaric.

For an isobaric process,

$$W = nR\Delta T$$

$$Q = nC_P\Delta T = n\left(\frac{5}{2}Right)\Delta T\quad	ext{for a monatomic gas}$$

$$\frac{W}{Q} = \frac{nR\Delta T}{\frac{5}{2}nR\Delta T} = \frac{2}{5}$$

Question 36

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is

Accepted Answer

(A) 13.2 cm

For a closed organ pipe, the third harmonic is $\dfrac{3v}{4l}$.

For an open organ pipe, the fundamental frequency is $\dfrac{v}{2l'}$.

Given that these are equal,

$$\frac{3v}{4l} = \frac{v}{2l'}$$

$$l' = \frac{4l}{3 	imes 2} = \frac{2l}{3} = \frac{2 	imes 20}{3} \approx 13.3\ 	ext{cm}$$

which matches the option 13.2 cm.

Question 37

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

Accepted Answer

(A) 26.8%

Efficiency of an ideal heat engine is

$$\eta = \left(1 - \frac{T_2}{T_1}ight)$$

where T$_2$ is the sink temperature and T$_1$ the source temperature.

$$\%\eta = \left(1 - \frac{273}{373}ight)	imes 100 = \left(\frac{100}{373}ight)	imes 100 = 26.8\%$$

Question 38

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to

Accepted Answer

(C) $\dfrac{5}{4}D$

As the track is frictionless, the total mechanical energy remains constant.

$$0 + mgh = \frac{1}{2}mv_L^{2} + 0 \Rightarrow h = \frac{v_L^{2}}{2g}$$

For completing the vertical circle, $v_L \geq \sqrt{5gR}$, so

$$h = \frac{5gR}{2g} = \frac{5}{2}R = \frac{5}{4}D$$

Question 39

Three objects, A : (a solid sphere), B : (a thin circular disk) and C : (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed $\omega$ about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation

Accepted Answer

(A) W$_C$ > W$_B$ > W$_A$

Work done required to bring them to rest is

$$\Delta W = \Delta KE = \frac{1}{2}I\omega^{2}$$

So $\Delta W \propto I$ for the same $\omega$.

$$W_A : W_B : W_C = \frac{2}{5}MR^{2} : \frac{1}{2}MR^{2} : MR^{2} = \frac{2}{5} : \frac{1}{2} : 1 = 4 : 5 : 10$$

$$W_C > W_B > W_A$$

Question 40

Which one of the following statements is incorrect?

Accepted Answer

(C) Coefficient of sliding friction has dimensions of length.

Coefficient of sliding friction has no dimensions.

$$f = \mu_s N \Rightarrow \mu_s = \frac{f}{N}$$

It is the ratio of two forces, so it is dimensionless — statement (3) is incorrect.

Question 41

A moving block having mass m, collides with another stationary block having mass 4m. The lighter block comes to rest after collision. When the initial velocity of the lighter block is v, then the value of coefficient of restitution (e) will be

Accepted Answer

(B) 0.25

According to the law of conservation of linear momentum,

$$mv + 4m 	imes 0 = 4mv' + 0 \Rightarrow v' = \frac{v}{4}$$

$$e = \frac{	ext{Relative velocity of separation}}{	ext{Relative velocity of approach}} = \frac{v/4}{v}$$

$$e = \frac{1}{4} = 0.25$$

Question 42

A block of mass m is placed on a smooth inclined wedge ABC of inclination $	heta$ as shown in the figure. The wedge is given an acceleration 'a' towards the right. The relation between a and $	heta$ for the block to remain stationary on the wedge is

Accepted Answer

(C) $a = g	an	heta$

Working in the frame of the wedge, a pseudo force $ma$ acts on the block towards the left.

Resolving along and perpendicular to the incline, for the block to stay put:

$$N\sin	heta = ma$$

$$N\cos	heta = mg$$

Dividing,

$$	an	heta = \frac{a}{g} \Rightarrow a = g	an	heta$$

Question 43

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field $\vec{E}$. Due to the force $q\vec{E}$, its velocity increases from 0 to 6 m/s in one second duration. At that instant the direction of the field is reversed. The car continues to

Accepted Answer

(B) 1 m/s, 3 m/s

Acceleration $a = \dfrac{6 - 0}{1} = 6$ ms$^{-2}$

For $t = 0$ to $t = 1$ s,

$$S_1 = \frac{1}{2}	imes 6(1)^{2} = 3\ 	ext{m}$$

For $t = 1$ s to $t = 2$ s (field reversed, car decelerates to rest),

$$S_2 = 6(1) - \frac{1}{2}	imes 6(1)^{2} = 3\ 	ext{m}$$

For $t = 2$ s to $t = 3$ s (car moves back),

$$S_3 = 0 - \frac{1}{2}	imes 6(1)^{2} = -3\ 	ext{m}$$

Total displacement $S = S_1 + S_2 + S_3 = 3$ m, so

$$	ext{Average velocity} = \frac{3}{3} = 1\ 	ext{ms}^{-1}$$

Total distance travelled $= 3 + 3 + 3 = 9$ m, so

$$	ext{Average speed} = \frac{9}{3} = 3\ 	ext{ms}^{-1}$$

Question 44

The moment of the force, $\vec{F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$ at (2, 0, –3), about the point (2, –2, –2), is given by

Accepted Answer

(C) $-7\hat{i} - 4\hat{j} - 8\hat{k}$

$$\vec{	au} = (\vec{r} - \vec{r}_0)	imes\vec{F}$$

$$\vec{r} - \vec{r}_0 = (2\hat{i} + 0\hat{j} - 3\hat{k}) - (2\hat{i} - 2\hat{j} - 2\hat{k}) = 0\hat{i} + 2\hat{j} - \hat{k}$$

Taking the cross product with $\vec{F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$,

$$\vec{	au} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\ 0 & 2 & -1\ 4 & 5 & -6\end{vmatrix} = -7\hat{i} - 4\hat{j} - 8\hat{k}$$

Question 45

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball

Accepted Answer

(C) 0.529 cm

Diameter of the ball $=$ MSR $+$ CSR $	imes$ (Least count) $-$ Zero error

$$= 0.5\ 	ext{cm} + 25 	imes 0.001 - (-0.004)$$

$$= 0.5 + 0.025 + 0.004 = 0.529\ 	ext{cm}$$

Question 46

The difference between spermiogenesis and spermiation is

Accepted Answer

(C) In spermiogenesis spermatozoa are formed, while in spermiation spermatozoa are released from sertoli cells into the cavity of seminiferous tubules.

Spermiogenesis is the transformation of spermatids into spermatozoa.

Spermiation is the release of the sperms from Sertoli cells into the lumen of the seminiferous tubule.

Question 47

The amnion of mammalian embryo is derived from

Accepted Answer

(A) ectoderm and mesoderm

The extraembryonic or foetal membranes are amnion, chorion, allantois and yolk sac.

Amnion is formed from mesoderm on the outer side and ectoderm on the inner side.

Chorion is formed from trophoectoderm and mesoderm, whereas the allantois and yolk sac membrane have mesoderm on the outer side and endoderm on the inner side.

Question 48

The contraceptive 'SAHELI'

Accepted Answer

(A) blocks estrogen receptors in the uterus, preventing eggs from getting implanted.

Saheli is the first non-steroidal, once a week pill.

It contains centchroman and its functioning is based upon selective estrogen receptor modulation, i.e. it blocks estrogen receptors in the uterus so that eggs cannot be implanted.

Question 49

Hormones secreted by the placenta to maintain pregnancy are

Accepted Answer

(D) hCG, hPL, progestogens, estrogens

The placenta releases human chorionic gonadotropic hormone (hCG), which stimulates the corpus luteum during pregnancy to release estrogen and progesterone and also rescues the corpus luteum from regression.

Human placental lactogen (hPL) is involved in growth of the body of the mother and the breast.

Progesterone maintains pregnancy and keeps the uterus silent by increasing the uterine threshold to contractile stimuli.

Question 50

Match the items given in Column I with those in Column II and select the correct option given below :

Column I:

(a) Proliferative Phase

(b) Secretory Phase

(c) Menstruation

Column II:

(i) Breakdown of endometrial lining

(ii) Follicular Phase

(iii) Luteal Phase

Accepted Answer

(D) (a)-(ii), (b)-(iii), (c)-(i)

During the proliferative phase the follicles start developing, hence it is called the follicular phase.

The secretory phase is also called the luteal phase, mainly controlled by progesterone secreted by the corpus luteum. Estrogen further thickens the endometrium maintained by progesterone.

Menstruation occurs due to a decline in progesterone level and involves breakdown of the overgrown endometrial lining.

Question 51

All of the following are part of an operon except

Accepted Answer

(D) an enhancer

The operon concept is for prokaryotes, and an operon consists of a promoter, an operator and structural genes.

Enhancer sequences are present in eukaryotes, so an enhancer is not part of an operon.

Question 52

A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by

Accepted Answer

(C) Both sons and daughters

The woman is a carrier, so either of her two X chromosomes can pass to any child.

Both son and daughter can inherit that X-chromosome, although only the son will be diseased if the condition is recessive.

Question 53

According to Hugo de Vries, the mechanism of evolution is

Accepted Answer

(B) Saltation

As per the mutation theory given by Hugo de Vries, evolution is a discontinuous phenomenon or saltatory phenomenon, i.e. saltation (single step large mutation).

Question 54

AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA?

Accepted Answer

(A) AGGUAUCGCAU

The coding strand and the mRNA have the same nucleotide sequence, except that 'T' – thymine is replaced by 'U' – uracil in mRNA.

So AGGTATCGCAT gives AGGUAUCGCAU.

Question 55

Among the following sets of examples for divergent evolution, select the incorrect option :

Accepted Answer

(C) Eye of octopus, bat and man

Divergent evolution occurs in the same structure, for example the forelimbs, heart and brain of vertebrates, which have developed along different directions due to adaptation to different needs.

Whereas the eye of octopus, bat and man are examples of analogous organs showing convergent evolution, so option (3) is the incorrect one.

Question 56

Conversion of milk to curd improves its nutritional value by increasing the amount of

Accepted Answer

(D) Vitamin B$_{12}$

Curd is more nourishing than milk.

It has an enriched presence of vitamins, especially vitamin B$_{12}$.

Question 57

Which of the following is not an autoimmune disease?

Accepted Answer

(D) Alzheimer's disease

Rheumatoid arthritis is an autoimmune disorder in which antibodies are produced against the synovial membrane and cartilage.

Vitiligo causes white patches on skin and is also characterised as an autoimmune disorder.

Psoriasis is a skin disease that causes itchy or sore patches of thick red skin and is also autoimmune.

Whereas Alzheimer's disease is due to deficiency of the neurotransmitter acetylcholine, so it is not autoimmune.

Question 58

The similarity of bone structure in the forelimbs of many vertebrates is an example of

Accepted Answer

(A) Homology

The forelimbs of many vertebrates have the same fundamental bone structure but perform different functions.

Such structures are homologous organs, so this is an example of homology, resulting from divergent evolution.

Question 59

Which of the following characteristics represent 'Inheritance of blood groups' in humans?

(a) Dominance

(b) Co-dominance

(c) Multiple allele

(d) Incomplete dominance

(e) Polygenic inheritance

Accepted Answer

(B) (a), (b) and (c)

I$^{A}$I$^{O}$ and I$^{B}$I$^{O}$ show a dominant-recessive relationship, i.e. dominance.

I$^{A}$I$^{B}$ shows codominance.

I$^{A}$, I$^{B}$ and I$^{O}$ are three different allelic forms of one gene, i.e. multiple allelism.

So (a), (b) and (c) are the characteristics shown.

Question 60

In which disease does mosquito transmitted pathogen cause chronic inflammation of lymphatic vessels?

Accepted Answer

(A) Elephantiasis

Elephantiasis is caused by the roundworm $Wuchereria$ $bancrofti$ and it is transmitted by the $Culex$ mosquito.

The worms cause chronic inflammation of the lymphatic vessels.

NEET (UG) 2018 — Code ZZ — Answer Key & Solutions

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