NEET (UG) 2021 — Code N1

Question 1

A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since

Accepted Answer

(D) All of the above

With a larger aperture of the objective lens, the light gathering power in the telescope is high.

Also, the resolving power or the ability to observe two objects distinctly also depends on the diameter of the objective. Thus an objective of large diameter is preferred.

Also, with large diameters fainter objects can be observed. Hence it also contributes to the better quality and visibility of images.

Hence, all options are correct.

Question 2

A thick current carrying cable of radius '$R$' carries current '$I$' uniformly distributed across its cross-section. The variation of magnetic field $B(r)$ due to the cable with the distance '$r$' from the axis of the cable is represented by

Accepted Answer

(B) A straight line rising to a peak at $r = R$, then falling away as a hyperbola

From Ampere's circuital law,

$$B = \frac{\mu_0 I}{2\pi R^{2}} \cdot r \qquad 	ext{if } r < R \ \Rightarrow\ B_{	ext{inside}} \propto r$$

$$B = \frac{\mu_0 I}{2\pi r} \qquad 	ext{if } r \geq R \ \Rightarrow\ B_{	ext{outside}} \propto \frac{1}{r}$$

Hence the correct plot of magnetic field $B$ with distance $r$ from the axis of the cable rises as a straight line up to $r = R$ and then falls away as a hyperbola.

Question 3

A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is

Accepted Answer

(C) 0.628 s

For a spring, $kx = F$.

Given $x = 5$ cm, $F = 10$ N:

$$\Rightarrow\ k(5 	imes 10^{-2}) = 10$$

$$\Rightarrow\ k = \frac{1000}{5} = 200\ 	ext{N/m}$$

Now, for a spring-mass system undergoing SHM,

$$T = 2\pi\sqrt{\frac{m}{k}}$$

Given $m = 2$ kg,

$$\Rightarrow\ T = 2\pi\sqrt{\frac{2}{200}} = \frac{2\pi}{10} = 0.628\ 	ext{s}$$

Question 4

Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine?

($g = 10$ m/s$^{2}$)

Accepted Answer

(A) 8.1 kW

Incident power on turbine

$$= \frac{d(mgh)}{dt} = gh\frac{dm}{dt}$$

$$= 10 	imes 60 	imes 15 = 9000\ 	ext{W}$$

Now, losses are 10%:

$$	herefore\ 	ext{power generated} = \left(1 - \frac{10}{100}ight) 	imes 9000$$

$$= 8100\ 	ext{W} = 8.1\ 	ext{kW}$$

Question 5

Polar molecules are the molecules

Accepted Answer

(C) Having a permanent electric dipole moment

In polar molecules, the centre of positive charges does not coincide with the centre of negative charges.

Hence, these molecules have a permanent electric dipole moment of their own.

Question 6

Consider the following statements (A) and (B) and identify the correct answer.

(A) A zener diode is connected in reverse bias, when used as a voltage regulator.

(B) The potential barrier of $p$-$n$ junction lies between 0.1 V to 0.3 V.

Accepted Answer

(B) (A) is correct and (B) is incorrect.

In reverse bias, after breakdown, the voltage across the zener diode becomes constant. Therefore a zener diode is connected in reverse bias when used as a voltage regulator.

The potential barrier of a silicon diode is nearly 0.7 V, not 0.1 V to 0.3 V.

So statement (A) is correct and statement (B) is incorrect.

Question 7

An inductor of inductance $L$, a capacitor of capacitance $C$ and a resistor of resistance '$R$' are connected in series to an ac source of potential difference '$V$' volts as shown in figure.

Potential difference across $L$, $C$ and $R$ is 40 V, 10 V and 40 V, respectively. The amplitude of curren

Accepted Answer

(C) $5\ \Omega$

$V_L = 40$ volt, $V_R = 40$ volt, $V_C = 10$ volt.

Now,

$$V_{RMS} = \sqrt{V_R^{2} + \left(V_L - V_Cight)^{2}}$$

$$= \sqrt{(40)^{2} + (40 - 10)^{2}} = 50\ 	ext{V}$$

$$I_{RMS} = \frac{I_0}{\sqrt{2}} = \frac{10\sqrt{2}}{\sqrt{2}} = 10\ 	ext{A}$$

Since $V_{RMS} = I_{RMS} 	imes Z$,

$$	herefore\ Z = \frac{V_{RMS}}{I_{RMS}} = \frac{50}{10} = 5\ \Omega$$

Question 8

The escape velocity from the Earth's surface is $v$. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is

Accepted Answer

(C) $4v$

Escape velocity from the Earth's surface

$$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2Gho\dfrac{4}{3}\pi R^{3}}{R}} = \sqrt{\frac{8Gho\pi}{3}R^{2}}$$

$$v_e \propto R \qquad 	ext{(for same density)}$$

$$\frac{v}{v_1} = \frac{R}{4R}$$

$$v_1 = 4v$$

Question 9

A screw gauge gives the following readings when used to measure the diameter of a wire

Main scale reading : 0 mm

Circular scale reading : 52 divisions

Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is

Accepted Answer

(C) 0.052 cm

Here, pitch of the screw gauge, $P = 1$ mm, and number of circular divisions, $n = 100$.

Thus least count

$$LC = \frac{P}{n} = \frac{1}{100} = 0.01\ 	ext{mm} = 0.001\ 	ext{cm}$$

So, diameter of the wire $= MSR + (CSR 	imes LC)$

$$= 0 + (52 	imes 0.001\ 	ext{cm}) = 0.052\ 	ext{cm}$$

Question 10

A cup of coffee cools from 90$^{\circ}$C to 80$^{\circ}$C in $t$ minutes, when the room temperature is 20$^{\circ}$C. The time taken by a similar cup of coffee to cool from 80$^{\circ}$C to 60$^{\circ}$C at a room temperature same at 20$^{\circ}$C is

Accepted Answer

(A) $\dfrac{13}{5}t$

From the average form of Newton's law of cooling,

$$-\left(\frac{T_1 + T_2}{2} - T_s\right)K = \frac{T_1 - T_2}{\Delta t}$$

where $T_1$ and $T_2$ are initial and final temperature and $T_s$ is the surrounding temperature.

$$\Rightarrow\ -K\left[\frac{(90 + 80)}{2} - 20\right] = \frac{90 - 80}{t}$$

$$\Rightarrow\ -K(65) = \frac{10}{t} \ \Rightarrow\ K = \frac{-2}{13t}$$

In the second case,

$$-K\left(\frac{80 + 60}{2} - 20\right) = \frac{(80 - 60)}{t_1}$$

$$\Rightarrow\ -K(50) = \frac{20}{t_1} \ \Rightarrow\ \frac{2}{13t}(50) = \frac{20}{t_1}$$

$$\Rightarrow\ t_1 = \frac{13t}{5}$$

Question 11

The half-life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be

Accepted Answer

(A) $\dfrac{1}{2\sqrt{2}}$

The activity of a radioactive substance is given as

$$A = A_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$$

Now,

$$\frac{A}{A_0} = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$$

$$\Rightarrow\ \frac{A}{A_0} = \left(\frac{1}{2}\right)^{\frac{150}{100}} = \left(\frac{1}{2}\right)^{\frac{3}{2}}$$

$$\Rightarrow\ \frac{A}{A_0} = \frac{1}{2\sqrt{2}}$$

Question 12

The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of $3.3 	imes 10^{-3}$ watt will be ($h = 6.6 	imes 10^{-34}$ J s)

Accepted Answer

(B) $10^{16}$

The power of a source is given as

$$P = \frac{E}{t} = \frac{n}{t}\left(\frac{hc}{\lambda}ight)$$

$$\Rightarrow\ \frac{n}{t} = \frac{P}{\left(\dfrac{hc}{\lambda}ight)}$$

(Here $\dfrac{n}{t}$ is the number of photons emitted per second)

$$\Rightarrow\ \frac{n}{t} = \frac{3.3 	imes 10^{-3} 	imes 6 	imes 10^{-7}}{6.6 	imes 10^{-34} 	imes 3 	imes 10^{8}}$$

$$= 10^{16}\ 	ext{photons per second}$$

Question 13

A parallel plate capacitor has a uniform electric field '$\vec{E}$' in the space between the plates. If the distance between the plates is '$d$' and the area of each plate is '$A$', the energy stored in the capacitor is

($\varepsilon_0$ = permittivity of free space)

Accepted Answer

(B) $\dfrac{1}{2}\varepsilon_0 E^{2} A d$

Energy density associated with the electric field is given by

$$u = \frac{dU}{dV} = \frac{1}{2}\varepsilon_0 E^{2}$$

$$\Rightarrow\ dU = \frac{1}{2}\varepsilon_0 E^{2}\,dV$$

Total energy stored in the space between the capacitor will be

$$U = \int dU = \int \frac{1}{2}\varepsilon_0 E^{2}\,dV$$

$$= \frac{1}{2}\varepsilon_0 E^{2}\int dV \qquad [E\ 	ext{is constant}]$$

$$= \frac{1}{2}\varepsilon_0 E^{2}V = \frac{1}{2}\varepsilon_0 E^{2}Ad \qquad [V = Ad]$$

Question 14

If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.

Accepted Answer

(A) $[	ext{F}][	ext{A}][	ext{T}^{2}]$

Energy, $E \propto 	ext{F}^{a}	ext{A}^{b}	ext{T}^{c}$

$$[E] = [	ext{F}^{a}][	ext{A}^{b}][	ext{T}^{c}]$$

$$\Rightarrow\ [	ext{ML}^{2}	ext{T}^{-2}] = [	ext{MLT}^{-2}]^{a}[	ext{LT}^{-2}]^{b}[	ext{T}]^{c}$$

$$[	ext{ML}^{2}	ext{T}^{-2}] = [	ext{M}^{a}	ext{L}^{a+b}	ext{T}^{-2a-2b+c}]$$

Comparing dimensions on both sides,

$$a = 1;\quad a + b = 2 \Rightarrow b = 1$$

$$-2 = -2a - 2b + c \Rightarrow -2 = -2 - 2 + c \Rightarrow c = 2$$

$$[E] = [	ext{FAT}^{2}]$$

Question 15

The equivalent capacitance of the combination shown in the figure is

Two capacitors of capacitance $C$ form the upper and lower branches between terminals $A$ and $B$, and a third capacitor $C$ is connected vertically between the right-hand ends of those two branches.

Accepted Answer

(A) $2C$

In the given circuit, the points at the right-hand ends of the two branches and the terminal $B$ are all at the same potential, as they are connected by a conducting wire.

So the third capacitor is short circuited. It does not store any charge.

The circuit can therefore be redrawn as the two remaining capacitors directly in parallel between $A$ and $B$:

$$C_{AB} = C + C = 2C \qquad 	ext{(Parallel combination)}$$

Question 16

A capacitor of capacitance '$C$', is connected across an ac source of voltage $V$, given by

$$V = V_0\sin\omega t$$

The displacement current between the plates of the capacitor, would then be given by

Accepted Answer

(D) $I_d = V_0\omega C\cos\omega t$

Given $V = V_0\sin\omega t$.

Now the displacement current $I_d$ is given by

$$I_d = C\frac{dV}{dt}$$

$$= C\frac{d}{dt}\left(V_0\sin\omega t\right)$$

$$= C\left(V_0\omega\right)\cos\omega t$$

$$I_d = V_0\omega C\cos\omega t$$

Question 17

In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?

Accepted Answer

(D) 60 cm

From the application of a potentiometer to compare two cells of emfs $E_1$ and $E_2$ by balancing lengths $\ell_1$ and $\ell_2$,

$$\frac{E_1}{E_2} = \frac{\ell_1}{\ell_2}$$

$$\Rightarrow\ \ell_2 = \ell_1\left(\frac{E_2}{E_1}ight) = (36\ 	ext{cm})\left(\frac{2.5\ 	ext{V}}{1.5\ 	ext{V}}ight)$$

$$= 60\ 	ext{cm}$$

Question 18

Column-I gives certain physical terms associated with flow of current through a metallic conductor.

Column-II gives some mathematical relations involving electrical quantities. Match Column-I and Column-II with appropriate relations.

Column-I:

(A) Drift Velocity

(B) Electrical Resistivity

(C) R

Accepted Answer

(D) (A)-(R), (B)-(S), (C)-(P), (D)-(Q)

Drift velocity, $v_d = \dfrac{eE	au}{m}$

Electrical resistivity, $ho = \dfrac{1}{\sigma} = \dfrac{E}{J}$

Relaxation period, $	au = \dfrac{m}{ne^{2}ho}$

Current density, $J = \dfrac{I}{A} = nev_d$

So (A)-(R), (B)-(S), (C)-(P), (D)-(Q).

Question 19

A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is

Accepted Answer

(C) 216 MeV

Mass number of reactant $= 240$, $BE$ per nucleon $= 7.6$ MeV.

Mass number of products $= 120$ each, $BE$ per nucleon of product $= 8.5$ MeV.

Total gain in $BE$ = ($BE$) of products $-$ ($BE$) of reactants

$$= [120 + 120] 	imes 8.5 - [240] 	imes 7.6$$

$$= (240) 	imes 8.5 - 240 	imes 7.6$$

$$= (2040 - 1824)\ 	ext{MeV}$$

Gain in $BE$ = 216 MeV

Question 20

Find the value of the angle of emergence from the prism. Refractive index of the glass is $\sqrt{3}$.

Accepted Answer

(D) 60$^{\circ}$

From the ray diagram shown in the figure, at point $P$ Snell's law gives

$$\sqrt{3}\sin 30^{\circ} = 1 	imes \sin e$$

$$\Rightarrow\ \sin e = \sqrt{3} 	imes \frac{1}{2} = \frac{\sqrt{3}}{2}$$

$$\Rightarrow\ e = 60^{\circ}$$

Question 21

If $E$ and $G$ respectively denote energy and gravitational constant, then $\dfrac{E}{G}$ has the dimensions of

Accepted Answer

(D) $[	ext{M}^{2}][	ext{L}^{-1}][	ext{T}^{0}]$

Dimensional formula of energy

$$[E] = [	ext{M}^{1}	ext{L}^{2}	ext{T}^{-2}] \qquad \ldots(	ext{I})$$

Dimensional formula of gravitational constant

$$[G] = [	ext{M}^{-1}	ext{L}^{3}	ext{T}^{-2}] \qquad \ldots(	ext{II})$$

From (I) and (II),

$$\frac{[E]}{[G]} = \frac{[	ext{M}^{1}	ext{L}^{2}	ext{T}^{-2}]}{[	ext{M}^{-1}	ext{L}^{3}	ext{T}^{-2}]} = [	ext{M}^{2}	ext{L}^{-1}	ext{T}^{0}]$$

Hence, dimensions of $\left[\dfrac{E}{G}ight] = [	ext{M}^{2}	ext{L}^{-1}	ext{T}^{0}]$.

Question 22

A small block slides down on a smooth inclined plane, starting from rest at time $t = 0$. Let $S_n$ be the distance travelled by the block in the interval $t = n - 1$ to $t = n$. Then, the ratio $\dfrac{S_n}{S_{n+1}}$ is

Accepted Answer

(A) $\dfrac{2n - 1}{2n + 1}$

Suppose $	heta$ is the inclination of the inclined plane, so the acceleration along the inclined plane is $a = g\sin	heta$.

$S_n$ = distance travelled by the object during the $n^{	ext{th}}$ second, with initial speed $u = 0$.

By the equation of uniformly accelerated motion,

$$S_n = u + \frac{a}{2}(2n - 1)$$

$$S_n = 0 + \frac{g\sin	heta}{2}(2n - 1) \qquad \ldots(i)$$

Distance travelled during the $(n + 1)^{	ext{th}}$ second,

$$S_{n+1} = 0 + \frac{g\sin	heta}{2}\left[2(n + 1) - 1ight] = \frac{g\sin	heta}{2}(2n + 1) \qquad \ldots(ii)$$

Dividing equations (i) and (ii),

$$\frac{S_n}{S_{n+1}} = \frac{(2n - 1)}{(2n + 1)}$$

Question 23

Match Column-I and Column-II and choose the correct match from the given choices.

Column-I:

(A) Root mean square speed of gas molecules

(B) Pressure exerted by ideal gas

(C) Average kinetic energy of a molecule

(D) Total internal energy of 1 mole of a diatomic gas

Column-II:

(P) $\dfrac{1}{3}

Accepted Answer

(B) (A)-(Q), (B)-(P), (C)-(S), (D)-(R)

Root mean square speed of a gas molecule $= \sqrt{\dfrac{3RT}{M}}$

Pressure exerted by an ideal gas $= \dfrac{1}{3}nm\bar{v}^{2}$

Average kinetic energy of a molecule $= \dfrac{3}{2}k_BT$

Total internal energy of a gas is $U = \dfrac{1}{2}nfRT$; here $n = 1$ and $f = 5$ for a diatomic gas, so $U = \dfrac{5}{2}RT$

Hence (A)-(Q), (B)-(P), (C)-(S), (D)-(R).

Question 24

A radioactive nucleus $^{A}_{Z}X$ undergoes spontaneous decay in the sequence

$$^{A}_{Z}X \rightarrow {}_{Z-1}B \rightarrow {}_{Z-3}C \rightarrow {}_{Z-2}D$$

where $Z$ is the atomic number of element $X$. The possible decay particles in the sequence are

Accepted Answer

(B) $\beta^{+}$, $\alpha$, $\beta^{-}$

On $\beta^{+}$ decay the atomic number decreases by 1.

On $\beta^{-}$ decay the atomic number increases by 1.

On $\alpha$ decay the atomic number decreases by 2.

So the sequence runs

$$^{A}_{Z}X \xrightarrow{\ \beta^{+}\ 	ext{decay}\ } {}_{Z-1}B \xrightarrow{\ \alpha\ 	ext{decay}\ } {}_{Z-3}C \xrightarrow{\ \beta^{-}\ 	ext{decay}\ } {}_{Z-2}D$$

Hence the correct order of decay is $\beta^{+}$, $\alpha$, $\beta^{-}$.

Question 25

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 $\Omega$. What will be the effective resistance if they are connected in series?

Accepted Answer

(C) 4 $\Omega$

All the wires are identical and of the same material so they will have the same value of resistance. Let it be $R$.

When these four are connected in parallel,

$$R_P = \frac{R}{4} \qquad \left(\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}ight)$$

Given $R_P = 0.25\ \Omega$:

$$	herefore\ 0.25 = \frac{R}{4} \ \Rightarrow\ R = 1\ \Omega$$

Now these four resistances are arranged in series:

$$R_S = R + R + R + R = 4R$$

$$	herefore\ R_S = 4 	imes 1 = 4\ \Omega$$

Question 26

A convex lens '$A$' of focal length 20 cm and a concave lens '$B$' of focal length 5 cm are kept along the same axis with a distance '$d$' between them. If a parallel beam of light falling on '$A$' leaves '$B$' as a parallel beam, then the distance '$d$' in cm will be

Accepted Answer

(A) 15

A parallel beam entering the convex lens converges towards its focus, a distance $f_A$ away.

For the beam to leave the concave lens parallel again, that convergence point must coincide with the focus of the concave lens, which lies a distance $f_B$ beyond it.

Hence

$$d = f_A - f_B = 20 - 5 = 15\ 	ext{cm}$$

Question 27

A particle is released from height $S$ from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively

Accepted Answer

(C) $\dfrac{S}{4}$, $\sqrt{\dfrac{3gS}{2}}$

Let the required height of the body be $y$.

When the body falls from rest through height $(S - y)$, then under constant acceleration

$$v^{2} = 0^{2} + 2g(S - y)$$

$$v = \sqrt{2g(S - y)} \qquad \ldots(1)$$

When the body is at height $y$ above ground, the potential energy of a body of mass $m$ is

$$U = mgy$$

As per the given condition, kinetic energy $K = 3U$:

$$\frac{1}{2}m v^{2} = 3 	imes mgy$$

$$\frac{1}{2} 	imes m 	imes 2g(S - y) = 3 	imes mgy \qquad 	ext{(using (1))}$$

$$S - y = 3y$$

$$	herefore\ y = \frac{S}{4} \qquad \ldots(2)$$

$$	herefore\ v = \sqrt{2 	imes g\left(S - \frac{S}{4}ight)} = \sqrt{\frac{3gS}{2}} \qquad \ldots(3)$$

Question 28

A dipole is placed in an electric field as shown. In which direction will it move?

Accepted Answer

(A) Towards the right as its potential energy will decrease.

Potential energy of an electric dipole in an external electric field is $U = -\vec{P} \cdot \vec{E}$.

The angle between the electric field and the electric dipole moment is $180^{\circ}$:

$$U = -PE\cos	heta = -PE\cos 180^{\circ} = +PE$$

On moving towards the right the electric field strength decreases, therefore the potential energy decreases.

The net force on the electric dipole is towards the right and the net torque acting on it is zero.

So it will move towards the right.

Question 29

Two charged spherical conductors of radius $R_1$ and $R_2$ are connected by a wire. Then the ratio of surface charge densities of the spheres ($\sigma_1/\sigma_2$) is

Accepted Answer

(A) $\dfrac{R_2}{R_1}$

When two conductors are connected by a conducting wire, then the two conductors should have the same potential.

So $V_1 = V_2$:

$$\frac{1}{4\pi\varepsilon_0}\frac{Q_1}{R_1} = \frac{1}{4\pi\varepsilon_0}\frac{Q_2}{R_2}$$

Multiplying each side by $\dfrac{R}{R}$ to bring in the surface area,

$$\Rightarrow\ \frac{Q_1R_1}{4\pi R_1^{2}\varepsilon_0} = \frac{Q_2R_2}{4\pi R_2^{2}\varepsilon_0}$$

$$\Rightarrow\ \frac{\sigma_1 R_1}{\varepsilon_0} = \frac{\sigma_2 R_2}{\varepsilon_0}$$

$$\Rightarrow\ \frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$$

Question 30

A body is executing simple harmonic motion with frequency '$n$', the frequency of its potential energy is

Accepted Answer

(A) $2n$

The equation of displacement of a particle executing SHM is given by $x = A\sin(\omega t + \phi)$.

Potential energy of a particle executing SHM is given by

$$U = \frac{1}{2}kx^{2} = \frac{1}{2}kA^{2}\sin^{2}(\omega t + \phi)$$

The time period of $x = A\sin(\omega t + \phi)$ is

$$T_1 = \frac{2\pi}{\omega} \ \Rightarrow\ 	ext{frequency } n_1 = \frac{\omega}{2\pi}$$

while the time period of $x^{2} = A^{2}\sin^{2}(\omega t + \phi)$ is

$$T_2 = \frac{\pi}{\omega} \ \Rightarrow\ 	ext{frequency } n_2 = \frac{\omega}{\pi}$$

Hence $n_2 = 2n_1$.

Question 31

The velocity of a small ball of mass $M$ and density $d$, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $\dfrac{d}{2}$, then the viscous force acting on the ball will be

Accepted Answer

(D) $\dfrac{Mg}{2}$

Let $F_v$ be the viscous force and $F_B$ be the buoyant force acting on the ball.

When the body moves with constant velocity, the acceleration is zero:

$$Mg = F_B + F_v$$

$$F_v = Mg - F_B$$

$$= dVg - \frac{d}{2}Vg \qquad (M = dVg,\ V = 	ext{volume of ball})$$

$$= \frac{d}{2}Vg$$

$$F_v = \frac{M}{2}g$$

Question 32

An electromagnetic wave of wavelength '$\lambda$' is incident on a photosensitive surface of negligible work function. If '$m$' mass is of photoelectron emitted from the surface has de-Broglie wavelength $\lambda_d$, then

Accepted Answer

(B) $\lambda = \left(\dfrac{2mc}{h}\right)\lambda_d^{2}$

For a surface of negligible work function, the whole photon energy becomes the kinetic energy of the photoelectron:

$$\frac{hc}{\lambda} = \frac{p^{2}}{2m}$$

The de-Broglie wavelength gives $p = \dfrac{h}{\lambda_d}$, so

$$\frac{hc}{\lambda} = \frac{h^{2}}{2m\lambda_d^{2}}$$

$$\Rightarrow\ \lambda = \frac{2mc\lambda_d^{2}}{h} = \left(\frac{2mc}{h}\right)\lambda_d^{2}$$

Question 33

For a plane electromagnetic wave propagating in $x$-direction, which one of the following combination gives the correct possible directions for electric field ($E$) and magnetic field ($B$) respectively?

Accepted Answer

(A) $-\hat{j} + \hat{k},\ -\hat{j} - \hat{k}$

The direction of propagation of an electromagnetic wave is along $\vec{E} 	imes \vec{B}$.

Given that the direction of propagation is along the $x$-axis, testing each option:

$$(1)\quad \left(-\hat{j} + \hat{k}ight) 	imes \left(-\hat{j} - \hat{k}ight) = 2\hat{i}$$

$$(2)\quad \left(\hat{j} + \hat{k}ight) 	imes \left[-\left(\hat{j} + \hat{k}ight)ight] = 0$$

$$(3)\quad \left(-\hat{j} + \hat{k}ight) 	imes \left(-\hat{j} + \hat{k}ight) = 0$$

$$(4)\quad \left(\hat{j} + \hat{k}ight) 	imes \left(\hat{j} + \hat{k}ight) = 0$$

Only option (1) gives a cross product along $+\hat{i}$.

Question 34

An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of $10^{5}$ m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the ele

Accepted Answer

(C) $8 	imes 10^{-20}$ N

Magnetic field produced due to the current carrying wire at the position of the electron:

$$B = \frac{\mu_0}{4\pi}\frac{2I}{r}$$

$$B = \frac{10^{-7} 	imes 2 	imes 5}{20 	imes 10^{-2}} = \frac{1}{2} 	imes 10^{-5}\ 	ext{(Tesla)}$$

directed upward to the plane of the paper.

Now, the force acting on the electron due to this field is

$$\vec{F} = q\left(\vec{v} 	imes \vec{B}ight)$$

$$\left|\vec{F}ight| = 1.6 	imes 10^{-19} 	imes 10^{5} 	imes \frac{1}{2} 	imes 10^{-5}$$

$$= 0.8 	imes 10^{-19}\ 	ext{N}$$

$$\left|\vec{F}ight| = 8 	imes 10^{-20}\ 	ext{N}$$

Question 35

The electron concentration in an $n$-type semiconductor is the same as hole concentration in a $p$-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.

Accepted Answer

(B) Current in $n$-type > current in $p$-type.

The current through a semiconductor is

$$I = neAv_d = neA\mu E$$

$$\frac{I_n}{I_p} = \frac{n_e e A \mu_e E}{n_h e A \mu_h E}$$

Since the concentrations are equal,

$$\frac{I_n}{I_p} = \frac{\mu_e}{\mu_h}$$

The mobility of electrons is greater than that of holes, $\mu_e > \mu_h$:

$$\Rightarrow\ I_n > I_p$$

Question 36

A particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle '$	heta$' to the horizontal, the maximum height attained by it equals $4R$. The angle of projection, $	heta$, is then given

Accepted Answer

(C) $	heta = \sin^{-1}\left(\dfrac{2gT^{2}}{\pi^{2}R}ight)^{\frac{1}{2}}$

To complete a circular path of radius $R$, the time period is $T$, so the speed of the particle is

$$U = \frac{2\pi R}{T} \qquad \ldots(1)$$

Now the particle is projected with the same speed at angle $	heta$ to the horizontal, so the maximum height is

$$H = \frac{U^{2}\sin^{2}	heta}{2g}$$

Given $H = 4R$:

$$\Rightarrow\ \frac{U^{2}\sin^{2}	heta}{2g} = 4R$$

$$\Rightarrow\ \sin^{2}	heta = \frac{8gR}{U^{2}} \qquad \ldots(2)$$

$$\Rightarrow\ \sin^{2}	heta = \frac{8gRT^{2}}{4\pi^{2}R^{2}} = \frac{2gT^{2}}{\pi^{2}R} \qquad 	ext{(using equation 1)}$$

$$\Rightarrow\ 	heta = \sin^{-1}\left(\frac{2gT^{2}}{\pi^{2}R}ight)^{\frac{1}{2}}$$

Question 37

Three resistors having resistances $r_1$, $r_2$ and $r_3$ are connected as shown in the given circuit — $r_1$ carries the whole current $i_1$ from $A$ and then splits into $r_2$ carrying $i_2$ and $r_3$ carrying $i_3$ in parallel, rejoining at $B$. The ratio $\dfrac{i_3}{i_1}$ of currents in terms o

Accepted Answer

(A) $\dfrac{r_2}{r_2 + r_3}$

In the parallel combination of resistances $r_2$ and $r_3$, the potential difference will be equal across both resistances.

So, $i_2r_2 = i_3r_3$:

$$\Rightarrow\ i_2 = \frac{i_3r_3}{r_2} \qquad \ldots(1)$$

As per Kirchhoff's first law,

$$\Rightarrow\ i_1 = i_2 + i_3$$

$$\Rightarrow\ i_1 = \left(\frac{r_3}{r_2} + 1ight)i_3 \qquad 	ext{(from equation 1)}$$

$$\Rightarrow\ \frac{i_3}{i_1} = \frac{r_2}{r_2 + r_3}$$

Question 38

A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of

Accepted Answer

(C) 20 cm from the plane mirror, it would be a virtual image

Using the lens formula for the first refraction from the convex lens,

$$\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f}$$

with $u = -60$ cm and $f = 30$ cm:

$$\Rightarrow\ \frac{1}{v_1} + \frac{1}{60} = \frac{1}{30} \ \Rightarrow\ v_1 = 60\ 	ext{cm}$$

This first image $I_1$ lies 60 cm beyond the lens, i.e. 20 cm behind the plane mirror.

The plane mirror will produce an image at a distance 20 cm to the left of it.

For the second refraction from the convex lens, $u = -20$ cm and $f = 30$ cm:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \ \Rightarrow\ \frac{1}{v} + \frac{1}{20} = \frac{1}{30}$$

$$\Rightarrow\ \frac{1}{v} = \frac{1}{30} - \frac{1}{20} \ \Rightarrow\ v = -60\ 	ext{cm}$$

Thus the final image is virtual and at a distance $60 - 40 = 20$ cm from the plane mirror.

Question 39

A uniform conducting wire of length $12a$ and resistance '$R$' is wound up as a current carrying coil in the shape of,

(i) an equilateral triangle of side '$a$'.

(ii) a square of side '$a$'.

The magnetic dipole moments of the coil in each case respectively are

Accepted Answer

(D) $\sqrt{3}Ia^{2}$ and $3Ia^{2}$

The current in the loop will be $\dfrac{V}{R} = I$, which is the same for both loops.

Now the magnetic moment of the triangular loop $= NIA$, where the wire of length $12a$ makes $\dfrac{12a}{3a}$ turns:

$$M_1 = \left(\frac{12a}{3a}\right) \cdot I \cdot \frac{\sqrt{3}}{4}a^{2} = \sqrt{3}Ia^{2}$$

And the magnetic moment of the square loop $= N'IA'$, where the wire makes $\dfrac{12a}{4a}$ turns:

$$M_2 = \left(\frac{12a}{4a}\right) \cdot I \cdot a^{2} = 3Ia^{2}$$

Question 40

For the given circuit, the input digital signals are applied at the terminals $A$, $B$ and $C$. What would be the output at the terminal $y$?

Accepted Answer

(C) A constant 5 V level throughout

The output of the combination of logic gates is given as

$$y = A \cdot B + \overline{B \cdot C}$$

Working through each time interval with the printed input waveforms:

$0$-$t_1$: $A$=0, $B$=0, $C$=1, so $AB$=0, $\overline{B \cdot C}$=1, $y$=1

$t_1$-$t_2$: $A$=1, $B$=0, $C$=1, so $AB$=0, $\overline{B \cdot C}$=1, $y$=1

$t_2$-$t_3$: $A$=0, $B$=1, $C$=0, so $AB$=0, $\overline{B \cdot C}$=1, $y$=1

$t_3$-$t_4$: $A$=1, $B$=1, $C$=0, so $AB$=1, $\overline{B \cdot C}$=1, $y$=1

$t_4$-$t_5$: $A$=0, $B$=0, $C$=1, so $y$=1

$t_5$-$t_6$: $A$=1, $B$=0, $C$=1, so $y$=1

$t_6$-$t_7$: $A$=0, $B$=0, $C$=1, so $y$=1

So the output $y$ is high (1) throughout, that is $v_0 = 5$ V.

Question 41

A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is ($g = 10$ m/s$^{2}$) nearly

Accepted Answer

(A) 4.2 kg m/s

Given, mass of ball $= 0.15$ kg and height from which the ball is dropped $= 10$ m.

Impulse, $\vec{I}$ = change in linear momentum $= \Delta\vec{P} = \vec{P}_f - \vec{P}_i$.

Velocity of ball at ground,

$$v = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 10} = 10\sqrt{2}\ 	ext{m/s}$$

$$\vec{I} = 0.15 	imes 10\sqrt{2}\left(-\hat{j}ight) - 0.15 	imes 10\sqrt{2}\left(\hat{j}ight)$$

$$\vec{I} = 2 	imes 0.15 	imes 10\sqrt{2}\left(-\hat{j}ight) = 4.2\left(-\hat{j}ight)$$

$$\Rightarrow\ 	ext{magnitude of impulse} = 4.2\ 	ext{kg m/s}$$

Question 42

A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?

Accepted Answer

(D) 0.2 A

In an ideal transformer, input power = output power:

$$\Rightarrow\ V_P I_P = V_S I_S = 	ext{Given power}$$

$$\Rightarrow\ 220 	imes I_P = 44$$

$$\Rightarrow\ I_P = 0.2\ 	ext{A}$$

Question 43

A series LCR circuit containing 5.0 H inductor, 80 $\mu$F capacitor and 40 $\Omega$ resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be

Accepted Answer

(B) 46 rad/s and 54 rad/s

The resonance frequency of a series LCR circuit is given as

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{5 	imes 80 	imes 10^{-6}}} = 50\ 	ext{rad/s}$$

Now the half power frequencies are given as

$$\omega = \omega_0 \pm \frac{R}{2L}$$

i.e.

$$\omega_L = 50 - \frac{40}{2 	imes 5} = 46\ 	ext{rad/s}$$

$$\omega_H = 50 + \frac{40}{2 	imes 5} = 54\ 	ext{rad/s}$$

Question 44

A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass '$m$' is suspended from the rod at 160 cm mark as shown in the figure. Find the value of '$m$' such that the rod is in equilibrium. ($

Accepted Answer

(C) $\dfrac{1}{12}$ kg

The rod is uniform, so its whole weight acts at its centre, the 100 cm mark.

Taking torques about the wedge at the 40 cm mark, the 2 kg mass sits 20 cm to its left while the rod's weight and the mass $m$ sit 60 cm and 120 cm to its right.

For equilibrium the anticlockwise torque equals the clockwise torque:

$$2 	imes 0.20 = 0.5 	imes 0.60 + m 	imes 1.20$$

$$0.4 = 0.3 + 1.2\,m$$

$$1.2\,m = 0.1$$

$$m = \frac{1}{12}\ 	ext{kg}$$

Question 45

A car starts from rest and accelerates at 5 m/s$^{2}$. At $t = 4$ s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t = 6$ s?

(Take $g = 10$ m/s$^{2}$)

Accepted Answer

(C) $20\sqrt{2}$ m/s, 10 m/s$^{2}$

Initial velocity of car $= 0$, acceleration of car $= 5$ m/s$^{2}$.

Velocity of car at $t = 4$ s: $v = u + at$

$$\Rightarrow\ v = 0 + 5 	imes 4 = 20\ 	ext{ms}^{-1}$$

At $t = 4$ s, a ball is dropped out of a window, so the velocity of the ball at this instant is 20 ms$^{-1}$ along the horizontal.

After 2 seconds of motion:

Horizontal velocity of ball $= 20$ ms$^{-1}$ (since $a_x = 0$)

Vertical velocity of ball, $v_y = u_y + a_y t = 0 + 10 	imes 2 = 20$ ms$^{-1}$ (since $a_y = g = 10$ m/s$^{2}$)

So the magnitude of velocity of the ball is

$$v = \sqrt{v_x^{2} + v_y^{2}} = 20\sqrt{2}\ 	ext{m/s}$$

Acceleration of the ball at $t = 6$ s is $g = 10$ m/s$^{2}$, as the ball is under free fall.

Question 46

Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

Accepted Answer

Electric potential due to a charged sphere $= \dfrac{kQ}{R}$, where $k = 9 	imes 10^{9}$ N m$^{2}$/C$^{2}$, $Q$ is the charge on the sphere and $R$ its radius.

Let the charge and radius of the smaller drop be $q$ and $r$ respectively. For the smaller drop,

$$V = \frac{kq}{r} = 220\ 	ext{V}$$

Let $R$ be the radius of the bigger drop. As the volume remains the same,

$$\left(\frac{4}{3}\pi r^{3}ight) 	imes 27 = \frac{4}{3}\pi R^{3}$$

$$\Rightarrow\ R = \sqrt[3]{27}\,r = 3r$$

Now, using charge conservation, $Q = 27q$:

$$V_{	ext{big drop}} = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9\left(\frac{kq}{r}ight)$$

$$= 9 	imes 220 = 1980\ 	ext{V}$$

Question 47

Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1 \gg R_2$, the mutual inductance $M$ between them will be directly proportional to

Accepted Answer

(C) $\dfrac{R_2^{2}}{R_1}$

Two concentric coils are of radius $R_1$ and $R_2$. Let the current in the outer loop be $i$.

Magnetic field at the centre

$$B = \frac{\mu_0 i}{2R_1}$$

Since $R_1 \gg R_2$, this field is effectively uniform over the inner coil, so the magnetic flux through the inner coil is

$$\phi = B 	imes \pi R_2^{2} = \frac{\mu_0 i}{2R_1} 	imes \pi R_2^{2} = \frac{\mu_0 i}{2} 	imes \frac{\pi R_2^{2}}{R_1}$$

As per definition, $\phi = Mi$:

$$\Rightarrow\ M = \left(\frac{\mu_0\pi}{2}ight)\frac{R_2^{2}}{R_1}$$

$$	herefore\ M \propto \frac{R_2^{2}}{R_1}$$

Question 48

A particle of mass '$m$' is projected with a velocity $v = kV_e$ ($k < 1$) from the surface of the earth.

($V_e$ = escape velocity)

The maximum height above the surface reached by the particle is

Accepted Answer

(C) $\dfrac{Rk^{2}}{1 - k^{2}}$

Given $v = kV_e$ where $k < 1$, thus $v < V_e$.

From conservation of mechanical energy,

$$\frac{1}{2}mv^{2} - \frac{GmM}{R} = -\frac{GmM}{(R + h)}$$

$$\Rightarrow\ \frac{v^{2}}{2} = \frac{GM}{R} - \frac{GM}{(R + h)} = \frac{h}{R(R + h)}GM$$

$$\Rightarrow\ \frac{1}{2}k^{2}V_e^{2} = \frac{GMh}{R(R + h)}$$

We know $V_e = \sqrt{\dfrac{2GM}{R}}$:

$$\Rightarrow\ \frac{1}{2}k^{2}\left(\frac{2GM}{R}\right) = \frac{GMh}{R(R + h)}$$

$$\Rightarrow\ k^{2}(R + h) = h$$

$$\Rightarrow\ h\left(1 - k^{2}\right) = k^{2}R$$

$$\Rightarrow\ h = \frac{Rk^{2}}{1 - k^{2}}$$

Question 49

In the product

$$\vec{F} = q\left(\vec{v} 	imes \vec{B}ight) = q\vec{v} 	imes \left(B\hat{i} + B\hat{j} + B_0\hat{k}ight)$$

For $q = 1$ and $\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}$ and $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}$

What will be the complete expression for $\vec{B}$ ?

Accepted Answer

(A) $-6\hat{i} - 6\hat{j} - 8\hat{k}$

Given $q = 1$, $\vec{v} = 2\hat{i} + 4\hat{j} + 6\hat{k}$ and $\vec{F} = 4\hat{i} - 20\hat{j} + 12\hat{k}$.

Expanding the cross product,

$$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & 6 \ B & B & B_0 \end{vmatrix}$$

$$\Rightarrow\ \hat{i}(4B_0 - 6B) - \hat{j}(2B_0 - 6B) + \hat{k}(2B - 4B)$$

Comparing L.H.S and R.H.S,

$$4B_0 - 6B = 4 \ \Rightarrow\ 2B_0 - 3B = 2 \qquad \ldots(1)$$

$$-(2B_0 - 6B) = -20 \ \Rightarrow\ B_0 - 3B = 10 \qquad \ldots(2)$$

$$2B - 4B = 12 \ \Rightarrow\ B = -6 \qquad \ldots(3)$$

From (2) and (3), $B = -6$ and $B_0 = -8$.

Hence, $\vec{B} = -6\hat{i} - 6\hat{j} - 8\hat{k}$.

Question 50

From a circular ring of mass '$M$' and radius '$R$' an arc corresponding to a 90$^{\circ}$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is '$K$' times '$MR^{2}$'. Then the va

Accepted Answer

(D) $\dfrac{3}{4}$

Given, mass of ring $= M$, radius of ring $= R$.

Now a 90$^{\circ}$ arc is removed from the circular ring, so the mass removed $= \dfrac{M}{4}$.

Mass of remaining portion $= \dfrac{3M}{4}$.

Moment of inertia of the remaining part

$$= \int dm\,r^{2}$$

$$\Rightarrow\ I = R^{2}\int dm \qquad (\because r = R)$$

$$\Rightarrow\ I = \frac{3MR^{2}}{4}$$

So the value of $K$ is $\dfrac{3}{4}$.

Question 51

Choose the correct option for graphical representation of Boyle's law, which shows a graph of pressure vs. volume of a gas at different temperatures :

Accepted Answer

(C) Three rectangular hyperbolas, the 600 K curve furthest from the origin and the 200 K curve nearest

According to Boyle's law,

$$P \propto \frac{1}{V} \ \Rightarrow\ P = \frac{k}{V} \ \Rightarrow\ PV = k$$

where $k$ is a proportionality constant and equal to $nRT$.

$$	herefore\ 	ext{Graph between } P 	ext{ vs. } V 	ext{ should be a rectangular hyperbola, and the product } PV 	ext{ increases with increase in temperature.}$$

So the isotherm for the highest temperature, 600 K, lies furthest from the origin.

Question 52

Statement I : Acid strength increases in the order given as HF $\ll$ HCl $\ll$ HBr $\ll$ HI.

Statement II : As the size of the elements F, Cl, Br, I increases down the group, the bond strength of HF, HCl, HBr and HI decreases and so the acid strength increases.

In the light of the above statements

Accepted Answer

(D) Both statement I and Statement II are true

In the modern periodic table, moving down the group as the size of the halogen atom increases, the H–X bond length also increases; as a result the bond enthalpy decreases. Hence, the acidic strength also increases.

So, the correct order of acidic strength is

$$	ext{HI} > 	ext{HBr} > 	ext{HCl} > 	ext{HF}$$

Both statements are therefore true, and the second correctly explains the first.

Question 53

Tritium, a radioactive isotope of hydrogen, emits which of the following particles?

Accepted Answer

(D) Beta ($\beta^{-}$)

Hydrogen has three isotopes: protium $^{1}_{1}$H, deuterium $^{2}_{1}$H or D, and tritium $^{3}_{1}$H or T.

Of these isotopes, only tritium is radioactive and emits low energy $\beta^{-}$ particles ($t_{1/2}$ = 12.33 years).

Question 54

The maximum temperature that can be achieved in blast furnace is :

Accepted Answer

(A) Upto 2200 K

The maximum temperature that can be achieved in a blast furnace is up to 2200 K.

(As per NCERT text, 2170 K maximum temperature is given in the figure of the blast furnace.)

Question 55

Noble gases are named because of their inertness towards reactivity. Identify an incorrect statement about them.

Accepted Answer

(A) Noble gases have very high melting and boiling points

Noble gases have weak dispersion forces, hence they have LOW melting and boiling points.

So the statement that noble gases have very high melting and boiling points is incorrect.

The other three statements are correct.

Question 56

Identify the compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali.

Accepted Answer

(B) CH$_3$CH$_2$–NH$_2$ (a primary amine)

Benzenesulphonyl chloride (C$_6$H$_5$SO$_2$Cl) is also known as Hinsberg's reagent.

The reaction of Hinsberg's reagent with a primary amine (CH$_3$CH$_2$NH$_2$) yields N-ethylbenzene sulphonamide, which still carries a hydrogen on nitrogen. That hydrogen is acidic because of the strongly electron-withdrawing sulphonyl group, so the solid dissolves in alkali.

The reaction of Hinsberg's reagent with a secondary amine (C$_2$H$_5$NHCH$_3$) gives N-ethyl-N-methyl benzene sulphonamide, which is insoluble in alkali due to the absence of an H-atom on nitrogen.

A 3$^{\circ}$ amine does not react with Hinsberg's reagent at all.

Question 57

Among the following alkaline earth metal halides, one which is covalent and soluble in organic solvents is :

Accepted Answer

(C) Beryllium chloride

Except for beryllium chloride, all the other chlorides of alkaline earth metals are ionic in nature.

Due to the small size of Be, beryllium chloride is essentially covalent and soluble in organic solvents.

Question 58

What is the IUPAC name of the organic compound formed in the following chemical reaction?

Acetone treated with (i) C$_2$H$_5$MgBr, dry ether; (ii) H$_2$O, H$^{+}$ gives the product.

Accepted Answer

(C) 2-methylbutan-2-ol

The ethyl carbanion of the Grignard reagent attacks the carbonyl carbon of acetone, giving the bromomagnesium alkoxide:

$$	ext{CH}_3	ext{—CO—CH}_3 \xrightarrow[	ext{Dry ether}]{	ext{C}_2	ext{H}_5	ext{MgBr}} 	ext{CH}_3	ext{—C(OMgBr)(C}_2	ext{H}_5	ext{)—CH}_3$$

On hydrolysis with H$_2$O/H$^{+}$ this becomes the tertiary alcohol:

$$	ext{CH}_3	ext{—C(OH)(CH}_3	ext{)—CH}_2	ext{—CH}_3$$

The product is 2-methylbutan-2-ol.

Question 59

BF$_3$ is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are :

Accepted Answer

(B) $sp^{2}$ and 6

In BF$_3$, boron forms three B–F bonds and has no lone pair.

Number of electrons around the boron atom is 6.

Hybridization of B is $sp^{2}$.

Shape is trigonal planar.

Question 60

The structures of beryllium chloride in solid state and vapour phase, are :

Accepted Answer

(D) Chain and dimer, respectively

Beryllium chloride has a chain structure in the solid state, with chlorine bridges linking the beryllium atoms.

In the vapour phase, beryllium chloride tends to form a chloro-bridged dimer, Be$_2$Cl$_4$, which dissociates into the linear monomer at high temperature.

So the answer is chain and dimer respectively.

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