Q: The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be

Elastic potential energy of a spring depends on the square of the extension:U = \tfrac{1}{2}kx^2 \;\Rightarrow\; U \propto x^2The extension goes from 2 cm to 8 cm — a factor of 4. Since energy scales as the square:\frac{U'}{U} = \left(\frac{x'}{x}\right)^2 = \left(\frac{8}{2}\right)^2 = 4^2 = 16U' = 16\,UThe trap is answering 4U by assuming energy scales linearly with extension.

Q: A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output?

Each component has a distinct job in the rectifier:• The centre-tapped transformer supplies two equal, opposite-phase voltages to the two diodes.• The diodes conduct on alternate half cycles, converting ac into a pulsating dc — this is rectification, not smoothing.• The load resistance is simply where the output is taken.• The capacitor is the filter. It charges up when the rectified voltage rises and discharges slowly through the load when the voltage falls, filling in the dips between pulses. This flattens the pulsating output and removes the ac ripple.So the capacitor is the answer. A large

Question 1

Let a wire be suspended from the ceiling (rigid support) and stretched by a weight W attached at its free end. The longitudinal stress at any point of cross-sectional area A of the wire is

Accepted Answer

Stress is defined as the internal restoring force set up per unit area of cross-section.

The wire is in equilibrium, so the tension at every point equals the weight hanging at the free end:
$T = W$

This tension acts across the cross-sectional area $A$, and since the force is along the length of the wire, the stress is longitudinal:
$$\text{Longitudinal stress} = \frac{F}{A} = \frac{W}{A}$$
Note that the stress does not double — the ceiling's reaction and the weight form an action–reaction pair across the same section, not two separate loads.

Question 2

The ratio of radius of gyration of a solid sphere of mass M and radius R about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is

Accepted Answer

Radius of gyration k is defined through I = Mk^2, so k = \sqrt{I/M}.Solid sphere about a diameter: I = \dfrac{2}{5}MR^2k_1 = \sqrt{\frac{2}{5}}\,RThin hollow sphere about a diameter: I = \dfrac{2}{3}MR^2k_2 = \sqrt{\frac{2}{3}}\,RTaking the ratio:\frac{k_1}{k_2} = \sqrt{\frac{2}{5}} 	imes \sqrt{\frac{3}{2}} = \sqrt{\frac{3}{5}}This value, \sqrt{3/5}, does not match any of the four options. Option (1) reads 3 : 5, which is the ratio inside the square root — the square root was not taken. This question was treated as a bonus.

Question 3

A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is

Accepted Answer

Force has the same direction as the change in momentum, not the direction of motion:\vec{F} = \frac{\Delta \vec{p}}{\Delta t}, \qquad \Delta\vec{p} = \vec{p}_f - \vec{p}_iTake east as \hat{i} and north as \hat{j}. The player moves south initially and east finally, both with speed u:\vec{p}_i = -mu\,\hat{j}    (southward)\vec{p}_f = mu\,\hat{i}    (eastward)\Delta\vec{p} = mu\,\hat{i} - (-mu\,\hat{j}) = mu(\hat{i} + \hat{j})The vector \hat{i}+\hat{j} has equal components along east and north, so it points north-east. The force is therefore along north-east.A common mistake here is answering "ea

Question 4

If \oint_s \vec{E} \cdot \overrightarrow{dS} = 0 over a surface, then

Accepted Answer

The integral \oint_s \vec{E}\cdot\overrightarrow{dS} is the net electric flux through the closed surface. Being zero means:\phi_{	ext{net}} = \phi_{	ext{out}} - \phi_{	ext{in}} = 0 \;\Rightarrow\; \phi_{	ext{out}} = \phi_{	ext{in}}So exactly as many field lines leave the surface as enter it. That is option (1).Why the others fail:• Zero net flux says nothing about the magnitude of \vec{E} at each point — the field can vary freely over the surface.• By Gauss's law \phi = q_{	ext{enclosed}}/\varepsilon_0, zero flux means the net enclosed charge is zero. Equal positive and negative charges

Question 5

The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm, potential energy stored in it will be

Accepted Answer

Elastic potential energy of a spring depends on the square of the extension:
$$U = \tfrac{1}{2}kx^2 \;\Rightarrow\; U \propto x^2$$
The extension goes from 2 cm to 8 cm — a factor of 4. Since energy scales as the square:
$$\frac{U'}{U} = \left(\frac{x'}{x}\right)^2 = \left(\frac{8}{2}\right)^2 = 4^2 = 16$$
$$U' = 16\,U$$
The trap is answering 4U by assuming energy scales linearly with extension.

Question 6

A 12 V, 60 W lamp is connected to the secondary of a step-down transformer, whose primary is connected to ac mains of 220 V. Assuming the transformer to be ideal, what is the current in the primary winding?

Accepted Answer

For an ideal transformer there are no losses, so power in equals power out:
$$P_{\text{primary}} = P_{\text{secondary}}$$
The lamp consumes 60 W, so the primary must draw 60 W as well:
$$V_p I_p = 60$$
$$220 \times I_p = 60$$
$$I_p = \frac{60}{220} \approx 0.27\ \text{A}$$
Note that the 12 V rating is not needed for the primary current — it only tells you the transformer steps down. The secondary current would be $60/12 = 5$ A, which is much larger, exactly as expected for a step-down transformer.

Question 7

A full wave rectifier circuit consists of two p-n junction diodes, a centre-tapped transformer, capacitor and a load resistance. Which of these components remove the ac ripple from the rectified output?

Accepted Answer

Each component has a distinct job in the rectifier:
• The centre-tapped transformer supplies two equal, opposite-phase voltages to the two diodes.
• The diodes conduct on alternate half cycles, converting ac into a pulsating dc — this is rectification, not smoothing.
• The load resistance is simply where the output is taken.
• The capacitor is the filter. It charges up when the rectified voltage rises and discharges slowly through the load when the voltage falls, filling in the dips between pulses. This flattens the pulsating output and removes the ac ripple.

So the capacitor is the answer. A larger capacitance gives a longer discharge time constant $RC$ and therefore smaller ripple.

Question 8

Light travels a distance x in time t1 in air and 10x in time t2 in another denser medium. What is the critical angle for this medium?

Accepted Answer

Step 1 — find the speeds.In air: v_1 = \dfrac{x}{t_1}     In the medium: v_2 = \dfrac{10x}{t_2}Step 2 — relate to refractive index.Since \mu = c/v, the refractive index is inversely proportional to speed:\frac{\mu_1}{\mu_2} = \frac{v_2}{v_1}Step 3 — apply the critical angle condition.At the critical angle i_c, light travelling from the denser medium (\mu_2) into the rarer one (\mu_1) refracts at 90°:\mu_2 \sin i_c = \mu_1 \sin 90° \;\Rightarrow\; \sin i_c = \frac{\mu_1}{\mu_2} = \frac{v_2}{v_1}\sin i_c = \frac{10x/t_2}{x/t_1} = \frac{10\,t_1}{t_2}i_c = \sin^{-1}\left(\frac{10\,t_1}{t_2}\right)

Question 9

Resistance of a carbon resistor determined from colour codes is (22000 ± 5%) Ω. The colour of third band must be

Accepted Answer

In the four-band colour code, the first two bands give the significant digits, the third band is the decimal multiplier, and the fourth is the tolerance.

Write the resistance in standard form:
$$22000\ \Omega = 22 \times 10^3\ \Omega$$
So the significant digits are 2 and 2, and the multiplier is $10^3$.

Reading the multiplier off the colour code, the digit 3 corresponds to orange.

(For reference: black 0, brown 1, red 2, orange 3, yellow 4, green 5. The first two bands here would both be red, and the ±5% tolerance band would be gold.)

Question 10

Given below are two statements:Statement I: Photovoltaic devices can convert optical radiation into electricity.Statement II: Zener diode is designed to operate under reverse bias in breakdown region.In the light of the above statements, choose the most appropriate answer from the options given belo

Accepted Answer

Statement I is correct. A photovoltaic device — a solar cell — is a p-n junction in which incident photons with energy greater than the band gap generate electron–hole pairs. The junction's built-in field separates these carriers, producing a photovoltage across the terminals. So optical radiation is indeed converted into electricity.Statement II is correct. An ordinary diode is destroyed if driven into reverse breakdown, but a Zener diode is heavily doped so that its depletion layer is very thin and breakdown occurs at a low, sharply defined reverse voltage. It is built to operate there safel

Question 11

The magnetic energy stored in an inductor of inductance 4 µH carrying a current of 2 A is

Accepted Answer

Energy stored in the magnetic field of an inductor:
$$U = \tfrac{1}{2}Li^2$$
Substituting $L = 4\ \mu\text{H} = 4 \times 10^{-6}$ H and $i = 2$ A:
$$U = \tfrac{1}{2} \times 4 \times 10^{-6} \times (2)^2$$
$$U = \tfrac{1}{2} \times 4 \times 10^{-6} \times 4 = 8 \times 10^{-6}\ \text{J}$$
$$U = 8\ \mu\text{J}$$
Watch the unit prefix — micro, not milli. The arithmetic gives 8, and the answer must stay in µJ because $L$ was in µH.

Question 12

The angular acceleration of a body, moving along the circumference of a circle, is

Accepted Answer

Angular quantities — angular velocity \vec{\omega} and angular acceleration \vec{\alpha} — are axial vectors. They are not directed along the circular path itself.Since\vec{\alpha} = \frac{d\vec{\omega}}{dt}and \vec{\omega} always points along the axis of rotation (its sense given by the right-hand rule), any change in \vec{\omega} for a body confined to a fixed circular path must also lie along that same axis. So \vec{\alpha} is along the axis of rotation.Don't confuse this with the linear quantities: centripetal acceleration points towards the centre, and tangential acceleration lies along t

Question 13

A Carnot engine has an efficiency of 50% when its source is at a temperature 327°C. The temperature of the sink is

Accepted Answer

Efficiency of a Carnot engine depends only on the two absolute temperatures:
$$\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$$
Convert to kelvin first — this is essential, the formula fails with Celsius:
$$T_{\text{source}} = 327 + 273 = 600\ \text{K}$$
Given $\eta = 50\% = \dfrac{1}{2}$:
$$\frac{1}{2} = 1 - \frac{T_{\text{sink}}}{600}$$
$$\frac{T_{\text{sink}}}{600} = \frac{1}{2} \;\Rightarrow\; T_{\text{sink}} = 300\ \text{K}$$
Convert back:
$$T_{\text{sink}} = 300 - 273 = 27°\text{C}$$

Question 14

Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (G = gravitational constant)

Accepted Answer

Step 1 — locate the null point.Let the field vanish at distance x from the mass m, so it is (R-x) from 9m. The two field contributions are opposite in direction and must be equal in magnitude:\frac{Gm}{x^2} = \frac{G(9m)}{(R-x)^2}\frac{(R-x)^2}{x^2} = 9 \;\Rightarrow\; \frac{R-x}{x} = 3R - x = 3x \;\Rightarrow\; x = \frac{R}{4}, \qquad R - x = \frac{3R}{4}Step 2 — find the potential there.Potential is a scalar, so the two contributions simply add (both negative — they do not cancel):V = -\frac{Gm}{x} - \frac{G(9m)}{R-x}V = -\frac{Gm}{R/4} - \frac{9Gm}{3R/4}V = -\frac{4Gm}{R} - \frac{12Gm}{R} =

Question 15

A vehicle travels half the distance with speed v and the remaining distance with speed 2v. Its average speed is

Accepted Answer

Average speed is total distance divided by total time — never the plain average of the two speeds.

Let the total distance be $2d$, so each half is $d$.
$$t_1 = \frac{d}{v}, \qquad t_2 = \frac{d}{2v}$$
$$t_{\text{total}} = \frac{d}{v} + \frac{d}{2v} = \frac{2d + d}{2v} = \frac{3d}{2v}$$
$$v_{\text{avg}} = \frac{2d}{3d/2v} = \frac{2d \times 2v}{3d} = \frac{4v}{3}$$

Equivalently, for equal distances the average speed is the harmonic mean:
$$v_{\text{avg}} = \frac{2v_1v_2}{v_1+v_2} = \frac{2 \cdot v \cdot 2v}{v + 2v} = \frac{4v}{3}$$
Note $4v/3 = 1.33v$, which is less than the arithmetic mean $1.5v$ — as it must be, since more time is spent at the slower speed.

Question 16

The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly (surface tension of soap solution = 0.03 N m−1)

Accepted Answer

Work done in creating a surface equals surface tension times the area created:
$$W = S \times \Delta A$$
The critical point: a soap bubble has two surfaces — an inner one and an outer one, both in contact with air. So the area created is twice the spherical area:
$$\Delta A = 2 \times 4\pi r^2$$
With $r = 2\ \text{cm} = 2 \times 10^{-2}$ m:
$$\Delta A = 2 \times 4\pi \times (2\times10^{-2})^2 = 2 \times 4\pi \times 4 \times 10^{-4}$$
$$W = 0.03 \times 2 \times 4\pi \times 4 \times 10^{-4}$$
$$W = 3.015 \times 10^{-4}\ \text{J} \approx 3.01 \times 10^{-4}\ \text{J}$$
If you forget the factor of 2, you get roughly $1.5 \times 10^{-4}$ J and no option matches. A liquid drop, by contrast, has only one surface.

Question 17

The minimum wavelength of X-rays produced by an electron accelerated through a potential difference of V volts is proportional to

Accepted Answer

The minimum wavelength (the cut-off of the continuous X-ray spectrum) corresponds to the extreme case where an electron loses all of its kinetic energy in a single collision, producing one photon.

Kinetic energy gained by the electron:
$$KE = eV$$
Setting this equal to the energy of the emitted photon:
$$eV = \frac{hc}{\lambda_{\min}}$$
$$\lambda_{\min} = \frac{hc}{eV}$$
Since $h$, $c$ and $e$ are all constants:
$$\lambda_{\min} \propto \frac{1}{V}$$
This is the Duane–Hunt law. Note that $\lambda_{\min}$ depends only on the accelerating voltage, not on the target material.

Question 18

The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to \left(\dfrac{1}{16}\right)^{th} of its initial value?

Accepted Answer

After n half-lives the activity falls by a factor of 2^n:\frac{A}{A_0} = \frac{1}{2^n}We need the activity to drop to \dfrac{1}{16}:\frac{1}{16} = \frac{1}{2^n} \;\Rightarrow\; 2^n = 16 = 2^4 \;\Rightarrow\; n = 4So four half-lives have elapsed:t = n 	imes T_{1/2} = 4 	imes 20 = 80\ 	ext{minutes}Tracking it step by step: 1 → 1/2 (20 min) → 1/4 (40 min) → 1/8 (60 min) → 1/16 (80 min).

Question 19

A metal wire has mass (0.4 ± 0.002) g, radius (0.3 ± 0.001) mm and length (5 ± 0.02) cm. The maximum possible percentage error in the measurement of density will nearly be

Accepted Answer

The wire is a cylinder, so its density is
$$\rho = \frac{M}{V} = \frac{M}{\pi r^2 \ell}$$
For a quantity built from products and powers, the fractional errors add, each weighted by its power. Here $r$ appears squared, so its error is counted twice:
$$\frac{\Delta\rho}{\rho} = \frac{\Delta M}{M} + 2\frac{\Delta r}{r} + \frac{\Delta \ell}{\ell}$$
Substituting (units cancel within each ratio, so no conversion is needed):
$$\frac{\Delta\rho}{\rho} = \frac{0.002}{0.4} + \frac{2 \times 0.001}{0.3} + \frac{0.02}{5}$$
$$= 0.005 + 0.00667 + 0.004 = 0.0157$$
As a percentage:
$$\frac{\Delta\rho}{\rho} \times 100 \approx 1.56\% \approx 1.6\%$$
The factor of 2 on the radius term is what separates 1.6% from the wrong answer — the radius contributes the largest share of the error.

Question 20

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ Hz and amplitude 48 V m⁻¹. Then the amplitude of oscillating magnetic field is

(Speed of light in free space = 3 × 10⁸ m s⁻¹)

Accepted Answer

In an electromagnetic wave in free space, the amplitudes of the electric and magnetic fields are locked together by the speed of light:
$$c = \frac{E_0}{B_0}$$
Rearranging for the magnetic field amplitude:
$$B_0 = \frac{E_0}{c} = \frac{48}{3 \times 10^8}$$
$$B_0 = 16 \times 10^{-8} = 1.6 \times 10^{-7}\ \text{T}$$
The frequency given (2.0 × 10¹⁰ Hz) is not needed — it is there as a distractor. The $E_0/B_0$ ratio holds regardless of frequency.

Question 21

The temperature of a gas is −50°C. To what temperature the gas should be heated so that the rms speed is increased by 3 times?

Accepted Answer

The rms speed of gas molecules:v_{	ext{rms}} = \sqrt{\frac{3RT}{M}} \;\Rightarrow\; v_{	ext{rms}} \propto \sqrt{T}Read the wording carefully. "Increased by 3 times" means the speed increases by 3v, so the final speed is v + 3v = 4v — not 3v.Initial temperature in kelvin:T_1 = -50 + 273 = 223\ 	ext{K}Using the proportionality:\frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \;\Rightarrow\; \frac{v}{4v} = \sqrt{\frac{223}{T_2}}\frac{1}{16} = \frac{223}{T_2} \;\Rightarrow\; T_2 = 223 	imes 16 = 3568\ 	ext{K}The options are in Celsius, so convert:T_2 = 3568 - 273 = 3295°	ext{C}

Question 22

An ac source is connected to a capacitor C. Due to decrease in its operating frequency

Accepted Answer

Capacitive reactance is inversely proportional to frequency:
$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$
So if $f$ decreases, $X_C$ increases. This rules out options (1) and (4).

With the source voltage unchanged, a larger reactance means a smaller current:
$$I = \frac{V}{X_C}$$
Inside the capacitor no charge actually crosses the gap between the plates — the current there is carried by the changing electric field, which Maxwell called the displacement current. Its magnitude always equals the conduction current in the connecting wires.

So as the conduction current falls, the displacement current falls with it. Displacement current decreases.

Question 23

For Young's double slit experiment, two statements are given below:

Statement I : If screen is moved away from the plane of slits, angular separation of the fringes remains constant.

Statement II : If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.

In the light of the above statements, choose the correct answer from the options given below:

Accepted Answer

The angular fringe width in a double slit experiment is
$$\theta = \frac{\lambda}{d}$$
where $d$ is the slit separation. Notice what is absent from this expression: the screen distance $D$.

Statement I: Moving the screen away changes the linear fringe width $\beta = \dfrac{\lambda D}{d}$, which grows with $D$. But the angular separation $\lambda/d$ has no $D$ in it, so it stays constant. Statement I is true.

Statement II: Since $\theta \propto \lambda$, a larger wavelength gives a larger angular separation, not a smaller one. Statement II is false.

So Statement I is true but Statement II is false.

Question 24

In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the Bracket series is

Accepted Answer

The Rydberg formula for hydrogen:\frac{1}{\lambda} = R\left[\frac{1}{n_2^2} - \frac{1}{n_1^2}\right]The shortest wavelength in any series is the series limit, obtained by taking n_1 = \infty.Balmer series (n_2 = 2):\frac{1}{\lambda} = R\left[\frac{1}{4} - 0\right] \;\Rightarrow\; \lambda = \frac{4}{R}Brackett series (n_2 = 4):\frac{1}{\lambda'} = R\left[\frac{1}{16} - 0\right] \;\Rightarrow\; \lambda' = \frac{16}{R}Dividing:\frac{\lambda'}{\lambda} = \frac{16/R}{4/R} = 4 \;\Rightarrow\; \lambda' = 4\lambdaThe ratio is simply (n_2')^2/(n_2)^2 = 16/4 = 4.

Question 25

The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14 eV, 2.30 eV and 2.75 eV respectively. If incident electromagnetic radiation has an incident energy of 2.20 eV, which of these photosensitive surfaces may emit photoelectrons?

Accepted Answer

Photoemission occurs only when the incident photon energy is at least equal to the work function of the surface:
$$E \geq \phi_0$$
Compare the incident energy of 2.20 eV against each:
• Caesium: $\phi_0 = 2.14$ eV. Since $2.20 > 2.14$, emission occurs. The excess $2.20 - 2.14 = 0.06$ eV appears as the maximum kinetic energy of the ejected electron.
• Potassium: $\phi_0 = 2.30$ eV. Since $2.20 < 2.30$, no emission.
• Sodium: $\phi_0 = 2.75$ eV. Since $2.20 < 2.75$, no emission.

Only caesium emits photoelectrons. Note that increasing the intensity of the radiation would not help K or Na — the threshold depends on photon energy, not on how many photons arrive.

Question 26

The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are

Accepted Answer

Random errors are those that cannot be traced to any systematic or constant cause. They arise from unpredictable fluctuations in experimental conditions — temperature drift, voltage supply variation, small changes in pressure — and from the observer's own inconsistency in judgement. They vary irregularly in both size and sign, which is exactly the behaviour described here.Why the others do not fit:• Instrumental errors come from a defect or faulty calibration in the apparatus. They are systematic — the same bias every time, not fluctuating.• Personal errors arise from the observer's carelessne

Question 27

In a series LCR circuit, the inductance L is 10 mH, capacitance C is 1 µF and resistance R is 100 Ω. The frequency at which resonance occurs is

Accepted Answer

At resonance the inductive and capacitive reactances cancel ($X_L = X_C$), giving
$$f = \frac{1}{2\pi\sqrt{LC}}$$
Substitute $L = 10\ \text{mH} = 10 \times 10^{-3}$ H and $C = 1\ \mu\text{F} = 1 \times 10^{-6}$ F:
$$\sqrt{LC} = \sqrt{10 \times 10^{-3} \times 1 \times 10^{-6}} = \sqrt{10^{-8}} = 10^{-4}$$
$$f = \frac{1}{2\pi \times 10^{-4}} = \frac{10^4}{2\pi}$$
$$f \approx 1591\ \text{Hz} = 1.59\ \text{kHz}$$
Two traps here: the resistance $R$ does not affect the resonant frequency at all (it only sets the sharpness of the peak), and the answer must be read in Hz — options (1) and (3) are in rad/s, which would be $\omega = 10^4$ rad/s.

Question 28

The venturi-meter works on

Accepted Answer

A venturi-meter measures the flow rate of a fluid through a pipe. It has a constricted throat: as the fluid enters the narrower section, the equation of continuity $(A_1v_1 = A_2v_2)$ forces its speed to increase.

By Bernoulli's principle, along a streamline
$$P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}$$
so where the speed rises, the pressure must fall. Measuring the pressure difference between the wide section and the throat therefore gives the flow speed.

The other options belong elsewhere: Huygens' principle concerns wave propagation, and the parallel and perpendicular axes theorems are used for moment of inertia.

Question 29

The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is

Accepted Answer

Open pipe (open at both ends): antinodes form at both ends, so the pipe length holds half a wavelength.\ell = \frac{\lambda}{2} \;\Rightarrow\; f_{	ext{open}} = \frac{v}{2\ell}Closed pipe (closed at one end): a node forms at the closed end and an antinode at the open end, so the pipe length holds a quarter wavelength.\ell = \frac{\lambda}{4} \;\Rightarrow\; f_{	ext{closed}} = \frac{v}{4\ell}Taking the ratio for the same length \ell and same speed of sound v:\frac{f_{	ext{open}}}{f_{	ext{closed}}} = \frac{v/2\ell}{v/4\ell} = \frac{4\ell}{2\ell} = \frac{2}{1}So the ratio is 2 : 1 — an open p

Question 30

An electric dipole is placed at an angle of 30° with an electric field of intensity 2 × 105 N C−1. It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.

Accepted Answer

Torque on a dipole in a uniform electric field:
$$\vec{\tau} = \vec{p} \times \vec{E} \;\Rightarrow\; \tau = pE\sin\theta$$
Step 1 — find the dipole moment.
$$4 = p \times 2 \times 10^5 \times \sin 30°$$
$$4 = p \times 2 \times 10^5 \times 0.5 = p \times 10^5$$
$$p = 4 \times 10^{-5}\ \text{C m}$$
Step 2 — extract the charge.
Since $p = q\ell$ with $\ell = 2\ \text{cm} = 0.02$ m:
$$q = \frac{p}{\ell} = \frac{4 \times 10^{-5}}{0.02} = 2 \times 10^{-3}\ \text{C}$$
$$q = 2\ \text{mC}$$

Question 31

The net magnetic flux through any closed surface is

Accepted Answer

This is Gauss's law for magnetism, one of Maxwell's four equations:
$$\oint \vec{B} \cdot \overrightarrow{ds} = 0$$
The physical reason is that magnetic monopoles do not exist. Unlike electric charges, which can be isolated as a lone positive or negative charge, magnetic poles always come in pairs — cut a bar magnet in half and you get two smaller magnets, each with its own north and south pole.

As a result, magnetic field lines never begin or end anywhere; they always form closed loops. Every line that enters a closed surface must also leave it, so the inward and outward contributions cancel exactly and the net flux is always zero — for any closed surface, whatever its shape or position.

Contrast this with the electric case, where $\oint \vec{E}\cdot\overrightarrow{dS} = q/\varepsilon_0$ can be non-zero because isolated charges do exist.

Question 32

A bullet is fired from a gun at the speed of 280 m s−1 in the direction 30° above the horizontal. The maximum height attained by the bullet is (g = 9.8 m s−2, sin30° = 0.5)

Accepted Answer

Only the vertical component of velocity matters for the height. At the highest point the vertical velocity is zero, giving
$$H = \frac{u^2\sin^2\theta}{2g}$$
Substituting $u = 280$ m s⁻¹, $\theta = 30°$, $g = 9.8$ m s⁻²:
$$H = \frac{(280)^2 \times (0.5)^2}{2 \times 9.8}$$
$$H = \frac{280 \times 280 \times 0.5 \times 0.5}{19.6}$$
$$H = \frac{19600}{19.6} = 1000\ \text{m}$$
Note that $\sin\theta$ is squared — using $\sin 30°$ once instead of twice gives 2000 m, which is option (2) and the intended trap.

Question 33

Two thin lenses are of same focal lengths (f), but one is convex and the other one is concave. When they are placed in contact with each other, the equivalent focal length of the combination will be

Accepted Answer

For two thin lenses in contact, the powers add:
$$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2}$$
By sign convention a convex lens has a positive focal length and a concave lens a negative one. Since their magnitudes are equal:
$$f_1 = +f, \qquad f_2 = -f$$
$$\frac{1}{f_{\text{eq}}} = \frac{1}{f} + \frac{1}{-f} = \frac{1}{f} - \frac{1}{f} = 0$$
$$f_{\text{eq}} = \infty$$
The combination has zero net power — it behaves like a plane glass slab, letting light pass without converging or diverging it. The two lenses cancel each other exactly.

Answering "zero" is the common error: $1/f_{\text{eq}} = 0$, so $f_{\text{eq}}$ is infinite, not zero.

Question 34

A bullet from a gun is fired on a rectangular wooden block with velocity u. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes \dfrac{u}{3}. Then it further penetrates into the block in the same direction before coming to rest exactly at the other e

Accepted Answer

The retardation a inside the wood is uniform throughout. Use v^2 = u^2 - 2as for each stage.Stage 1 — first 24 cm: velocity drops from u to u/3.\left(\frac{u}{3}ight)^2 = u^2 - 2a(24)\frac{u^2}{9} = u^2 - 48a \;\Rightarrow\; 48a = u^2 - \frac{u^2}{9} = \frac{8u^2}{9}a = \frac{u^2}{54} \qquad \ldots(	ext{i})Stage 2 — remaining distance s: velocity drops from u/3 to 0.0 = \left(\frac{u}{3}ight)^2 - 2as \;\Rightarrow\; 2as = \frac{u^2}{9}Substituting a from (i):2 	imes \frac{u^2}{54} 	imes s = \frac{u^2}{9}\frac{s}{27} = \frac{1}{9} \;\Rightarrow\; s = 3\ 	ext{cm}Total length:L = 24 + 3 =

Question 35

A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s−1. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take g = 10 m s−2)

Accepted Answer

Take upward as positive, with the bridge as the origin. The ball starts with $u = +4$ m s⁻¹ and undergoes acceleration $-g$ throughout.

Using $s = ut - \tfrac{1}{2}gt^2$ with $t = 4$ s:
$$s = (4)(4) - \tfrac{1}{2}(10)(4)^2$$
$$s = 16 - \tfrac{1}{2}(10)(16) = 16 - 80 = -64\ \text{m}$$
The negative sign simply means the displacement is downward from the bridge. So the water surface lies 64 m below:
$$h = 64\ \text{m}$$
The single equation handles the whole journey — upward flight, turnaround, and fall — because the acceleration never changes. There is no need to split it into separate stages.

Question 36

10 resistors, each of resistance R are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased n times. The value of n is

Accepted Answer

Series combination: resistances add.R_{	ext{eq}} = 10R \;\Rightarrow\; i = \frac{E}{10R}Parallel combination: for n identical resistors the equivalent is R/n.R_{	ext{eq}}' = \frac{R}{10} \;\Rightarrow\; i' = \frac{E}{R/10} = \frac{10E}{R}Taking the ratio:\frac{i'}{i} = \frac{10E/R}{E/10R} = \frac{10E}{R} 	imes \frac{10R}{E} = 100n = 100The factor is 100, not 10, because the change works twice over: parallel connection lowers the resistance by a factor of 10 compared to a single resistor, while series raised it by a factor of 10 — a swing of 10 × 10.

Question 37

A wire carrying a current I along the positive x-axis has length L. It is kept in a magnetic field \vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k}) T. The magnitude of the magnetic force acting on the wire is

Accepted Answer

Force on a current-carrying wire in a magnetic field:
$$\vec{F} = I\vec{L} \times \vec{B}$$
The wire lies along the positive $x$-axis, so $\vec{L} = L\hat{i}$:
$$\vec{F} = IL\,\hat{i} \times (2\hat{i} + 3\hat{j} - 4\hat{k})$$
Evaluate term by term using $\hat{i}\times\hat{i} = 0$, $\hat{i}\times\hat{j} = \hat{k}$, $\hat{i}\times\hat{k} = -\hat{j}$:
$$\vec{F} = IL\left[2(0) + 3\hat{k} - 4(-\hat{j})\right] = IL(3\hat{k} + 4\hat{j})$$
Magnitude:
$$|\vec{F}| = IL\sqrt{3^2 + 4^2} = IL\sqrt{25} = 5IL$$
The $2\hat{i}$ component of $\vec{B}$ contributes nothing — it is parallel to the current, and a magnetic field parallel to the current exerts no force.

Question 38

A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity \dfrac{3\pi}{Gd} represents

Accepted Answer

For a satellite orbiting just above the surface, the orbital radius equals the earth's radius $R$. From Kepler's third law:
$$T = 2\pi\sqrt{\frac{R^3}{GM}}$$
Express the mass in terms of density. Treating the earth as a uniform sphere:
$$M = d \times \frac{4}{3}\pi R^3$$
Substituting:
$$T = 2\pi\sqrt{\frac{R^3}{G \cdot d \cdot \frac{4}{3}\pi R^3}}$$
The $R^3$ cancels completely — the period does not depend on the earth's size at all, only on its density:
$$T = 2\pi\sqrt{\frac{3}{4\pi G d}}$$
Squaring both sides:
$$T^2 = 4\pi^2 \times \frac{3}{4\pi G d} = \frac{3\pi}{Gd}$$
So $\dfrac{3\pi}{Gd}$ represents $T^2$.

Question 39

Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 (g = 10 m s−2).

Accepted Answer

For the body to stay put on the floor, it must accelerate along with the car. The only horizontal force available to push it forward is static friction from the floor.

Applying Newton's second law to the body:
$$f = ma$$
Static friction has an upper limit:
$$f \leq f_{\max} = \mu_s N = \mu_s mg$$
Combining, the largest acceleration the friction can sustain is
$$ma_{\max} = \mu_s mg$$
The mass cancels — the answer is independent of how heavy the body is:
$$a_{\max} = \mu_s g = 0.15 \times 10 = 1.5\ \text{m s}^{-2}$$
Beyond this, friction cannot supply the required force and the body slides backwards relative to the car.

Question 40

The resistance of platinum wire at 0°C is 2 Ω and 6.8 Ω at 80°C. The temperature coefficient of resistance of the wire is

Accepted Answer

Resistance varies with temperature asR = R_0(1 + \alpha \Delta T)where R_0 is the resistance at 0°C and \alpha is the temperature coefficient.Substituting R_0 = 2\ \Omega, R = 6.8\ \Omega and \Delta T = 80 - 0 = 80°C:6.8 = 2\left[1 + \alpha(80)ight]\frac{6.8}{2} = 1 + 80\alpha \;\Rightarrow\; 3.4 = 1 + 80\alpha80\alpha = 2.4\alpha = \frac{2.4}{80} = 0.03 = 3 	imes 10^{-2}\ °	ext{C}^{-1}Because the reference is 0°C, \Delta T is numerically the same in Celsius or kelvin, so no conversion is needed here.

Question 41

The radius of inner most orbit of hydrogen atom is 5.3 × 10−11 m. What is the radius of third allowed orbit of hydrogen atom?

Accepted Answer

In the Bohr model the orbital radius grows as the square of the principal quantum number:
$$r_n = r_1 \times \frac{n^2}{Z}$$
For hydrogen $Z = 1$, so $r_n \propto n^2$.

For the third orbit, $n = 3$:
$$r_3 = r_1 \times 3^2 = 9 r_1$$
$$r_3 = 9 \times 5.3 \times 10^{-11} = 47.7 \times 10^{-11}\ \text{m}$$
Convert to ångström using $1\ \text{Å} = 10^{-10}$ m:
$$r_3 = 4.77 \times 10^{-10}\ \text{m} = 4.77\ \text{Å}$$
The dependence is on $n^2$, not $n$ — multiplying by 3 instead of 9 gives 1.59 Å, which is option (3) and the intended trap.

Question 42

Which of the following reactions will NOT give primary amine as the product?

Accepted Answer

Look carefully at the difference between a nitrile (CN) and an isonitrile (NC) — the point of attachment to carbon changes the product.

• Option 1: An amide with Br₂/KOH is the Hoffmann bromamide degradation. It removes one carbon and gives a primary amine (CH₃NH₂).
• Option 2: A nitrile CH₃–C≡N reduced by LiAlH₄ adds hydrogens to the carbon end, giving CH₃CH₂NH₂ — a primary amine.
• Option 4: An amide reduced by LiAlH₄ also gives a primary amine.
• Option 3: An isocyanide CH₃–N≡C is attached to nitrogen. On reduction the two hydrogens add to the terminal carbon, giving CH₃–NH–CH₃ — a secondary amine, not primary.

So option 3 is the one that does not give a primary amine.

Question 43

Match List-I with List-II.List-I: (A) Coke   (B) Diamond   (C) Fullerene   (D) GraphiteList-II: (I) Carbon atoms are sp3 hybridised   (II) Used as a dry lubricant   (III) Used as a reducing agent   (IV) Cage like molecules

Accepted Answer

Match each form of carbon to its defining property:

• Coke is nearly pure carbon and is widely used to reduce metal oxides to metals in metallurgy — a reducing agent (III).
• Diamond has each carbon bonded tetrahedrally to four others, so every carbon is sp³ hybridised (I).
• Fullerene (buckminsterfullerene, C₆₀) is built from fused five- and six-membered rings that close into a hollow sphere — a cage-like molecule (IV).
• Graphite has layers that slide over one another easily, making it soft and slippery — used as a dry lubricant (II).

So A-III, B-I, C-IV, D-II.

Question 44

Given below are two statements: one is labelled as Assertion A and the other as Reason R.

Assertion A: Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.

Reason R: The deep blue solution is due to the formation of amide.

In the light of the above statements, choose the correct answer.

Accepted Answer

Assertion is true. When an alkali metal like sodium dissolves in liquid ammonia, the metal releases electrons which become surrounded by ammonia molecules — these are called ammoniated electrons:	ext{M} + (x+y)	ext{NH}_3 ightarrow [	ext{M(NH}_3)_x]^+ + [	ext{e(NH}_3)_y]^-These trapped electrons absorb light in the visible region, giving the deep blue colour. Because the electrons are unpaired, the solution is paramagnetic.Reason is false. The blue colour comes from the ammoniated (solvated) electron, not from any amide. Amide formation happens only slowly and is a separate process.So A i

Question 45

In Lassaigne's extract of an organic compound, both nitrogen and sulphur are present, which gives blood red colour with Fe3+ due to the formation of

Accepted Answer

When both nitrogen and sulphur are present together in the organic compound, fusion with sodium produces sodium thiocyanate rather than sodium cyanide:
$$\text{Na} + \text{C} + \text{N} + \text{S} \rightarrow \text{NaSCN}$$
The thiocyanate ion then reacts with ferric ions to give a blood-red complex:
$$\text{Fe}^{3+} + \text{SCN}^- \rightarrow [\text{Fe(SCN)}]^{2+} \;\; (\text{blood red})$$
Note there is no Prussian blue here — that would require free cyanide ions, but with sulphur present the cyanide is locked up as thiocyanate. The blood-red colour is the confirmatory test for N and S together.

Question 46

The conductivity of centimolar solution of KCl at 25°C is 0.0210 ohm−1 cm−1 and the resistance of the cell containing the solution at 25°C is 60 ohm. The value of cell constant is

Accepted Answer

Conductivity, conductance and cell constant are related by
$$\kappa = G \times G^*$$
where $\kappa$ is conductivity, $G$ is conductance and $G^*$ is the cell constant. Conductance is the reciprocal of resistance, $G = 1/R$, so
$$\kappa = \frac{G^*}{R} \;\Rightarrow\; G^* = \kappa \times R$$
Substituting the values:
$$G^* = 0.0210 \times 60 = 1.26\ \text{cm}^{-1}$$
The molarity of the solution is not needed — it is extra information. The cell constant depends only on the geometry of the cell (electrode area and separation).

Question 47

Given below are two statements: one is labelled as Assertion A and the other as Reason R.

Assertion A: Metallic sodium dissolves in liquid ammonia giving a deep blue solution, which is paramagnetic.

Reason R: The deep blue solution is due to the formation of amide.

In the light of the above statements, choose the correct answer.

Accepted Answer

Assertion is true. A few reactions — certain radical combination reactions, for example — proceed with essentially no energy barrier, so their activation energy is effectively zero.Reason is true. Activation energy is correctly defined as the minimum extra energy that reactant molecules must absorb for their energy to reach the threshold value needed to react.But R does not explain A. The definition of activation energy tells you what it is; it says nothing about why a particular reaction could have that value equal to zero. The two statements are both correct but independent.So both are true,

Question 48

Which one is an example of heterogenous catalysis?

Accepted Answer

The distinction is about phases: heterogeneous catalysis has the catalyst in a different phase from the reactants; homogeneous catalysis has them in the same phase.• Option 4 is the Haber process:	ext{N}_2(g) + 3	ext{H}_2(g) \xrightarrow{	ext{Fe}(s)} 2	ext{NH}_3(g)The iron catalyst is a solid while the reactants are gases — different phases, so this is heterogeneous.The other three are all homogeneous — catalyst and reactants share the same phase:• SO2 oxidation with gaseous NO — all gases.• Sugar hydrolysis with H+ — all in aqueous solution.• Ozone decomposition with NO — all gases.So opt

Question 49

The given compoundC6H5–CH=CH–CH(X)–CH2–CH3(a phenyl group attached to CH=CH–CHX–CH2–CH3) is an example of ______.

Accepted Answer

The classification depends on what kind of carbon bears the halogen X.

Here X sits on a carbon that is sp³ hybridised and is directly attached to a carbon–carbon double bond (C=C). A saturated carbon adjacent to a double bond is called an allylic position, so this is an allylic halide.

Ruling out the others:
• Vinylic halide would need X directly on a doubly-bonded (sp²) carbon — it is not.
• Aryl halide would need X directly on the benzene ring — it is not.
• Benzylic halide would need X on the carbon directly attached to the ring — here that carbon is part of the C=C, and X is one carbon further along.

So it is an allylic halide.

Question 50

Given below are two statements: one is labelled as Assertion A and the other as Reason R.Assertion A: Helium is used to dilute oxygen in diving apparatus.Reason R: Helium has high solubility in O2.

Accepted Answer

Assertion is true. Divers breathe a helium–oxygen mixture instead of nitrogen–oxygen. Helium's very low solubility in blood means it does not dissolve into the bloodstream under the high pressures at depth, avoiding the dangerous bubbles ("the bends") that nitrogen would form on ascent.On the reason: the official key treats R as true (helium and oxygen mix/diffuse readily), giving option 2 — both true, but R not the correct explanation of the assertion. Note, however, that the reason as worded ("high solubility in O2") is chemically loose: helium is chosen for its low solubility in blood, not

Question 51

A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is AxBy, then the value of x + y is

Accepted Answer

Start by counting the atoms and voids in a close-packed lattice.

Let there be $N$ atoms of B forming the ccp arrangement. In any close-packed structure the number of tetrahedral voids is twice the number of packing atoms, so there are $2N$ tetrahedral voids.

Atoms of A fill 1/3 of these voids:
$$\text{A} = \frac{1}{3} \times 2N = \frac{2N}{3}$$
The ratio A : B is therefore
$$\frac{2N/3}{N} = \frac{2}{3}$$
giving the formula
$$\text{A}_2\text{B}_3$$
So $x = 2$, $y = 3$, and
$$x + y = 5$$

Question 52

Given below are two statements:Statement I: A unit formed by the attachment of a base to 1′ position of sugar is known as nucleoside.Statement II: When nucleoside is linked to phosphorous acid at 5′-position of sugar moiety, we get nucleotide.In the light of the above statements, choose the correct

Accepted Answer

Statement I is true. A nucleoside is exactly this: a nitrogenous base joined to the 1′ carbon of the sugar. (The sugar carbons are numbered 1′, 2′, 3′... with primes to distinguish them from the atoms in the base.)Statement II is false because of one word. A nucleotide is formed when the nucleoside is linked to phosphoric acid at the 5′ position — not phosphorous acid. Phosphoric acid (H3PO4) and phosphorous acid (H3PO3) are different compounds, and it is the phosphate (from phosphoric acid) that forms the backbone of DNA and RNA.So Statement I is true but Statement II is false.

Question 53

The relation between nm (nm = the number of permissible values of magnetic quantum number (m)) for a given value of azimuthal quantum number (l), is

Accepted Answer

For a given azimuthal quantum number l, the magnetic quantum number m runs over all integers from -l to +l, including zero:m = -l, \ldots, -1, 0, +1, \ldots, +lCounting these values givesn_m = 2l + 1The question asks for l in terms of n_m, so rearrange:2l = n_m - 1 \;\Rightarrow\; l = \frac{n_m - 1}{2}That is option (1). For example, for a d-subshell l = 2, so n_m = 5 (the five d-orbitals), and indeed l = (5-1)/2 = 2.

Question 54

Amongst the following the total number of species NOT having eight electrons around central atom in its outermost shell, isNH3, AlCl3, BeCl2, CCl4, PCl5

Accepted Answer

Count the electrons around the central atom in each — the octet rule is satisfied when there are exactly eight.

• NH₃: nitrogen has three bond pairs and one lone pair = 8 electrons. Octet complete.
• AlCl₃: aluminium forms three bonds = 6 electrons. Short of octet.
• BeCl₂: beryllium forms two bonds = 4 electrons. Short of octet.
• CCl₄: carbon forms four bonds = 8 electrons. Octet complete.
• PCl₅: phosphorus forms five bonds = 10 electrons. Expanded octet.

Three species — AlCl₃, BeCl₂ and PCl₅ — do not have exactly eight electrons around the central atom.

So the answer is 3.

Question 55

The correct order of energies of molecular orbitals of N2 molecule, is

Accepted Answer

For the lighter second-period molecules — B₂, C₂ and N₂ — there is significant s–p mixing, which pushes the σ2p_z orbital above the π2p orbitals. This gives the order:
$$\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^*2p_x = \pi^*2p_y) < \sigma^*2p_z$$
The key feature to remember: for N₂ and lighter, the two π bonding orbitals come before the σ2p_z. Only from O₂ onwards does the σ2p_z drop below the π orbitals (that is what option 2 describes, which is wrong for N₂).

So option 1 is correct.

Question 56

The number of σ bonds, π bonds and lone pair of electrons in pyridine, respectively are:

Accepted Answer

Pyridine is a six-membered aromatic ring, C₅H₅N, with one nitrogen replacing a CH group of benzene.

Sigma bonds: The ring has 6 C–C/C–N sigma bonds around it. Five carbons each carry one hydrogen, adding 5 C–H sigma bonds. Total = 6 + 5 = 11 sigma bonds.

Pi bonds: Like benzene, pyridine has three alternating double bonds in the ring, contributing 3 pi bonds.

Lone pairs: The nitrogen has one lone pair — and importantly this lone pair lies in an sp² orbital in the plane of the ring, not in the aromatic π system, which is why pyridine is basic. That is 1 lone pair.

So 11, 3, 1.

Question 57

Intermolecular forces are forces of attraction and repulsion between interacting particles that will include:

(A) dipole-dipole forces (B) dipole-induced dipole forces (C) hydrogen bonding (D) covalent bonding (E) dispersion forces

Accepted Answer

Intermolecular forces act between molecules. The key is to spot the one item that is an intramolecular force.Covalent bonding (D) is the force that holds atoms together within a single molecule — it is a chemical bond, not an intermolecular force. So D must be excluded.The genuine intermolecular forces (all van der Waals type or hydrogen bonding) are:• A dipole–dipole forces• B dipole–induced dipole forces• C hydrogen bonding• E dispersion (London) forcesSo the correct set is A, B, C, E.

Question 58

Which of the following statements are NOT correct?(A) Hydrogen is used to reduce heavy metal oxides to metals.(B) Heavy water is used to study reaction mechanism.(C) Hydrogen is used to make saturated fats from oils.(D) The H–H bond dissociation enthalpy is lowest as compared to a single bond betwee

Accepted Answer

Check each statement for correctness — the question asks which are wrong.

Correct statements: (A) hydrogen does reduce heavy metal oxides to metals; (B) heavy water (D₂O) is used to trace reaction mechanisms; (C) hydrogenation of oils produces saturated fats (vanaspati).

The two incorrect statements:
• (D) is wrong. The H–H bond enthalpy is actually the highest — not lowest — among single bonds between two atoms of any element (about 435 kJ/mol), because the tiny hydrogen atoms allow very close, strong overlap.
• (E) is wrong. Hydrogen can only reduce oxides of metals that are less active than iron (that lie below it in the reactivity series). Oxides of metals more active than iron cannot be reduced by hydrogen.

So D and E are the incorrect statements.

Question 59

Which amongst the following molecules on polymerization produces neoprene?

Accepted Answer

Neoprene is a synthetic rubber made by the free-radical polymerisation of chloroprene (2-chloro-1,3-butadiene):
$$n\,\text{CH}_2{=}\text{C(Cl)}{-}\text{CH}{=}\text{CH}_2 \xrightarrow{\text{polymerisation}} [{-}\text{CH}_2{-}\text{C(Cl)}{=}\text{CH}{-}\text{CH}_2{-}]_n$$
The monomer is a butadiene skeleton carrying a chlorine on the second carbon — that is option 2.

The others give different polymers: option 1 (buta-1,3-diene) gives Buna rubber, and option 4 (isoprene, with a methyl group instead of chlorine) gives natural rubber. Neoprene needs the chlorine, which is what makes it oil- and weather-resistant.

Question 60

Some tranquilizers are listed below. Which one from the following belongs to barbiturates?

Accepted Answer

Barbiturates are a specific class of tranquilizers derived from barbituric acid.

Among the options, Veronal (also called barbital) is a derivative of barbituric acid and is therefore a barbiturate.

The other three — chlordiazepoxide, meprobamate and valium — are also tranquilizers but belong to different chemical classes (the first and third are benzodiazepines), not barbiturates.

So the answer is Veronal.

Question 61

The element expected to form largest ion to achieve the nearest noble gas configuration is

Accepted Answer

To reach the nearest noble gas (neon) configuration, each of these gains or loses electrons to become isoelectronic — all end up with 10 electrons:
• O → O²⁻ (gains 2)
• F → F⁻ (gains 1)
• N → N³⁻ (gains 3)
• Na → Na⁺ (loses 1)

For an isoelectronic series, the ionic size depends on the nuclear charge pulling on those 10 electrons: fewer protons means a weaker pull and a larger ion. The order of nuclear charge is Na(11) > O(8) > F(9)... wait, arrange by protons: N has only 7 protons, the fewest of the four.

With the smallest nuclear charge holding the 10 electrons, N³⁻ has the largest ionic radius. Also, as the negative charge on an anion increases, the ion swells — and N³⁻ carries the highest negative charge here.

So nitrogen forms the largest ion.

Question 62

Select the correct statements from the following:

(A) Atoms of all elements are composed of two fundamental particles.
(B) The mass of the electron is 9.10939 × 10⁻³¹ kg.
(C) All the isotopes of a given element show same chemical properties.
(D) Protons and electrons are collectively known as nucleons.
(E) Dalton's atomic theory regarded the atom as an ultimate particle of matter.

Accepted Answer

Test each statement:• (A) Wrong. Atoms are made of three fundamental particles — electrons, protons and neutrons — not two.• (B) Correct. The electron mass is indeed 9.10939 × 10−31 kg.• (C) Correct. Isotopes differ only in neutron number, so they have the same electronic configuration and hence identical chemical properties.• (D) Wrong. Nucleons are the particles in the nucleus — protons and neutrons, not protons and electrons.• (E) Correct. Dalton's theory treated the atom as the ultimate, indivisible particle of matter.So the correct statements are B, C and E only.

Question 63

The stability of Cu2+ is more than Cu+ salts in aqueous solution due to

Accepted Answer

At first glance Cu+ should be favoured, because forming Cu2+ requires the large second ionisation enthalpy. So why is Cu2+ more stable in water?The answer is the hydration energy. The Cu2+ ion is smaller and doubly charged, so it attracts water molecules far more strongly than the singly-charged Cu+. Its hydration enthalpy is very large and negative (about −2121 kJ/mol).This enormous release of energy on hydration more than compensates for the extra second ionisation enthalpy needed to reach Cu2+. On balance, Cu2+(aq) sits lower in energy and is the more stable species in aqueous solution.So t

Question 64

Which one of the following statements is correct?

Accepted Answer

Check each biological statement:• Option 1 — correct. The daily requirement of both Mg and Ca is around 200–300 mg, which is 0.2–0.3 g.• Option 2 — wrong. Enzymes using ATP in phosphate transfer require Mg as the cofactor, not Ca.• Option 3 — wrong. Bone is not inert; it is constantly being dissolved and redeposited throughout life.• Option 4 — wrong. It is Ca, not Mg, that plays the key role in neuromuscular function, interneuronal transmission, cell membrane integrity and blood clotting.So only option 1 is correct.

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