Q&A: practice questions with answers

Concept: Rutherford bombarded a thin gold foil with fast α-particles and studied how they scattered. This led to the discovery of the nucleus (1911).

1. Most α-particles passed straight → most of the atom is empty space.
2. A few were deflected → positive charge occupies a very small volume.
3. Very few (about 1 in 20,000) bounced back → a tiny, dense, positively charged nucleus at the centre.

Yaad rakhein: Electron – J.J. Thomson; Proton – Goldstein; Nucleus – Rutherford; Neutron – Chadwick.

Answer: (C) Nucleus

Concept: The sky looks blue because tiny air molecules scatter blue light (shorter wavelength) much more than red light.

1. Scattering needs particles of the atmosphere.
2. With no atmosphere, there would be no scattering of sunlight.
3. No scattered light reaches the eye from the sky, so it appears dark (black).

Yaad rakhein: Astronauts and the Moon's sky see a black sky – no atmosphere, no scattering.

Answer: (A) black

Concept: Speed of light in a medium $v=\frac{c}{n}$; a greater refractive index means a smaller speed.

1. Refractive indices: air $\approx 1.0003$, water $=1.33$, glass $\approx 1.5$.
2. So speed: air $>$ water $>$ glass.
3. Hence light is faster in water than in glass.

Trick: Optically denser medium → light slows down (higher $n$, lower $v$).

Answer: (D) greater in water than in glass

Concept: Alkanes are saturated hydrocarbons with only single C–C bonds. Their general formula is ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}+2}$.

1. Methane ($n=1$): $\mathrm{C}{\mathrm{H}}_4$; ethane ($n=2$): ${\mathrm{C}}_2{\mathrm{H}}_6$.
2. Alkenes (one double bond): ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}}$.
3. Alkynes (one triple bond): ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2\mathrm{n}-2}$.

Trick: Alkane +2, alkene 0, alkyne −2 (H count relative to $2n$).

Answer: (D) ${\text{C}}_n{\text{H}}_{2n+2}$

Concept: Non-biodegradable substances (plastics, glass, metals) are not broken down by microbes and stay in the environment for a long time.

1. Reduce – use less of such materials.
2. Reuse – use items again; Recycle – convert them into new products.
3. Throwing them openly pollutes soil and water, blocks drains and harms animals – not a suitable method.

Yaad rakhein: 3 R's – Reduce, Reuse, Recycle. Open dumping is never a solution.

Answer: (C) Throw them openly

Concept: Work done by a constant force: $W=F\times s$, where $s$ is the displacement in the direction of the force.

1. $F=2\text{ N}$, $s=0$.
2. $W=2\times 0=0$.
3. No displacement along the force → no work, however large the force.

Yaad rakhein: Pushing a wall or holding a load still → work done $=0$.

Answer: (D) 0

Concept: Heat produced per second (power) in a resistor: $P=I^{2}R$.

1. $P=100\text{ J/s}=100\text{ W}$, $R=4\Omega$.
2. $I^2=\frac{P}{R}=\frac{100}{4}=25$.
3. $I=\sqrt{25}=5\text{ A}$.

Yaad rakhein: Heat per second $=I^{2}R$; total heat $H=I^{2}Rt$.

Answer: (A) 5 A

Concept: A homogeneous mixture has a uniform composition throughout and no visible boundaries between its components.

1. Air is a mixture of gases – mainly ${\mathrm{N}}_2$ (~78%) and ${\mathrm{O}}_2$ (~21%), plus Ar, $\mathrm{C}{\mathrm{O}}_2$, water vapour.
2. The gases mix uniformly, so any sample has the same composition.
3. Hence air is a homogeneous mixture (a gaseous solution).

Yaad rakhein: Air, sugar solution, brass – homogeneous; muddy water, oil + water – heterogeneous.

Answer: (D) Homogeneous mixture

Concept: Ozone (${\mathrm{O}}_3$) in the upper atmosphere shields the earth from harmful ultraviolet (UV) radiation of the sun.

1. (i) UV radiation is highly damaging (e.g. skin cancer) – incorrect.
2. (ii) UV splits ${\mathrm{O}}_2$ into O atoms, which combine with ${\mathrm{O}}_2$ to form ${\mathrm{O}}_3$ – correct.
3. (iii) Ozone levels began to drop sharply in the 1980s, linked to CFCs – correct.

Yaad rakhein: ${\mathrm{O}}_2\stackrel{\mathrm{U}\mathrm{V}}{\to }\mathrm{O}+\mathrm{O}$, $\mathrm{O}+{\mathrm{O}}_2\to {\mathrm{O}}_3$; CFCs deplete ozone (Montreal Protocol, 1987).

Answer: (B) Only (ii) and (iii)

Concept: Each part of the female reproductive system has a specific role from making eggs to nourishing the embryo.

1. Ovary → makes female gametes (eggs): (i)-(d).
2. Oviduct (fallopian tube) → site of fertilisation: (ii)-(a).
3. Uterus → implantation of the embryo: (iii)-(b).
4. Placenta → nourishes the growing embryo: (iv)-(c).

Trick: Egg made in ovary → meets sperm in oviduct → sticks in uterus → fed by placenta.

Answer: (C) (i)-(d), (ii)-(a), (iii)-(b), (iv)-(c)

Concept: In a displacement reaction, a more reactive metal pushes out a less reactive metal from its salt solution.

1. Example: $\mathrm{F}\mathrm{e}+\mathrm{C}\mathrm{u}\mathrm{S}{\mathrm{O}}_4\to \mathrm{F}\mathrm{e}\mathrm{S}{\mathrm{O}}_4+\mathrm{C}\mathrm{u}$ – iron is above copper in the series.
2. $\mathrm{C}\mathrm{u}+\mathrm{F}\mathrm{e}\mathrm{S}{\mathrm{O}}_4$ → no reaction, as copper is below iron.
3. So the deciding factor is the relative position of the metals in the reactivity series.

Yaad rakhein: Higher in the reactivity series displaces lower – never the reverse.

Answer: (D) Position of the metals in the reactivity series

Concept: Covalent compounds form by sharing of electrons, usually between non-metals. Ionic compounds form by transfer of electrons from a metal to a non-metal.

1. $\mathrm{M}\mathrm{g}\mathrm{C}{\mathrm{l}}_2$, NaCl, MgO – metal + non-metal → ionic.
2. $\mathrm{C}{\mathrm{H}}_4$ – carbon shares its 4 valence electrons with 4 H atoms.
3. So $\mathrm{C}{\mathrm{H}}_4$ is covalent (4 single C–H bonds).

Trick: Metal + non-metal → ionic; non-metal + non-metal → covalent.

Answer: (D) ${\text{CH}}_4$

Concept: In multiple fission, the nucleus of a single-celled organism divides many times and the cell splits into many daughter cells at once.

1. Plasmodium (malaria parasite) reproduces by multiple fission.
2. Leishmania – binary fission (in a definite plane).
3. Yeast and Hydra – budding.

Yaad rakhein: Plasmodium → multiple; Amoeba, Leishmania → binary; Yeast, Hydra → budding.

Answer: (A) Plasmodium

Concept: A solution is a homogeneous mixture of two or more substances. The component present in larger amount is the solvent; the one in smaller amount is the solute.

1. (A) Wrong – a solution is homogeneous.
2. (B) Wrong – concentration can also be mass by volume, volume by volume, etc.
3. (D) Wrong – solute particles are too small to be separated by filtration.
4. (C) Correct – solute is present in a smaller amount than the solvent.

Yaad rakhein: Solution particles $<1\text{ nm}$ – they pass through filter paper and do not settle.

Answer: (C) The solute is present in a smaller amount than the solvent.

Concept: Rhizopus (bread mould) reproduces by spore formation. The thread-like hyphae form the vegetative body, not the reproductive part.

1. Sporangiophores – vertical hyphae carrying the sporangia (reproductive).
2. Sporangium – knob-like sac containing spores (reproductive).
3. Spores – develop into new Rhizopus (reproductive).
4. Hyphae – fine threads that spread over bread for nutrition, i.e. vegetative.

Yaad rakhein: Rhizopus: hyphae = body; blob on top (sporangium) = spore bag.

Answer: (D) Hyphae

Concept: The reaction between an acid and a base to give salt and water is called a neutralisation reaction.

1. General form: acid + base → salt + water.
2. Example: $\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}+\mathrm{H}\mathrm{C}\mathrm{l}\to \mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}+{\mathrm{H}}_2\mathrm{O}$.
3. The ${\mathrm{H}}^{+}$ of the acid and $\mathrm{O}{\mathrm{H}}^{-}$ of the base combine to form ${\mathrm{H}}_2\mathrm{O}$.

Yaad rakhein: Antacid for acidity, baking soda for ant/bee sting, lime for acidic soil – all neutralisation.

Answer: (A) Neutralisation reaction

Concept: Work done $W=F\times s$. When force and displacement are in the same direction, the work is positive.

1. $F=1\text{ N}$, $s=1\text{ m}$ along the force.
2. $W=1\times 1=1\text{ J}$.
3. Same direction → positive, so $W=+1\text{ J}$.

Yaad rakhein: 1 joule = work done when a force of 1 N moves an object by 1 m along the force.

Answer: (C) $+1$ J

Concept: Tides (periodic rise and fall of sea water) are caused mainly by the gravitational pull of the moon on the earth's oceans.

1. The moon pulls the ocean water facing it more strongly, causing a bulge (high tide).
2. As the earth rotates, places pass through bulges, giving high and low tides.
3. Blue sky and rainbow are due to scattering/dispersion of light; thunderstorms are weather phenomena.

Yaad rakhein: Gravitation explains: tides, the moon orbiting the earth, planets orbiting the sun.

Answer: (C) Tides

Concept: For a spherical mirror, the distance between the pole (P) and the principal focus (F) is called the focal length $f$.

1. Radius of curvature $R$ = distance from pole to centre of curvature C.
2. Aperture = diameter of the reflecting surface.
3. Relation: $R=2f$, i.e. $f=\frac{R}{2}$.

Trick: PF = $f$, PC = $R$ = $2f$.

Answer: (B) focal length

Concept: A concave (diverging) lens always forms a virtual, erect and diminished image between F and the optical centre, on the same side as the object – wherever the object is placed.

1. Ray parallel to the axis diverges as if coming from F.
2. Ray through the optical centre O goes straight.
3. Their backward extensions meet between F and O → virtual, erect, smaller image.

Trick: Concave lens and convex mirror: image always virtual, erect, diminished.

Answer: (B) Virtual and diminished