Concept: The force per unit length on a current-carrying wire in a magnetic field is given by \[ \frac{F}{L} = B I \sin \theta \]
where \(B\) is the magnetic field, \(I\) is the current, and \(\theta\) is the angle between the wire and the field.


Calculation: \[ \frac{F}{L} = 0.5 \times 2 \times \sin 30^\circ = 1 \times 0.5 = 0.5~N/m \]

Explanation: The wire experiences a magnetic force of 0.5 N/m due to its orientation and the given current.
Quick Tip: Use the formula \(F = BIL \sin\theta\) carefully and make sure to use the angle between current and magnetic field.

Concept: At maximum height, the vertical velocity becomes zero. Using \[ v = u - g t \]
where \(u = 20\) m/s, \(g = 9.8\) m/s\(^2\).


Calculation: \[ 0 = 20 - 9.8 t \implies t = \frac{20}{9.8} \approx 2.04~s \approx 2~s \]

Explanation: It takes approximately 2 seconds for the ball to reach its peak.
Quick Tip: Always set final velocity to zero for maximum height calculation.

Concept: Work done against gravity is given by \[ W = m g h \]
where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is height.


Calculation: \[ W = 10 \times 9.8 \times 5 = 490 \approx 500~J \]

Explanation: Lifting a 10 kg object to 5 m requires 500 J of work.
Quick Tip: Remember to multiply mass, gravity, and height to calculate work.

Concept: Work done at constant pressure is \[ W = P (V_f - V_i) \]

Calculation: \[ W = 1 \times 10^5 (2 - 5) = -3 \times 10^5~J \]

Explanation: Negative sign indicates work is done on the gas during compression.
Quick Tip: Pay attention to the sign of work: compression \(\rightarrow\) negative, expansion \(\rightarrow\) positive.

Concept: The mirror formula is \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
where \(f\) is the focal length, \(u\) is the object distance, and \(v\) is the image distance.


Calculation: \[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30} \] \[ v = 30~cm \]

Explanation: The positive image distance indicates a real and inverted image.
Quick Tip: Always apply the mirror formula carefully and keep track of signs: concave mirror real image distance is positive.

Concept: Magnetic field due to a long straight wire: \[ B = \frac{\mu_0 I}{2 \pi r} \]
Magnetic force on moving charge: \[ F = q v B \sin \theta \]
Since proton moves parallel to current, \(\theta = 90^\circ\) for perpendicular field contribution.


Calculation: \[ B = \frac{4 \pi \times 10^{-7} \times 10}{2 \pi \times 0.05} = 4 \times 10^{-5}~T \] \[ F = 1.6 \times 10^{-19} \times 2 \times 10^5 \times 4 \times 10^{-5} = 1.28 \times 10^{-19}~N \]

Explanation: The proton experiences a tiny magnetic force due to the current-carrying wire.
Quick Tip: Remember to convert units correctly and use \(\theta = 90^\circ\) if velocity is parallel to wire for perpendicular magnetic field.

Concept: Power dissipated in resistor: \[ P = I^2 R \]

Calculation: \[ P = (2)^2 \times 5 = 4 \times 5 = 20~W \]

Explanation: The resistor dissipates 20 W as heat.
Quick Tip: Use \(P = I^2 R\) or \(P = V I\) depending on given quantities.

Concept: Volume flow rate \(Q = A v\), cross-sectional area \(A = \pi r^2\)


Calculation: \[ A = \pi (0.1)^2 = 0.0314~m^2 \] \[ Q = 0.0314 \times 2 = 0.0628~m^3/s \]

Explanation: Water flows through the pipe at a rate of 0.0628 m\(^3\)/s.
Quick Tip: Always calculate cross-sectional area first: \(A = \pi r^2\).

Concept: From the stoichiometry of the reaction, \(2\) moles of \(H_2\) produce \(2\) moles of \(H_2O\).


Calculation: \[ 2~mol H_2 \rightarrow 2~mol H_2O \] \[ 4~mol H_2 \rightarrow 4~mol H_2O \]

Explanation: Doubling hydrogen doubles the water produced.
Quick Tip: Always check the stoichiometric coefficients in chemical equations.

Concept: Energy of a photon: \[ E = \frac{hc}{\lambda} \]
where \(h = 6.626 \times 10^{-34}\) Js, \(c = 3 \times 10^8\) m/s, \(\lambda = 656.3 \times 10^{-9}\) m.


Calculation: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{656.3 \times 10^{-9}} \approx 3.03 \times 10^{-19}~J \]

Explanation: Photon energy corresponds to the given wavelength.
Quick Tip: Convert wavelength to meters before calculation and use \(E = hc/\lambda\).

Concept: Rate law: \(r = k[A]^2[B]\)


Calculation: \[ [A] \to 2[A],\quad [B] \to \frac{1}{2}[B] \] \[ r_new = k(2[A])^2 (\frac{1}{2}[B]) = k \cdot 4[A]^2 \cdot 0.5[B] = 2 k[A]^2[B] = 2 r_old \]

Explanation: Rate doubles due to changes in concentrations.
Quick Tip: Apply the rate law with new concentrations carefully and simplify step by step.

Concept: First law of thermodynamics: \[ \Delta U = Q - W \]

Calculation: \[ \Delta U = 100 - 40 = 60~J \]

Explanation: Internal energy increases by 60 J.
Quick Tip: Keep track of signs: work done by gas is subtracted.

Concept: Parent chain = butane (4 carbons), with methyl group on C2.


Explanation: Number the chain from end closest to branch; correct name is 2-Methylbutane.
Quick Tip: Always identify the longest carbon chain as parent before naming.

Concept: Let change in \([N_2] = -x\), then \([H_2]\) decreases by \(3x\), \([NH_3]\) increases by \(2x\).


Calculation: \[ K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(2x)^2}{(1-x)(1-3x)^3} = 0.5 \]
Solving, \(x = 0.25\)


Equilibrium concentrations: \[ [N_2] = 1 - x = 0.75~M, \quad [H_2] = 1 - 3x = 0.25~M, \quad [NH_3] = 2x = 0.5~M \]

Explanation: Concentrations satisfy \(K_c = 0.5\).
Quick Tip: Use ICE table method for equilibrium calculations.

Concept: Zinc loses 2 electrons to become Zn\(^{2+}\).


Explanation: Oxidation state of zinc in products is +2.
Quick Tip: Remember oxidation = loss of electrons, reduction = gain.

Woman: X\(^c\)X\(^c\), Man: X\(^N\)Y. Son inherits Y from father and X\(^c\) from mother \(\rightarrow\) 100% color-blind.


Explanation: All sons will inherit the affected X from the mother.
Quick Tip: For X-linked recessive traits, sons inherit X from mother, Y from father.

Man: AB, Woman: OO

Gametes: Man → A, B; Woman → O


Child genotypes: \[ AO, BO \]
Probability of blood group A = 1 out of 4 = 25%
Quick Tip: Use Punnett square for blood group inheritance.

Concept: Amylase catalyzes hydrolysis of starch (polysaccharide) to maltose (disaccharide).


Explanation: Salivary and pancreatic amylase are responsible for starch digestion.
Quick Tip: Identify enzyme function based on substrate: starch → maltose = amylase.

Parent genotypes: TT × tt


F1 genotype: All Tt (heterozygous)


Phenotype: All Tall (dominant trait)
Quick Tip: Cross homozygous dominant with homozygous recessive → F1 all heterozygous, show dominant trait.

Concept: Alveoli have thin walls and large surface area for diffusion of O\(_2\) and CO\(_2\).


Explanation: Gas exchange occurs across alveolar and capillary walls.
Quick Tip: Remember alveoli = primary site due to large surface area and thin membrane.

Concept: Dihybrid cross of heterozygotes → 16 offspring


Phenotypic ratio:
- 9 Round Yellow (\(R Y\))
- 3 Round Green (\(R y y\))
- 3 Wrinkled Yellow (\(r r Y\))
- 1 Wrinkled Green (\(r r y y\)) Quick Tip: Use Punnett square or forked-line method to determine dihybrid phenotypic ratio.

TSH stimulates thyroid to secrete thyroxine, regulating metabolism.
Quick Tip: Remember anterior pituitary secretes tropic hormones that act on other glands.

Concept: Helicase breaks hydrogen bonds between DNA strands, unwinding the double helix.


Explanation: DNA polymerase synthesizes new strands; helicase opens the template.
Quick Tip: Identify enzyme function: unwinding = helicase, synthesis = polymerase.

Concept: Hardy-Weinberg principle: \(q^2\) = frequency of homozygous recessive


Calculation: \[ q^2 = 0.36 \implies q = \sqrt{0.36} = 0.6 \]

Explanation: Frequency of recessive allele \(r\) is 0.6.
Quick Tip: Use \(q^2 = fraction of recessive phenotype\) to find allele frequency.