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No expanded working has been attached for this question yet. The verified answer is Option C.

Efficiency = \( \frac{W(120N)INPUT}{WORK OUTPUT} \times \frac{100}{1}\) And for incline plane V.R =\( \frac{1}{Sin\theta}\) = V.R = \(\frac{1}{Sin30} =\frac{1}{0.5}\) = 2 \(\text{therefore } \frac{60}{120} = \frac{M.A}{2}\) M.A = \(\frac{120}{100}\) = 1.2 \(\text{but M.A } = \frac{load}{effort} = \frac{120}{\text{E}}\) \(\text{therefore} \frac{120}{\text{E}}\) = 1.2 E = \(\frac{120}{1.2 }\) = 100N Therefore, Effort up the plane = 100N

from hooke's law F=Ke \( \frac{F_1}{e_1} - \frac{F_2}{e_2} \text{therefore} \frac{4}{0.02} = \frac{15}{e_2} \) e1 = \( \frac{0.02 x 15}{4} = 0.075M \) The new length of the string is 4 + 0.075 = 4.075 = 4.08

Mathematically Boyle's law = PV = K Where P = Pressure and V = Volume By definition Pressure = \( \frac{F}{A} = \frac{F}{M \times M} \) Again Workdone = force x displacement therefore \( PV = \frac{F}{A} \times V = \frac{F}{M^2} \times \frac{M^3}{1} \\ PV = \frac{F}{x\times x} \times \frac{x \times x \times x}{1} = F \times x \) = force x displacement

Spaces and Forces: The differences observed in solids, liquids, and gases can primarily be attributed to the arrangement, spacing, and intermolecular forces acting between the molecules. Solids: Tightly packed molecules with strong intermolecular forces, giving a definite shape and volume. Liquids: Close molecules that can move past each other, resulting in a definite volume but no definite shape. Gases: Widely spaced molecules with weak intermolecular forces, filling the entire container.

Workdone = product of the charge Q, and the p.d V. i.e Work done = QV = 600 * 10-6 *2.1 * 103 =1.8j

\( R = \frac{\rho L}{A}\) \(\implies \rho\) = \(\frac{R.A}{L}\) = \(\frac{0.1 \times 3.14}{2} \left ( \frac{10^{-3}}{2} \right)^2\) \(\therefore\) \(\rho\) = \(\frac{0.1 \times 3.14 \times 10^{-6}}{2 \times 4}\) = 0.3925 x 10\(^{-7}\) OR 3.930 x 10 \(^{-8}\)\Omegam

In general, Power = IV; \( \implies 40KW = IV \\ \text{Therefore } 40000 = 1 \times 800\\ \implies I = \frac{40000}{800} = 50A,\text{tune the current through} \) Resistor = 50A power loss= I2R = 502 x 2 = 2500 x 2 = 5.0 x 103W

A moving iron ammeter can measure both alternating current (AC) and direct current (DC) due to its design, which allows it to respond to the magnetic field created by both types of current.

since the wires are parallel to each other, the forces which they exert on each other are equal and opposite, since they must agree with newton's law of action an reaction. Hence the force exerted by the wire is equally 5 x 10-5

Lenz's lew of electrical magnetic induction is usefully referred to as the law of conservation of energy

Power of the lamp = 60W; \(I = {\frac{P}{V}}\) \(I = {\frac{60}{120}}\) = \(0.5A\) \(R = {\frac{V}{I}}\) \(R_mains\) = \(\frac{220}{0.5}\) = \(440\Omega\) \(R_lamp\) = \(\frac{120}{0.5}\) = \(240\Omega\) \(\therefore\) The non-inductive resistance to keep the lamp = \((440 - 240)\Omega\) = \(200\Omega\)

80 days = 4 half lives (i.e 4 x 20dqs = 80dq) But the fraction always left by any radioactive material after 4 half lives is \( \left ( \frac{1}{2} \right)^4 = \frac{1}{16} \)

\(\lambda\) = \(\frac{h}{mv}\), then, h = mv\(\lambda\) m = \(\frac{h}{v \lambda}\) = \(\frac{6.63 \times 10^{-34}}{6.0 \times 10^{-7} \times 3 \times 10^8}\) = 3.68 x 10\(^{-36}\) kg OPtion A is the closest.

Nitric acid is not used in the preparation of hydrogen because it is a strong oxidizing agent, meaning it readily oxidizes the hydrogen gas produced during the reaction with a metal, converting it to water and thus preventing the collection of hydrogen gas; instead, acids like dilute hydrochloric acid or sulphuric acid are preferred for hydrogen production as they do not oxidize hydrogen.