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The colour which an object presents to the eye is the colour of light it reflects. Red rose flower will therefore absorb any other colour and reflect red. Therefore when viewed in blue light, it has no colur to reflect and will therefore appear black. Although mixture of red light and blue light = magenta .

Dielectric constant = (c)/c° = (60)/20 = 3

\(\frac{1}{\text{f}}\) = \(\frac{1}{\text{u}}\) + \(\frac{1}{\text{v}}\) \(\frac{1}{\text{f}}\) = \(\frac{1}{25}\) + \(\frac{1}{-50}\) u = 25cm, v = - 50cm \(\frac{1}{\text{f}}\) = \(\frac{1}{u}\) - \(\frac{1}{50}\) = \(\frac{2 - 1}{50}\) f = 50cm P = \(\frac{1}{\text{f}}\) = \(\frac{1}{50}\) = 0.02cm. ( in dioptres = 2D) 1D = 1m\(^{-1}\)

Inductance (L) in series: 4 + 8 = 12 Therefore Energy stored = W = \(\frac{1}{2} LI^2\) = \(\frac{1}{2} \times 12 \times 10^2\) = 600J

Resistance in metals are proportional to temperature

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

T (1)/2 = (0.693)/\lambda = (0.693)/10-5 = 0.693 x 105 = 6.93 x 104 sec

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

Work function, w = hfo 6.6 x 10-34 x 1.3 x 1015 = 8.58 x 10-19

Metals and hydrogen are deposited at the cathode, while non-metals and oxygen are deposited on the anode ∴ metals Y will form the anode which is deposited at the cathode made of x

Power in Series Arrangement Total Resistance (R_series): R\(_{series}\) = R\(_1\) + R\(_2\) = 5Ω + 10Ω = 15Ω Total Power (P_series): P = \(\frac{V^2}{\text{R}}\) = \(\frac{24^2}{15}\) = 38.4W Power in parallel Arrangement Total Resistance(R\(_{parallel}\)): R\(_{parallel}\) = \(\frac{R_1 \times R_2}{R_1 + R_2}\) = \(\frac{5 \times 10}{5 + 10}\) = 3.33 Total Power(P_{parallel}\): P = \(\frac{V^2}{\text{R}}\) = \(\frac{24^2}{3.33}\) = 172.8W Ratio of Powers Power ratio = \(\frac{\text{Pseries}}{\text{Pparallel}}\) = \(\frac{38.4}{172.8}\) = \(\frac{1}{4.5}\) = \(\frac{2}{9}\) Therefore, there is no correct answer.

The resultant, R is given by pythagoras rule, R2 = P2 + Q2 = 62 + 82 = 36 + 64 = 100 R = √100 = 10.0N

Under neutral equilibrium, the body rolls and the position of the center of gravity , C. G is neither raised nor lowered

Electric potential is work done in taking a unit mass or charge from infinity to that point V = \(\frac{\text{W}}{\text{Q}}\) = \(\frac{\text{Fr}}{\text{Q}}\) = \(\frac{Q_1Q_2}{4\piεor^2}\) x \(\frac{\text{r}}{Q_1}\) = \(\frac{- q}{4\piεor}\)

m = v/u = -3 v = -3u f = 15cm 1/v + 1/U = 1/f 1/-3U + 1/u = 1/15 2/3u = 1/15 3U = 15 x 2 U = 30/3 = 10 cm