Question Library

Total energy in parred connection is equal to emt of one cell, = 2v but total internal resistance is given by (1)/r = (1)/r1 + (1)/r2 + (1)/r3 = (1)/2 + (1)/2 + (1)/2 = (3)/2 ∴ r = (2)/3 = 0.67Ω

presbyopia is loss of accommodation

\(\frac{1}{\text{f}}\) = \(\frac{1}{\text{u}}\) + \(\frac{1}{\text{v}}\) \(\frac{1}{\text{f}}\) = \(\frac{1}{25}\) + \(\frac{1}{-50}\) 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

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

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 \(\frac{\text{r}}{Q_1}\) = \(\frac{- q}{4\piεor}\)

f1 = 225Hz, T1 = 225N f2 = ? T2 = 324N But f ∝ \sqrtT => f = K\sqrtT => K = f/\sqrtT ∴ f1/\sqrtT1 = f2/\sqrtT2 225/\sqrt225 = f2/\sqrt324 f2 = 225\sqrt324 \sqrt225 f2 = 270Hz

p = \(\frac{M}{V} = \frac{M}{\frac{4}{3} \pi (\frac{d}{\pi})^3}\) = \(\frac{6m}{\pi d^3}\)

P = \(\frac{M}{AL}\) L = \(\frac{M}{CA}\) = \(\frac{1.76}{7000 \times 0.001257}\) = 0.2m

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

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

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

Expansion and Contraction: The density of mercury affects how it expands and contracts with temperature changes. A liquid with a consistent and predictable expansion rate is essential for accurate temperature readings. Calibration: The density of the liquid must be known to calibrate the thermometer accurately. This ensures that the scale on the thermometer corresponds correctly to temperature changes. Thermal Conductivity: While conductivity can influence how quickly a liquid reaches thermal equilibrium with its surroundings, it is not as critical as density for the overall accuracy of the thermometer. But in choosing mercury as a thermometric liquid, the density of mercury is not usually considered.

K.E = P.E = mgh = 0.3 x 10 x 100 - 0.3 x 10 x 75 = 75J