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

N = N\(_o\)(\(\frac{1}{2}\))\(^{\frac{t}{T_{\frac{1}{2}}}}\) N = 2g, N\(_o\) = 128g, t =24days and \(T_{\frac{1}{2}}\) = ? 2 = 128(\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\) \(\frac{2}{128}\) = (\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\) \(\frac{1}{64}\) = (\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\) (\(\frac{1}{2}\))\(^6\) = (\(\frac{1}{2}\))\(^{\frac{24}{T_{\frac{1}{2}}}}\) 6 = \(\frac{24}{T_{\frac{1}{2}}}\) \(T_{\frac{1}{2}}\) = \(\frac{24}{6}\) = 4 days.

V\(_1\) = U - gt U = V + gt = 0 + 10t h = ut\(\frac{1}{2}\) - gt\(^2\), 50 = 10t\(^2\) \(\frac{1}{2}\) x 10 t\(^2\) 50 = 10t\(^2\) - 5t\(^2\) 50 = 5t\(^2\) = 10 t\(^2\) = 10 t = \(\sqrt{10}\) = 3.16s (ii) V = U - gt, O = U - 10 x 3.16 U = 31.16

Explanation: This is due to the accumulation of hydrogen bubbles in the positive copper plate, thereby insulating the electrodes. The hydrogen-zinc pair sets up opposing e.m.f. that reduces the e.m.f. originally established. Correction: It can be corrected by introducing a suitable depolarises.

Fresh water is less dense than the salt water, hence when the ship sails in fresh water, it will sink more, thus the fraction of the ship above the water surface will decrease.

\(\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\) Given \(V_{2} = 2V_{1}\) \(T_{1} = 27^\circC + 273 = 300k\) \(\frac{V_{1}}{300} = \frac{2V_{1}}{T_{2}}\) \(\frac{1}{300} = \frac{2}{T_{2}} \implies T_{2} = 600k or 327^\circC\)

Weight of handbag(W) = 162N acting vertically downward Angle of pull (\(\theta\)) = 60º to the vertical Since the handbag is in equilibrium, the total vertical force must equal the weight of the handbag. Let F be the force exerted by each student on the handle. The vertical component of the force from each student is given by: Fv = F cos 60º but there are two students 2Fcos60º = 162 2F(0.5) = 162 ( cos60º = 0.5) F = 162 ( since 2 x 0.5 = 1) Therefore, the force exerted by each student on the handbag is: 162N ​

Refractive Index = Real Depth/App. Depth 4/3 = 6/app. depth App Depth = (3 x 6)/4 = 4.5m

1 2 3 Dry 30^\circC 29^\circC 25^\circC Wet 22^\circC 22^\circC 21^\circC Diff 8^\circC 7^\circC 4^\circC \(Relative Humidity \propto \frac{1}{temp}\) Hence, There is a steady increase in the relative humidity between Day 1 and Day 3

The airtight stopper prevents heat loss by convection because it is tightly sealed.

For a simple pendulum, the period T = 2π \sqrt 1/g Since g, 2 and π are constant, => T ∝ √l ∴ T = K√l Let the original length = l thus for the length l, period T = 17s ∴ 17 = K√1 .................. [1] Again, for the new length [1 - 1.5] the period T = 8.5s ∴ 8.5 = k \sqrt [1-1.5] ...............[2] thus dividing eqn [1] by eqn[2] we have 17/8.5 = √(l/l-1.5) ∴2 = √(l/l-1.5) =>4 = (l/l-1.5) ∴l = 4[l-1.5] = 4l - 6.0 ∴4l - l = 6.0 3l = 6.0 l = 2.0m

given: u = 15m/s, a = 4m/s\(^2\), t = 10secs, S = ? Using: S = ut + \(\frac{1}{2}\)at\(^2\) S = 15 x 10 + 0.5 x 4 x 100 S = 350m.

Taking moment about the 40cm mark Total clockwise moments = total anti-clockwise moment m x 30 = 90 x 10 + 72 x 30 30m = 900 + 2160 = 3060 m = \(\frac{3060}{30}\) = 102g.

period = 0.25 x 4 = 1.0sec, time taken to make one complete circle.

V\(_{max}\) = \omegar \omega = 2\(\pi\)f but f = \(\frac{1}{T}\) \omega = \(\frac{2\pi}{T}\) = \(\frac{2\pi}{4\pi}\) ( since T = 4\(\pi\) given) \omega = 0.5rads/secs V\(_{max}\) = \omegar = 0.5 x 3 = 1.5m/s ( r = 3cm given)