Author: mrchuakh

24P1Q30

30 D

Mass of constituents
$\displaystyle \begin{aligned}&=30{{m}_{n}}+26{{m}_{p}}+26{{m}_{e}} \\ &=30(1.00867u)+26(1.00728u)+26(0.00055u) \\ &=56.46368u \end{aligned}$

Mass defect
$\displaystyle \begin{aligned}&=56.4638u-55.93494u \\ &=0.52875u \\ &=0.52875\times 1.66\times {{10}^{-27}} \\ &=8.78\times {{10}^{28}}\text{ kg} \end{aligned}$

24P1Q21

21 A

$\displaystyle \begin{aligned}\varepsilon &={{V}_{PQ}} \\ &=\frac{2.0}{2.0+1.0}\times 0.60 \\ &=0.40\text{ V} \end{aligned}$

The unknown cell must be oriented in the same direction as the driver cell (0.60 V). If it is oriented the other way, a null deflection is not possible.

24P1Q29

29 B

Option B is the only option that kind of put Δx Δp ⪆ h into words.

24P1Q28

28 D

Momentum of photon, $\displaystyle p=\frac{h}{\lambda }$

Energy of photon, $\displaystyle E=\frac{hc}{\lambda }=pc$

24P1Q27

27 A

First calculate the current in the transmission cables.

$\displaystyle \begin{aligned}(P&=VI) \\ 100.00\times {{10}^{6}}&=400\times {{10}^{3}}I \\ I&=250\text{ A} \end{aligned}$

Total resistance of the cables, $\displaystyle r=2\times 100\times 0.030=6.0\text{ }\Omega\text{ }$

Power lost in the cables $\displaystyle ={{I}^{2}}r={{250}^{2}}\times 6.0=375000\text{ W}$

Power input to the sub-station $\displaystyle =100.00\times {{10}^{6}}-375000=99.625\text{ MW}$

24P1Q26

26 B

The AC voltage changes from +V₀ to −V₀ in half a cycle.

24P1Q25

25 D

In one minute, distance flown $\displaystyle =150\times 60=9000\text{ m}$

Flux cut $\displaystyle \phi =BA=(1.0\times {{10}^{-5}})(9000\times 80)=7.2\text{ Wb}$

24P1Q24

24 B

Since the magnetic force provides the required centripetal force, $\displaystyle Bqv=\frac{m{{v}^{2}}}{r}\text{ }\Rightarrow \text{ }r=\frac{mv}{Bq}$

$\displaystyle T=\frac{2\pi r}{v}=\frac{2\pi }{v}\frac{mv}{Bq}=\frac{2\pi m}{Bq}$

T is directly proportional to m, inversely proportional to B and q, independent of v and r.

24P1Q23

23 C

Option A: F_E is outward. F_B is inward. If the two forces cancel out, net force will be zero. This allows the ion to move along a straight line at a constant speed.

Option B: F_E is in the downward and rightward direction. F_B is upward. The vertical components cancel out, net force will be rightward. This allows the ion to move straight rightward at increasing speed.

Option C: F_E is inward. F_B is downward. The net force will be inward and downward. It is not possible for the ion to travel along a straight line.

Option D: F_E is rightward. F_B is zero. The net force will be rightward. This allows the ion to move straight rightward at increasing speed.

24P1Q22

22 A

As temperature rises, thermistor’s resistance decreases.

Reading on A₁ must increase since the total resistance of the circuit has decreased.

As the thermistor’s resistance decreases, the effect resistance across the thermistor also decreases (Think PDP).

Reading on A₂ must decrease since p.d. across the resistor has decreased.