Author: mrchuakh

23P1Q21

21 B

Option A: Would have been correct for “P.d. across the internal resistance of a cell”.

Option B: Correct.

Option C: Would have been correct if “total energy dissipated in it over a period of time” is replaced by “Rate of energy dissipation in the resistor (power)”.

Option D: The statement is a bit vague and unconventional but it is not totally wrong actually. Energy is converted from electrical to non-electrical in the cell due to the internal resistance of the cell. Perhaps would be more correct if “to move a unit charge” is replaced by “per unit charge moving”.

23P1Q20

20 B

$\displaystyle \begin{aligned} (V&=k\frac{Q}{r}) \\ \frac{{{V}_{P}}}{{{V}_{R}}}&=\frac{1}{SP}\div \frac{1}{SR} \\ \frac{{{V}_{P}}}{636}&=\frac{2\sqrt{2}}{3} \\ {{V}_{P}}&=599.6\text{ V} \end{aligned}$

23P1Q19

19 D

Because the protons experience a constant downward force, and with an initial rightward velocity, it travels along a parabolic path towards plate Y.

23P1Q18

18 A

Fringe width x (which is equivalent to fringe separation for the double slit) is directly proportional to screen distance D. $\displaystyle (\Delta y=\frac{L\lambda }{D})$

As t progresses and D increases at a constant rate, so should x.

23P1Q16

16 D

$\displaystyle (\sin {{\theta }_{1}}=\frac{\lambda }{b})$

Assume small diffraction angles, so first minimal angle can be written as $\displaystyle {{\theta }_{1}}=\frac{\lambda }{b}$ .

In addition, width of the central maximum x can be written as the arc length subtended by angle $\displaystyle 2{{\theta }_{1}}$ .

In other words, $\displaystyle x=D\times 2{{\theta }_{1}}$

Substitute $\displaystyle {{\theta }_{1}}=\frac{\lambda }{b}$ : $\displaystyle \begin{aligned} & x=D\times 2\frac{\lambda }{b} \\ & bx=D\times 2\frac{c}{f} \\ & b=\frac{2cD}{fx} \end{aligned}$

23P1Q17

17 A

$\displaystyle \begin{aligned} d\sin \theta &=n\lambda \\ \frac{0.001}{N}\sin 30{}^\circ &=2(500\times {{10}^{-9}}) \\ N&=500 \end{aligned}$

23P1Q15

15 B

Intensity after the horizontally polarising filter $\displaystyle ={{I}_{0}}$

Intensity after the 45° polarising filter $\displaystyle ={{I}_{0}}{{(\cos 45{}^\circ )}^{2}}=0.5{{I}_{0}}$

Intensity after the vertically polarising filter $\displaystyle =(0.5{{I}_{0}}){{(\cos 45{}^\circ )}^{2}}=0.25{{I}_{0}}$

23P1Q14

14 D

Wave X leads wave Y by $\displaystyle \frac{2\text{ units}}{6\text{ units}}=\frac{1}{3}$ cycle, corresponding to 120°.

Intensity is proportional to amplitude-square.

Ratio of intensities $\displaystyle ={{(\frac{{{A}_{x}}}{{{A}_{y}}})}^{2}}={{(\frac{3}{2})}^{2}}=2.25$

23P1Q13

13 B

Mass of each molecule, $\displaystyle \begin{aligned} & m=(2.9\times {{10}^{-2}})\div {{N}_{A}} \\ & =(2.9\times {{10}^{-2}})\div (6.02\times {{10}^{23}}) \\ & =4.817\times {{10}^{-26}}\text{ kg} \end{aligned}$

Maximum speed of air molecule, $\displaystyle \begin{aligned} & {{v}_{\max }}=\omega {{x}_{0}} \\ & =(\frac{2\pi }{0.003})(2\times {{10}^{-6}}) \\ & =4.189\times {{10}^{-3}}\text{ m }{{\text{s}}^{-1}} \end{aligned}$

Maximum KE of air molecule, $\displaystyle \begin{aligned} & K{{E}_{\max }}=\frac{1}{2}(4.817\times {{10}^{-26}}){{(4.189\times {{10}^{-3}})}^{2}} \\ & =4.23\times {{10}^{-31}}\text{ J} \end{aligned}$

23P1Q12

12 C

P’s temperature rises more steeply. So P has a smaller specific heat capacity. (Think $\displaystyle P=mc\frac{dT}{dt}$ )

P takes longer to melt completely. So P has a larger specific latent heat of fusion. ( $\displaystyle P\Delta t=mL$ )