MDCAT Physics Numericals with Solutions 2026 (g = 10)

Two forces of $3\mathrm{N}$ and $4\mathrm{N}$ act at right angles. Find the magnitude of the resultant.

1. $R=\sqrt{A^2+B^2+2AB\cos \theta }.Here\theta =90^{\circ },so\cos \theta =0.$
2. $R=\sqrt{9+16}=\sqrt{25}=5\mathrm{N}.$
3. $7\mathrm{N}$ is the parallel sum. $1\mathrm{N}$ is the anti-parallel difference. Neither applies.

$5\mathrm{N}$ — a 3–4–5 triangle. If you wrote $7\mathrm{N}$, you added magnitudes as scalars.

A car starts from rest and accelerates uniformly at $4m/s$ ² for $5\mathrm{s}$. Distance covered?

1. $u=0$, so $s=ut$ + ½ $a{t}^2=\frac{1}{2}(4)(25)=50\mathrm{m}$.
2. at² without the ½ is $100\mathrm{m}$. at is $20m/s$, a change in speed, not a distance.

$50\mathrm{m}$. If your option was $100\mathrm{m}$, you dropped the half.

$u=20$ m/s, $\theta =45^{\circ },g=10$ m/s². Horizontal range on level ground?

1. $R=u^2$ sin2 $\theta$ / g. At $45^{\circ },2\theta =90^{\circ },\sin =1.$
2. $R=(400)/10=40\mathrm{m}$ — the maximum range on level ground.

$40\mathrm{m}$. Complementary angles share range; $45^{\circ }$ is the peak.

A $50\mathrm{kg}$ student stands on a scale in a lift accelerating upward at $2m/s$ ². $g=10.$ Scale reading?

1. $N-mg=ma\Rightarrow N=m(g+a)=50(12)=600\mathrm{N}.$
2. $500\mathrm{N}$ is true weight ($a=0).400\mathrm{N}$ is m(g − a), the down case.

$600\mathrm{N}$. Apparent weight rises when the lift accelerates up.

A $10\mathrm{N}$ force displaces a body $4\mathrm{m}$ at $60^{\circ }$ to the force. Work done?

1. $W=Fs\cos \theta =10\times 4\times \frac{1}{2}=20\mathrm{J}.$
2. $40\mathrm{J}$ is Fs with the angle forgotten. $40\sqrt{3}$ used sin instead of cos.

$20\mathrm{J}$. Only the component of F along s does work.

Water in a pipe of radius $4\mathrm{cm}$ flows at $1m/s$. The pipe narrows to $2\mathrm{cm}$. Speed in the throat?

1. $A\propto r^2,so{A}_2/A_1=(2/4{)}^2=\frac{1}{4}.$
2. $A_1{v}_1=A_2{v}_2\Rightarrow v_2=4v_1=4m/s.$

$4m/s$. Half the radius is a quarter the area, four times the speed.

$m=0.25$ kg on $k=25$ N/m. Period? Does doubling amplitude change T?

1. $T=2\pi \sqrt{m/k}=2\pi \sqrt{0.25/25}=2\pi \sqrt{0.01}=2\pi (0.1)=0.2\pi s.$
2. Amplitude does not appear. Double A, T stays.

$T=0.2\pi$ s, independent of amplitude (ideal SHM).

A capacitor charged to $Q_0$ discharges through R. Charge at $t=RC?$

1. $Q=Q_0{e}^{-t/RC}=Q_0{e}^{-1}\approx 0.37Q_0.$
2. 0.63 $Q_0$ is how much has left (or how much a charging capacitor has gained).

About 37% remains. One time constant is not ‘empty’.

A cell $E=12$ V, $r=2\Omega$. Load for maximum power, and that power?

1. $R=r=2\Omega .$
2. $I=E/(R+r)=12/4=3\mathrm{A}.P=I^{2}R=9\times 2=18\mathrm{W}.$

Match R to r. $18\mathrm{W}$ in the load; an equal $18\mathrm{W}$ is wasted in r.

Same peak Vm. Compare $V_{dc}$ for half-wave and full-wave.

1. Half-wave: $V_{dc}$ = Vm/ $\pi$.
2. Full-wave: $V_{dc}$ = 2Vm/ $\pi$ — twice the humps, twice the average.
3. $f_{out}$ is f versus 2f. That is a separate item.

Full-wave has the higher DC output and double the ripple frequency.

A photon of $\lambda =400\mathrm{nm}$. Energy in eV? Use hc ≈ $1240\mathrm{eV}$ ·nm.

1. $E=1240/400=3.1\mathrm{eV}.$
2. An $800\mathrm{nm}$ photon is $1.55\mathrm{eV}$ — half the energy, twice the wavelength.

$3.1\mathrm{eV}$. Shorter $\lambda$, larger E. Brightness does not enter.

Wavelength of the 3 → 2 jump in hydrogen?

1. $E_3=-13.6/9\approx -1.51\mathrm{eV},E_2=-13.6/4=-3.4\mathrm{eV}.$
2. $\Delta E\approx 1.89\mathrm{eV}.\lambda \approx 1240/1.89\approx 656\mathrm{nm}.$
3. That is H- $\alpha$, Balmer, visible red.

$656\mathrm{nm}$. 2→1 is Lyman- $\alpha$ at $122\mathrm{nm}$, UV, not visible.

A sample has $T_{1/2}$ = 2 h. Fraction remaining after 6 h?

1. $n=6/2$ = 3 half-lives.
2. $Remaining=(1/2{)}^3=1/8.$
3. 1/3 is the trap that treats half-life as a third.

1/8 remains. Activity is also 1/8 of the original, same clock.