MDCAT PAST PAPERS AND SYLLABUS
MDCAT Past Paper
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چار آپشن میں سے کسی ایک پر کلک کرنے سے جواب سرخ ہو جائے گا۔
1/2
1/4
1/8
1/16
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Explanation
After 3 half-lives, the remaining fraction is:
(1/2)³ = 1/8
4 m/s
2 m/s
3 m/s
Depends on masses
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Explanation
m₁ = 2 kg, u₁ = 5 m/s
- m₂ = 3 kg, u₂ = 1 m/s
- Both moving in same direction
- Relative speed of approach = u₁ − u₂ = 4 m/s
Two times
Four times
Eight times
Sixteen times
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Explanation
F = k * (q₁q₂) / r²
Changes:
- q₁ and q₂ → doubled → (2q)² = 4 times
- r → halved → (½r)² = ¼
So:
F’ = k * (4q²) / (¼r²) = 16F
Point of projection
Highest point
Between launching and highest point
All the points
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Explanation
- At launch point, both vertical and horizontal components of velocity exist.
- At the highest point, vertical component is zero.
- So, total velocity is maximum at launch.
Midway between them
Along the perpendicular bisector
Close to either charge
At Infinity
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Explanation
- At the midpoint, the electric fields due to each charge are equal in magnitude but opposite in direction, so they partially cancel out, making the net field weakest there.
Slightly less than critical velocity
Equal to critical velocity
Greater than critical velocity
Increasing gradually but less than critical velocity
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Explanation
- Critical velocity is the maximum velocity at which fluid flows in laminar form.
- Turbulent flow begins when the fluid exceeds this critical velocity .
9 J
3 J
6 J
8 J
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Explanation
KE = ½ mv² = ½ × 0.5 × 6² = 0.25 × 36 = 9 J
Path difference = λ/2
Path difference = λ
Path difference = λ/4
Path difference = 3λ/4
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Explanation
Constructive interference (max loudness) occurs when path difference = nλ
(λ, 2λ, etc.)
Always positive
Always negative
Zero
Depends on the path taken
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Explanation
Gravitational force acts downward, so moving upward means work is done against gravity → negative work.
1/°C
1/K
1/A
1/Ω
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Explanation
Temperature coefficient α is defined as (Δρ / ρΔT)
Unit = (Ω·m) / (Ω·m·°C) = 1/°C