{"questions":[{"id":"mdcat-phy-kin-01","v":1,"subject":"physics","chapter":"kinematics","seed":true,"difficulty":1,"stem":"The slope of a velocity–time graph gives:","options":["Displacement","Acceleration","Distance","Speed"],"answer":1,"explain":{"en":"Slope = change in velocity ÷ change in time = acceleration. The area under a velocity–time graph gives displacement.","ur":"ڈھلوان = رفتار میں تبدیلی ÷ وقت میں تبدیلی = اسراع۔ ولاسٹی ٹائم گراف کے نیچے کا رقبہ ہٹاؤ (ڈسپلیسمنٹ) دیتا ہے۔"},"source":{"kind":"parcha-original"},"status":"sample","exam":"mdcat"},{"id":"mdcat-phy-kin-02","v":1,"subject":"physics","chapter":"kinematics","difficulty":1,"stem":"The area under a velocity–time graph represents:","options":["Acceleration","Force","Displacement","Momentum"],"answer":2,"explain":{"en":"Velocity × time = displacement, so the area between the graph and the time axis is the displacement (area below the axis counts as negative).","ur":"رفتار × وقت = ہٹاؤ، اس لیے گراف اور وقت کے محور کے درمیان رقبہ ہٹاؤ ہے (محور سے نیچے کا رقبہ منفی گنا جاتا ہے)۔"},"source":{"kind":"parcha-original"},"status":"sample","exam":"mdcat"},{"id":"mdcat-phy-kin-03","v":1,"subject":"physics","chapter":"kinematics","difficulty":1,"stem":"A car starts from rest with a uniform acceleration of 2 m/s². Its velocity after 5 s is:","options":["10 m/s","2.5 m/s","7 m/s","25 m/s"],"answer":0,"explain":{"en":"v = u + at = 0 + 2 × 5 = 10 m/s.","ur":"v = u + at = 0 + 2 × 5 = 10 m/s۔"},"source":{"kind":"parcha-original"},"status":"sample","exam":"mdcat"},{"id":"mdcat-phy-kin-04","v":1,"subject":"physics","chapter":"kinematics","difficulty":2,"stem":"The same car (from rest, 2 m/s²) covers what distance in those 5 s?","options":["10 m","50 m","20 m","25 m"],"answer":3,"explain":{"en":"s = ut + ½at² = 0 + ½ × 2 × 5² = 25 m. Check: average velocity (0 + 10)/2 = 5 m/s, times 5 s = 25 m.","ur":"s = ut + ½at² = 0 + ½ × 2 × 5² = 25 m۔ تصدیق: اوسط رفتار (0 + 10)/2 = 5 m/s، ضرب 5 s = 25 m۔"},"source":{"kind":"parcha-original"},"status":"sample","exam":"mdcat"},{"id":"mdcat-phy-kin-05","v":1,"subject":"physics","chapter":"kinematics","difficulty":2,"stem":"Ignoring air resistance, a projectile has its maximum horizontal range when launched at:","options":["30°","60°","45°","90°"],"answer":2,"explain":{"en":"Range R = v² sin 2θ / g, which is greatest when sin 2θ = 1, i.e. 2θ = 90°, θ = 45°. Angles of 30° and 60° give equal (smaller) ranges.","ur":"رینج R = v² sin 2θ / g، جو sin 2θ = 1 پر سب سے زیادہ ہے، یعنی θ = 45°۔ 30° اور 60° پر رینج برابر مگر کم ہوتی ہے۔"},"source":{"kind":"parcha-original"},"status":"sample","exam":"mdcat"},{"id":"mdcat-phy-kin-06","v":1,"subject":"physics","chapter":"kinematics","difficulty":1,"stem":"A stone is dropped from rest. Taking g = 10 m/s², its speed after 2 s is:","options":["5 m/s","20 m/s","10 m/s","40 m/s"],"answer":1,"explain":{"en":"v = u + gt = 0 + 10 × 2 = 20 m/s (air resistance ignored).","ur":"v = u + gt = 0 + 10 × 2 = 20 m/s (ہوا کی مزاحمت نظرانداز)۔"},"source":{"kind":"parcha-original"},"status":"sample","exam":"mdcat"}]}