12th Grade > Physics
NEWTON S LAWS OF MOTION MCQs
Total Questions : 30
| Page 1 of 3 pages
Answer: Option C. -> 2×106 N
:
C
Thrust on the rocket F=udmdt=5×104(40)=2×106N
:
C
Thrust on the rocket F=udmdt=5×104(40)=2×106N
Answer: Option C. -> 1.5×10−3N−s
:
C
Impulse=force×time=50×10−5×3=1.5×10−3N−s
:
C
Impulse=force×time=50×10−5×3=1.5×10−3N−s
Answer: Option A. -> 40 cm
:
A
Distance covered by m2=50cm/s×0.4s=20cmSince the m1 mass cover double distance therefore S=2×20=40cm
:
A
Distance covered by m2=50cm/s×0.4s=20cmSince the m1 mass cover double distance therefore S=2×20=40cm
Answer: Option A. -> 480 N
:
A
v1=−12 m\s and v2=+20m/s [because direction is reversed]
m=150gm=0.15kg,t=0.01sec
Force exerted by the bat on the ball F=m[v2−v1]t=0.15[20−(−12)0.01]=480Newton
:
A
v1=−12 m\s and v2=+20m/s [because direction is reversed]
m=150gm=0.15kg,t=0.01sec
Force exerted by the bat on the ball F=m[v2−v1]t=0.15[20−(−12)0.01]=480Newton
Answer: Option C. -> All will be at the same distance
:
C
For W, 2W, 3W apparent weight will be zero because the system is falling freely. So there will be no extension in any spring i.e. the distances of the weight from the rod will be same.
:
C
For W, 2W, 3W apparent weight will be zero because the system is falling freely. So there will be no extension in any spring i.e. the distances of the weight from the rod will be same.
Answer: Option C. -> 40 km/hour
:
C
Initially the wagon of mass 1000 kg is moving with velocity of 50 km/h
So its momentum =1000×50kg×kmh
When a mass 250kg is dropped into it. New mass of the system = 1000+250 = 1250 kg
Let v is the velocity of the system.
By the conservation of linear momentum : Initial momentum = Final momentum
1000×50=1250×v
v=50,0001250=40kmh.
:
C
Initially the wagon of mass 1000 kg is moving with velocity of 50 km/h
So its momentum =1000×50kg×kmh
When a mass 250kg is dropped into it. New mass of the system = 1000+250 = 1250 kg
Let v is the velocity of the system.
By the conservation of linear momentum : Initial momentum = Final momentum
1000×50=1250×v
v=50,0001250=40kmh.
Question 9. Three masses of 15 kg. 10 kg and 5 kg are suspended vertically as shown in the fig. If the string attached to the support breaks and the system falls freely, what will be the tension in the string between 10 kg and 5 kg masses. Take g=10ms−2 . It is assumed that the string remains tight during the motion
Answer: Option D. -> Zero
:
D
In the condition of free fall, tension becomes zero.
:
D
In the condition of free fall, tension becomes zero.
Answer: Option C. -> 8N
:
C
By comparing the above problem with general expression. T1=(m2+m3)m1+m2+m2=(3+5)102+3+5=8Newton
:
C
By comparing the above problem with general expression. T1=(m2+m3)m1+m2+m2=(3+5)102+3+5=8Newton