ques 1.......a boy walks from his home to his school at an average speed of 10 km per hour and returns to his home at an average speed of 15 km per hour . what is his average speed in km per hour during the entire round trip ?
ques 2........a thief running at a speed of 8 km per hour is chased by a police man whose speed is 10 km per hour if the thief is 100 metre ahead of the policemen what is the time required by the police men to catch the thief ?
ques 3....a man of mass 60 kg jumps at a speed of 4 meters per second from a stationary boat and as a result the boat moves off with a speed of 0.8 meters per second. what is the mass of the boat?
ques 4.....two persons start walking at a speed of 3 km per hour from a road intersection along two road that make an angle of 60 degree with each other . what will be the shortest distance separating them at the end of 20 minutes?
ques 5....a car starts at rest and moves on a straight road with a uniform acceleration of 10 m/s square for the first 10 seconds . during the next 10 seconds , the car moves with the velocity it has attained . what is the total distance covered by car in 20 seconds ?
ques 6....a crane can raise a load of mass 120kg vertically upward with a speed of 5 metres per second . if the acceleration due to gravity is 10 m/s square what is the power ( in kilowatts ) of the engine ?
ques 7.....water in a river is flowing from west to east at a speed of 5 metres per minute. a man on the south bank of the river , capable of swimming at a speed of 10 mete per minute , wishes to cross the river in the shortest possible time . in which direction must be swim ?
ques 8....a string can withstand a maximum tension of 25N . what is the greatest speed at which a body of mass 1kg can be whriled in a horizontal circle using 1 metre length of the string?
ques 9.....the escape velocity for a body projected vertically upwards from the surface of the earth is ve . if the body is projected in a direction making at angle of 30 degree with the vertical, what is the escape velocity ?
ques 10....the work done in a raising a body of mass 2kg to a height of 1 metre above the ground is ? (
the accelration due to gravity is to m/s square.
ques 11 ...a sound generator placed under water has a frequency of 580 hz. it produces waves of wavelength 2.5 m. what is the speed of sound in water ?
ques 12 ...a man standing in front of a high wall fires a gun. he hears the echo 2 seconds after the gun is fired . if the speed of sound is 340m/s . how far is the wall from the man. ?
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answers
1 ----12
2----- 3min
3 --------300kg
4-----1km
5-----1500m
6-----6kilowatt
7---5m/s
8---11.2km/s
9-----20 joule
10--------1450m/s
11------340m
1. Avg = (10+15) / 2 = 12.5km/h.
2. 100m = 0.1km.
d1 = V*t + 0.1 = 8t + 0.1 = The thief's distance at end of chase.
d2 = 10t = Policeman's distance at end of chase.
10t = 8t + 0.1,
10t - 8t = 0.1,
2t = 0.1,
t = 0.05h = 3min.
3. The velocity is inversly proportional to the mass:
Mass = (4/0.8) * 60kg = 300kg.
4. d = 3km/h + (20/60)h = 1km.
5. d1 = Vo*t + 0.5at^2,
d1 = 0 + 0.5*10*(10)^2 = 500m.
V1 = at = 10 * 10 = 100m/s.
d2 = Vo*t = 100m/s * 10s = 1000m/2.
Dt = d1 + d2 = 500 + 1000 = 1500m.
6. F = mg = 120kg * 10N/kg = 1200N.
Po = = FV = 1200 * 5m/s = 6000 Watts =
6 Kilowatts.
10. F = mg = 2kg * 10N/kg = 20N.
W = Fd = 20 * 1 = 20 Joules.
11. L = VT = V*(1/f) = 2.5m,
V*(1/580) = 2.5,
V = 2.5 * 580 = 1450m/s.
12. T = 2s / 2 = 1s = Time to reach the wall.
d = VT = 340m/s * 1s = 340m.
Question 1:
To find the average speed during the entire round trip, we need to calculate the total distance traveled and the total time taken.
The distance from home to school is the same as the distance from school to home. Let's call this distance "d".
To calculate the total distance, we add the distance from home to school and the distance from school to home: 2d.
The time taken to travel from home to school is given by: time = distance / speed = d / 10 km/h.
Similarly, the time taken to travel from school to home is given by: time = distance / speed = d / 15 km/h.
The total time for the round trip is the sum of these two times: d / 10 km/h + d / 15 km/h = (3d + 2d) / 30 km/h = 5d / 30 km/h.
Finally, we can find the average speed by dividing the total distance by the total time: Average speed = total distance / total time = 2d / (5d / 30 km/h) = 2d / (d / 6) = 12 km/h.
Therefore, the average speed during the entire round trip is 12 km/h.
Answer: The average speed during the entire round trip is 12 km/h.
Question 2:
To find the time required for the policeman to catch the thief, we need to calculate the time it takes for the policeman to cover the distance between them.
The thief is initially 100 meters ahead of the policeman. Let's call this distance "d".
The relative speed of the policeman with respect to the thief is the difference in their speeds: relative speed = policeman's speed - thief's speed = 10 km/h - 8 km/h = 2 km/h.
To convert this speed into meters per hour, we multiply by 1000: relative speed = 2 km/h * 1000 m/km = 2000 m/h.
Now we can calculate the time required for the policeman to cover the distance: time = distance / speed = d / 2000 m/h.
Plugging in the given value of 100 meters for the distance, we get: time = 100 m / 2000 m/h = 1/20 h.
To convert this to minutes, we multiply by 60: time = (1/20 h) * 60 min/h = 3 min.
Therefore, it takes the police 3 minutes to catch the thief.
Answer: The time required for the police to catch the thief is 3 minutes.
Question 3:
To find the mass of the boat, we can use the principle of conservation of momentum.
When the man jumps from the boat, the total momentum of the system (man + boat) is conserved.
The momentum of an object is given by the product of its mass and velocity: momentum = mass * velocity.
Before the man jumps, the momentum of the system is zero because both the man and the boat are stationary.
After the man jumps, the boat starts moving with a speed of 0.8 m/s in the opposite direction to the man's jump.
Let's assume the mass of the man is "m" kg, and the mass of the boat is "M" kg.
The initial momentum of the system is 0, and the final momentum is given by: momentum of man + momentum of boat = 0.
Since the momentum of the man is given by: momentum = mass * velocity = m * 4 m/s = 4m.
And the momentum of the boat is given by: momentum = mass * velocity = M * (-0.8 m/s) = -0.8M.
Setting the initial and final momentum equal to each other, we get: 4m + (-0.8M) = 0.
Simplifying the equation, we have: 4m = 0.8M.
Dividing both sides of the equation by 0.8, we get: 5m = M.
Therefore, the mass of the boat is 5 times the mass of the man, which is 5 * 60 kg = 300 kg.
Answer: The mass of the boat is 300 kg.
Question 4:
To find the shortest distance separating the two persons at the end of 20 minutes, we can use the concept of vector addition.
Since the two persons are walking along roads that make an angle of 60 degrees with each other, the distance between them can be found using the formula: distance = √(a^2 + b^2 - 2abcosθ).
Let's assume the speed of each person is 3 km/h. The distance traveled by each person in 20 minutes is given by: distance = speed * time = (3 km/h) * (20/60) h = 1 km.
Let's represent the distance between the two persons by "c" km.
Using the given angle of 60 degrees, and the distances a = b = 1 km, we can plug these values into the distance formula to find "c": c = √(1^2 + 1^2 - 2 * 1 * 1 * cos60°).
Simplifying the equation, we have: c = √(1 + 1 - 2 * 1 * 1 * 0.5) = √(2 - 1) = √1 = 1 km.
Therefore, the shortest distance separating the two persons at the end of 20 minutes is 1 km.
Answer: The shortest distance separating the two persons at the end of 20 minutes is 1 km.