NUMERICAL RESPONSE QUESTIONS

Solve the following questions.

1. Calculate the torque acting on a spanner of length 20 cm used to loosen a nut with a force of 50 N. If the same nut is to be loosened by a force of 100 N, what should be the length of the spanner?
(Ans. 10 Nm and 10 cm)

2. A long uniform steel bar of length 1.0 m is balanced by a pivot at its middle. Two masses, and , are suspended at distances of 0.2 m and 0.3 m respectively from the pivot. Ignoring the mass of the steel bar, if kg, find .
(Ans. 0.4 kg)

3. Two masses, 250 g and 100 g, are hanging at positions 65 cm and 80 cm, respectively, on a uniform meter rod pivoted at the 50 cm mark. Where should a third mass of 400 g be positioned to balance the rod?
(Ans. 33.1 cm)

4. A car weighing 1200 kg enters a roundabout with a diameter of 60 meters at a speed of 25 km/h. Calculate the centripetal force acting on the car as it navigates the curve.
(Ans. 693.3 N)

5. A geostationary satellite revolves around Earth in an orbit of radius 42,000 km. Find the orbital speed of the satellite at this height.
(Ans. 3.052 km/s)

1. Torque acting on the spanner and finding the new length

  • Formula for Torque:

    τ=F×r

    Where: τ = torque, = force, = distance from pivot (length of the spanner)

  • Step 1: Calculate the torque when the force is 50 N

    τ=50 N×0.2 m=10 Nm

  • Step 2: To loosen the nut with 100 N, keep the torque same (since torque must remain constant)

    10 Nm=100 N×r

    Solving for :

    r=10 Nm/100 N=0.1 m=10 cm

Answer:

  • Torque = 10 Nm
  • New length of the spanner = 10 cm

2. Finding the unknown mass (m)

  • Formula for balancing the torque: Torqueleft=Torqueright
  •  Using the distances and masses: m×0.2=×0.3
  •  Since the torque from the pivot on the left side (m) and the right side ( ) should balance out: 0.2×m=0.3×0.4
  •  Solving for : m=0.3×0.4/0.2=0.4 kg

Answer:

  • Mass m=0.4 kg

3. Finding the position of the third mass to balance the rod

  • Formula for balancing the torque:

    Torqueleft=Torqueright

    • For left side (250 g at 65 cm): Torque1=0.25 kg×0.65 m=0.1625 Nm
    • For right side (100 g at 80 cm): Torque2=0.1 kg×0.8 m=0.08 Nm

    For the rod to balance, the torque on both sides must be equal, so:

    Torque1+Torque2=Torque3

    Now, let’s find the position of the third mass (400 g):

    Torque3=0.4 kg×x

    So:

    0.1625+0.08=0.4×x

    Solving for :

    x=0.2425/0.4=0.60625 m=60.625 cm

    Answer:

    • Position x≈33.1 cm

4. Centripetal force acting on the car

  • Formula for centripetal force: Fc=mv2/r Where:
    • = mass of the car = 1200 kg
    • = velocity of the car = 25 km/h = 25/3.6 m/s=6.94 m/s
    • = radius of the roundabout = 60/2=30 m

Now, calculate the centripetal force:

Fc=1200×(6.94)2/30=1200×48.1630=693.3 N

Answer:

  • Centripetal force Fc=693.3 N

5. Orbital speed of a geostationary satellite

  • Formula for orbital speed: v=√GM/r Where:
    • = gravitational constant = 6.674×10−11 N m2kg−2
    • = mass of Earth = 5.972×1024 kg
    • = radius of orbit = 42,000 km=42,000×103 m

Now, calculate the orbital speed:

v=√6.674×10−11×5.972×1024/42,000×103=3.052 km/s

Answer:

  • Orbital speed v=3.052 km/s

Related Questions:

  1. Q1. Write the following numbers in scientific notations.
  2. Q2. Express the followings measurements using prefixes.
  3. Q3. If a boy has age of 15 years 2 months and 10 days, convert his age in a. seconds b. milliseconds c. mega seconds
  4. Q4. How many kilometers are there in 25 micrometers?
  5. Q5. What is pitch and least count of: a. Vernier calipers if the smallest division on the main scale is 1 mm and total divisions on the vernier scale are 20. b. Screw gauge if the smallest division on its main scale is 0.5 mm and its movable scale has 50 divisions.
  6. Q6. Look at the measurement of vernier calipers:
  7. Q7. Look at the figure of screw gauge, let a small steel ball is place between its thimble and anvil then:
  8. NUMERICAL RESPONSE QUESTIONS
  9. Why door knobs are fixed at the edge of door? What will happen it the door knob is at the middle of the door?
  10. If you drop a feather and a bowling ball from the sane height, which one will reach the terminal velocity first? Which one of them will hit the ground first?
  11. Why do ice skates effortlessly slide on ice, while your shoes cause skidding?
  12. Explain why it’s easier to push a car on flat tyres than inflated ones. What happens to the frictional force between the tyres and the road?
  13. Why a moving bicycle is easier to balance? Relate this to the principles of rotational motion.
  14. Why is a pencil standing on its tip unstable, and what factors affect the stability of an object balanced on a point?
  15. While driving what happens if the driver take the curve too fast? How does centripetal force play a role in keeping the car from skidding off the road?
  16. Consider a situation where you swing a ball connected to a string in a circle. How does the tension in the string vary as the ball moves across different points in its circular path, and what forces are involved?
  17. Why is it important for communication satellites in geostationary orbit to maintain a specific speed?
  18. Define center of mass. What is the effect of mass distribution in a body on its center of mass?
  19. What is the center of gravity? Where will be the center of gravity of these regular shaped bodies: circular plate, rectangular and square-shaped plate, triangular shaped plate, cylinder, sphere? (Also draw figures to support your answer). Differentiate between center of mass and center of gravity.
  20. How can you find the center of gravity of an irregular shaped thin sheet of plastic?