NUMERICAL RESPONSE QUESTIONS

Solve the following numerical questions.

1. Consider a spring with a spring constant of 8000 N/m. If a force of 500 N is applied to the spring, what will be the displacement of the spring?
(Ans. 6.25 cm)

2. In a force multiplier, the small piston has a diameter of 15 cm and the large piston has a diameter of 30 cm. If 2500 N force is applied to the small piston, how much force will be produced on the large piston?
(Ans. 10000 N)

3. A hydraulic car lift lifts a car of mass 1000 kg when we apply a force of 50 N on the small piston. The radius of its small piston is 20 cm. Find the radius of its large piston.
(Ans. 78.4 cm)

4. Water column in a beaker is 70 cm. Find the pressure of water in the beaker. Take the density of water4. as 1000 kg/m³.
(Ans. 6.86 kPa)

5. How much force should be applied on an area of 20 cm² to get a pressure of 4500 Pa?
(Ans. 9 N)

1. Spring Displacement

We use Hooke’s Law for this:

F=k×x

Where:

  • F is the force applied (500 N)
  • is the spring constant (8000 N/m)
  • x is the displacement (what we need to find)

Rearrange the formula to solve for xx:

x=F/k

Substitute the values:

x=500/8000=0.0625 m

So the displacement is:

x=6.25 cm

Answer: 6.25 cm


2. Force Multiplier

In a force multiplier (hydraulic press), the force is related to the areas of the pistons and the force applied using the principle of Pascal’s Law:

F1/A1=F2/A2

Where:

  • F1 is the force applied on the small piston (2500 N)
  • A1 and A2 are the areas of the small and large pistons
  • F is the force on the large piston (what we need to find)

The area of a piston is given by the formula for the area of a circle:

A=πr2

We can express the ratio of the forces in terms of the ratio of the areas:

F1/F2=A1/A2=(r1/r2)2

Since the diameters are given:

  • Small piston diameter d1=15 cm, so the radius r1=7.5 cm
  • Large piston diameter d2=30 cm, so the radius r2=15 cm

Now, we find the ratio of the forces:

F2/F1=(r2/r1)2=(157.5)2=4

Thus:

F2=4×F1=4×2500=10000 N

Answer: 10000 N


3. Radius of Large Piston in Hydraulic Lift

In this case, we use Pascal’s Law again, which states that the pressure is the same in both pistons. Thus:

F1/A1=F2/A2

Where:

  • F1=50 N (force on small piston)
  • F2=m×g=1000×9.8=9800 N (force on large piston, mass of car)
  • r1=20 cm=0.2 m (radius of small piston)
  • r2 (radius of large piston, which we need to find)

Using the formula for the area of a circle, A=πr2, the equation becomes:

F1/πr12=F2/πr22

Simplifying:

F1/r12=F2/r22

Rearranging to solve for r2:

r2=√F2r12/F1

Substitute the values:

r2=√9800×(0.2)2/50      =√9800×0.04/50      =√392/50         =√7.84    ≈2.8 m

So, the radius of the large piston is approximately:

r2≈78.4 cm

Answer: 78.4 cm


4. Pressure of Water in Beaker

The pressure at the bottom of a water column is given by:

P=ρgh

Where:

  • ρ=1000 kg/m3 (density of water)
  • g=9.8 m/s2 (acceleration due to gravity)
  • h=0.7 m (height of the water column)

Substitute the values:

P=1000×9.8×0.7=6860 Pa=6.86 kPa

Answer: 6.86 kPa


5. Force to Create a Given Pressure

The pressure is given by the formula:

P=F/A

Where:

  • P=4500 Pa
  • A=20 cm2=20×10−4 m2

Rearranging to solve for the force:

F=P×A

Substitute the values:

F=4500×20×10−4=9 N

Answer: 9 N