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