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An object of 5kg mass is placed on a cylindrical stand having cross section . the radius of the base of the stand is 4cm . if the same object is placed on cylindrical stand with radius the base of stand 2cm, the pressure becomes : a. one fourth b. two times c. half d. four times and explain your answer as well?
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An object of 5kg mass is placed on a cylindrical stand having cross se...
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An object of 5kg mass is placed on a cylindrical stand having cross se...
Given:
Mass of the object, m = 5 kg
Radius of the first cylindrical stand, r1 = 4 cm = 0.04 m
Radius of the second cylindrical stand, r2 = 2 cm = 0.02 m
To Find:
The change in pressure when the object is placed on the second cylindrical stand.
Solution:
1. Pressure:
Pressure is defined as the force per unit area. Mathematically, it can be written as:
Pressure = Force/Area
2. Area of the Base:
The area of the base of a cylinder can be calculated using the formula:
Area = π * (radius)^2
3. Force on the Base:
The force on the base of the cylindrical stand is the weight of the object, which can be calculated using the formula:
Force = mass * acceleration due to gravity
Force = m * g
4. Calculation:
Let's calculate the pressure exerted by the object on the first cylindrical stand and the second cylindrical stand.
For the first cylindrical stand:
Area1 = π * r1^2
Force1 = m * g
Pressure1 = Force1 / Area1
For the second cylindrical stand:
Area2 = π * r2^2
Force2 = m * g
Pressure2 = Force2 / Area2
5. Comparison:
Let's compare the pressures on the two cylindrical stands.
Pressure2 / Pressure1 = (Force2 / Area2) / (Force1 / Area1)
Pressure2 / Pressure1 = (m * g) / (Area2 / Area1)
Substituting the values:
Pressure2 / Pressure1 = (5 kg * 9.8 m/s^2) / ((π * (0.02 m)^2) / (π * (0.04 m)^2))
Pressure2 / Pressure1 = (5 * 9.8) / ((0.02)^2 / (0.04)^2)
Pressure2 / Pressure1 = (5 * 9.8) / (0.04 / 0.16)
Pressure2 / Pressure1 = 49 / 0.25
Pressure2 / Pressure1 = 196
6. Answer:
The pressure on the second cylindrical stand is four times the pressure on the first cylindrical stand. Therefore, the correct option is d. four times.
Explanation:
When the radius of the base of a cylindrical stand decreases, the area of the base decreases. As pressure is inversely proportional to the area, a decrease in area leads to an increase in pressure. In this case, when the radius is halved, the area becomes one-fourth of the original area. Thus, the pressure becomes four times the original pressure.
Mass of the object, m = 5 kg
Radius of the first cylindrical stand, r1 = 4 cm = 0.04 m
Radius of the second cylindrical stand, r2 = 2 cm = 0.02 m
To Find:
The change in pressure when the object is placed on the second cylindrical stand.
Solution:
1. Pressure:
Pressure is defined as the force per unit area. Mathematically, it can be written as:
Pressure = Force/Area
2. Area of the Base:
The area of the base of a cylinder can be calculated using the formula:
Area = π * (radius)^2
3. Force on the Base:
The force on the base of the cylindrical stand is the weight of the object, which can be calculated using the formula:
Force = mass * acceleration due to gravity
Force = m * g
4. Calculation:
Let's calculate the pressure exerted by the object on the first cylindrical stand and the second cylindrical stand.
For the first cylindrical stand:
Area1 = π * r1^2
Force1 = m * g
Pressure1 = Force1 / Area1
For the second cylindrical stand:
Area2 = π * r2^2
Force2 = m * g
Pressure2 = Force2 / Area2
5. Comparison:
Let's compare the pressures on the two cylindrical stands.
Pressure2 / Pressure1 = (Force2 / Area2) / (Force1 / Area1)
Pressure2 / Pressure1 = (m * g) / (Area2 / Area1)
Substituting the values:
Pressure2 / Pressure1 = (5 kg * 9.8 m/s^2) / ((π * (0.02 m)^2) / (π * (0.04 m)^2))
Pressure2 / Pressure1 = (5 * 9.8) / ((0.02)^2 / (0.04)^2)
Pressure2 / Pressure1 = (5 * 9.8) / (0.04 / 0.16)
Pressure2 / Pressure1 = 49 / 0.25
Pressure2 / Pressure1 = 196
6. Answer:
The pressure on the second cylindrical stand is four times the pressure on the first cylindrical stand. Therefore, the correct option is d. four times.
Explanation:
When the radius of the base of a cylindrical stand decreases, the area of the base decreases. As pressure is inversely proportional to the area, a decrease in area leads to an increase in pressure. In this case, when the radius is halved, the area becomes one-fourth of the original area. Thus, the pressure becomes four times the original pressure.
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An object of 5kg mass is placed on a cylindrical stand having cross section . the radius of the base of the stand is 4cm . if the same object is placed on cylindrical stand with radius the base of stand 2cm, the pressure becomes : a. one fourth b. two times c. half d. four times and explain your answer as well?
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An object of 5kg mass is placed on a cylindrical stand having cross section . the radius of the base of the stand is 4cm . if the same object is placed on cylindrical stand with radius the base of stand 2cm, the pressure becomes : a. one fourth b. two times c. half d. four times and explain your answer as well? for Class 8 2024 is part of Class 8 preparation. The Question and answers have been prepared according to the Class 8 exam syllabus. Information about An object of 5kg mass is placed on a cylindrical stand having cross section . the radius of the base of the stand is 4cm . if the same object is placed on cylindrical stand with radius the base of stand 2cm, the pressure becomes : a. one fourth b. two times c. half d. four times and explain your answer as well? covers all topics & solutions for Class 8 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An object of 5kg mass is placed on a cylindrical stand having cross section . the radius of the base of the stand is 4cm . if the same object is placed on cylindrical stand with radius the base of stand 2cm, the pressure becomes : a. one fourth b. two times c. half d. four times and explain your answer as well?.
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