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Find the height of the cylinder whose vo...

Find the height of the cylinder whose volume is `1.54m^(3)` and diameter of the base is 140 cm?

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To find the height of the cylinder given its volume and diameter, we can follow these steps: ### Step 1: Convert the diameter to meters The diameter of the base is given as 140 cm. To convert this to meters, we use the conversion factor \(1 \text{ cm} = 0.01 \text{ m}\). \[ \text{Diameter in meters} = 140 \text{ cm} \times 0.01 \text{ m/cm} = 1.4 \text{ m} \] **Hint:** Remember that to convert centimeters to meters, you divide by 100. ### Step 2: Calculate the radius The radius \(r\) of the cylinder is half of the diameter. \[ r = \frac{\text{Diameter}}{2} = \frac{1.4 \text{ m}}{2} = 0.7 \text{ m} \] **Hint:** The radius is always half of the diameter. ### Step 3: Use the volume formula for a cylinder The formula for the volume \(V\) of a cylinder is given by: \[ V = \pi r^2 h \] Where: - \(V\) is the volume, - \(r\) is the radius, - \(h\) is the height, - \(\pi\) is a constant approximately equal to \(3.14\) or can be used as \(\frac{22}{7}\). ### Step 4: Substitute the known values into the formula We know the volume \(V = 1.54 \text{ m}^3\) and the radius \(r = 0.7 \text{ m}\). We will use \(\pi \approx \frac{22}{7}\). \[ 1.54 = \frac{22}{7} \times (0.7)^2 \times h \] ### Step 5: Calculate \( (0.7)^2 \) First, calculate \( (0.7)^2 \): \[ (0.7)^2 = 0.49 \] ### Step 6: Substitute back into the equation Now substitute \(0.49\) back into the equation: \[ 1.54 = \frac{22}{7} \times 0.49 \times h \] ### Step 7: Solve for \(h\) Rearranging the equation to solve for \(h\): \[ h = \frac{1.54 \times 7}{22 \times 0.49} \] ### Step 8: Calculate the right-hand side Calculating the numerator and denominator: 1. **Numerator:** \[ 1.54 \times 7 = 10.78 \] 2. **Denominator:** \[ 22 \times 0.49 = 10.78 \] Now substituting these values back into the equation for \(h\): \[ h = \frac{10.78}{10.78} = 1 \text{ m} \] ### Final Answer The height of the cylinder is \(1 \text{ meter}\). ---
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RS AGGARWAL-VOLUME AND SURFACE AREA OF SOLIDS-EXERCISE 20 B
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