What is the percentage error in the measurement of volume of sphere?

The percentage error in the measurement of the volume of a sphere is 6\%.

What is the percent error for the density measurement?

Step 1: Subtract the accepted value from the experimental value. Step 3: Divide that answer by the accepted value. Step 4: Multiply that answer by 100 and add the \% symbol to express the answer as a percentage. The percent error of your density calculation is 2\%.

What will be the percentage error in measurement of density if the error in measurement of the radius length and mass of a wire are 3 4 & 5 respectively?

Formula is delta of (density) = (delta of (mass) /mass} + { 3 × delta of (length)/ length) . So, Maximum Error = 13\%.

What is percentage error in volume of a sphere when error in measuring its radius is 2\% *?

Hence,,the error in volume of sphere is 6\%.

When measuring the radius of a sphere an error of 0.6 percent takes place in the measurement What will be the permissible error while measuring the surface area?

The permissible error in its surface area is 0.6\%. Reason: The permissible error is calculated by the formula ΔAA=4Δrr. The error in the measured of radius of the sphere is 0.3\%.

What is the fractional error in G calculated from?

The fractional error in g calculated from T = 2pi√(l/g) ,it fractional error in l and T are p and q respectively is.

How do you calculate the percentage of error?

Percent Error Calculation Steps

1. Subtract one value from another.
2. Divide the error by the exact or ideal value (not your experimental or measured value).
3. Convert the decimal number into a percentage by multiplying it by 100.
4. Add a percent or \% symbol to report your percent error value.

What is the maximum percentage error in density of a body if percentage?

Answer: Thus, the maximum possible percentage error in the relative density of the material of the body is 11\%.

What is maximum percentage error in density of Cube?

This measurement is made by taking the measurements of the cube’s mass and length of its edge. The maximum errors in these measurements are found to be 0.3\% and 0.2\% respectively.

What is percentage error in surface area of sphere if the diameter of sphere measured by vernier calipers is 1.71 cm?

Therefore, the percentage error in surface area is \[1.169\\% \].

What is the percentage error in the measurement of volume of sphere if percentage error in its measurement of its radius is 3 \%?

Here we will use the formulae of volume of sphere to solve this question and we know that volume of sphere = 43πr3 , from this formula we can clearly observe that v is directly proportional to r. Hence, the error in volume of the sphere will be 9\% .