In an experiment , the following observations were recorded:
`L = 2.820 m , M = 3.00 kg , l = 0.087 cm , diameter , D = 0.041 cm`. Taking ` g = 9.81 m s^(-2)` and using the formula , `Y = ( 4MgL)/( pi D^(2) l)` , find the maximum permissible error in `Y`.

In an experiment for determining the value of acceleration due to gravity (g) using a simpe pendulum , the following observations were recorded: Length of the string (l) = 98.0 cm Diameter of the bob (d) = 2.56 cm Time for 10 oscillations (T) = 20.0 s Calculate the value of g with maximum permissible absolute error and the percentage relative error.

If L = 2.01 m +- 0.01 m , what is 3L?

In an experiment for measurement of young's modulus, following readings are taken. Load = 3.00 kg, length = 2.820 m, diameter = 0.041 cm and extension = 0.87. We shell determine the percentage error in measurement of Y.

In resonance tube experiment , the velocity of sound is given by v = 2 f_(0) ( l_(2) - l_(1)) . We found l_(1)=25.0 cm and l_(2)=75.0 cm . If there is no error in frequency, what will be the maximum permissible errror in the speed of sound ? (Take f_(0) = 325 Hz )

If the measured value of resistance R = 1.05 Omega , wire diameter d = 0.60 mm , and length l = 75.3 cm , then find the maximum permissible error in resistivity , rho = ( R ( pi d^(2) // 4)) /( l) .

How many electrons in a given atom can have the following quantum numbers ? (a) n = 3, l = 1 (b) n = 3, l = 2, m_l = 0 (c ) n = 3, l = 2, m_l = +2, m_s = + 1/2 (d) n = 3 .

If l_(1) = (10.0 pm 0.1) cm and l_(2) = (9.0 pm 0.1) cm , find the their sum , difference and error in each .

In Searle's exp to find Young's modulus, the diameter of wire is mesured as D=0.05cm length of wire is L=125cm , and when a weight, m=20.0 kg is put, extension in wire was found to be 0.100cm . Find maximum permissible percentage error in Young's modulus ( Y )

Describe the orbital with the following quantum numbers : (i) n = 1, l = 0 " " (ii) n = 2, l = 1, m = 0 (iii) n = 3, l = 2 " " (iv) n = 4, l = 1 (v) n = 3, l = 0, m = 0 " " (vi) n = 3, l = 1 .

If measured value of resistance R = 1.05Omega wire diameter d = 0.60 mm and length l = 75.3 cm If maximum error in resistance measurment is 0.01 Omega and least count of diameter and length measuring device are 0.01 mm and 0.1cm respectively then find max permissible error in resistiviry rho=(R((pid^(2))/(4)))/(l) .