Young's modulus is determined by the equation given by Y=49000 m/l dyne/cm² where M is the mass and l is the extension of wire used in the experiment. Now error in Young modules (Y) is estimated by taking data from M-l plot in graph paper. The smallest scale divisions are 5 g and 0.02 cm along load axis and extension axis respectively. If the value of M and l are 500 g and 2 cm respectively then percentage error of Y is :

Solution

Physics sir · 3 steps

Y goes as M over l, so just add the percentage errors in M and l.

1. Errors from the graph. On graph paper, the error in each reading is the smallest scale division on that axis. So Δ M = 5 g and Δ l = 0.02 cm. The constant 49000 carries no error.

   (Δ Y)/Y × 100 = ((Δ M)/M + (Δ l)/l) × 100

2. Put in numbers. (Δ M)/M = 5/500 = 0.01, which is 1%. (Δ l)/l = 0.02/2 = 0.01, which is also 1%.
3. So. The percentage error in Y is 1% + 1% = 2%, option (a).

The idea to keep. For a quantity like M/l, add the fractional errors whether the variable multiplies or divides. On a graph, each reading's error is the least count of its axis.

Answer from the NTA final answer key. Solution by PrepWiser.