GATE ME 2015 Set 1 — Question 40

From the assembly diagram, the relationship between the dimensions is:
$L_1 = L_2 + L_3 + L_4$
Therefore, the required dimension $L_4$ is given by:
$L_4 = L_1 - L_2 - L_3$First, we calculate the nominal value of $L_4$:
$L_{4, \text{nominal}} = L_{1, \text{nominal}} - L_{2, \text{nominal}} - L_{3, \text{nominal}}$
Given:
$L_{1, \text{nominal}} = 22.0$ mm
$L_{2, \text{nominal}} = 10.0$ mm
$L_{3, \text{nominal}} = 10.0$ mm$L_{4, \text{nominal}} = 22.0 - 10.0 - 10.0 = 2.0$ mm
All options have a nominal value of 2.0 mm.Next, we calculate the tolerance for $L_4$. The question states to assume a normal distribution of part dimensions, which implies using statistical tolerancing, specifically the Root Sum Squares (RSS) method.The tolerances for the given parts are:
Tolerance of $L_1$, $\delta_1 = \pm 0.01$ mm
Tolerance of $L_2$, $\delta_2 = \pm 0.005$ mm
Tolerance of $L_3$, $\delta_3 = \pm 0.005$ mmFor addition or subtraction of dimensions, the combined tolerance using the RSS method is the square root of the sum of the squares of the individual tolerances.$\delta_4 = \sqrt{\delta_1^2 + \delta_2^2 + \delta_3^2}$
$\delta_4 = \sqrt{(0.01)^2 + (0.005)^2 + (0.005)^2}$
$\delta_4 = \sqrt{0.0001 + 0.000025 + 0.000025}$
$\delta_4 = \sqrt{0.00015}$
$\delta_4 \approx 0.012247$ mmRounding this to three decimal places, we get $\delta_4 \approx 0.012$ mm.So, the dimension $L_4$ is $2.0 \pm 0.012$ mm.This matches option (B).Note: If we had used the worst-case (arithmetic) stack-up, the tolerance would be $\delta_4 = \delta_1 + \delta_2 + \delta_3 = 0.01 + 0.005 + 0.005 = 0.020$ mm, which corresponds to option (D). However, the mention of "normal distribution" specifically directs the use of the statistical method.