GATE 2016 ME (ME2) – Question 63

Metrology and Inspection · Linear and angular measurements, comparators · 2 marks · Multiple choice

For the situation shown in the figure below the expression for H in terms of r, R and D is

Diagram for GATE 2016 ME (ME2) question 63

$H=D+\sqrt{r^2+R^2}$
$H=(R+r)+(D+r)$
$H=(R+r)+\sqrt{D^2-R^2}$
$H=(R+r)+\sqrt{2D(R+r)-D^2}$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (D) $H=(R+r)+\sqrt{2D(R+r)-D^2}$

Explanation

The circle centres are separated by R+r. Their horizontal separation is D−R−r, so vertical centre separation is sqrt[(R+r)²−(D−R−r)²]=sqrt[2D(R+r)−D²]. Add the large-circle centre height R and small-circle top offset r. Thus **H=R+r+sqrt[2D(R+r)−D²], D**.