• # question_answer Three blocks ${{B}_{1}},$ ${{B}_{2}}$ and ${{B}_{3}}$ are to inserted in cannel of width S maintaining a minimum gap of width $T=0.125\,\,m,$ as shown in figure. For $=\,18.75\pm 0.08\,;\,Q=25.00\pm 0.12\,;\,R=28.125\pm 0.1$and $S=72.35+X,$ (where all dimensions are in mm), the tolerance X is: A) $+0.38$                      B) $-\,0.38$C) $+0.05$                      D) $-\,0.05$

Tolerance build up, $P+Q+R+T=(18.75\pm 0.08)+(25.00\pm 0.12)+$ $(28.125\pm 0.10)+0.125$ $=72.000\pm 0.30$ $72.35+X=72.00\pm 0.30$ $X=72.30-72.35=-\,0.05\,mm$