Q A lamina is in the shape of an isosceles trapezium The two parallel sides are 5 m apart - Physics -

Question

Q A lamina is in the shape of an isosceles trapezium The two
parallel sides are 5 m apart and their lengths are 3 m and 2 m The
distance of the centre of mass of the trapezium form the longer of
the parallel sides is :

(a) 8/3 m (b) 8/15 m (c) 6 m (d) 7/3 m

Sanjay Singh · Accepted Answer

Dear student

Y coordinate of centre of mass is ycm=A1y1+A2y2+A3y3A1+A2+A3Area of triangle ABC(A1)=Area of triangle DEF(A2)=12×12×5=54m2Area of rectangle ACDE (A3) is =2×5=10m2y cordinate of centre of mass of a triangle is =h3  (h is the of perpendicular dropped from one of its vertex on the opposite base)h=5my1=y2=53 and y3=52ycm=A1y1+A2y2+A3y3A1+A2+A3=54×53+54×53+10×5254+54 +10  =256+25104+10=256+2552+10multiplying numerator and the denometor by 617515+60=17575=73mRegards

Q A lamina is in the shape of an isosceles trapezium. The two parallel sides are 5 m apart and their lengths are 3 m and 2 m . The distance of the centre of mass of the trapezium form the longer of the parallel sides is : (a) 8/3 m (b) 8/15 m (c) 6 m (d) 7/3 m

Q A lamina is in the shape of an isosceles trapezium. The two parallel sides are 5 m apart and their lengths are 3 m and 2 m . The distance of the centre of mass of the trapezium form the longer of the parallel sides is :

(a) 8/3 m (b) 8/15 m (c) 6 m (d) 7/3 m