An ornament weighing 36 g in air weighs only 34 g in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it.Specific gravity of gold is 19.3 and that of copper is 8.9.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

2.2 g

4.4 g

1.1 g

3.6 g

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given that mreal =36g,mapp =34g . Density of gold ρAu=19.3g/cc Density of copper ρCu=8.9g/cc We know that loss of weight = weight of displaced water =36−34=2g= Buoyant force =B Here, mreal =mAu +mCu=36g ..........(i) Let v be the volume of the ornament in cm3. Then B=v×ρw×g=2×g ⇒ mAuρAu+mCuρCuρw×g=2×g⇒mAuρCu+mCuρAu=2ρAuρCu⇒8.9mAu+19.3mCu=2×19.3×8.9=343.54.........(ii) From Eqs. (i) and (ii), 8.9mAu+19.3mCu=343.54⇒ 8.9mAu+mCu+10.4mCu=343.54⇒ 8.9×36+10.4mCu=343.54⇒mCu=2.225g So the amount of copper in the ornament is 2.2g .

Watch 3-min video & get full concept clarity