A piece of cork whose weight is 19 gm is attached to a bar of silver weighing 63 gm. The two together just not in water. The specific gravity of silver is 10-5. Find the specific gravity of the cork. Density of water = 1 gm cm ^(-3)

A piece of cork whose weight is 19 gm is attached to a bar of silver w

1.	A piece of cork whose weight is 19 gm is attached to a bar of silver weighing 63 gm. The two together just not in water. The specific gravity of silver is 10-5. Find the specific gravity of the cork. Density of water = 1 gm cm ^(-3)
Answer» SOLUTION :Density of SILVER ` = 10.5 gm cm ^(-3)` `"" ` MASS of cork = `m_1= 19 gm ` `""`Mass of silver `= m_2 = 63 gm` Total mass of FLOATING body = ` m_1+ m_2 = 82 gm ` Mass of floating body= mass of water DISPLACED ` therefore ` Mass of displaced = ` ("mass " )/( " density")= (82)/(1) = 82 cm ^(3)` Volume of silver `= (" mass")/(" density")= ( 63)/( 10.5) = 6 cm ^(3)` Volume of cork = Total Volume - Volume of silver ` "" = 82- 6= 7^(2)cm ^(3)` Density of cork = ` (" mass ")/( " volume")= (19)/(76) = 0.25 gm cm ^(-3)` Specific gravity = ` (" density of cork")/( " density of water ")= (0.25 gm cm ^(-3))/( 1 gm cm ^(-3))= 0.25`