Q.
A cubical block of wood $10\ cm$ on a side floats at the interface between oil and water with its lower surface horizontal and 4 cm below the interface. The density of oil is $0.6gcm^{- 3}$. The mass of block is
A) $706\ g$
B) $607\ g$
C) $760\ g$
D) $670\ g$
Explanation
(c) Weight of block
= Weight of displaced oil + Weight of displaced water
$\Rightarrow mg = V_{1}\rho_{0}g + V_{2}\rho_{w}g$
$\Rightarrow m = (10 \times 10 \times 6) \times 0.6 + (10 \times 10 \times 4) \times 1 = 760\ gm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.