Question
Download Solution PDFA cubical block of wood of 1 m side and density 500 kg/m3 floats in seawater of density 1000 kg/m3. The depth of immersion of the block is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Buoyancy Principle:
Buoyant Force (F) = Weight of displaced volume
Buoyant Force = Weight of displaced volume = (A × h × ρ w × g) N
A = Base area of a cube, h = Depth of cube immersed in the seawater
Floating principle,
Buoyant force (F) = Weight of cube (W)
Calculation:
Weight of the cube (W) = Density × Volume × g = 500 × 13 = (500 × g) N ...(1)
Buoyant Force (F) = (A × h × ρw × g) N = (12 × h × 1000 × g) = (1000 × g × h) N ...(2)
Buoyant force (F) = Weight of cube (W)
From equation 1 and 2,
500 × g = 1000 × g × h
h = 0.5 m
Last updated on Jul 1, 2025
-> SSC JE ME Notification 2025 has been released on June 30.
-> The SSC JE Mechanical engineering application form are activated from June 30 to July 21.
-> SSC JE 2025 CBT 1 exam for Mechanical Engineering will be conducted from October 2 to 31.
-> SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.