A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle’s apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8.
The new value of increase in length of the steel wire is:
(A) 3.0 mm (B) 4.0 mm
(C) 5.0 mm (D) zero
Answers
Answer:
A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle’s apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8.
The new value of increase in length of the steel wire is:
(A) 3.0 mm (B) 4.0 mm
(C) 5.0 mm (D) zero
Answer: (A) 3.0 mm
Explanation:
given that ΔL = 4.0 mm,
As we know that
F / A = Y (ΔL / L)
⇒ Mg / A = Y (ΔL / L)
⇒ (8ρv)g / A = Y ( ΔL / L) -------- (1)
and
(8ρvg-2ρvg) / A = Y ( ΔL' / L)
6ρvg / A = Y ( ΔL' / L) ............(2)
On solving equation (1) & (2)
dividing (1) by (2)
4 / 3 = ΔL / ΔL'
⇒ ΔL' = ( 3 / 4) Δl = ( 3 / 4 ) x 4 = 3 mm
Hence, the new value of increase in length of the steel wire is: 3.0 mm
Answer:
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