An object suspended by a wire stretches it by 10 mm. When object is immersed in a liquid the elongation in wire reduces by (10/3) mm. The ratio of relative densities of the object and liquid is

Question

An object suspended by a wire stretches it by 10 mm. When object is immersed in a liquid the elongation in wire reduces by (10/3) mm. The ratio of relative densities of the object and liquid is (A) 3: 1 (B) 1: 3 (C) 1: 2 (D) 2: 1

Tardigrade Admin · Accepted Answer

Δ L=(F L/A Y) ⇒ Elongation propto force and force is due to weight So elongation propto weight Δ L1 propto text weight ...(1) When not submerged in liquid Δ L2 propto text apparant weight ...(2) When submerged in liquid Dividing (1) by (2) (10/10-(10)3)=(M g/M g-(M g ρ)σ) ⇒ (1/1-(1)3)=(1/1-(ρ)σ) Solving this we get (ρ/σ)=(1/3) So relative densities of object (σ) and liquid (ρ) is 3: 1 Let density of liquid =ρ Let density of object =σ Mass of object =M

Q. An object suspended by a wire stretches it by $10 mm$ . When object is immersed in a liquid the elongation in wire reduces by $\frac{10}{3} mm$ . The ratio of relative densities of the object and liquid is

$3 : 1$

$1 : 3$

$1 : 2$

$2 : 1$

Q. An object suspended by a wire stretches it by 10mm. When object is immersed in a liquid the elongation in wire reduces by 310​mm. The ratio of relative densities of the object and liquid is

3:1

1:3

1:2

2:1

Q. An object suspended by a wire stretches it by $10 mm$ . When object is immersed in a liquid the elongation in wire reduces by $\frac{10}{3} mm$ . The ratio of relative densities of the object and liquid is

$3 : 1$

$1 : 3$

$1 : 2$

$2 : 1$