An iceberg is floating partially immersed in sea water. The density of sea water is 1.03 g cm -3 and that of ice is 0.92 g cm -3. The approximate percentage of total volume of iceberg above the level of sea water is

Question

An iceberg is floating partially immersed in sea water. The density of sea water is 1.03 g cm -3 and that of ice is 0.92 g cm -3. The approximate percentage of total volume of iceberg above the level of sea water is (A) 8 (B) 11 (C) 34 (D) 89

Tardigrade Admin · Accepted Answer

(V-Δ V) × 1.03 × g=V × 0.92 × g or (V-Δ V/V)=(0.92/1.03) or (Δ V/V)=1-(0.92/1.03)=(0.11/1.03) or (Δ V/V) × 100=(11/1.03) ≈ 11

Q. An iceberg is floating partially immersed in sea water. The density of sea water is $1.03 g c m^{- 3}$ and that of ice is $0.92 g c m^{- 3}$ . The approximate percentage of total volume of iceberg above the level of sea water is

8

11

34

89

$(V - Δ V) \times 1.03 \times g = V \times 0.92 \times g$
or $\frac{V - Δ V}{V} = \frac{0.92}{1.03}$
or $\frac{Δ V}{V} = 1 - \frac{0.92}{1.03} = \frac{0.11}{1.03}$
or $\frac{Δ V}{V} \times 100 = \frac{11}{1.03} \approx 11$

Q. An iceberg is floating partially immersed in sea water. The density of sea water is 1.03gcm−3 and that of ice is 0.92gcm−3. The approximate percentage of total volume of iceberg above the level of sea water is

8

11

34

89

(V−ΔV)×1.03×g=V×0.92×g or VV−ΔV​=1.030.92​ or VΔV​=1−1.030.92​=1.030.11​ or VΔV​×100=1.0311​≈11

Q. An iceberg is floating partially immersed in sea water. The density of sea water is $1.03 g c m^{- 3}$ and that of ice is $0.92 g c m^{- 3}$ . The approximate percentage of total volume of iceberg above the level of sea water is

$(V - Δ V) \times 1.03 \times g = V \times 0.92 \times g$
or $\frac{V - Δ V}{V} = \frac{0.92}{1.03}$
or $\frac{Δ V}{V} = 1 - \frac{0.92}{1.03} = \frac{0.11}{1.03}$
or $\frac{Δ V}{V} \times 100 = \frac{11}{1.03} \approx 11$