A body of density 1.2 × 103 kg / m 3 is dropped from rest from a height 1 m into a liquid of density 2.4 × 103 kg / m 3. Neglecting all dissipative effects, the maximum depth to which the body sinks before returning to float on the surface is

Question

A body of density 1.2 × 103 kg / m 3 is dropped from rest from a height 1 m into a liquid of density 2.4 × 103 kg / m 3. Neglecting all dissipative effects, the maximum depth to which the body sinks before returning to float on the surface is (A) 0.1 m (B) 1 m (C) 0.01 m (D) 2 m

Tardigrade Admin · Accepted Answer

d=1.2 × 103, h=1 m, ρ=2.4 × 103 By work Energy Theorem, M g(h+d)=B d V(1.2 × 10-3) g(1+d)=V(2.4 × 10-3) g d ∴ d=1 m

Q. A body of density $1.2 \times 1 0^{3} k g / m^{3}$ is dropped from rest from a height $1 m$ into a liquid of density $2.4 \times 1 0^{3} k g / m^{3}$ . Neglecting all dissipative effects, the maximum depth to which the body sinks before returning to float on the surface is

0.1 m

1 m

0.01 m

2 m

$d = 1.2 \times 1 0^{3}, h = 1 m, ρ = 2.4 \times 1 0^{3}$
By work Energy Theorem,
$M g (h + d) = B d$
$V (1.2 \times 1 0^{- 3}) g (1 + d) = V (2.4 \times 1 0^{- 3}) g d$
$∴ d = 1 m$

Q. A body of density 1.2×103kg/m3 is dropped from rest from a height 1m into a liquid of density 2.4×103kg/m3. Neglecting all dissipative effects, the maximum depth to which the body sinks before returning to float on the surface is

0.1 m

1 m

0.01 m

2 m

d=1.2×103,h=1m,ρ=2.4×103 By work Energy Theorem, Mg(h+d)=Bd V(1.2×10−3)g(1+d)=V(2.4×10−3)gd ∴d=1m

Q. A body of density $1.2 \times 1 0^{3} k g / m^{3}$ is dropped from rest from a height $1 m$ into a liquid of density $2.4 \times 1 0^{3} k g / m^{3}$ . Neglecting all dissipative effects, the maximum depth to which the body sinks before returning to float on the surface is

$d = 1.2 \times 1 0^{3}, h = 1 m, ρ = 2.4 \times 1 0^{3}$
By work Energy Theorem,
$M g (h + d) = B d$
$V (1.2 \times 1 0^{- 3}) g (1 + d) = V (2.4 \times 1 0^{- 3}) g d$
$∴ d = 1 m$