Question

A horse drinks water from a cubical container of side 1 m. The level of the stomach of horse is at 2 m from the ground. Assume that all the water drunk by the horse is at a level of 2m from the ground. Then minimum work done by the horse in drinking the entire water of the container is (Take $\rho_{water}$ = 1000 kg/m³ and g = 10 m/s²) –(a) 10 kJ(b) 15 kJ(c) 20 kJ(d) zero

Answer 1

$\boxed{\sf {\green{Answer}}}$

Work done = 20KJ

$\boxed{\sf {\green{Explanation}}}$

Given:

• side of cubical container(a) = 1m
• h = 2m
• $\rho_{water}$ = 1000 kg/m³
• g = 10 m/s²

To Find

work done by the horse

Let us find volume of container

Volume of cube = a³

V = (1m)³ = 1m³

Now, let us find mass

$\boxed{\bold {\pink{m = V \times \rho}}}$

$\mathsf{m = 1{m}^{3} \times 1000\dfrac{kg}{{m}^{3}}}$

$\mathsf{m = 1000kg}$

Work done by horse (W)

$\boxed{\bold {\red{W= mgh}}}$

$\mathsf{W = 1000kg \times 10m{s}^{-2} \times 2m}$

$\mathsf{W = 20000kg {m}^{2}{s}^{-2} }$

$\mathsf{W = 20000J }$

$\mathsf{W = 20KJ }$ ______option(c)

$\texttt{\blue{Aravind}\:\red{Reddy}....!}$

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

Side of cube(S) = 1m.

Height (h) = 2m.

g = 10 m/s².

${ \rho_{water} = 1000kg/m^3}$

$\huge{\bold{\underline{Explanation:-}}}$

$\large{Density = \frac{Mass}{Volume}}$

$\large{ \rho = \frac{M}{V}}$

$\large{ M = \rho \times V ------(1)}$

Volume of cuboid = $\bold{(S)^3}$

Substituting the values

$\large{V = (1m)^3}$

$\large{V = 1 \: m^3 ------(2)}$

Here work done will be.

$\large{\bold{W = Mgh}}$

From equation (1) and (2)

$\large{\bold{W = \rho Vgh}}$

Substituting the values.

$\large{ W = 1000 \times 1 \times 10 \times 2}$

$\large{\bold{W = 20,000 \: J}}$

Now,

$\huge{\boxed{\boxed{W = 20 \: KJ}}}$

So, the work done by the horse is 20 KJ (option --- (c)).