Skip to main content

Visit the website manishverma.site for latest posts, courses, admission & more.

Due to cold weather a 1 m water pipe of cross-sectional area ....

Due to cold weather a 1 m water pipe of cross-sectional area $1 cm^2$ is filled with ice at $-10^\circ C $. Resistive heating is used to melt the ice. Current of 0.5 A is passed through $4 k\Omega $ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required?
(Given latent heat of fusion for water/ice $=3.33 \times 10^5 J Kg^{-1}$, specific heat of ice $=2\times 10^3 J Kg^{-1}$ and density of ice $=10^3 Kg m^{-3} $ )

(A) 3.53 s
(B) 0.353 s
(C) 35.3 s
(D) 70.65 s

Solution

Electrical Energy = Thermal Energy

$\therefore I^2 Rt = Cm\Delta T + mL = m (C\Delta T + L) $

$\Rightarrow 0.5^2 \times 4000 \times t = \rho V (2 \times 10^3 \times 10 + 3.33 \times 10^5 )$

$\Rightarrow 1000 t = 10^3 \times Al (2 \times 10^4 + 33.3 \times 10^4 )$

$\Rightarrow t = 1 \times 10^{-4} \times 1 (35.3 \times 10^4 ) = 35.3 s $

Answer: (C)

Popular posts from this blog

A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1 $ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$