A) remain at the same height
B) fall at the rate of 1 cm/h
C) fall at the rate of 2 cm/h
D) go up at the rate of 1 cm/h
Correct Answer: B
Solution :
Weight of candle is equal to weight of liquid displaced. From Archimedes? principle when a body is immersed in a liquid completely or partly, then there is an apparent loss in its weight. This apparent loss in weight is equal to the weight of the liquid displaced by the body. Also, \[\operatorname{volume} of candle = Area \times length\] \[=\,\,\pi {{\left( \frac{d}{2} \right)}^{2}}\times 2L\] \[\operatorname{Weight} of candle= weight of liquid displaced\] \[V\rho g=V'\,\rho 'g\] \[\Rightarrow \,\,\,\,\,\,\,\left( \pi \frac{{{d}^{2}}}{4}\times 2L \right)\rho \,\,=\,\,\left( \pi \frac{{{d}^{2}}}{4}\,\times L \right)\rho '\] \[\Rightarrow \,\,\,\,\,\,\frac{\rho }{\rho '}=\frac{1}{2}\] Since candle is burning at the rate of 2 cm/h, then after an hour, candle length is \[2L - 2\] \[\therefore \,\,~~\left( 2L-2 \right)\rho =\left( L-x \right)\rho '\] \[\therefore \,\,~~\frac{\rho }{\rho '}=\frac{L-x}{2(L-1)}\] \[\Rightarrow \,\,\,\,\,\,\,\frac{1}{2}=\frac{L-x}{2(L-1)}\] \[\Rightarrow \,\,\,\,\,\,\,x=1\,cm\] Hence, in one hour it melts 1 cm and so it falls the rate of 1 cm/h. .
You need to login to perform this action.
You will be redirected in
3 sec
